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Let the eigenvalues of 2 x 2 matrix A be 1, -2 with eigenvectors x1 and x2 respectively. Then the eigenvalues and eigenvectors of the matrix A^2 - 3A+4I would respectively, be
(a) 2,14; x1,x2
(b) 2,14; x1+x2:x1-x2
(c) 2,0; x1, x2
(d) 2,0; x1+x2,x1-x2
Let λ be an eigenvalue of A and X be the eigenvector of A.
Then according to the definition of eigenvalues and eigenvector we get
AX = λ X ........................(1)
Now, multiplying both sides of equation 1 with A we get
A2X = A λ X = λ AX = λ λ X = λ2X ............... (2)
So, the matrix A2 has the same eigenvector as in A but the eigenvalue is square of the eigenvalue of A.
Now, multiplying both sides of equation 1 with -3 we get
-3 AX = -3 λX ..........................(3)
Now, adding equation (2) and (3) we get
(A2 - 3A)X = ( λ2 - 3 λ )X ........................(4)
Now, we add 4$I$ X to the both sides of equation 4
(A2 - 3A) X + 4$I$ X = ( λ2 - 3 λ )X + 4$I$X
=> (A2 - 3A + 4$I$)X = ( λ2 - 3 λ +4 )X [ since $I$X = X]
So, the eigenvalue of the matrix (A2 - 3A +4$I$) is ( λ2 - 3 λ + 4 ) and the eigenvector remains same as A.
Now, the value of λ is -2 and 1
So, putting those values in the expression of the eigenvalue we get the eigenvalues of the given matrix which are 2 and 14
So, Option A is the correct answer.
$\rightarrow$Matrix A
$\rightarrow$Eigen values 1 , -2
$\rightarrow$So, Matrix A$^{2}$ - 3A + 4
$\Rightarrow$ 1-3(1)+4 and (-2)$^{2}$ -3(-2) +4
$\Rightarrow$ Eigen values are 2 and 14 respectively.
$\therefore$ A & P(A) = a$_{0}$ I + a$_{1}$ A + a$_{2}$A have same eigen vectors.
So, option A is the correct answer.
My doubt is:
Say 'e' is eigen value of A , then eigen value of A^2 is e^2. Also the eigen value of pA will be p*e.
But this rules doesn't apply on + and - then how can we say that eigen value of matrix Matrix A^2 - 3A + 4I will be
eigen of (A^2) - eigen of(3A) +4
edited
If λ$_{i}$ is any eigenvalue of A and φ$_{i}$ is an eigenvector corresponding to λ$_{i}$ ,then for any polynomial p(λ) we have p(A).φ$_{i}$ = p(λ$_{i}$).φ$_{i}$.
Better u should read the properties of eigen value and eigen vector, u will get it.
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Solve the following system of equations by substitution method. Equation 2) -x + 5y + 3z = 2. By using this website, you agree to our Cookie Policy. The graph of this linear system follows: Figure \(\PageIndex{2}\) The substitution method for solving systems is a completely algebraic method. Check the solution in both original equations. In two variables ( x and y ) , the graph of a system of two equations is a pair of lines in the plane. Question 9 : Solve the following system of equations by substitution method. The solution is x = 1, y = –2. -4x + y = 6 and -5x - y = 21. There are three possibilities: There is no need to graph the lines unless you are asked to. Question 8 : Solve the following system of equations by substitution method. Solve for x in the second equation. Substitute for x in the other equation. Learn vocabulary, terms, and more with flashcards, games, and other study tools. When solving linear systems, you have two methods — substitution or elimination — at your disposal, and which one you choose depends on the problem. Equation 3) 3x - 2y – 4z = 18 2x + y = 20 and 6x - 5y = 12. This Solver (SOLVE linear system by SUBSTITUTION) was created by by ichudov(507) : View Source, Show, Put on YOUR site About ichudov: I am not a paid tutor, I am the owner of this web site. Solve this system of equations by using substitution. Solution: Step 1: Solve for either variable in either equation. If the coefficient of any variable is 1, which means you can easily solve for it in terms of the other variable, then substitution is a very good bet. The substitution method involves algebraic substitution of one equation into a variable of the other. Solving Systems of Linear Equations Using Substitution Systems of Linear equations: A system of linear equations is just a set of two or more linear equations. Substitute the value found for y into any equation involving both variables. Thus … Solving Linear Systems by Substitution The substitution method for solving linear systems is a completely algebraic technique. Solve this new equation. Free system of equations substitution calculator - solve system of equations unsing substitution method step-by-step This website uses cookies to ensure you get the best experience. This is called the substitution method A means of solving a linear system by solving for one of the variables and substituting the result into the other equation., and the steps are outlined in the following example. *** Solving systems of liner equations using the substitution method in Algebra 1. Solving a Linear System of Linear Equations in Three Variables by Substitution . ***Class video lesson created for my Algebra 1 classes. This method is fairly straight forward and always works, the steps are listed below. This will be the sample equation used through out the instructions: Equation 1) x – 6y – 2z = -8. Start studying Solving Systems of Linear Equations: Substitution (6.2.2). Example 1: Solve by substitution: {2 x + y = 7 3 x − 2 y = − 7. y = -2 and 4x - …
## solving linear systems by substitution
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# Lesson Video: Dividing Polynomials by Binomials Using Factorization Mathematics • 10th Grade
In this video, we will learn how to divide polynomials by binomials using factorization.
14:26
### Video Transcript
In this video, weβll learn how to divide polynomials by binomials using factorization. There are a number of ways to divide algebraic expressions. One of these has used a process called polynomial long division. But this can be quite a slow method. And so we should always check to see whether there are other ways that we can perform such division. One of these methods is to use factorization. Essentially, we begin by writing our division as a fraction. And then we can write both the numerator and denominator where possible in factored form.
Once weβve done this, we simplify as we w ould with any other numeric fraction by dividing by a common factor. Now, before we go any further, itβs worth noting that this process relies on us being confident factoring algebraic expressions. These will primarily be of order two. We will also consider polynomials of order three, which can be expressed as a product of two binomials. So please ensure that you are able to factor expressions of this kind before moving any further. Weβll now consider a simple example.
Simplify two π₯ squared plus five π₯ minus three over π₯ plus three.
First, we recall that this fraction line means divide. And so when we simplify, weβre actually saying how can we divide two π₯ squared plus five π₯ minus three by π₯ plus three. Well, once our division problem is written as a fraction, we look to factor where possible. Now, itβs not possible to factor the expression on the denominator. But we can factor the numerator. Letβs look to factor two π₯ squared plus five π₯ minus three. There are a number of ways of doing this. One way is called the AC method. Itβs called the AC method because given a quadratic equation of the form ππ₯ squared plus ππ₯ plus π, we begin by multiplying the value of π and π. In our equation, π, which is the coefficient of π₯ squared, is two and π is negative three. Two multiplied by negative three is negative six.
Our next step, is just like when we factor a quadratic equation where the coefficient of π₯ squared is one. We look for two numbers that multiply to make negative six and add to make five. Well, six multiplied by negative one is negative six. And six plus negative one is five. And so weβre going to look to split this middle term up into six π₯ and negative one π₯. We now write our quadratic as two π₯ squared plus six π₯ minus one equals three. Now, weβve not done anything mind-blowing here. Weβve just rewritten our original expression. If we were to now simplify the expression on the right-hand side, that would take us back to the expression on the left.
The next step is to consider the two pairs of terms. Weβre going to factor each pair. We see that two π₯ squared and six π₯ have a highest common factor or a greatest common factor of two π₯. And so two π₯ squared plus six π₯ can be written as two π₯ times π₯ plus three. Similarly, negative one π₯ minus three have a common factor of negative one. So when we factor this expression, we get negative one π₯ plus three.
Notice now that each term contains a factor of π₯ plus three. And so we can factor by π₯ plus three. Two π₯ times π₯ plus three divided by π₯ plus three gives us two π₯. Then negative one times π₯ plus three divided by π₯ plus three gives us negative one. And so we fully factored our quadratic. Itβs π₯ plus three times two π₯ minus one. And this is great because we could now rewrite our fraction. Weβve replaced the quadratic with its factored form. And we see itβs equal to π₯ plus three times two π₯ minus one all over π₯ plus three.
Now that itβs in this form, we can simplify our fraction as we would in numerical fraction by dividing through by any common factors. In this case, we can divide through by π₯ plus three. When we do, we see that our expression fully simplifies to two π₯ minus one over one or simply two π₯ minus one. And so the answer to our question and, in fact, the answer to two π₯ squared plus five π₯ minus three divided by π₯ plus three is two π₯ minus one. Now, we did use something called the AC method to factor our quadratic expression. You may be used to using some other method. And thatβs absolutely fine as long as you do indeed end up with π₯ plus three times two π₯ minus one.
Weβll now look at how to find the denominator of a fraction when dividing a polynomial by a binomial using factorization.
Find the denominator of the fraction in this equation: Three π₯ squared plus 11π₯ plus eight over what is equal to π₯ plus one.
Letβs call the denominator of our fraction π¦ for now, where π¦ is going to be some function of π₯. Then weβre going to recall the relationship between fractions and division. A fraction is just another way of writing a division. So what this question is also asking us is, what is the value of π¦ such that three π₯ squared plus 11π₯ plus eight divided by π¦ equals π₯ plus one? Now, we could rearrange this to make π¦ the subject. Or we can quote that if π divided by π is equal to π, then π divided by π must be equal to π. This makes a lot of sense because if we rearrange each equation, we find that π is equal to π times π. We can think of π and π as a factor pair of π. We can therefore say that three π₯ squared plus 11π₯ plus eight divided by π₯ plus one must be equal to π¦.
But how do we work out this division on the lef-hand side? Well, we have a number of methods, but factorization is generally the most straightforward. Weβre going to look to factor the expression three π₯ squared plus 11π₯ plus eight. There are a number of ways to do this. One method is kind of observation. Itβs a quadratic equation, and there are no common factors apart from one in each of our terms. And so we know we can write it as the product of two binomials. The first term in each binomial must be three π₯ and π₯ because three π₯ times π₯ gives us the three π₯ squared we need.
And then we need to look for factor pairs of eight bearing in mind that one of these is going to be multiplied by three. And then once that happens, when we add our numbers together, weβre going to get 11. Well, a factor pair we could use is eight and one. And if we multiply three by one, we get three. Then three plus eight is 11. For this to work, both eight and one need to be positive. And so we factored our expression. Itβs three π₯ plus eight times π₯ plus one. And so we can now rewrite our equation as three π₯ plus eight times π₯ plus one all over π₯ plus one equals π¦.
Now, our next step since itβs written as a fraction is to simplify like we would any other fraction by dividing through by a common factor. Here, we see we have a common factor of π₯ plus one. π₯ plus one divided by π₯ plus one is simply one. And so we see that π¦ is equal to three π₯ plus eight? And since we said π¦ was the denominator of our fraction, then the denominator is three π₯ plus eight. Now, a really quick way to check this answer is to check that the product of π₯ plus one and our denominator is indeed equal to three π₯ squared plus 11π₯ plus eight. And in fact, if we multiply these two binomials, we do indeed get three π₯ squared plus 11π₯ plus eight.
In our next example, weβll look at how we can use a similar method to help us find the value of an unknown.
Find the value of π that makes the expression π₯ squared minus ππ₯ plus 30 divisible by π₯ minus five.
When we divide polynomials by binomials, we look to begin by writing them as a fraction and then simplifying as far as possible. And so to divide π₯ squared minus ππ₯ plus 30 by π₯ minus five, we begin by writing it as π₯ squared minus ππ₯ plus 30 over π₯ minus five. And so the implication is that for our expression to be divisible by π₯ minus five, this fraction can be simplified. And we simplify, of course, by dividing through by a common factor. On the denominator of our fraction, we have π₯ minus five. So that must indicate to us that π₯ minus five is a factor of π₯ squared minus ππ₯ plus 30. So we should be able to write π₯ squared minus ππ₯ plus 30 as some binomial β Iβve called that π₯ plus π, where π is a constant β times π₯ minus five.
But how do we decide what π needs to be? We think about how we factor quadratic expressions where the coefficient of π₯ squared is equal to one. We have an π₯ at the front of each binomial. And then we look for two numbers whose product is the constant β so here, that would be 30 β and whose sum is the coefficient of π₯. So here, thatβs negative π. These two numbers give us the numerical parts of our binomials. And so whilst we donβt know what π is, we can say that π multiplied by negative five must be 30. And so weβll solve for π by dividing through by negative five. 30 divided by negative five is negative six. So we say that π is equal to negative six.
If we replace π with negative six β and now forget about the denominators because theyβre of course equal β we see that the numerators must be equal. We see that π₯ squared minus ππ₯ plus 30 must be equal to π₯ minus six times π₯ minus five. If we distribute these parentheses and simplify, that will tell us the value of π. So we begin by multiplying the first term in each binomial. π₯ times π₯ is π₯ squared. We then multiply the outer terms to give us negative five π₯, then multiply the inner terms to give us negative six π₯. Finally, we multiply the last terms. Negative six times negative five is 30. So we get π₯ squared minus five π₯ minus six π₯ plus 30. And since negative five π₯ minus six π₯ is negative 11π₯, this becomes π₯ squared minus 11π₯ plus 30.
Letβs compare both sides of this equation. We have π₯ squared on both sides, and we have plus 30. And so we can say that these two terms must be equal. Negative ππ₯ must be equal to negative 11π₯. Well, for this to be true, π must therefore be equal to 11. The value of π that makes the expression π₯ squared minus ππ₯ plus 30 divisible by π₯ minus five is 11.
Weβll now look at how this process can help us solve problems involving geometry.
A rectangle has an area of π¦ cubed plus two π¦ squared plus five π¦ plus 10 square centimeters and a width of π¦ plus two centimeters. Find its length in terms of π¦ and its perimeter when π¦ equals four.
Our rectangle could look a little something like this. We know it has a width of π¦ plus two centimeters, and weβre trying to find its length in terms of π¦. For now, weβll call its length π₯ centimeters, where π₯ is going to be some function in π¦. Now, we know that the area of a rectangle is given by multiplying its width by its length. So in this case, the area of the rectangle would be π¦ plus two times π₯. Letβs write that as π₯ times π¦ plus two. But weβre actually given an expression for the area. Itβs π¦ cubed plus two π¦ squared plus five π¦ plus 10. And so we actually have an equation that we can look to solve or at least make π₯ the subject.
Weβre going to make π₯ the subject by dividing both sides by π¦ plus two. On the right-hand side, that leaves us with π₯. But what happens on the left-hand side? Well, for now, weβll write it as a fraction, since a fraction line simply means divide. And one of the best ways that we have to simplify a fraction, which is the same as dividing, is to begin by factoring where possible. Letβs factor the expression π¦ cubed plus two π¦ squared plus five π¦ plus 10. To do this, we begin by factoring the pairs of terms. When we factor π¦ cubed plus two π¦ squared, we get π¦ squared times π¦ plus two.
Similarly, when we factor five π¦ plus 10, we get five times π¦ plus two. We now know that there is a common factor of π¦ plus two in our two terms. And so we can factor by π¦ plus two. π¦ squared times π¦ plus two divided by π¦ plus two is π¦ squared. Then when we divide our second term, five times π¦ plus two, by π¦ plus two, weβre simply left with five. This means we can rewrite our fraction as π¦ plus two times π¦ squared plus five over π¦ plus two. And now, we see that there is a common factor of π¦ plus two on both the numerator and denominator of our fraction. And so weβre going to divide through by π¦ plus two. And when we do, we find that π₯ is equal to π¦ squared plus five.
Remember, we said that the length of our rectangle was π₯ centimeters. So in fact, weβve discovered that the length in terms of π¦ is π¦ squared plus five centimeters. Now, weβre not quite finished. The question wants us to find the perimeter of this rectangle when π¦ is equal to four. So weβre going to substitute π¦ equals four into the expressions for the width and length. The width becomes four plus two, which is six centimeters. And the length becomes four squared plus five, which is 21 centimeters. The perimeter is the entire distance around the outside. So weβre going to add 21 and 6 and then multiply that by two to represent the other two sides. 21 plus six is 27, and 27 times two is 54. The perimeter of our rectangle is therefore 54 centimeters.
In this video, we saw that to divide a polynomial by a binomial, one of the most efficient methods is to use factorization. When using this method, the first thing that we do is we write our division as a fraction. The dividend, thatβs the expression that weβre dividing into, is the numerator of our fraction. And the divisor, which is what weβre dividing by, is the denominator. We then factor where possible. And in this video, weβve mainly factored the numerators. But there will be instances where we need to factor the denominator too. Once weβve done this, we look for any common factors and we divide through. This is the same as simplifying any normal fraction. And once itβs fully simplified, weβve performed our division.
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New Zealand
Level 8 - NCEA Level 3
# Key features of cot, sec and cosec curves
Lesson
The functions cotangent, secant and cosecant are defined as the reciprocal functions of tangent, cosine and sine respectively.
$\cot\left(x\right)$cot(x) $=$= $\frac{1}{\tan\left(x\right)}$1tan(x) $\sec\left(x\right)$sec(x) $=$= $\frac{1}{\cos\left(x\right)}$1cos(x) $\csc\left(x\right)$csc(x) $=$= $\frac{1}{\sin\left(x\right)}$1sin(x)
We can use the properties of $\tan\left(x\right)$tan(x), $\cos\left(x\right)$cos(x) and $\sin\left(x\right)$sin(x) to deduce the properties of their reciprocals.
## Asymptotes
The sine and cosine functions vary continuously between $-1$1 and $1$1, passing through zero twice in every period. When $\sin\left(x\right)=0$sin(x)=0 we should have $\csc\left(x\right)=\frac{1}{0}$csc(x)=10 which is undefined. Similarly, when $\cos\left(x\right)=0$cos(x)=0, the definition requires the impossible expression $\frac{1}{0}$10 for $\sec\left(x\right)$sec(x).
So, there must be discontinuities in the $\sec\left(x\right)$sec(x) and $\csc\left(x\right)$csc(x) functions corresponding to the points at which $\cos\left(x\right)$cos(x) and $\sin\left(x\right)$sin(x) are zero.
We see also that when sine or cosine are close to but not quite equal to zero, the corresponding values of cosecant and secant can be made very large in the positive or negative direction, depending on which side of zero the variable $x$x is.
We say that the secant function has vertical asymptotes at the points where the cosine function is zero. That is, $\sec\left(x\right)$sec(x) is asymptotic at $x=\frac{\pi}{2}\pm n\pi$x=π2±nπ, where $n$n is an integer.
Similarly, $\csc\left(x\right)$csc(x) has vertical asymptotes wherever $\sin\left(x\right)=0$sin(x)=0. That is, at $x=0,\pm\pi,\pm2\pi,...$x=0,±π,±2π,... and so on.
The range of the tangent function is the whole of the real numbers: $-\infty<y<. It has vertical asymptotes wherever the cosine function is zero due to the fact that$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$tan(x)=sin(x)cos(x). The reciprocal of$\tan\left(x\right)$tan(x) must also have the range$-\infty<y< but its vertical asymptotes occur where $\sin\left(x\right)=0$sin(x)=0, because $\cot\left(x\right)=\frac{\cos\left(x\right)}{\sin\left(x\right)}$cot(x)=cos(x)sin(x).
The following graph illustrates the asymptotes. Notice that both $\cot\left(x\right)$cot(x) and $\csc\left(x\right)$csc(x) share the same asymptotes because they both have a denominator of $\sin\left(x\right)$sin(x).
Comparing the location of the asymptotes of each reciprocal trigonometric function.
## Maxima and minima
### Outcomes
#### M8-2
Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions
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Everyday maths 2 (Wales)
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# 7.1 Simplifying fractions
Watch the video below which looks at how to simplify fractions before having a go yourself in Activity 14.
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#### Transcript
In this video, you'll look at how to simplify fractions. You'll be used to seeing the results of company surveys given as fractions: '7/8 people say they were satisfied with our customer service'. This is an example of a fraction in its simplest form. It could be that 184 people were surveyed and that 161 of them responded to say they were satisfied. Although they are equivalent fractions, you can see that the fraction '7/8 people' is a lot more user friendly than saying 161/184 people responded to say they were satisfied.
If you're asked to give an answer as a fraction in its simplest form, you firstly find a number that you can divide both parts of the fraction by, and divide it. For example, to simplify the fraction 12/18, you can divide both the top and the bottom number by 2, to give 6/9.
Keep going until you can't find a number that you can divide both parts of the fraction by. The fraction 6/9 can be divided again, this time by 3, to get 2/3. This is now the fraction in its simplest form.
Now let's simplify the fraction 40/120. Remember, find a number that you can divide both parts of the fraction by, and keep going until there is no longer a number that works. So, divide the top and bottom by 10. Divide the top and bottom by 2. Divide the top and bottom by 2. This is the fraction in its simplest form.
End transcript
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## Activity 14: Simplifying fractions
Show the following fractions in simplest form, where possible:
• a.
• b.
• c.
• d.
• e.
• f.
• a. =
• b. =
• c. =
• d. =
• e. can’t be simplified
• f. =
Next you’ll look at expressing a quantity of an amount as a fraction.
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# The Law of Sines
We will discuss here about the law of sines or the sine rule which is required for solving the problems on triangle.
In any triangle the sides of a triangle are proportional to the sines of the angles opposite to them.
That is in any triangle ABC,
$$\frac{a}{sin A}$$ = $$\frac{b}{sin B}$$ = $$\frac{c}{sin C}$$
Proof:
Let ABC be a triangle.
Now will derive the three different cases:
Case I: Acute angled triangle (three angles are acute): The triangle ABC is acute-angled.
Now, draw AD from A which is perpendicular to BC. Clearly, D lies on BC
Now from the triangle ABD, we have,
⇒ sin B = AD/c, [Since, AB = c]
⇒ AD= c sin B ……………………………………. (1)
Again from the triangle ACD we have,
⇒ sin C = AD/b, [Since, AC = b]
⇒ AD = b sin C ...………………………………….. (2)
Now, from (1) and (2) we get,
c sin B = b sin C
⇒ b/sin B = c/sin c………………………………….(3)
Similarly, if we draw a perpendicular to AC from B, we will get
a/sin A = c/sin c………………………………….(4)
Therefore, from (3) and (4) we get,
$$\frac{a}{sin A}$$ = $$\frac{b}{sin B}$$ = $$\frac{c}{sin C}$$
Case II: Obtuse angled triangle (one angle is obtuse): The triangle ABC is obtuse angled.
Now, draw AD from A which is perpendicular to produced BC. Clearly, D lies on produced BC.
Now from the triangle ABD, we have,
⇒ sin (180 - B) = AD/c, [Since ∠ABD = 180 - B and AB = c]
⇒ sin B = AD/c, [Since sin (180 - θ) = sin θ]
⇒ AD = c sin B ……………………………………. (5)
Again, from the triangle ACD, we have,
⇒ sin C = AD/b, [Since, AC = b]
⇒ AD = b sin C ……………………………………. (6)
Now, from (5) and (6) we get,
c sin B = b sin C
b/sin B = c/sin C ……………………………………. (7)
Similarly, if we draw a perpendicular to AC from B, we will get
a/sin A = b/sin B ……………………………………. (8)
Therefore, from (7) and (8) we get,
$$\frac{a}{sin A}$$ = $$\frac{b}{sin B}$$ = $$\frac{c}{sin C}$$
Case III: Right angled triangle (one angle is right angle): The triangle ABC is right angled. The angle C is a right angle.
Now from triangle ABC, we have,
sin C = sin π/2
⇒ sin C = 1, [Since, sin π/2 = 1], ……………………………………. (9)
sin A = BC/AB
⇒ sin A = a/c, [Since, BC = a and AB = c]
⇒ c = a/sin A ……………………………………. (10)
and sin B = AC/AB
⇒ sin B = b/c, [Since, AC = b and AB = c]
⇒ c = b/sin B ……………………………………. (11)
Now from (10) and (11) we get,
a/sin A = b/sin B = c
⇒ a/sin A = b/sin B = c/1
Now from (9) we get,
⇒ $$\frac{a}{sin A}$$ = $$\frac{b}{sin B}$$ = $$\frac{c}{sin C}$$
Therefore, from all three cases, we get,
$$\frac{a}{sin A}$$ = $$\frac{b}{sin B}$$ = $$\frac{c}{sin C}$$. Proved.
Note:
1. The sine rule or the law of sines can be expressed as
$$\frac{sin A}{a}$$ = $$\frac{sin B}{b}$$ = $$\frac{sin C}{c}$$
2. The sine rule or the law of sines is a very useful rule to express sides of a triangle in terms of the sines of angles and vice-versa in the following manner.
We have $$\frac{a}{sin A}$$ = $$\frac{b}{sin B}$$ = $$\frac{c}{sin C}$$ = k$$_{1}$$ (say)
⇒ a = k$$_{1}$$ sin A, b = k$$_{1}$$ sin B and c = k$$_{1}$$ sin C
Similarly, sin A/a = sin B/b = sin C/c = k$$_{2}$$ (say)
⇒ sin A = k$$_{2}$$ a, sin B = k$$_{2}$$ b and sin C = k$$_{2}$$ c
Solved problem using the law of sines:
The triangle ABC is isosceles; if ∠A = 108°, find the value of a : b.
Solution:
Since the triangle ABC is isosceles and A = 108°, A + B + C = 180°, hence it is evident that B = C.
Now, B + C = 180° - A = 180° - 108°
⇒ 2B = 72° [Since, C = B]
⇒ B = 36°
Again, we have, $$\frac{a}{sin A}$$ = $$\frac{b}{sin B}$$
Therefore, $$\frac{a}{b}$$ = $$\frac{sin A}{sin B}$$ = $$\frac{sin 108°}{sin 36°}$$ = $$\frac{cos 18°}{sin 36°}$$
Now, cos 18° = $$\sqrt{1 - sin^{2} 18°}$$
= $$\sqrt{1 - (\frac{\sqrt{5} - 1}{4})^{2}}$$
= ¼$$\sqrt{10 + 2\sqrt{5}}$$
and sin 36° = $$\sqrt{1 - cos^{2} 36°}$$
= $$\sqrt{1 - (\frac{\sqrt{5} + 1}{4})^{2}}$$
= ¼$$\sqrt{10 - 2\sqrt{5}}$$
Therefore, a/b = $$\frac{\frac{1}{4}\sqrt{10 + 2\sqrt{5}}}{\frac{1}{4}\sqrt{10 - 2\sqrt{5}}}$$
= $$\frac{\sqrt{10 + 2\sqrt{5}}}{\sqrt{10 - 2\sqrt{5}}}$$
= $$\sqrt{\frac{(10 + 2\sqrt{5})^{2}}{10^{2} - (2\sqrt{5})^{2}}}$$
= $$\frac{10 + 2\sqrt{5}}{\sqrt{80}}$$
⇒ $$\frac{a}{b}$$ = $$\frac{2√5(√5 + 1)}{4 √5}$$
$$\frac{a}{b}$$ = $$\frac{√5 + 1}{2}$$
Therefore, a : b = (√5 + 1) : 2
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## Recent Articles
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# 2005 AIME II Problems/Problem 14
## Problem
In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$
## Solution 1
$[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("A",A,N); label("B",B,SW); label("D",D,SE); label("E",E,S); label("C",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$
By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$, we have
\begin{align*} \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\ &= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\ &= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2} \end{align*}
Substituting our knowns, we have $\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}$. The answer is $q = \boxed{463}$.
## Solution 2 (Similar Triangles)
Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.
From here, we can use Heron's Formula to find the altitude. The area of the triangle is $\sqrt{21*6*7*8} = 84$. We can then use similar triangles with triangle $AQC$ and triangle $DSC$ to find $DS=\frac{24}{5}$. Consequently, from Pythagorean theorem, $SC = \frac{18}{5}$ and $AS = 14-SC = \frac{52}{5}$. We can also use the Pythagorean theorem on triangle $AQB$ to determine that $BQ = \frac{33}{5}$.
Label $AR$ as $y$ and $RE$ as $x$. $RB$ then equals $13-y$. Then, we have two similar triangles.
Firstly: $\triangle ARE \sim \triangle ASD$. From there, we have $\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}$.
Next: $\triangle BRE \sim \triangle BQA$. From there, we have $\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}$.
Solve the system to get $x = \frac{2184}{463}$ and $y = \frac{4732}{463}$. Notice that 463 is prime, so even though we use the Pythagorean theorem on $x$ and $13-y$, the denominator won't change. The answer we desire is $\boxed{463}$.
## Solution 3 (LoC and LoS bash)
Let $\angle CAD = \angle BAE = \theta$. Note by Law of Sines on $\triangle BEA$ we have $$\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}$$ As a result, our goal is to find $\sin{\angle BEA}$ and $\sin{\theta}$ (we already know $AB$).
Let the foot of the altitude from $A$ to $BC$ be $H$. By law of cosines on $\triangle ABC$ we have $$169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}$$ It follows that $AH = \frac{56}{5}$ and $HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}$.
Note that by PT on $\triangle AHD$ we have that $AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}$. By Law of Sines on $\triangle ADC$ (where we square everything to avoid taking the square root) we see $$\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.$$ How are we going to find $\sin{\angle BEA}$ though? $\angle BEA$ and $\theta$ are in the same triangle. Applying Law of Sines on $\triangle ABC$ we see that $$\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.$$ $\theta$, $\angle B$, and $\angle BEA$ are all in the same triangle. We know they add up to $180^{\circ}$. There's a good chance we can exploit this using the identity $\sin{p} = \sin{180^{\circ}-p}$.
We have that $\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}$. Success! We know $\sin{\theta}$ and $\sin{\angle B}$ already. Applying the $\sin$ addition formula we see $$\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.$$ This is the last stretch! Applying Law of Sines a final time on $\triangle BEA$ we see $$\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.$$ It follows that the answer is $\boxed{463}$.
## Solution 4 (Ratio Lemma and Angle Bisector Theorem)
Let $AK$ be the angle bisector of $\angle A$ such that $K$ is on $BC$.
Then $\angle KAB = \angle KAC$, and thus $\angle KAE = \angle KAD$.
By the Ratio Lemma, $\frac{BE}{KE} = \frac{BA}{KA} * \frac{\sin{BAE}}{\sin{KAE}}$ and $\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}$.
This implies that $\frac{BE}{KE*BA} = \frac{CD}{KD*CA}$.
Thus, $\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}$.
$DK = CK - 6 = 14*15/27 - 6 = 16/9$. Thus, $\frac{BE}{KE} = \frac{13*54}{14*16}$.
Additionally, $BE + KE = 9$. Solving gives that $q = 463.$
Alternate: By the ratio lemma, $BD/DC = (13/14)*(\sin BAD/\sin DAC)$ $EC/EB = (14/13)*(\sin EAC/\sin BAE)$
Combining these, we get $(BD/DC)(14/13) = (EC/EB)(13/14)$ $(3/2)(14/13)(14/13) = (15-x)(x)$
$x = 2535/463$ Thus, $q = 463$
## Solution 5 (Isogonal lines with respect to A angle bisesector)
Since $AE$ and $AD$ are isogonal with respect to the $A$ angle bisector, we have $$\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.$$ To prove this, let $\angle BAE=\angle DAC=x$ and $\angle BAD=\angle CAE=y.$ Then, by the Ratio Lemma, we have $$\frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}$$ $$\frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}$$ and multiplying these together proves the formula for isogonal lines. Hence, we have $$\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}$$ so our desired answer is $\boxed{463}.$
## Solution 6 (Tangent subtraction formulas)
Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle.
$[asy] import olympiad; import cse5; import geometry; size(300); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("A",A,N); label("B",B,SW); label("D",D,SE); label("E",E,S); label("C",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); pair G = foot(E,C,A); pair F = foot(D,C,A); draw(D--F); draw(E--G); label("G",G,N); label("F",F,N); [/asy]$
Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling $\angle DAG = \alpha$. Now we know that $\overline{DG} = \frac{24}{5}$ and $\overline{GC} = \frac{18}{5}$. Therefore, $\overline{AG} = \frac{52}{5}$, so $\tan{(\alpha)} = \frac{6}{13}$. Our goal now is to use tangent $\angle EAG$ in triangle $AEG$. We set $\overline{BE}$ to $x$, so $\overline{ED} = 9 - x$ and $\overline{EC} = 15 - x$, so $\overline{EG} = \frac{4}{5}(15-x)$ and $\overline{GC} = \frac{3}{5}(15-x)$ so $\overline{AG} = \frac{3x+25}{5}$. Now we just need tangent of $\angle EAG$.
We find this using $\tan{(EAG)} = \tan{(A - \alpha)} = \frac{\tan{A} - \tan{\alpha}}{1 + \tan{A}\tan{\alpha}}$, which is $\frac{\frac{12}{5} - \frac{6}{13}}{1 + \frac{12}{5}\frac{6}{13}}$ or $\frac{126}{137}$. Now we solve the equation $\tan{\angle EAG} = \frac{126}{137} = \frac{\frac{60-4x}{5}}{\frac{3x+25}{5}}$, so $x = \frac{2535}{463}$
## Solution 7 (Super fast solution, 2 billion IQ)
Let $\angle BAE = \angle CAD = \theta, \angle EAD = \alpha, BE = x \rightarrow ED = 9-x$
Via ratio lemma, we have $\frac{BD}{DC} = \frac{AB}{AC} \cdot \frac{\sin (\theta + \alpha)}{\sin (\theta)}$ and $\frac{BE}{EC} = \frac{AB}{AC} \cdot \frac{\sin (\theta)}{\sin (\theta + \alpha)}$
Multiplying the two equations, we have $\left (\frac{AB}{AC} \right)^2 = \frac{BD \cdot BE}{DC \cdot EC}$. Plugging in the values we know, we have $\frac{169}{196} = \frac{3x}{30-2x}$. Solving for $x$, we get $x = \frac{2535}{463} \rightarrow \boxed{463}$
Solution by hiker.
## Solution 8(literally 2 minute solution)
Let $ED = x$, such that $BE = 9-x$. Since $\overline{AE}$ and $\overline{AD}$ are isogonal, we get $\frac{9-x}{6+x} \cdot \frac{9}{6} = \frac{13^2}{14^2} \Rightarrow 588(9 - x) = 338(6 + x)$, and we can solve to get $x = \frac{1632}{463}$(and $BE = \frac{1146}{463}$). Hence, our answer is $\boxed{463}$. - Spacesam
## Solution 9 (Long but no IQ Required Altogether Bash)
Diagram borrowed from Solution 1.
$[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("A",A,N); label("B",B,SW); label("D",D,SE); label("E",E,S); label("C",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$
Applying Law of Cosines on $\bigtriangleup ABC$ with respect to $\angle C$ we have $$AB^2=AC^2+BC^2-2(AC)(BC)\cos C$$ Solving gets $\cos C=\frac{3}{5}$, which implies that $$\sin C=\sqrt{1-\cos C}=\frac{4}{5}$$ Applying Stewart's Theorem with cevian $AD$ we have $$(BC)(AD)^2+(BC)(BD)(CD)=(CD)(AB)^2+(BD)(AC)^2$$ Solving gets $AD=\frac{4\sqrt{205}}{5}$.
Applying Law of Sines on $\bigtriangleup ACD$ to solve for $\sin CAD$ we have $$\frac{AD}{\sin C}=\frac{CD}{\sin CAD}$$ Solving gets $\sin CAD=\frac{6\sqrt{205}}{205}$. Thus $\sin BAE=\sin CAD=\frac{6\sqrt{205}}{205}$.
Applying Law of Sines on $\bigtriangleup ABC$ we have $$\frac{AC}{\sin B}=\frac{AB}{\sin C}$$ Solving gets $\sin B=\frac{56}{65}$.
Applying Stewart's Theorem with cevian $AE$ we have $$(BC)(AE)^2+(BC)(BE)(CE)=(CE)(AB)^2+(BE)(AC)^2$$ $$(BC)(AE)^2+(BC)(BE)(BC-BE)=(BC-BE)(AB)^2+(BE)(AC)^2$$ Solving gets $AE=\sqrt{\frac{15BE^2-198BE+2535}{15}}$
Finally, applying Law of Sines on $\bigtriangleup BAE$ we have $$\frac{AE}{\sin B}=\frac{BE}{\sin BAE}$$ $$\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}$$ $$7605BE^2-32342BE+2535=0$$ Solving
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## 3
P (1 + 2i)
Pi
P (1+ 2i) + Pi = P (1 + 3i)
n
P (1 + (n-1)i)
Pi
P (1+ (n-1)i) + Pi = P (1 + ni)
earns P*i dollars (\$8) of interest since under simple compounding, interest is paid only on the original principal amount P (\$100). Thus at the end of year 2, the balance in the account is obtained by adding P dollars (the original principal) plus P*i (the interest from year 1) plus P*i (the interest from year 2) to obtain P+P*i+P*i (\$100+\$8+\$8=\$116). After some algebraic manipulation, this can be written conveniently mathematically as P*(1+2*i) dollars (\$100*1.16=\$116).
Table 4.4 extends the above logic to year 3 and then generalizes the approach for year n. If we return our attention to our original goal of developing a formula for Fn which is expressed only in terms of the present amount P, the annual interest rate i, and the number of years n, the above development and Table 4.4 results can be summarized as follows:
Example 3
Determine the balance which will accumulate at the end of year 4 in an account which pays 10%/yr simple interest if a deposit of \$500 is made today.
Fn = P * (1 + n*i) F4 = 500 * (1 + 4*0.10) F4 = 500 * (1 + 0.40) F4 = 500 * (1.40) F4 = \$700
### 4.6.4 Compound Interest
For compound interest, interest is earned (charged) on the original principal amount plus any accumulated interest from previous years at the rate of i% per year (i%/ yr). Table 4.5 illustrates the annual calculation of compound interest. In the Table 4.5 and the formulas which follow, i is expressed as a decimal amount (i.e., 8% interest is expressed as 0.08).
At the beginning of year 1 (end of year 0), P dollars (e.g., \$100) are deposited in an account earning i%/yr (e.g., 8%/yr or 0.08) compound interest. Under compound interest, during year 1 the P dollars (\$100) earn P*i dollars (\$100*0.08 = \$8) of interest. Notice that this the same as the amount earned under simple compounding. This result is expected since the interest earned in previous years is zero for year 1. At the end of the year 1 the balance in the account is obtain by adding P dollars (the original principal, \$100) plus P*i (the interest earned during year 1, \$8) to obtain P+P*i (\$100+\$8=\$108). Through algebraic manipulation, the end of year 1 balance can be expressed mathematically as P*(1+i) dollars (\$100*1.08=\$108).
During year 2 and subsequent years, we begin to see the power (if you are a lender) or penalty (if you are a borrower) of compound interest over simple interest. The beginning of year 2 is the same point in time as the end of year 1 so the balance in the account is P*(1+i) dollars (\$108). During year 2 the account earns i% interest on the original principal, P dollars (\$100), and it earns i% interest on the accumulated interest from year 1, P*i dollars (\$8). Thus the interest earned in year 2 is [P+P*i]*i dollars ([\$100+\$8]*0.08=\$8.64). The balance at the end of year 2 is obtained by adding P dollars (the original principal) plus P*i (the interest from year 1) plus [P+P*i]*i (the interest from year 2) to obtain P+P*i+[P+P*i]*i dollars (\$100+\$8+\$8.64=\$116.64). After some algebraic ma
Table 4.5 The Mathematics of Compound Interest
Year (t)
Amount At Beginning Of Year
Interest Earned During Year
Amount At End Of Year
(Ft)
0 0
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Square Root
# What Is The Square Root of 4? How To Find The Square Root of 4?
Written by Prerit Jain
Updated on: 17 Aug 2023
### What Is The Square Root of 4? How To Find The Square Root of 4?
The square root of 4 is 2. The inversion of subtraction is addition, and the inversion of multiplication is division in the same way the inversion of the square root of a value is squaring the number. The value of any number that when multiplied by itself gives the original number. The number 2 when they are squared 22 gives 4, and when the number is multiplied by itself 2*2 it gives the original value 4.
For any given number before finding its square root, we must know whether the given number is a perfect square number or not. A perfect square number is one that when squared has a whole number and was easy to find. A non-perfect square number when squared has only integers and not a whole number and is found using the long division method.
## Method to find the square root of 4
• Prime factorization method
• Repeated subtraction method
• Long division method
### Prime factorization method
Prime factorization is one of the easiest methods to find the square root of any number. Let us take n as the prime number, by grouping the similar numbers together we get n2, by multiplying the similar n2. The value we get is the square root of the number.
The prime factor of 4 = 2*2
By squaring them we get 22, since two is the only similar square we take the common one.
Here, the common square is 2.
Hence, the √4 = 2
### Repeated subtraction method
For the repeated subtraction method we should start to multiply the given number with only odd numbers. The step in which zero is obtained is the square root of the given number.
4 – 1 = 3
3 – 3 = 0
Here, the value zero is obtained in the second step.
Hence, the √4 = 2
### Long division method
The divisor for the perfect square is easy to find. The number when multiplied twice should give the value of the original number. In this case 2 when multiplied gives 4. Hence 2 is used as a divisor.
Hence, the √4 = 2
## Solved examples
Example 1: What is the sum of the square of 4 and the square root of 4?
Square of 4= 16
The square root of 4 = 2
The sum of the square of 4 and the square root of 4 = 16+2 = 18
Example 2: Joe wants to find the square root value of 81/9 help him solve the sum
The 9 = 3.
The 81 = 9
Therefore, the 4/81 = 2/9.
Example 3: Solve 44 + 24
The 4 = 2.
Hence 4 (2) + 2 (2).
= 8+4
=12.
Example 4: what is the area of the square whose each side is equal to 2cm
Area of the square = x2
Here x= 2
Therefore, 22 = 2 2 = 4
Hence the area of the square with 2cm = 4 sq. cm
What is the square root of 4?
The square root of 4 = 2.
Explain why the square root of 4 is a rational number.
When finding the prime factor of 4 we get 22 which is entirely an even power and hence all the numbers are positive integers which makes them rational numbers.
What is the value of 4 and the square root of 4?
The square root of 4 is 2, therefore 44 = 4*2 = 8
What is the real value for the square root of 4?
The actual value for the 4 = 2
What is the square root of 25?
The square root of 25 = 5
What is the square root of (-4)?
The square root of minus digits is imaginary numbers or imaginary units which really don’t have a square root.
Is 4 a perfect square number?
Yes, 4 is a perfect square number.
Written by by
Prerit Jain
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# 8. Equations Involving Fractions
In this section, we can find the solution easily by multiplying throughout by the lowest common denominator (LCD). This will simplify our equation and make it easier to solve.
We need to remember to multiply all terms in the equation (both sides of the equal sign) by the LCD, otherwise the final answer won't be correct.
## a. Algebraic types
### Example 1
Solve for x:
x/5+3/10 = 1/2
We first look at the denominators of the fractions and determine the lowest common denominator. In this case, it will be 10.
We multiply throughout by 10 and the result is:
10xx(x/5+3/10) = 10xx(1/2)
2x+3=5
Then, we just solve the simpler equation we've found.
Subtract 3 from both sides:
2x=2
Divide boths sides by 2, giving us the final answer:
x=1
### Example 2
Solve for x:
(2x)/3+2/5 = 8+x/2
Once again, we look at the denominators of the fractions and determine the lowest common denominator. In this case, it will be 30.
We multiply throughout by 30 and the result is:
30xx((2x)/3+2/5) = 30xx(8+x/2)
20x+12=240+15x
Subtracting 12 from both sides:
20x=228+15x
Substract 15x from both sides:
5x=228
Divide both sides by 5, giving us the final answer:
x=228/5=45.6
### Example 3
This next one has the variable in the denominator. We'll still need to find the LCD as before.
Solve for x:
4/x+1/3 = 7/(5x)
In this case, the lowest common denominator is 15x.
We multiply throughout by 15x and the result is:
15x xx(4/x+1/3) = 15x xx(7/(5x))
60+5x=21
Subtracting 60 from both sides:
5x=-39
Divide both sides by 5, giving us the final answer:
x=-39/5=-7.8
Continues below
## b. Word problems
### Example 4
An aquarium can be filled by one hose in 7 minutes and a second thinner hose in 10 minutes.
How long will it take to fill the tank if both hoses operate together?
Since each hose makes the filling time less, we have to add the reciprocals together and take the reciprocal of the result.
We need to use:
1/T=1/T_1+1/T_2
So we have:
1/T=1/7+1/10
1/T=(10+7)/70=17/70
So
T=70/17=4.1176
So it will take 4.1 minutes to fill the tank with both hoses operating together.
### Related pages
You may find these useful:
For 2 resistors with resistances R1and R2 in parallel, the combined resistance R is given by:
1/R=1/R_1+1/R_2
For a particular circuit, the combined resistance R was found to be 4 ohms (Ω), and R1= 10 Ω. Find R2.
We have:
1/4=1/10+1/R_2
We need to multiply throughout by the lowest common denominator: 20R2
1/4=1/10+1/R_2
(20R_2)/4=(20R_2)/10+(20R_2)/(R_2)
5R_2=2R_2+20
3R_2=20
R_2=20/3=6 2/3Omega
"Omega" is the symbol for "ohms", the unit of resistance.
### Example 6
A car averaged 30 km/h going from home to work and 40 km/h on the return journey. If the total time for the two journeys is 50 minutes, how far is it from home to work?
Let the length of the journey from home to work be x km.
Recall that
text(speed) = text(distance)/text(time)
So
text(time) = text(distance)/text(speed)
We must use the same time units throughout. We will use hours.
Now
50\ text(minutes)=50/60=5/6text(hours)
In the forward journey, the car's time was x/30 hours.
For the return journey, the time was x/40 hours.
The total time was x/30+x/40=5/6\ text(hours)
So (4x+3x)/120=(7x)/120=5/6
This gives us 7x=(5xx120)/6=100
That is x=100/7=14.286
So the distance from home to work is 14.3 km.
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Mathematics Grade 4
Educational Links
Strand: NUMBER AND OPERATIONS IN BASE TEN (4.NBT)
Generalize place value understanding for multi-digit whole numbers by analyzing patterns, writing whole numbers in a variety of ways, making comparisons, and rounding (Standards 4.NBT.1–3). Use place value understanding and properties of operations to perform multidigit addition, subtraction, multiplication, and division using a one-digit divisor (Standards 4.NBT.4–6). Expectations in this strand are limited to whole numbers less than or equal to 1,000,000.
• Arithmetic Four
In this game for two players, students solve problems involving whole number and integer addition, subtraction, multiplication, and division.
• Clock Arithmetic
In this activity students experiment with clock arithmetic and use long division skills.
• Grade 4 Mathematics Module 1: Place Value, Rounding, and Algorithms for Addition and Subtraction
In this 25-day module of Grade 4, students extend their work with whole numbers. They begin with large numbers using familiar units (hundreds and thousands) and develop their understanding of millions by building knowledge of the pattern of times ten in the base ten system on the place value chart (4.NBT.1). They recognize that each sequence of three digits is read as hundreds, tens, and ones followed by the naming of the corresponding base thousand unit (thousand, million, billion).
• IXL Game: Mixed operations: Addition, subtraction, multiplication, and division word problems
This game helps fourth graders understand how to read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. This is just one of many online games that supports the Utah Math core. Note: The IXL site requires subscription for unlimited use.
• Mental Division Strategy
The task presents the scenario where Jillian says "I know that 20 times 7 is 140 and if I take away 2 sevens that leaves 126. So 126 ÷ 7 = 18." Students must then respond to this statement by answering if she's correct, drawing a picture showing her reasoning and using that method to solve another problem.
• Order of Operations Four
This applet pits two players against each other in solving expressions using the four basic operations.
• Ordering 4-digit Numbers
This task includes three-digit numbers with large hundreds digits and four-digit numbers with small thousands digits so that students must infer the presence of a 0 in the thousands place in order to compare. It also includes numbers with strategically placed zeros and an unusual request to order them from greatest to least in addition to the more traditional least to greatest.
• Reasoning About Division
The Common Core allows students to get a better understanding of Math concepts and skills, including division. This Teaching Channel video shows how you can teach reasoning about division in your classroom. (7 min.)
• Rounding to the Nearest 100 and 1000
Part (a) of this task fits squarely within third grade when students "use place value understanding to round whole numbers to the nearest 10 or 100". Part (b) is a first step in rounding beyond tens and hundreds.
• Rounding to the Nearest 1000
The purpose of this task is for students to estimate the position of numbers less than 10,000 on the number line, to practice rounding, and to make the connection between rounding and location on the number line.
• Sample Assessment Task: Numbers of Stadium Seats
This sample three-part assessment task calls for students to demonstrate reasoning skills and a deep conceptual knowledge of place value in atypical ways. Use the navigation at the upper right of this page to access the task.
• Spy Game
Encrypted spy messages are the starting point for this lesson plan on modular arithmetic.
• Thousands and Millions of Fourth Graders
Students are given facts about the number of fourth graders in different places and asked to make calculations comparing the numbers. The purpose of this task is to help students understand the multiplicative relationship between commonly used large numbers (thousands and millions) by using their understanding of place value.
• Threatened and Endangered
Given facts about two endangered and threatened animals students are asked a question requiring them to understand multiplying and dividing whole numbers by powers of 10.
• To Regroup or Not to Regroup
This task presents an incomplete problem and asks students to choose numbers to subtract (subtrahends) so that the resulting problem requires different types of regrouping.
• USBE Core Guide (4NBT1)
This core guide was developed by USBE and Utah educators.
• USBE Core Guide (4NBT3)
This core guide was developed by USBE and Utah educators.
• USBE Core Guide (4NBT6)
This core guide was developed by USBE and Utah educators.
• USBE Core Guide(4NBT2)
This core guide was developed by USBE and Utah educators.
• What's My Number?
The purpose of this task is for students to reason about base-ten numbers on the number line in a way that will require them to use what they know about place value.
http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialist - Shannon Ference and see the Mathematics - Elementary website. For general questions about Utah's Core Standards contact the Director - DIANA SUDDRETH .
Email: diana.suddreth@schools.utah.gov
These materials have been produced by and for the teachers of the State of Utah. Copies of these materials may be freely reproduced for teacher and classroom use. When distributing these materials, credit should be given to Utah State Board of Education. These materials may not be published, in whole or part, or in any other format, without the written permission of the Utah State Board of Education, 250 East 500 South, PO Box 144200, Salt Lake City, Utah 84114-4200.
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# Calculate derivative of function y=2x^2-10x+13 +2(x^2-5x+6) using 2 methods.
justaguide | Certified Educator
You cannot ask more than one question at a time. So I am only providing one basic method.
We need to find the derivative of the function y=2x^2-10x+13 +2(x^2-5x+6).
y= 2x^2-10x+13 +2(x^2-5x+6)
=> y = 2x^2 - 10x + 13 + 2x^2 - 10x + 12
=> y = 4x^2 - 20x + 25
If f(x) = x^n, f'(x) = n*x^(n - 1)
y' = 2*4x - 20
=> y' = 8x - 20
The required derivative of the function is y' = 8x - 20
giorgiana1976 | Student
Either we can differentiate each term of the sum, or we can notice that the expression is a complete square and we'll differentiate the complete square.
We'll re-write the expression of the function to emphasize the fact that it represents a complet square.
y=2x^2-10x+13 + 2(x^2-5x+6)
y = (x^2 - 4x + 4) + (x^2 - 6x + 9) + 2(x^2-5x+6)
y = (x-2)^2 + (x-3)^2 + 2(x^2-5x+6)
If we'll put (x-2) as a and (x-3) as b, and we'll re-write the expresison, we'll get a perfect square:
y=a^2 + 2ab + b^2
y = (a+b)^2
y = (x-2+x-3)^2
We'll combine like terms:
y = (2x-5)^2
Now, we'll differentiate both sides, with respect to x, using chain rule:
dy/dx = 2(2x-5)*(2x-5)'
dy/dx = 2(2x-5)*2
dy/dx = 4(2x-5)
We'll remove the brackets:
dy/dx = 8x - 20
The other method is to differentiate each term of the sum, with respect to x.
dy/dx = d(x-2)^2/dx + 2d[(x-2)(x-3)]/dx + d(x-3)^2/dx
dy/dx = 2(x-2)+ 2d(x^2-5x+6)/dx + 2(x-3)
dy/dx = 2x - 4 + 2(2x-5) + 2x -6
dy/dx = 4x -10 + 4x -10
dy/dx = 8x - 20
Both methods yields the same result: dy/dx = 8x - 20.
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# Lesson 6
Estimating Probabilities Using Simulation
Let’s simulate real-world situations.
### 6.1: Which One Doesn’t Belong: Spinners
Which spinner doesn't belong?
### 6.2: Diego’s Walk
Your teacher will give your group the supplies for one of the three different simulations. Follow these instructions to simulate 15 days of Diego’s walk. The first 3 days have been done for you.
• Simulate one day:
• If your group gets a bag of papers, reach into the bag, and select one paper without looking inside.
• If your group gets a spinner, spin the spinner, and see where it stops.
• If your group gets two number cubes, roll both cubes, and add the numbers that land face up. A sum of 2–8 means Diego has to wait.
• Record in the table whether or not Diego had to wait more than 1 minute.
• Calculate the total number of days and the cumulative fraction of days that Diego has had to wait so far.
• On the graph, plot the number of days and the fraction that Diego has had to wait. Connect each point by a line.
• If your group has the bag of papers, put the paper back into the bag, and shake the bag to mix up the papers.
• Pass the supplies to the next person in the group.
day Does Diego have
to wait more
than 1 minute?
total number
of days Diego
fraction
of days Diego
1 no 0 $$\frac{0}{1} =$$ 0.00
2 yes 1 $$\frac{1}{2} =$$ 0.50
3 yes 2 $$\frac{2}{3} \approx$$ 0.67
4
5
6
7
8
9
10
11
12
13
14
15
1. Based on the data you have collected, do you think the fraction of days Diego has to wait after the 16th day will be closer to 0.9 or 0.7? Explain or show your reasoning.
2. Continue the simulation for 10 more days. Record your results in this table and on the graph from earlier.
day Does Diego have
to wait more
than 1 minute?
total number
of days Diego
fraction
of days Diego
16
17
18
19
20
21
22
23
24
25
3. What do you notice about the graph?
4. Based on the graph, estimate the probability that Diego will have to wait more than 1 minute to cross the crosswalk.
Let's look at why the values tend to not change much after doing the simulation many times.
1. After doing the simulation 4 times, a group finds that Diego had to wait 3 times. What is an estimate for the probability Diego has to wait based on these results?
1. If this group does the simulation 1 more time, what are the two possible outcomes for the fifth simulation?
2. For each possibility, estimate the probability Diego has to wait.
3. What are the differences between the possible estimates after 5 simulations and the estimate after 4 simulations?
2. After doing the simulation 20 times, this group finds that Diego had to wait 15 times. What is an estimate for the probability Diego has to wait based on these results?
1. If this group does the simulation 1 more time, what are the two possible outcomes for the twenty-first simulation?
2. For each possibility, estimate the probability Diego has to wait.
3. What are the differences between the possible estimates after 21 simulations and the estimate after 20 simulations?
3. Use these results to explain why a single result after many simulations does not affect the estimate as much as a single result after only a few simulations.
### 6.3: Designing Experiments
For each situation, describe a chance experiment that would fairly represent it.
1. Six people are going out to lunch together. One of them will be selected at random to choose which restaurant to go to. Who gets to choose?
2. After a robot stands up, it is equally likely to step forward with its left foot or its right foot. Which foot will it use for its first step?
3. In a computer game, there are three tunnels. Each time the level loads, the computer randomly selects one of the tunnels to lead to the castle. Which tunnel is it?
4. Your school is taking 4 buses of students on a field trip. Will you be assigned to the same bus that your math teacher is riding on?
### Summary
Sometimes it is easier to estimate a probability by doing a simulation. A simulation is an experiment that approximates a situation in the real world. Simulations are useful when it is hard or time-consuming to gather enough information to estimate the probability of some event.
For example, imagine Andre has to transfer from one bus to another on the way to his music lesson. Most of the time he makes the transfer just fine, but sometimes the first bus is late and he misses the second bus. We could set up a simulation with slips of paper in a bag. Each paper is marked with a time when the first bus arrives at the transfer point. We select slips at random from the bag. After many trials, we calculate the fraction of the times that he missed the bus to estimate the probability that he will miss the bus on a given day.
### Glossary Entries
• chance experiment
A chance experiment is something you can do over and over again, and you don’t know what will happen each time.
For example, each time you spin the spinner, it could land on red, yellow, blue, or green.
• event
An event is a set of one or more outcomes in a chance experiment. For example, if we roll a number cube, there are six possible outcomes.
Examples of events are “rolling a number less than 3,” “rolling an even number,” or “rolling a 5.”
• outcome
An outcome of a chance experiment is one of the things that can happen when you do the experiment. For example, the possible outcomes of tossing a coin are heads and tails.
• probability
The probability of an event is a number that tells how likely it is to happen. A probability of 1 means the event will always happen. A probability of 0 means the event will never happen.
For example, the probability of selecting a moon block at random from this bag is $$\frac45$$.
• random
Outcomes of a chance experiment are random if they are all equally likely to happen.
• sample space
The sample space is the list of every possible outcome for a chance experiment.
For example, the sample space for tossing two coins is:
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# Trigonometry Identity proving
If $\sin(x-y) =\cos y$ prove that $\tan y = \frac{1+ \sin y}{\cos y}$.
Is there an error with the question? I don't seem to be able to get the answer. Should it be $\tan x$ instead of $\tan y = \frac{1+\sin y}{\cos y}$ ?
$$\tan x\times \cos y = 1 + \sin y$$ $$\frac{\sin x}{\cos x} \times \cos y = \frac{\sin x\times\cos y}{\cos x}$$ $$\frac{\sin x\times\cos y}{\cos x}= 1 + \sin y$$ $$\frac{\sin x\times\cos y}{\sin(x-y)}= 1 + \sin y$$ I am stuck at this step.
• That should be x on the RHS instead of y. – MonK May 28 '15 at 9:29
$$\cos y=\sin(x-y)=\sin x\cos y-\cos x\sin y$$
$$\iff\cos x\sin y=\cos y(\sin x-1)\implies\dfrac{\sin y}{\cos y}=\dfrac{\sin x-1}{\cos x}$$
Now $\cos^2x=(1-\sin x)(1+\sin x)\implies\dfrac{1-\sin x}{\cos x}=\dfrac{\cos x}{1+\sin x}$
Since $\sin(x-y)= \cos(y)$, expanding it gives \begin{equation*} \sin (x) \cos(y)-\cos(x) \sin(y)=\cos(y). \end{equation*} Dividing both sides by $\cos(y)$ we have \begin{equation*} \sin(x)-\cos(x)\tan(y)=1. \end{equation*} So from here, we get
• To obtain $\sin x$, $\cos x$, and $\tan x$, type \sin x, \cos x, and \tan x, respectively, in math mode. – N. F. Taussig May 28 '15 at 10:59
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# Graph Transformations
A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022
## Graph Transformations
You should have seen some graph transformations before, such as translations and reflections – recall that reflections in the $x$-axis flip $f(x)$ vertically and reflections in the $y$-axis flip $f(x)$ horizontally. Here, we will also look at stretches.
There are 4 main types of graph transformation that we will cover. Each transformation has the same effect on all functions.
Make sure you are happy with the following topics before continuing.
A Level
## Type 1: $y = f(x+k)$
For the transformation $y=f(x+k)$, for $k>0$:
• $f(x+k)$ is $f(x)$ moved $k$ to the left
• $f(x-k)$ is $f(x)$ moved $k$ to the right
In this example, we have $f(x) = x^2 - 4$ and $y=f(x+2)$
So, subtract $2$ from the $x$-coordinates of $f(x)$ to get $y=f(x+2)$
A Level
## Type 2: $y = f(x)+k$
For the transformation $y=f(x)+k$, for $k>0$:
• $f(x)+k$ is $f(x)$ moved $k$ upwards
• $f(x)-k$ is $f(x)$ moved $k$ downwards
In this example, we have $f(x) = x^2 - 4$ and $y=f(x)+3$
So, add $3$ to the $y$-coordinates of $f(x)$ to get $y=f(x)+3$
A Level
## Type 3: $y = af(x)$
For the transformation $y=af(x)$:
• If $|a|>1$ then $af(x)$ is $f(x)$ stretched vertically by a factor of $a$
• If $0<|a|<1$ then $f(x)$ is squashed vertically
• If $a<0$ then is $f(x)$ also reflected in the $x$-axis
In this example, we have $f(x) = x^2 - 4$ and $y=2f(x)$
This is a stretch vertically, so multiply the $y$-coordinates of $f(x)$ by $2$ to get $y=2f(x)$
A Level
## Type 4: $y = f(ax)$
For the transformation $y=f(ax)$:
• If $|a|>1$ then $f(ax)$ is $f(x)$ squashed horizontally by a factor of $a$
• If $0<|a|<1$ then $f(x)$ is stretched horizontally
• If $a<0$ then is $f(x)$ also reflected in the $y$-axis
In this example, we have $f(x) = x^2 - 4$ and $y=f(2x)$
This is a squash horizontally, so divide the $x$-coordinates of $f(x)$ by $2$ (or multiply by $\dfrac{1}{2}$) to get $y=f(2x)$
A Level
## Note:
• For these transformations, any asymptotes need to be moved correspondingly.
• A squash by a factor of $a$ is equivalent to a stretch by a factor of $\dfrac{1}{a}$
• When drawing graph transformation, only a sketch including important points is necessary.
## Combinations of Transformations
For combinations of transformations, it is easy to break them up and do them one step at a time (do the bit in the brackets first). You can sketch the graph at each step to help you visualise the whole transformation.
e.g. for $f(x) = x^2 - 4$ and $y=2f(x+2)$, draw the graph of
$y=f(x+2)$ first, and then use this graph to draw the graph of
$y=2f(x+2)$
Note: These transformations can also be combined with modulus functions.
A Level
## Example Questions
$y = f(4x)$ means that the graph of $f(x)$ is squashed horizontally by a factor of $4$.
Hence, the graph will look like:
Firstly, since the coefficient before $f(x)$ is negative, we need to reflect $f(x)$ in the $x$-axis.
The coefficient of $\dfrac{1}{2}$ before $-f(x)$ means that the graph of $-f(x)$ is squashed vertically by a factor of $2$.
Hence, the graph will look like:
Split the transformation up into $2$ parts – firstly sketch $y=3f(x)$ which is a stretch vertically by a scale factor of $3$ (multiply the $y$-coordinates by $3$:
Then, do the second transformation – $y=3f(x)-1$ means that we need to move the graph down by $1$ (subtract $1$ from the $y$-coordinates):
A Level
A Level
A Level
A Level
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Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Table of 429
The Table of 429 provides the results gained by repeated addition of number 429 for a certain number of times. Multiplication tables are the basics in Mathematics. Solving the problems with the help of multiplication tables boosts problem-solving and time management skills, which are essential from an exam point of view. Get the PDF of table 429 at BYJU’S, for a better academic performance.
## What is the 429 Times Table?
The repeated addition of the number 429 represents the multiplication table of 429. The table of 429 is given here for ease of learning. Students can learn the table of 429 to solve the problems in a quick way.
429×1 = 429 429 429×2 = 858 429 + 429 = 858 429×3 = 1287 429 + 429 + 429 = 1287 429×4 = 1716 429 + 429 + 429 + 429 = 1716 429×5 = 2145 429 + 429 + 429 + 429 + 429 = 2145 429×6 = 2574 429 + 429 + 429 + 429 + 429 + 429 = 2574 429×7 = 3003 429 + 429 + 429 + 429 + 429 + 429 + 429 = 3003 429×8 = 3432 429 + 429 + 429 + 429 + 429 + 429 + 429 + 429 = 3432 429×9 = 3861 429 + 429 + 429 + 429 + 429 + 429 + 429 + 429 + 429 = 3861 429×10 = 4290 429 + 429 + 429 + 429 + 429 + 429 + 429 + 429 + 429 + 429 = 4290
## Multiplication Table of 429
A multiplication table of 429 is given here for students who are in search of 429 multiples. This table will help students to solve the problems quickly.
429 × 1 = 429 429 × 2 = 858 429 × 3 = 1287 429 × 4 = 1716 429 × 5 = 2145 429 × 6 = 2574 429 × 7 = 3003 429 × 8 = 3432 429 × 9 = 3861 429 × 10 = 4290 429 × 11 = 4719 429 × 12 = 5148 429 × 13 = 5577 429 × 14 = 6006 429 × 15 = 6435 429 × 16 = 6864 429 × 17 = 7293 429 × 18 = 7722 429 × 19 = 8151 429 × 20 = 8580
## Solved Example on the Table of 429
Example 1:
Find the value of 429 times 15 plus 50.
Solution:
429 times 15 plus 50
= 429 x 15 + 50
= 6435 + 50
= 6485
## Frequently Asked Questions on Table of 429
### What are the prime factors of 429?
3, 11, 13 are the prime factors of 429.
### Evaluate 429 times 11 minus 50 plus 70.
429 times 11 minus 50 plus 70
= 429 x 11 – 50 + 70
= 4719 – 50 + 70
= 4739
### List the first five multiples of 429.
The first five multiples of 429 are 429, 858, 1287, 1716, and 2145.
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# L'Hôpital's Rule
(Redirected from L'Hopital's Rule)
L'Hopital's Rule is a theorem dealing with limits that is very important to calculus.
## Theorem
The theorem states that for real functions $f(x),g(x)$, if $\lim f(x),g(x)\in \{0,\pm \infty\}$ $$\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$$ Note that this implies that $$\lim\frac{f(x)}{g(x)}=\lim\frac{f^{(n)}(x)}{g^{(n)}(x)}=\lim\frac{f^{(-n)}(x)}{g^{(-n)}(x)}$$
## Proof
One can prove using linear approximation: The definition of a derivative is $f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$ which can be rewritten as $f'(x) = \frac{f(x+h)-f(x)}{h} + \eta(h)$. Just so all of us know $\eta(h)$ is a function that is both continuous and has a limit of $0$ as the $h$ in the derivative function approaches $0$. After multiplying the equation above by $h$, we get $f(x+h) = f(x) +f'(x)h+h\cdot \eta(h)$.
We have already assumed by the hypothesis that the derivative equals zero. Hence, we can rewrite the function as $\frac{f(x_0+h)}{g(x_0+h)}=\frac{f'(x_0)h + h\cdot \eta(h)}{g'(x_0)h+h\cdot \epsilon(h)}$, which would hence prove our lemma for L'Hospital's rule.
Text explanation:
Let $z(x) = \frac{f(x)}{g(x)}$, where $f(x)$ and $g(x)$ are both nonzero functions with value $0$ at $x = a$.
(For example, $g(x) = \cos\left(\frac{\pi}{2} x\right)$, $f(x) = 1-x$, and $a = 1$.)
Note that the points surrounding $z(a)$ aren't approaching infinity, as a function like $f(x) = 1/x-1$ might at $f(a)$.
The points infinitely close to $z(a)$ will be equal to $\lim_{b\to 0} \frac{f(a+b)}{g(a+b)}$.
Note that $\lim_{b\to 0} f(a+b)$ and $\lim_{b\to 0} g(a+b)$ are equal to $f'(a)$ and $g'(a)$.
As a recap, this means that the points approaching $\frac{f(a)}{g(a)}$, where $a$ is a number such that $f(a)$ and $g(a)$ are both equal to $0$, are going to approach $\frac{f'(x)}{g'(x)}$.
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In geometry, a triangle is a kind of two-dimensional polygon, which has actually three sides. As soon as the 2 sides space joined finish to end, the is referred to as the peak of the triangle. An edge is formed between two sides. This is among the important facets of geometry. Triangle possess different properties, and also each of this properties can be studied at different levels the education. In this article, you will know what is the incenter of a triangle, formula, properties and examples.
You are watching: What are the properties of the incenter of a triangle
## Incenter of a triangle Meaning
The incenter the a triangle is the intersection point of all the three inner angle bisectors that the triangle. In various other words, it deserve to be identified as the suggest where the internal angle bisectors of the triangle cross. This allude will it is in equidistant native the political parties of a triangle, as the main axis’s junction suggest is the centre suggest of the triangle’s enrolled circle.
The incenter of a triangle is the center of its enrolled circle i m sorry is the largest circle that will fit within the triangle. This circle is additionally called an incircle of a triangle. This can be it was observed from the listed below figure.
## Incenter that a Triangle Formula
Suppose (x1, y1), (x2, y2) and also (x3, y3) room the works with of vertices that a triangle ABC and also a, b and also c room the lengths of its sides, then the triangle’s incenter deserve to be calculated utilizing the formula:
(LARGE (fracax_1+bx_2+cx_3a + b + c, fracay_1+by_2+cy_3a + b + c))
The above formula help in resolving the troubles like How to discover the incenter that a triangle v 3 coordinates.
To deal with such problems, we have the right to just instead of the works with in the formula after ~ finding the lengths of political parties of a triangle making use of the street formula in coordinate geometry.
### Incenter the a Triangle edge Formula
Let E, F and G it is in the points where the angle bisectors of C, A and also B cross the political parties AB, AC and also BC, respectively.
Using the edge sum residential property of a triangle, we have the right to calculate the incenter that a triangle angle.
In the above figure,
∠AIB = 180° – (∠A + ∠B)/2
Where ns is the incenter that the offered triangle.
### Incenter of a Triangle Properties
Below are the few important nature of triangles’ incenter.
If i is the incenter of the triangle alphabet (as shown in the over figure), then line segments AE and also AG, CG and also CF, BF and BE room equal in length, i.e. AE = AG, CG = CF and also BF = BE.If ns is the incenter of the triangle ABC, then ∠BAI = ∠CAI, ∠BCI = ∠ACI and also ∠ABI = ∠CBI (using angle bisector theorem).The sides of the triangle are tangents to the circle, and thus, EI = FI = GI = r recognized as the inradii of the circle or radius the incircle.If s is the semiperimeter of the triangle and also r is the inradius the the triangle, then the area of the triangle is same to the product that s and r, i.e. A = sr.The triangle’s incenter constantly lies inside the triangle.
### How to discover Incenter the a Triangle
There room two different instances in which we have actually to discover the triangles’ incenter. In construction, us can find the incenter, by drawing the edge bisectors that the triangle. However, in coordinate geometry, we can use the formula to obtain the incenter.
Let’s understand this v the aid of the below examples.
Example 1: Find the works with of the incenter the a triangle who vertices are provided as A(20, 15), B(0, 0) and also C(-36, 15).
Solution:
Let the given points be:
A(20, 15) = (x1, y1)
B(0, 0) = (x2, y2)
C(-36, 15) = (x3, y3)
Using distance formula we can discover the length of political parties AB, BC and also CA as:
AB = c = √<(20 – 0)2 +(15 – 0)2> = 25
BC = a = √<(0 + 36)2 + (0 – 15)2> = 39
CA = b = √<(20 + 36)2 + (15 – 15)2> = 56
Substituting these values in incenter formula,
((fracax_1+bx_2+cx_3a + b + c, fracay_1+by_2+cy_3a + b + c))
= <(780 + 0 -900)/(39 + 56 + 25), (585 + 0 + 375)/(39 + 56 + 25)>= (-120/120, 960/120)
= (-1, 8)
Example 2: Using ruler and compasses only, draw an it is intended triangle of side 5 cm and also its incircle.
See more: How Long To Walk 1 Km ? How Long Does It Take To Walk 1 Km
Solution:
### Incenter the a right Triangle
The incenter of a right angled triangle is given in the listed below diagram.
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# Difference Between GCF and LCM (With Table)
The number system is one of the most basic and integral parts of mathematics from basic to advanced level. In mathematical operations, Greatest Common Factor (GCF) and Lowest Common Multiple (LCM) are the most useful to simplify a fraction. These mathematical methodologies help us to find the solutions of fractions, ratios, and numerous operations. Be it adding or simplifying fractions, a basic knowledge of GFC and LCM is all we need.
## GCF vs LCM
The basic difference between GCF and LCM is that GCF finds the greatest factor which is common to a given set of numbers. The factor means a number that divides other numbers and leaves zero (0) as its remainder. And as for LCM, it is the lowest multiple, common to a set of numbers. Multiple is something that is divided by another number with no remainder.
GCF is a widely used maths technique that is mostly learned in elementary school and continued to be used constantly after then. GFC helps to reduce a set of bigger numbers into a form that is smaller and simpler. Also during the process of factorization in the case of algebraic expressions, a GFC is found which is used to simplify the question.
LCM is yet another most important technique developed by mathematicians. It is also learned at the elementary level as soon as the teaching of fractions starts. LCM is used to add or subtract fractions that do not have a common denominator (these kinds of fractions are also called unlike fractions). An LCM is taken off the denominators concerned and the fractions are thus added.
## What is GCF?
GCF whose full form is Greatest common factor is one of the most widely used methods in the field of mathematics. Students learn it at an early age and apply it to solve their mathematical problems. Problems dealing with Simplification include breaking down a bigger number into its simplest and smallest form.
Problems relating to algebra include. simplification of an equation by putting the GFC outside the bracket. And finally, it can be used to solve various word problems too. GFC as its name suggests deals in factors. Factors are the numbers that can divide a bigger number into smaller parts with zero (0) as a reminder.
For example, two (2) is 6 because two divided by six don’t leave any remainder. GFC results are much smaller than the results of LCM as it finds factors. For example, we can take the numbers six (6) and eight (8). If we find the list of factors of these two numbers, the factors of six(6) are two(2) and three(3), i.e. 2×3. And the factors of 8 are two(2), two(2), and two(2) i.e. 2×2×2. So, the factors which appear as common in both six(6) and eight (8) are two (2). Therefore GCF of the numbers 6 and 8 equals 2.
While finding the GCF which is also known as HCF (Highest Common Factor) we take the numbers concerned separately to make the calculation easier, rather than taking it altogether. Prime numbers (numbers that have 1 or themselves as a factor) are used as factors.
## What is LCM?
LCM whose full form is Lowest Common Multiple is another widely used mathematical device invented to help us add fractions that do not have a common denominator (unlike fractions). It is also taught at the elementary level along with GFC as soon as the concepts of fractions kick into the course. They are also used to find out when certain events happening on the loop will coincide. And this helps in solving many word problems.
Euclid who developed or rather found out these two concepts of LCM and GCF wanted to make the study of mathematics easier. LCM as the name shows deals in Multiples. Multiples are numbers which when divided by smaller numbers, have no remainder left.
For example, we can take the numbers six (6) and eight (8). If we find the list of factors of these two numbers- The factors of six(6) are two(2) and three(3), i.e. 2×3. And the factors of 8 are two(2), two(2), and two(2) i.e. 2×2×2. So, the Lowest Common Multiple of these two numbers is 2×2×2×3 which equals 48. So, The number to which 6 and 8 can divide leaving no remainder is 48.
We can find the Lowest Common Multiple of a set of numbers together and use prime numbers (numbers with no factors except itself and one) to find the Lowest Common Multiple.
## MainDifferences Between GCF and LCM
1. GCF is used in the simplification of a bigger number into its smaller form for easier calculation.Whereas LCM is used to add fractions with different denominator(unlike fractions).
2. GCF deals in factors which are numbers that divide other bigger numbers and leave nothing as remainder. However, LCM deals with multiples,which are numbers that are divided by smaller numbers with no remainder.
3. GCF results are smaller than the results of LCM as it considers factors. LCM results are greater than GFC as if considering multiples.
4. For ease in finding GCF when the numbers are taken separately. But LCM can be found more easily if a table containing all numbers is taken at once.
5. While calculating the results, factors that are only common to every number in the set are taken in the case of GCF. Whereas while calculating LCM, every factor which appears is taken.
## Conclusion
Mathematics as a subject provides us with various techniques to make it easier for us to solve a particular mathematical calculation. GCF and LCM being two of the most important tools which were developed long ago are still fully functioning and very useful even in the present day. Students mostly get confused while reading these two terms, but the difference lies in their names themselves.
Learning the proper use of GCF and LCM helps us understand the basic concepts. And thus with eternal importance that’s attached to these two terms, we can solve, simplify and add fractions, equations, etc. Before indulging in the concept of factorization, tutors help us to understand these terms. In some special type problems, both may look similar. We often get confused about which to use and when to use. No doubt, this is the basis of many complex problems you may face ahead.
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Mathematics
# One to One Function
3.8k views
1 Introduction 2 One to one Function 3 What is a one to one function 4 Number of One to one Functions 5 How to determine if a function is one to one 6 Inverse of one to one function 7 One to One Function Examples 8 Summary
12 October 2020
## Introduction
Before we discuss types of Functions and some of them in more detail. Lets quickly recap what is a Function?
We can say Functions are like mathematical objects that takes X values as Input to the machine and then gives the output as Y values or we can f(x) that is the function of X which is in one to one correspondence. We can describe this input numbers "x" as being the domain of the function, while the output numbers f(x) are the range of the function.
It is a special type of relationship where every Domain value will have only one Range Value.
There is a different type of functions, One to One function, Many One functions, Onto function, Into Function. But here we will discuss One to One function and Many to one function in detail.
The type of function depends on the mapping.
## One to One Function
Referring to the above diagram C, it is a one to one function because every value of x (in Domain) when plugged into a function will get us a different single value of y (in co-domain).
## What is a one to one function?
As we have seen above as the name suggests one to one function is a function in which each of the element in the Domain will have unique mapping only with one of the elements in Codomain or also called Range/Image.
This means in a one-one function given any Y value there is only one X that can be paired with the given Y. We also call this Injective.
Example if x1≠ x2 in the domain then f(x1) ≠ f(x2) in Co- Domain too.
Let’s first understand this function with an example
Suppose there is function f(x) defined as Real to Real wherein Domain has Real numbers and Range (Co-domain/Image) also comes from Real numbers.
i.e. f(x): X (Real Numbers) → Y (Real Numbers)
Let’s define f(x) = 3x +4
Here Domain X and Range Y are as follows:
In the above diagram if you see when we input 1 in the function f(x) in place of x we get the output of 7 as the Range or Image.
Similarly, when we input 2 and 3 in place of x in the above function f(x) = 3x +4, we will get the output of 10 and 13 respectively.
So here if we see for every one element of input there is only one output or image in the range section. We will never see that one element of input will have two or more outputs. For reference from the above example input of just 2 in the above function cannot have 7 and 10 both as output.
If that happens then it is not a one-one function or a function at all.
This is what is called one to one function.
### Many to One Function
When we have one or the same value as output for two or more input of real number then it is called Many to One Function.
So, let's say if we consider f(x)= x2 then if we replace x with 1, we get the output as (1)2 =1.
Similarly, when we replace x with -1 then also, we get the output as (-1) 2 =1. This would mean for two different Input values we are getting the same output value/number.
Referring to the above diagram and function we see that with more than one input in the function we get only one output and is called Many to One Function i.e. many elements have only one image or value.
Example: In a classroom, many students are mapped to a single teacher. This is a common many to one function example.
## Number of One to One Functions
The number of one to one functions would mean the number of possible or ways in which each of the elements in Domain can be mapped uniquely or exclusively to elements in Co-Domain.
Let’s understand this with an
Example 1
Suppose we have x and y in Domain. 1 and 2 in Co -Domain.
How many possible one-to-one function you can think of from x, y to 1,2.
To be honest its nothing but unique permutations and combinations we can have. Say with the above example the only two possible one to one functions are
So, we can say there are not infinite but finite many.
Example 2
How many possible one-to-one function can you think of with the Given {a, b, c,} {1, 2, 3,}?
So typically, we will write this in below 6 ways
So here we can say there is 3! Or 6 Of one to one functions for {a, b, c,} {1, 2, 3,}
Now if I ask you how many 6 people can get together and decide to take a photograph having 5 people at a time?
Answer: We would write the number of one-to-one function as 6P5
## How to determine if a function is one to one
Some of the most frequently asked questions are based on how to tell if a function is one to one.
So, How to determine if a function is one to one?
There are 3 ways to check whether a function is a one-to-one function or not- Algebraic, Graphical, and Calculus.
### 1. Calculus Method
Suppose we are given a function y=f(x)
First We need to find the derivative of f(x) i.e. f'(x)
Now we have to observe that f'(x) is > 0 for the entire given interval or < 0 for the entire given interval. In both the case, we can say our given function f(x) is a one to one function.
However, if we find that within the given interval f'(x) is at time <0 or =0 and at times > 0 then it is not one to one function.
We can also see if the one-to-one function is monotonic that is strictly increasing throughout the interval or strictly decreasing throughout the interval.
Example
Here Domain is very important.
$$f: (-1,1) \rightarrow R$$
$$f(x) = x^2+2x$$
Then f’(x) =2x +2 = 2(x+1)
So, if we put x=-1/2 or 1/2 we always get value > 0
This would mean the given function is a one-to-one function.
### 2. Algebraic Method
Algebraic one to one function would mean if f(a) = f(b) which mean a=b
Now Let’s say we have a function f(x) = (3x+5)/5
Then by plugging in a and b in place of x we get
\begin{align}\frac{(3a+5)}{5} &= \frac{(3b+5)}{5}\\3a+5&= 3b+5\\3a &=3b\\a&=b \end{align}
We can say they are not 2 different values of x but the same value of x.
### 3. Graphical Method:
We already know that if a function with the input of different x value has an output of different y value then it is a one to one function.
By using the Horizontal Line Test, we can determine if the given one to one function graph represents a one-to-one function. So, if any horizontal line is going to intersect the graph of the function in exactly one point then the function is a one to one function.
If some horizontal line intersects the one to one function graph more than once, then the function is not one-to-one.
## The Inverse of one to one function
The inverse function would mean the inverse of the parent function or any other function.
Let's say we have a function f(x) then the inverse function would be f-1(x).
Now let’s talk about the Inverse of one to one function.
If we have an inverse of one to one function that would mean domain of our original function f(x) = Range of Inverse f-1(x). If we have Domain of inverse of one to one function i.e. f-1(x) then that would be the Range of our original function f(x).
There is a special property to inverse functions. If we are given two one to one functions and we think they are inverse one to one functions to one another, we can test this by plugging one function into another.
Let’s say my function is f(x) = 5x
And inverse function is f-1(x) = 1/5x then take this and plug it into original function f(x) we get
\begin{align}f(f-1(x)) &= 5\times\frac{1}{5x}\\&=x \end{align}
That would mean these are indeed inverse of each other.
### Inverse One to One Function Graph
The one to one function graph of an inverse one to one function is the reflection of the original graph over the line y = x.
The original function is = 2x + 1.
The new red line is our inverse of y = 2x + 1.
Note: Not all graphs will be a function that produces inverse.
If the one to one function passes the Horizontal Line Test, its inverse will pass the Vertical Line Test for functions. In the next section let us look at some one to one function examples.
## One to One Function Examples
Example 1
Show that f(y)=1/y is a one to one function.
Assume that there exists some y1 and y2, such that f(y1) = f(y2). i.e. 1/ 1/y1= 1/ y2
But, $$\frac{1}{y_1}= \frac{1}{y_2}$$
\begin{align}&⇒\frac{1}{y_1} - \frac{1}{y_2}=0\\&⇒ \frac{(y_1 - y_2)}{ y_1 y_2} = 0\\&⇒ \text{the numerator must equal zero! i.e. }y_2− y_1=0\\&⇒ y_2 = y_1 \end{align}
Thus it is a one-to-one function.
Example 2
Which of the following functions are one to one functions?
(a) f (x) = 3x-2 (b) f (x) = x2+3
Solution:
If above is the one to one function then it must satisfy the condition
f(x1) = f(x2)
a. Here in 1s equation by replacing x we get,
\begin{align}3 x_1-2 &= 3 x_2 - 2\\ 3 x_1 &= 3 x_2 \qquad \text{ Or}\\x_1 &= x_2 \end{align}
Therefore f(x) is one to one function
b. Here in 2nd equation by replacing x we get,
\begin{align}X_1^2+3 &= X_2^2+3\\ X_1^2& = X_2^2\\X_1 &≠ X_2 \end{align}
Therefore f(x) is not one to one function.
## Summary
Functions are very important as they can be applied in algebra, calculus, science, and engineering. It can also be applied to our day to day life.
If we look around, we can find many such examples of one to one relationship and real-life examples one to one function examples:
• One family stay in one house, and the house has one family.
• One person has one passport, and that passport belongs to one person.
• One person has one name, and that name belongs to one person.
• Each student gets one desk which can only be used by one student.
• A company creates one product, and that product is only made by that company.
• A CEO hires only one personal assistant, and that assistant only works with that CEO.
• People have unique fingerprints, which only belong to that individual.
And thus, it makes it important for us to study one to one functions, How to determine if a function is one to one, and various properties related to it.
Written by Rashi Murarka
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# SOLUTION USING THEVENIN’S THEOREM
## Presentation on theme: "SOLUTION USING THEVENIN’S THEOREM"— Presentation transcript:
SOLUTION USING THEVENIN’S THEOREM
Working out resistor currents in unbalanced bridge circuits can be frustrating, especially when depending entirely on Kirchoff laws to do so. See Fig-1 below. However, one can employ other mathematical tools like Thevenin and Norton equivalents called “Source Transformations” to make the task easier. For example, Thevenin’s Theorem states that any complex linear circuit can be simplified to an equivalent voltage source and series resistor connected to a load resistor. (See Fig-1A below) Left click on this slide to see how this is done.
Procedure: Call Rx the Load Resistor (See Fig-1 below) Remove Rx and calculate the Voltage potential between A & B This is the equivalent Voltage source. (See Fig-2 below) Place a short across 100V and calculate resistance between A & B This result is the equivalent series resistance. (See fig-3 below) Draw the series equivalent circuit and calculate the Rx current (See Fig.4 below) This result, 0.98A, is the current through Rx. Since the current through Rx (0.98A) is now known, the remaining currents in Fig-1 can be easily calculated using Kirchoff Laws.
KIRCHOFF LAWS Kirchoff’s Current Law: The sum of all currents entering a node must be equal currents leaving the node. Kirchoff Voltage Law: The sum of the voltages around a closed loop must equal to zero. To set up the problem draw the circuit and make notations as follows: Draw the circuit and enter Volts and resistance values (see Fig-5) Using conventional current flow (+ to -) enter arrows indicating direction of current flow in all resistors Label unknown currents I1, I2, & I3 through R1, R2, & R3 respectively Label R4 current as (I1 - I3 ) & R5 current as (I2 + I3)
KIRCHOFF LAWS Since there are 3 unknowns currents (I1, I2 & I3), select 3 loops in Fig-5 and from these, develop 3 equations to be used for working out I1, I2 & I3. Loop 1: 100V, R1 & R4 Loop 2: 100V, R2 & R5 Loop 3: R1, R2 & R3 Eq. 1: – R1 I1 – R4(I1 – I3) = 0 Eq. 2: – R2 I2 – R5(I2 + I3) = 0 Eq. 3: R1 I1 + R3 I3 –R2 I2 = 0 Substitute resistor values for R1, R2 & R3 Eq. 1A: – 12 I1 – 24(I1 – I3) = 0 Eq. 2A: – 36 I2 – 18(I2 + I3) = 0 Eq. 3A: I I3 –36 I2 = 0
KIRCHOFF LAWS From previous slide
Eq. 1A: – 12 I1 – 24(I1 – I3) = 0 Eq. 2A: – 36 I2 – 18(I2 + I3) = 0 Eq. 3A: I I3 –36 I2 = 0 Expand to get rid of parentheses Eq. 1B: – 12 I1 – 24I1 + 24I3 = 0 Eq. 2B: – 36 I2 – 18I2 – 18I3 = 0 Eq. 3B: I I3 –36 I2 = 0 Consolidating we get: Eq. 1C: – 36 I I3 = Eq. 2C: – 54 I2 – 18 I3 = Eq. 3C: I I3 –36 I2 = 0 We have 3 simultaneous equations with 3 unknowns to find I1, I2 & I3. Now the fun begins.
From previous slide Eq. 1C: – 36 I I3 = Eq. 2C: – 54 I2 – 18 I3 = Eq. 3C: I I3 –36 I2 = 0 These 3 equations can be reduced to two equations with two unknown currents using traditional algebra. The procedure is as follows: Multiply Eq. 1C through by 18: 18(100 – 36 I I3 = 0) to get: Eq. 1D: – 648 I I3 =0 Multiply Eq. 2C through by 24: 24(100 – 54 I2 – 18 I3 = 0) to get D: – 1296 I2 – 432 I3 =0 By Adding Eq. 1D to Eq. 2D we get: – 648 I1 – 1296 I2 or Eq. 4A: 4200 = 648 I I2 (One of 2 equations with 2 unknowns) Repeat procedure using eq. 2C & 3C to get the other equation with 2 unknowns
KIRCHOFF LAWS Eq. 2C: – 54 I2 – 18 I3 = Eq. 3C: I I3 –36 I2 = 0 Multiply Eq. 2C by 14 to get 14 (100 – 54 I2 – 18 I3 = 0) or: Eq. 2E: – 756 I2 – 252 I3 = 0 Multiply Eq. 3C by 18 to get 18(12 I I3 –36 I2 = 0) or: Eq. 3E: I I3 – 648 I2 = 0 Add Eq. 3E to Eq. 2E to get I1 – 1404 I2 or Eq. 4B: 1400 = – 216 I I2 (The 2nd Equation with 2 unknowns So now we have: Eq. 4A: 4200 = 648 I I2 Eq. 4B: = – 216 I I2 Multiply Eq. 4A by 1404 to get: Eq. 5A = – I I2 Multiply Eq. 4B by 1296 to get: Eq: 5B = I I2 Subtract Eq.5A from Eq. 5B to get: = I1 or I1 = 3.43A
KIRCHOFF LAWS We calculated I1 = 3.43A going through R1. I3 Should be easy to find. Using the 100V, R1 R4 loop we get: 100 = (12 x 3.43) + 24( I3) or 100 = –24 I3 or I3 = I3 = / 24 = A ~ 0.98A As you can see from the above application of Kirchoff Laws, it takes a good working knowledge of Algebra and a close watch on the details. Even with these qualities, there are many pitfalls on the way to a solution: getting the signs wrong, arithmetic errors, misplacement of variables etc. I don’t think I want to ever repeat such an exercise again. When ever possible, choose the Thev enin Voltage equivalents or the Norton Current equivalents.
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Difference between revisions of "2014 AMC 12B Problems/Problem 15"
Problem
When $p = \sum\limits_{k=1}^{6} k \ln{k}$, the number $e^p$ is an integer. What is the largest power of 2 that is a factor of $e^p$?
$\textbf{(A)}\ 2^{12}\qquad\textbf{(B)}\ 2^{14}\qquad\textbf{(C)}\ 2^{16}\qquad\textbf{(D)}\ 2^{18}\qquad\textbf{(E)}\ 2^{20}$
Solution
Let's write out the sum. Our sum is equal to $$1 \ln{1} + 2 \ln{2} + 3 \ln{3} + 4 \ln{4} + 5 \ln{5} + 6 \ln{6} =$$ $$\ln{1^1} + \ln{2^2} + \ln{3^3} + \ln {4^4} + \ln{5^5} + \ln {6^6} =$$ $$\ln{(1^1\times2^2\times3^3\times4^4\times5^5\times6^6)}$$ Raising $e$ to the power of this quantity eliminates the natural logarithm, which leaves us with $$e^p = 1^1\times2^2\times3^3\times4^4\times5^5\times6^6$$ This product has $2$ powers of $2$ in the $2^2$ factor, $4*2=8$ powers of $2$ in the $4^4$ factor, and $6$ powers of $2$ in the $6^6$ factor. This adds up to $2+8+6=16$ powers of two which divide into our quantity, so our answer is $\boxed{\textbf{(C)}\ 2^{16}}$
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# Ex.9.1 Q7 Rational-Numbers Solution - NCERT Maths Class 7
Go back to 'Ex.9.1'
## Question
Rewrite the following rational numbers in the simplest form:
(i) \begin{align}\frac{ - 8}{6}\end{align}
(ii) \begin{align}\frac{25}{45}\end{align}
(iii) \begin{align}\frac{ -44}{72}\end{align}
(iv) \begin{align}\frac{ -8}{10}\end{align}
Video Solution
Rational Numbers
Ex 9.1 | Question 7
## Text Solution
What is known?
Rational numbers.
What is unknown?
Simplest form of the given rational numbers.
Reasoning:
While solving such type of questions find the \begin{align}H.C.F\end{align} of numerator and denominator and then divide both numerator and denominator by the \begin{align}H.C.F\end{align}. After dividing it you will get the simplest form of the rational number.
Steps:
(i) \begin{align}\frac{ - 8}{6}\end{align}
\begin{align}H.C.F\end{align}.of \begin{align}8\end{align} and \begin{align}6\end{align} is two. Dividing the numerator and denominator by \begin{align}H.C.F\end{align}., we get
\begin{align}\frac{{ - 8 \div 2}}{{6 \div 2}} = \frac{{ - 4}}{3}\end{align}
(ii) \begin{align}\frac{25}{45}\end{align}
\begin{align}H.C.F\end{align} of \begin{align}25\end{align} and \begin{align}45\end{align} is \begin{align} 5\end{align}. Dividing the numerator and denominator by \begin{align}H.C.F\end{align}., we get,
\begin{align}\frac{{25 \div 5}}{{45 \div 25}} = \frac{5}{9}\end{align}
(iii) \begin{align}\frac{ -44}{72}\end{align}
\begin{align}H.C.F\end{align} of \begin{align}44\end{align} and \begin{align}72\end{align} is \begin{align}4\end{align}. Dividing the numerator and denominator by \begin{align}H.C.F\end{align}., we get,
\begin{align}\frac{{ - 44 \div 4}}{{72 \div 4}} = \frac{{ - 11}}{{18}}\end{align}
(iv) \begin{align}\frac{ -8}{10}\end{align}
\begin{align}H.C.F\end{align} of \begin{align}8\end{align} and \begin{align}10\end{align} is \begin{align}2\end{align}. Dividing the numerator and denominator by \begin{align}H.C.F\end{align}., we get,
\begin{align}\frac{{ - 8 \div 2}}{{10 \div 2}} = \frac{{ - 4}}{5}\end{align}
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Question Video: Determining Which Sampling Method Is Simple Random Sampling | Nagwa Question Video: Determining Which Sampling Method Is Simple Random Sampling | Nagwa
# Question Video: Determining Which Sampling Method Is Simple Random Sampling Mathematics • First Year of Preparatory School
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A restaurant owner wants to know if his guests enjoyed the food served in one evening. He decides to ask some of his guests. He chooses a sample. Which of the following methods is considered a random sampling method? [A] Choosing those who sit at tables that have even numbers [B] Choosing female guests [C] Choosing those who order chicken [D] Choosing male guests [E] Choosing those who enter the restaurant after 11 PM
03:56
### Video Transcript
A restaurant owner wants to know if his guests enjoyed the food served in one evening. He decides to ask some of his guests. He chooses a sample. Which of the following methods is considered a random sampling method? Option (A) choosing those who sit at tables that have even numbers. Option (B) choosing female guests. Option (C) choosing those who order chicken. Option (D) choosing male guests. Or option (E) choosing those who enter the restaurant after 11 PM.
We’re given five different ways of taking a sample of the guests in a restaurant, that is, five sampling methods. And we’re asked which of these can be considered a random sampling method. So let’s begin by defining what we mean by a random sampling method. We remind ourselves first that a population is defined as the entire set of objects we’re analyzing or concerned with. In our case then, we can say that the population is all of the restaurant’s guests on the evening in question. We then define a sample as a smaller subset or a selection of the population. A random sample, sometimes classed as a simple random sample, is a sample where every member of the population has an equal chance, or probability, of selection. The sampling method is the method we use to select the sample members from the population.
Let’s consider each of the methods given in options (A) to (E), starting with option (A). This sampling method selects guests who sit at tables that have even numbers. Assuming that guests were seated randomly at the tables in the restaurant on arrival, there should be nothing to distinguish between guests who sit at even-numbered tables and those who sit at odd-numbered tables. Since guests are equally likely to have been allocated an even-numbered table as an odd one to sit at, the method described in option (A) can be considered a random sampling method.
Now let’s consider option (B), where the sampling method is to choose female guests. If our sampling method selects only female guests, then any male guests will have zero chance of selection. The selection is therefore not random, since we’re excluding any male guests completely from the sample. Therefore, option (B) cannot be considered a random sampling method. Noting that the same logic applies in option (D), where instead of female guests, only male guests are chosen, we can also discount option (D). Option (D) cannot be considered a random sampling method.
Now let’s go back and look at option (C). In option (C), only guests who order chicken are selected. Now if this were a restaurant where the only thing served was chicken, we could then say that taking a sample of those guests who order chicken would be considered a random sample. But we’re not given any information as to what’s on the menu. So we can’t assume that only chicken is served. Then choosing only guests who order chicken means that anyone who does not order chicken has no chance of being selected. All guests do not have an equal chance of being selected, so the sampling method is not random and we can discount option (C).
Finally, considering option (E), where only guests entering the restaurant after 11 PM are chosen, this excludes any guests who enter the restaurant before 11 PM. All of those guests have zero chance of being selected, and the chance of being selected is not equal for every member of the population of guests. This sampling method cannot therefore be considered as a random sampling method. And so we can eliminate option (E).
Hence, option (A), that is, choosing those guests who sit at tables that have even numbers, is the only sampling method that is considered random.
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3 7 MATHEMATICS Therefore, their product = (x 5) (40 x) =40x x x = x + 45x 00 So, x + 45x 00 = 4 (Given that product = 4) i.e., x + 45x 34 = 0 i.e., x 45x + 34 = 0 Therefore, the number of marbles John had, satisfies the quadratic equation x 45x + 34 = 0 which is the required representation of the problem mathematically. (ii) Let the number of toys produced on that day be x. Therefore, the cost of production (in rupees) of each toy that day = 55 x So, the total cost of production (in rupees) that day = x (55 x) Therefore, x (55 x) = 750 i.e., 55x x = 750 i.e., x + 55x 750 = 0 i.e., x 55x = 0 Therefore, the number of toys produced that day satisfies the quadratic equation x 55x = 0 which is the required representation of the problem mathematically. Example : Check whether the following are quadratic equations: (i) (x ) + = x 3 (ii) x(x + ) + 8 = (x + ) (x ) (iii) x (x + 3) = x + (iv) (x + ) 3 = x 3 4 Solution : (i) LHS = (x ) + = x 4x = x 4x + 5 Therefore, (x ) + = x 3 can be rewritten as x 4x + 5 = x 3 i.e., x 6x + 8 = 0 It is of the form ax + bx + c = 0. Therefore, the given equation is a quadratic equation (--04)
6 QUADRATIC EQUATIONS 75 Note that we have found the roots of x 5x + 3 = 0 by factorising x 5x + 3 into two linear factors and equating each factor to zero. Example 4 : Find the roots of the quadratic equation 6x x = 0. Solution : We have 6x x = 6x + 3x 4x =3x (x + ) (x + ) =(3x )(x + ) The roots of 6x x = 0 are the values of x for which (3x )(x + ) = 0 Therefore, 3x = 0 or x + = 0, i.e., x = 3 or x = Therefore, the roots of 6x x = 0 are and. 3 We verify the roots, by checking that and satisfy 6x x = 0. 3 Example 5 : Find the roots of the quadratic equation 3x 6x 0. Solution : 3x 6x = 3x 6x 6x = 3x 3x 3x = 3x 3x So, the roots of the equation are the values of x for which Now, 3x 0 for x x x. 3 So, this root is repeated twice, one for each repeated factor 3x. Therefore, the roots of 3x 6x 0 are 3, (--04)
7 76 MATHEMATICS Example 6 : Find the dimensions of the prayer hall discussed in Section 4.. Solution : In Section 4., we found that if the breadth of the hall is x m, then x satisfies the equation x + x 300 = 0. Applying the factorisation method, we write this equation as x 4x + 5x 300 = 0 x (x ) + 5 (x ) = 0 i.e., (x )(x + 5) = 0 So, the roots of the given equation are x = or x =.5. Since x is the breadth of the hall, it cannot be negative. Thus, the breadth of the hall is m. Its length = x + = 5 m. EXERCISE 4.. Find the roots of the following quadratic equations by factorisation: (i) x 3x 0 = 0 (ii) x + x 6 = 0 (iii) x 7x 5 0 (iv) x x + 8 = 0 (v) 00 x 0x + = 0. Solve the problems given in Example. 3. Find two numbers whose sum is 7 and product is Find two consecutive positive integers, sum of whose squares is The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 3 cm, find the other two sides. 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ` 90, find the number of articles produced and the cost of each article. 4.4 Solution of a Quadratic Equation by Completing the Square In the previous section, you have learnt one method of obtaining the roots of a quadratic equation. In this section, we shall study another method. Consider the following situation: The product of Sunita s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age? To answer this, let her present age (in years) be x. Then the product of her ages two years ago and four years from now is (x )(x + 4) (--04)
8 QUADRATIC EQUATIONS 77 Therefore, (x )(x + 4) = x + i.e., x + x 8 = x + i.e., x 9 = 0 So, Sunita s present age satisfies the quadratic equation x 9 = 0. We can write this as x = 9. Taking square roots, we get x = 3 or x = 3. Since the age is a positive number, x = 3. So, Sunita s present age is 3 years. Now consider the quadratic equation (x + ) 9 = 0. To solve it, we can write it as (x + ) = 9. Taking square roots, we get x + = 3 or x + = 3. Therefore, x = or x = 5 So, the roots of the equation (x + ) 9 = 0 are and 5. In both the examples above, the term containing x is completely inside a square, and we found the roots easily by taking the square roots. But, what happens if we are asked to solve the equation x + 4x 5 = 0? We would probably apply factorisation to do so, unless we realise (somehow!) that x + 4x 5 = (x + ) 9. So, solving x + 4x 5 = 0 is equivalent to solving (x + ) 9 = 0, which we have seen is very quick to do. In fact, we can convert any quadratic equation to the form (x + a) b = 0 and then we can easily find its roots. Let us see if this is possible. Look at Fig. 4.. In this figure, we can see how x + 4x is being converted to (x + ) 4. Fig (--04)
9 78 MATHEMATICS The process is as follows: x + 4x =(x + 4 x ) + 4 x =x + x + x = (x + ) x + x =(x + ) x + x + =(x + ) x + (x + ) =(x + ) (x + ) =(x + ) 4 So, x + 4x 5 = (x + ) 4 5 = (x + ) 9 So, x + 4x 5 = 0 can be written as (x + ) 9 = 0 by this process of completing the square. This is known as the method of completing the square. In brief, this can be shown as follows: So, x + 4x = x x 4 x + 4x 5 = 0 can be rewritten as 4 x 45 =0 i.e., (x + ) 9 = 0 Consider now the equation 3x 5x + = 0. Note that the coefficient of x is not a perfect square. So, we multiply the equation throughout by 3 to get 9x 5x + 6 = 0 Now, 9x 5x + 6 = = 5 (3 x) 3x (3 x) 3x 6 = 5 5 3x 6 4 = 5 3x (--04)
10 QUADRATIC EQUATIONS 79 So, 9x 5x + 6 = 0 can be written as i.e., 5 3x 4 5 3x =0 = 4 So, the solutions of 9x 5x + 6 = 0 are the same as those of 5 3x. 4 i.e., 3x 5 = or 5 3x = (We can also write this as 5 3x, where ± denotes plus minus.) Thus, 3x = 5 or So, x = 5 or x 5 x 6 6 Therefore, x = or x = 4 6 i.e., x = or x = 3 Therefore, the roots of the given equation are and. 3 Remark : Another way of showing this process is as follows : The equation 3x 5x + = 0 is the same as 5 x x =0 3 3 Now, x 5 x = x (--04)
11 80 MATHEMATICS = = 5 5 x x x So, the solutions of 3x 5x + = 0 are the same as those of 5 x 0, 6 6 which are x 5 6 = ± 6, i.e., x = 5 = and x = 5 = Let us consider some examples to illustrate the above process. Example 7 : Solve the equation given in Example 3 by the method of completing the square. Solution : The equation x 5 3 5x + 3 = 0 is the same as x x 0. Now, 5 3 x x = x = x Therefore, x 5x + 3 = 0 can be written as x So, the roots of the equation x 5x + 3 = 0 are exactly the same as those of 5 5 x 0. Now, x =0 is the same as Therefore, 5 x = 4 4 i.e., x = i.e., x = 5 or x i.e., x = 3 or x = 5 x (--04)
12 QUADRATIC EQUATIONS 8 Therefore, the solutions of the equations are Let us verify our solutions. 3 x and. 3 Putting x 3 3 in x 5x + 3 = 0, we get 5 3 0, which is correct. Similarly, you can verify that x = also satisfies the given equation. In Example 7, we divided the equation x 5x + 3 = 0 throughout by to get x 5 x 3 = 0 to make the first term a perfect square and then completed the square. Instead, we can multiply throughout by to make the first term as 4x = (x) and then complete the square. This method is illustrated in the next example. Example 8 : Find the roots of the equation 5x 6x = 0 by the method of completing the square. Solution : Multiplying the equation throughout by 5, we get 5x 30x 0 = 0 This is the same as (5x) (5x) = 0 i.e., (5x 3) 9 0 = 0 i.e., (5x 3) 9 = 0 i.e., (5x 3) =9 i.e., 5x 3 = 9 i.e., 5x = 3 9 So, x = Therefore, the roots are Verify that the roots are and and (--04)
13 8 MATHEMATICS Example 9 : Find the roots of 4x + 3x + 5 = 0 by the method of completing the square. Solution : Note that 4x + 3x + 5 = 0 is the same as (x) + (x) =0 i.e., 3 9 x =0 i.e., x =0 i.e., x 3 4 = But x cannot be negative for any real value of x (Why?). So, there is 4 no real value of x satisfying the given equation. Therefore, the given equation has no real roots. Now, you have seen several examples of the use of the method of completing the square. So, let us give this method in general. Consider the quadratic equation ax + bx + c = 0 (a 0). Dividing throughout by b c a, we get x x 0 a a This is the same as b b c x 0 a a a i.e., b b 4ac x = 0 a 4a So, the roots of the given equation are the same as those of b b 4ac x 0, a 4a i.e., those of b b 4ac x a 4a () 05-6 (--04)
14 QUADRATIC EQUATIONS 83 If b 4ac 0, then by taking the square roots in (), we get b x = a b 4ac a Therefore, x = b b 4ac a b b 4ac b b 4ac So, the roots of ax + bx + c = 0 are and, if a a b 4ac 0. If b 4ac < 0, the equation will have no real roots. (Why?) Thus, if b 4ac 0, then the roots of the quadratic equation ax + bx + c = 0 are given by b± b 4ac a This formula for finding the roots of a quadratic equation is known as the quadratic formula. Let us consider some examples for illustrating the use of the quadratic formula. Example 0 : Solve Q. (i) of Exercise 4. by using the quadratic formula. Solution : Let the breadth of the plot be x metres. Then the length is (x + ) metres. Then we are given that x(x + ) = 58, i.e., x + x 58 = 0. This is of the form ax + bx + c = 0, where a =, b =, c = 58. So, the quadratic formula gives us the solution as x = i.e., x = 4()(58) or x i.e., x = 6 or x = Since x cannot be negative, being a dimension, the breadth of the plot is 6 metres and hence, the length of the plot is 33m. You should verify that these values satisfy the conditions of the problem (--04)
15 84 MATHEMATICS Example : Find two consecutive odd positive integers, sum of whose squares is 90. Solution : Let the smaller of the two consecutive odd positive integers be x. Then, the second integer will be x +. According to the question, x + (x + ) = 90 i.e., x + x + 4x + 4 = 90 i.e., x + 4x 86 = 0 i.e., x + x 43 = 0 which is a quadratic equation in x. Using the quadratic formula, we get x = i.e., x = or x = 3 But x is given to be an odd positive integer. Therefore, x 3, x =. Thus, the two consecutive odd integers are and 3. Check : + 3 = + 69 = 90. Example : A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude m (see Fig. 4.3). Find its length and breadth. Solution : Let the breadth of the rectangular park be x m. So, its length = (x + 3) m. Therefore, the area of the rectangular park = x(x + 3) m = (x + 3x) m. Now, base of the isosceles triangle = x m. Therefore, its area = x = 6 x m. According to our requirements, x + 3x =6x + 4 i.e., x 3x 4 = 0 Using the quadratic formula, we get Fig (--04)
16 QUADRATIC EQUATIONS 85 x = 3 5 But x (Why?). Therefore, x = 4. = 3 5 = 4 or So, the breadth of the park = 4m and its length will be 7m. Verification : Area of rectangular park = 8 m, area of triangular park = 4 m = (8 4) m Example 3 : Find the roots of the following quadratic equations, if they exist, using the quadratic formula: (i) 3x 5x + = 0 (ii) x + 4x + 5 = 0 (iii) x x + = 0 Solution : (i) 3x 5x + = 0. Here, a = 3, b = 5, c =. So, b 4ac = 5 4 = > 0. Therefore, x = 5 5, i.e., x = or x = So, the roots are 3 and. (ii) x + 4x + 5 = 0. Here, a =, b = 4, c = 5. So, b 4ac = 6 0 = 4 < 0. Since the square of a real number cannot be negative, therefore b 4ac will not have any real value. So, there are no real roots for the given equation. (iii) x x + = 0. Here, a =, b =, c =. So, b 4ac = 8 8 = 0 Therefore, x = 0 0, 4 i.e., x. So, the roots are, (--04)
17 86 MATHEMATICS Example 4 : Find the roots of the following equations: (i) Solution : x 3, x 0 (ii) 3, x 0, x x x (i) x 3. Multiplying throughout by x, we get x x + = 3x i.e., x 3x + = 0, which is a quadratic equation. Here, a =, b = 3, c = So, b 4ac = 9 4 = 5 > 0 Therefore, x = 3 5 (Why?) So, the roots are 3 5 and 3 5. (ii) 3, x 0,. x x As x 0,, multiplying the equation by x (x ), we get (x ) x =3x (x ) =3x 6x So, the given equation reduces to 3x 6x + = 0, which is a quadratic equation. Here, a = 3, b = 6, c =. So, b 4ac = 36 4 = > 0 Therefore, x = So, the roots are 3 3 and (--04)
18 QUADRATIC EQUATIONS 87 Example 5 : A motor boat whose speed is 8 km/h in still water takes hour more to go 4 km upstream than to return downstream to the same spot. Find the speed of the stream. Solution : Let the speed of the stream be x km/h. Therefore, the speed of the boat upstream = (8 x) km/h and the speed of the boat downstream = (8 + x) km/h. The time taken to go upstream = distance 4 speed 8 x hours. 4 Similarly, the time taken to go downstream = 8 x According to the question, hours x 8 x = i.e., 4(8 + x) 4(8 x) = (8 x) (8 + x) i.e., x + 48x 34 = 0 Using the quadratic formula, we get x = = = 4860 = 6 or Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = 54. Therefore, x = 6 gives the speed of the stream as 6 km/h. EXERCISE 4.3. Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) x 7x + 3 = 0 (ii) x + x 4 = 0 (iii) 4x 4 3x 3 0 (iv) x + x + 4 = 0. Find the roots of the quadratic equations given in Q. above by applying the quadratic formula (--04)
19 88 MATHEMATICS 3. Find the roots of the following equations: (i) x 3, x 0 (ii), x 4, 7 x x4 x The sum of the reciprocals of Rehman s ages, (in years) 3 years ago and 5 years from now is. Find his present age In a class test, the sum of Shefali s marks in Mathematics and English is 30. Had she got marks more in Mathematics and 3 marks less in English, the product of their marks would have been 0. Find her marks in the two subjects. 6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field. 7. The difference of squares of two numbers is 80. The square of the smaller number is 8 times the larger number. Find the two numbers. 8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken hour less for the same journey. Find the speed of the train. 9. Two water taps together can fill a tank in 9 3 hours. The tap of larger diameter takes 0 8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. 0. An express train takes hour less than a passenger train to travel 3 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is km/h more than that of the passenger train, find the average speed of the two trains.. Sum of the areas of two squares is 468 m. If the difference of their perimeters is 4 m, find the sides of the two squares. 4.5 Nature of Roots In the previous section, you have seen that the roots of the equation ax + bx + c = 0 are given by x = b b 4ac a If b 4ac > 0, we get two distinct real roots b b 4ac. a a b b 4ac and a a 05-6 (--04)
20 QUADRATIC EQUATIONS 89 b b b If b 4ac = 0, then x = 0, i.e., x or a a a b So, the roots of the equation ax + bx + c = 0 are both a Therefore, we say that the quadratic equation ax + bx + c = 0 has two equal real roots in this case. If b 4ac < 0, then there is no real number whose square is b 4ac. Therefore, there are no real roots for the given quadratic equation in this case. Since b 4ac determines whether the quadratic equation ax + bx + c = 0 has real roots or not, b 4ac is called the discriminant of this quadratic equation. So, a quadratic equation ax + bx + c = 0 has (i) two distinct real roots, if b 4ac > 0, (ii) two equal real roots, if b 4ac = 0, (iii) no real roots, if b 4ac < 0. Let us consider some examples. Example 6 : Find the discriminant of the quadratic equation x 4x + 3 = 0, and hence find the nature of its roots. Solution : The given equation is of the form ax + bx + c = 0, where a =, b = 4 and c = 3. Therefore, the discriminant b 4ac = ( 4) (4 3) = 6 4 = 8 < 0 So, the given equation has no real roots. Example 7 : A pole has to be erected at a point on the boundary of a circular park of diameter 3 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected? Solution : Let us first draw the diagram (see Fig. 4.4). Let P be the required location of the pole. Let the distance of the pole from the gate B be x m, i.e., BP = x m. Now the difference of the distances of the pole from the two gates = AP BP (or, BP AP) = 7 m. Therefore, AP = (x + 7) m. Fig (--04)
21 90 MATHEMATICS Now, AB = 3m, and since AB is a diameter, APB = 90 (Why?) Therefore, AP + PB =AB (By Pythagoras theorem) i.e., (x + 7) + x =3 i.e., x + 4x x = 69 i.e., x + 4x 0 = 0 So, the distance x of the pole from gate B satisfies the equation x + 7x 60 = 0 So, it would be possible to place the pole if this equation has real roots. To see if this is so or not, let us consider its discriminant. The discriminant is b 4ac = 7 4 ( 60) = 89 > 0. So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park. Solving the quadratic equation x + 7x 60 = 0, by the quadratic formula, we get x = = Therefore, x = 5 or. Since x is the distance between the pole and the gate B, it must be positive. Therefore, x = will have to be ignored. So, x = 5. Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and m from the gate A. Example 8 : Find the discriminant of the equation 3x x + 3 the nature of its roots. Find them, if they are real. Solution : Here a = 3, b = and c. 3 = 0 and hence find Therefore, discriminant b 4ac = ( ) = 4 4 = 0. Hence, the given quadratic equation has two equal real roots. The roots are b, b,,, i.e.,, i.e.,. a a (--04)
23 9 MATHEMATICS A NOTE TO THE READER In case of word problems, the obtained solutions should always be verified with the conditions of the original problem and not in the equations formed (see Examples, 3, 9 of Chapter 3 and Examples 0,, of Chapter 4) (--04)
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CHAPTER THE INTEGERS In golf tournaments, a player s standing after each hole is often recorded on the leaderboard as the number of strokes above or below a standard for that hole called a par. A player
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# What is harmonic mean with example?
## What is harmonic mean with example?
The harmonic mean is a type of numerical average. It is calculated by dividing the number of observations by the reciprocal of each number in the series. Thus, the harmonic mean is the reciprocal of the arithmetic mean of the reciprocals. The harmonic mean of 1, 4, and 4 is: 3 ( 1 1 + 1 4 + 1 4 ) = 3 1 .
## What is a harmonic sequence give example?
Harmonic Sequence Definition The sum of harmonic sequences is known as harmonic series. It is an infinite series that never converges to a limit. For example, let’s take an arithmetic sequence as 5, 10, 15, 20, 25,… with the common difference of 5. Then its harmonic sequence is: 1/5, 1/10, 1/15,1/20,1/25….
What are the example of harmonic?
The pendulum oscillating back and forth from the mean position is an example of simple harmonic motion. Bungee Jumping is an example of simple harmonic motion. The jumper oscillating up and down is undergoing SHM due to the elasticity of the bungee cord.
### Which of the following is the best example of simple harmonic motion?
The pendulum oscillating back and forth from the mean position is an example of simple harmonic motion. Bungee Jumping is an example of simple harmonic motion.
### How do you find the harmonic series of a function?
Enter the harmonic series. The harmonic series is the sum from n = 1 to infinity with terms 1/ n. If you write out the first few terms, the series unfolds as follows: 1 + 1/2 + 1/3 + 1/4 + 1/5 +. . .etc. As n tends to infinity, 1/ n tends to 0. However, the series actually diverges.
What is a harmonic series?
A harmonic series is a series that contains the sum of terms that are the reciprocals of an arithmetic series’ terms. This article will explore this unique series and understand how they behave as an infinite series.
## What is the sum of the harmonic sequence?
Harmonic sequence mathematics can be defined as The reciprocal form of the Arithmetic Sequence with numbers that can never be 0. The sum of harmonic sequence is known as Harmonic Series. In Mathematics, we can define progression as a series of numbers arranged in a predictable pattern.
## What is the difference between harmonic numbers?
The finite partial sums of the diverging harmonic series, = ∑ =, are called harmonic numbers. The difference between H n and ln n converges to the Euler–Mascheroni constant. The difference between any two harmonic numbers is never an integer. No harmonic numbers are integers, except for H 1 = 1.
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# 3.3: Combining Decimals- Addition and Subtraction with Decimals
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Addition of decimal numbers is quite similar to addition of whole numbers. For example, suppose that we are asked to add 2.34 and 5.25. We could change these decimal numbers to mixed fractions and add.
\begin{aligned} 2.34 + 5.25 & = 2 \frac{34}{100} + 5 \frac{25}{100} \\ & = 7 \frac{59}{100} \end{aligned}\nonumber
However, we can also line the decimal numbers on their decimal points and add vertically, as follows.
$\begin{array}{r} 2.34 \\ + 5.25 \\ \hline 7.59 \end{array}\nonumber$
Note that this alignment procedure produces the same result, “seven and fifty nine hundredths.” This motivates the following procedure for adding decimal numbers.
To add decimal numbers, proceed as follows:
1. Place the numbers to be added in vertical format, aligning the decimal points.
2. Add the numbers as if they were whole numbers.
3. Place the decimal point in the answer in the same column as the decimal points above it.
Example 1
Solution
Place the numbers in vertical format, aligning on their decimal points. Add, then place the decimal point in the answer in the same column as the decimal points that appear above the answer.
$\begin{array}{r} 3.125 \\ +4.814 \\ \hline 7.939 \end{array}\nonumber$
Thus, 3.125 + 4.814 = 7.939.
Exercise
5.893
Example 2
Jane has $4.35 in her purse. Jim has$5.62 in his wallet. If they sum their money, what is the total?
Solution
Arrange the numbers in vertical format, aligning decimal points, then add.
$\begin{array}{r} \ 4.35 \\ + \ 5.62 \\ \hline \ 9.97 \end{array}\nonumber$
Exercise
Alice has $8.63 in her purse and Joanna has$2.29. If they combine sum their money, what is the total?
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# Lesson 18Scaling Two Dimensions
Let’s change more dimensions of shapes.
### Learning Targets:
• I can create a graph representing the relationship between volume and radius for all cylinders (or cones) with a fixed height.
• I can explain in my own words why changing the radius by a scale factor changes the volume by the scale factor squared.
## 18.1Tripling Statements
, , , , and all represent positive integers. Consider these two equations:
1. Which of these statements are true? Select all that apply.
1. If is tripled, is tripled.
2. If , , and are all tripled, then is tripled.
3. If is tripled, is tripled.
4. If , , and are all tripled, then is tripled.
2. Create a true statement of your own about one of the equations.
## 18.2A Square Base
Clare sketches a rectangular prism with a height of 11 and a square base and labels the edges of the base . She asks Han what he thinks will happen to the volume of the rectangular prism if she triples .
Han says the volume will be 9 times bigger. Is he right? Explain or show your reasoning.
### Are you ready for more?
A cylinder can be constructed from a piece of paper by curling it so that you can glue together two opposite edges (the dashed edges in the figure).
1. If you wanted to increase the volume inside the resulting cylinder, would it make more sense to double , , or does it not matter?
2. If you wanted to increase the surface area of the resulting cylinder, would it make more sense to double , , or does it not matter?
3. How would your answers to these questions change if we made a cylinder by gluing together the solid lines instead of the dashed lines?
## 18.3Playing with Cones
There are many cones with a height of 7 units. Let represent the radius and represent the volume of these cones.
1. Write an equation that expresses the relationship between and . Use 3.14 as an approximation for .
2. Predict what will happen to the volume if you triple the value of .
3. Graph this equation.
4. What happens to the volume if you triple ? Where do you see this in the graph? How can you see it algebraically?
## Lesson 18 Summary
There are many rectangular prisms that have a length of 4 units and width of 5 units but differing heights. If represents the height, then the volume of such a prism is
The equation shows us that the volume of a prism with a base area of 20 square units is a linear function of the height. Because this is a proportional relationship, if the height gets multiplied by a factor of , then the volume is also multiplied by a factor of :
What happens if we scale two dimensions of a prism by a factor of ? In this case, the volume gets multiplied by a factor of twice, or .
For example, think about a prism with a length of 4 units, width of 5 units, and height of 6 units. Its volume is 120 cubic units since . Now imagine the length and width each get scaled by a factor of , meaning the new prism has a length of , width of , and a height of 6. The new volume is cubic units since .
A similar relationship holds for cylinders. Think of a cylinder with a height of 6 and a radius of 5. The volume would be cubic units since . Now, imagine the radius is scaled by a factor of . Then the new volume is or cubic units. So scaling the radius by a factor of has the effect of multiplying the volume by !
Why does the volume multiply by when only the radius changes? This makes sense if we imagine how scaling the radius changes the base area of the cylinder. As the radius increases, the base area gets larger in two dimensions (the circle gets wider and also taller), while the third dimension of the cylinder, height, stays the same.
## Lesson 18 Practice Problems
1. There are many cylinders with a height of 18 meters. Let represent the radius in meters and represent the volume in cubic meters.
1. Write an equation that represents the volume as a function of the radius .
2. Complete this table, giving three possible examples.
1
3. If the radius of a cylinder is doubled, does the volume double? Explain how you know.
4. Is the graph of this function a line? Explain how you know.
2. As part of a competition, Diego must spin around in a circle 6 times and then run to a tree. The time he spends on each spin is represented by and the time he spends running is . He gets to the tree 21 seconds after he starts spinning.
1. Write an equation showing the relationship between and .
2. Rearrange the equation so that it shows as a function of .
3. If it takes Diego 1.2 seconds to spin around each time, how many seconds did he spend running?
3. The table and graph represent two functions. Use the table and graph to answer the questions.
x y 1 2 3 4 5 6 3 -1 0 4 5 -1
1. For which values of is the output from the table less than the output from the graph?
2. In the graphed function, which values of give an output of 0?
4. A cone has a radius of 3 units and a height of 4 units.
1. What is this volume of this cone?
2. Another cone has quadruple the radius, and the same height. How many times larger is the new cone’s volume?
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# Solve for z 1/(6z)+1 1/6=-48
16z+116=-48
Convert 116 to an improper fraction.
A mixed number is an addition of its whole and fractional parts.
16z+1+16=-48
Add 1 and 16.
Write 1 as a fraction with a common denominator.
16z+66+16=-48
Combine the numerators over the common denominator.
16z+6+16=-48
Add 6 and 1.
16z+76=-48
16z+76=-48
16z+76=-48
Move all terms not containing z to the right side of the equation.
Subtract 76 from both sides of the equation.
16z=-48-76
To write -48 as a fraction with a common denominator, multiply by 66.
16z=-48⋅66-76
Combine -48 and 66.
16z=-48⋅66-76
Combine the numerators over the common denominator.
16z=-48⋅6-76
Simplify the numerator.
Multiply -48 by 6.
16z=-288-76
Subtract 7 from -288.
16z=-2956
16z=-2956
Move the negative in front of the fraction.
16z=-2956
16z=-2956
Set up the rational expression with the same denominator over the entire equation.
Multiply each term by a factor of 1 that will equate all the denominators. In this case, all terms need a denominator of 6z. The -2956 expression needs to be multiplied by zz to make the denominator 6z.
16z=-2956⋅zz
Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 6z.
295(z)
Multiply 295 by z.
16z=-295z6z
16z=-295z6z
Since the expression on each side of the equation has the same denominator, the numerators must be equal.
1=-(295z)
Rewrite the equation as -(295z)=1.
-(295z)=1
Multiply 295 by -1.
-295z=1
Divide each term by -295 and simplify.
Divide each term in -295z=1 by -295.
-295z-295=1-295
Cancel the common factor of -295.
Cancel the common factor.
-295z-295=1-295
Divide z by 1.
z=1-295
z=1-295
Move the negative in front of the fraction.
z=-1295
z=-1295
The result can be shown in multiple forms.
Exact Form:
z=-1295
Decimal Form:
z=-0.00338983…
Solve for z 1/(6z)+1 1/6=-48
### Solving MATH problems
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 4 * 2/5 = 8/5 = 1 3/5 = 1.6
Spelled result in words is eight fifths (or one and three fifths).
### How do you solve fractions step by step?
1. Multiple: 4 * 2/5 = 4 · 2/1 · 5 = 8/5
Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(8, 5) = 1. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - four multiplied by two fifths = eight fifths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Fraction to decimal
Write the fraction 3/22 as a decimal.
• Fraction and a decimal
Write as a fraction and a decimal. One and two plus three and five hundredths
• Unknown number
I think the number - its sixth is 3 smaller than its third.
• Rolls
Mom bought 13 rolls. Dad ate 3.5 rolls. How many rolls left when Peter yet put two at dinner?
• Product of two fractions
Product of two fractions is 9 3/5 . If one of the fraction is 9 3/7. Find the other fraction.
• Cupcakes 2
Susi has 25 cupcakes. She gives 4/5. How much does she have left?
• Withdrawal
If I withdrew 2/5 of my total savings and spent 7/10 of that amount. What fraction do I have in left in my savings?
• Simple equation 6
Solve equation with one variable: X/2+X/3+X/4=X+4
• One-third 2
One-third of the people in a barangay petitioned the council to allow them to plant in vacant lots, and another 1/5 of the people petitioned to have a regular garbage collection. What FRACTION of the barangay population made the petition?
• Pizza 5
You have 2/4 of a pizza, and you want to share it equally between 2 people. How much pizza does each person get?
• New bridge
Thanks to the new bridge, the road between A and B has been cut to one third and is now 10km long. How much did the road between A and B measure before?
• Cakes
On the bowl were a few cakes. Jane ate one-third of them, and Dana ate a quarter of those cakes that remained. a) What part (of the original number of cakes) Dana ate? b) At least how many cakes could be (initially) on the bowl?
• Passenger boat
Two-fifths of the passengers in the passenger boat were boys. 1/3 of them were girls and the rest were adult. If there were 60 passengers in the boat, how many more boys than adult were there?
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App that does your homework
App that does your homework is a software program that helps students solve math problems. We can solve math word problems.
The Best App that does your homework
Here, we debate how App that does your homework can help students learn Algebra. definite integrals are used for finding the value of a function at a specific point. There are two types: definite integrals of first and second order. The definite integral of the first order is sometimes called the definite integral from the left to evaluate an area under a curve, whereas the definite integral of the second order is used to find an area under a curve between two values. Definite integrals can be solved by using integration by parts. This equation says that you can break your integral into two parts, one on each side of the equals sign, which will cancel out giving you just the value of your integral. You can also use complex numbers in the denominator to simplify things even more! If you want to solve definite integrals by hand, following these steps should get you going: Step 1: Find your area under the graph by drawing small rectangles where you want to find your answer. Step 2: Evaluate your integral by plugging in numbers into each rectangle. Step 3: Add up all your rectangles' areas and divide by n (where n is the number of rectangles). This will tell you how much area you evaluated for this particular function.
Differential equations are equations that describe the relationship between a quantity and a change in that quantity. There are many types of differential equations, which can be classified into two main categories: linear and nonlinear. One example of a differential equation is the equation y = x2, which describes the relationship between the height and the width of a rectangle. In this case, x represents height and y represents width. If we want to find out how high or how wide a rectangle will be, we can find the height or width by solving this equation. For example, if we want to know what the height of a rectangle will be, we simply plug in an x value and solve for y. This process is called “back substitution” because it makes use of back-substitution. For example, if we want to know what the width of a rectangle will be, we plug in an x value and solve for y. Because differential equations describe how one quantity changes when another quantity changes, solving them can often be used to predict what will happen to one variable if another variable changes or is kept constant. In addition to predicting what will happen in the future, differential equations can also be used to simulate how systems behave in the past or present. Because these simulations involve using estimates of past values as inputs into models instead of actual values from the past, they are often referred as
Partial fraction decomposition (PFD) is a method for solving simultaneous equations. It gives the solution of A * B = C in terms of A and B, and C = A * B. If we have two equations, A * B = C and A + B = C, then PFD gives us an equation of the form (A * B) - (A + B) = 0. The PFD algorithm solves the system by finding a solution to the following equation: A(B - C) = 0 This can be expressed as a simpler equation in terms of partial fractions as: B - C / A(B - C) = 0 This solution is called a "mixed" or "mixed-order" solution. Mixed-order solutions typically have less accuracy than higher-order solutions, but are much faster to compute. The PFD solver computes mixed-order solutions based on an interpolation scheme that interpolates between values of a function at points where it crosses zero. This scheme makes the second derivative zero on these points, and therefore the interpolant will be quadratic on these points. These points are computed iteratively so that they become increasingly accurate while computing time is reduced. Typically, linear systems like this are solved by double-differencing or Taylor's series expansion to approximate the second derivative term at
The trigonomic equation solver is a tool that can be used to calculate the value of a trigonomic system. This is done by adding all of the elements together and then calculating how many calories are required to break them down. It works in a very similar way to the general formula for calories, where it takes into account different sources of calories.
Graph an equation in a table or graph to show how two values change over time. Graphs are a great way to show cause and effect. To solve an equation by graphing, first find the set of values that you want to represent your answer. Usually, you will want to plot one value against time and see how it changes over time. If you are solving a rate problem, you will plot the rate of change against time. You can also plot other quantities against time, such as distance and volume. For each pair of values that you plot on your graph, consider what is changing (the independent variable) and what is staying the same (the constant). You can then use your graph to see if there is a pattern or relationship between the two variables. If there is a pattern, then you can use that information to solve for one of the values.
App is great since it not only solves a plethora of problems by just taking a picture, it even explains step by how to solve the problems down to the nitty gritty (if you played the 10 dollars for plus that is). Only down side is that plus costs 10 dollars but that’s probably only a problem since I’m broke so 5 stars
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Online Variable Way App that solves word math problems Radical equations solver Algebra problems to solve How to solve an absolute value equation Algebra solving website
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# Single Variable Expressions
## Use symbols and operations to understand and define variables.
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Single Variable Expressions
License: CC BY-NC 3.0
Shelly is making bracelets to sell at her town's market this summer. She spent $150 on supplies and will make$4 for every bracelet she sells. Her profit for selling b\begin{align*}b\end{align*} bracelets is given by the expression 4b150\begin{align*}4b-150\end{align*}. How can Shelly calculate her profit if she sells 50 bracelets this summer?
In this concept, you will learn how to evaluate single variable expressions.
### Expressions
An expression is a mathematical phrase that contains numbers and operations.
Here are some examples of expressions:
• 3x+10\begin{align*}3x+10\end{align*}
• 15+71\begin{align*}-15+7-1\end{align*}
• 521\begin{align*}5^2-1\end{align*}
• 15r+2\begin{align*}15r+2\end{align*}
A variable is a symbol or letter (such as x,m,R,y,P\begin{align*}x,m,R,y,P\end{align*}, or a\begin{align*}a\end{align*}) that is used to represent a quantity that might change in value. A variable expression is an expression that includes variables. Another name for a variable expression is an algebraic expression.
Here are some examples of variable expressions:
• 3x+10\begin{align*}3x+10\end{align*}
• 10r\begin{align*}10r\end{align*}
• b3+2\begin{align*}b^3+2\end{align*}
• mx3\begin{align*}mx-3\end{align*}
A single variable expression is a variable expression with just one variable in it.
You can use a variable expression to describe a real world situation where one or more quantities has an unknown value or can change in value.
To evaluate a variable expression means to find the value of the expression for a given value of the variable. To evaluate, substitute the given value for the variable in the expression and simplify using the order of operations. To follow the order of operations, you always need to do any multiplication/division first before any addition/subtraction.
Here is an example.
Evaluate the expression 10k44\begin{align*}10k-44\end{align*} for k=12\begin{align*}k=12\end{align*}.
First, remember that when you see a number next to a letter, like “10k\begin{align*}10k\end{align*}”, it means to multiply.
Next, substitute 12 in for the letter k\begin{align*}k\end{align*} in the expression.
10(12)44\begin{align*}10(12)-44\end{align*}
Notice that you can put parentheses around the 12 to keep it separate from the number 10.
Now, simplify the expression using the order of operations. You will need to multiply first and then subtract.
10(12)44==1204476\begin{align*}\begin{array}{rcl} 10(12)-44 &=& 120-44 \\ &=& 76 \end{array}\end{align*}
The answer is 76.
Here is another example that involves division.
Evaluate the expression x3+2\begin{align*}\frac{x}{3}+2\end{align*} for x=24\begin{align*}x=24\end{align*}.
First, remember that a fraction bar is like a division sign. x3\begin{align*}\frac{x}{3}\end{align*} is the same as x÷3\begin{align*}x \div 3\end{align*}.
Next, substitute 24 in for the letter x\begin{align*}x\end{align*} in the expression.
243+2\begin{align*}\frac{24}{3}+2\end{align*}
Now, simplify the expression using the order of operations. You will need to divide first and then add.
243+2==8+210\begin{align*}\begin{array}{rcl} \frac{24}{3}+2 &=&8+2\\ &=& 10 \end{array} \end{align*}
The answer is 10.
### Examples
#### Example 1
Earlier, you were given a problem about Shelly and her bracelet business. Shelly is selling bracelets this summer and her profit for selling b\begin{align*}b\end{align*} bracelets is given by the expression 4b150\begin{align*}4b-150\end{align*}. Shelly wants to calculate what her profit will be if she sells 50 bracelets.
To calculate her profit from selling 50 bracelets, Shelly needs to evaluate the expression 4b150\begin{align*}4b-150\end{align*} for b=50\begin{align*}b=50\end{align*}.
First, substitute 50 in for the letter b\begin{align*}b\end{align*} in the expression.
4(50)150\begin{align*}4(50)-150\end{align*}
Now, simplify the expression using the order of operations. You will need to multiply first and then subtract.
4(50)150==20015050\begin{align*}\begin{array}{rcl} 4(50)-150 &=& 200-150\\ &=& 50 \end{array}\end{align*}
Shelly's profit from selling 50 bracelets would be \$50.
#### Example 2
Evaluate x75\begin{align*}\frac{x}{7}-5\end{align*} if x\begin{align*}x\end{align*} is 49.
First, substitute 49 in for the letter x\begin{align*}x\end{align*} in the expression.
4975\begin{align*}\frac{49}{7}-5\end{align*}
Now, simplify the expression using the order of operations. You will need to divide first and then subtract.
4975==752\begin{align*}\begin{array}{rcl} \frac{49}{7}-5 &=& 7-5 \\ &=& 2 \end{array}\end{align*}
The answer is 2.
#### Example 3
Evaluate 4x9\begin{align*}4x-9\end{align*} if x\begin{align*}x\end{align*} is 20.
First, substitute 20 in for the letter x\begin{align*}x\end{align*} in the expression.
4(20)9\begin{align*}4(20)-9\end{align*}
Now, simplify the expression using the order of operations. You will need to multiply first and then subtract.
4(20)9==80971\begin{align*}\begin{array}{rcl} 4(20)-9 &=& 80-9 \\ &=& 71 \end{array}\end{align*}
The answer is 71.
#### Example 4
Evaluate 5y+6\begin{align*}5y+6\end{align*} if y\begin{align*}y\end{align*} is 9.
First, substitute 9 in for the letter y\begin{align*}y\end{align*} in the expression.
5(9)+6\begin{align*}5(9)+6\end{align*}
Now, simplify the expression using the order of operations. You will need to multiply first and then add.
5(9)+6==45+651\begin{align*}\begin{array}{rcl} 5(9)+6 &=& 45+6\\ &=& 51 \end{array}\end{align*}
The answer is 51.
#### Example 5
Evaluate a48\begin{align*}\frac{a}{4}-8\end{align*} if a\begin{align*}a\end{align*} is 36.
First, substitute 36 in for the letter a\begin{align*}a\end{align*} in the expression.
3648\begin{align*}\frac{36}{4}-8\end{align*}
Now, simplify the expression using the order of operations. You will need to divide first and then subtract.
3648==981\begin{align*}\begin{array}{rcl} \frac{36}{4}-8 &=& 9-8\\ &=& 1 \end{array}\end{align*}
The answer is 1.
### Review
Evaluate each expression if the given value of r\begin{align*}r\end{align*} is 9.
1. r3\begin{align*}\frac{r}{3}\end{align*}
2. 63r\begin{align*}63-r\end{align*}
3. 11r\begin{align*}11r\end{align*}
4. 2r+7\begin{align*}2r+7\end{align*}
5. 3r+r\begin{align*}3r+r\end{align*}
6. 4r2r\begin{align*}4r-2r\end{align*}
7. r+5r\begin{align*}r+5r\end{align*}
8. 12r1\begin{align*}12r-1\end{align*}
Evaluate each expression for h=12\begin{align*}h=12\end{align*}.
9. 703h\begin{align*}70-3h\end{align*}
10. 6h+6\begin{align*}6h+6\end{align*}
11. 4h9\begin{align*}4h-9\end{align*}
12. 11+h4\begin{align*}11+\frac{h}{4}\end{align*}
13. 3h+h\begin{align*}3h+h\end{align*}
14. 2h+5h\begin{align*}2h+5h\end{align*}
15. 6h2h\begin{align*}6h-2h\end{align*}
### Answers for Review Problems
To see the Review answers, open this PDF file and look for section 1.4.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
Algebraic Expression
An expression that has numbers, operations and variables, but no equals sign.
Evaluate
To evaluate an expression or equation means to perform the included operations, commonly in order to find a specific value.
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
### Image Attributions
1. [1]^ License: CC BY-NC 3.0
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## Want to keep learning?
This content is taken from the University of Padova's online course, Precalculus: the Mathematics of Numbers, Functions and Equations. Join the course to learn more.
3.15
## Precalculus
Skip to 0 minutes and 10 seconds We’re going to develop a general strategy for finding the roots of a polynomial. It will be based on the factor theorem. And it will involve finding the root factorization of the polynomial. Now in order to reveal, fully, the structure of a polynomial, we would really like to factor it into the following form– a product of linear factors of the form x minus r and then a final term Q of x where Q is irreducible, has no roots. A general strategy for doing this is the following. First, we find a root r of the polynomial P by some means. We then use polynomial division and the factor theorem to divide P by x minus r.
Skip to 1 minute and 1 second And we know that it will be divisible, so we’ll have P equal x minus r times another polynomial P1. P1 will be of lower degree than p. We now repeat the process. We look at P1. We find a root of P1. We take out a linear factor. And we go on until we have the root factorization. So you see that we’re using so-called order reduction. At each step, the new polynomial will have degree less than the previous one. Let’s illustrate this with the following example. We want to find the root factorization of a certain polynomial of degree 4. First step, we observe that P(1) equals 0, as you can easily check. It follows that 1 is a root of P.
Skip to 1 minute and 48 seconds Therefore, by the factor theorem, P will be divisible by x minus 1. Now we can perform this long division. I will spare you the gruesome details. But here’s the answer and a certain polynomial of degree 3 is the quotient. We now proceed to study that polynomial Q of degree 3. We observe that Q vanishes at minus 3. That is Q of minus 3 is 0. That is minus 3 is a root. That means that x plus 3 will divide Q by the factor theorem. When we do that long division, then a certain quadratic x squared plus 2 appears. That quadratic is clearly irreducible. It can never be 0. It has no roots. So the procedure has ended.
Skip to 2 minutes and 36 seconds And now we get the root factorization of P. Now when you look at this example and the way we’ve solved it, surely that comes to mind to ask but how did we find the root 1 for the polynomial P? How did we find the root minus 3 for the polynomial Q? To be perfectly honest, we guessed. However, we guessed in an educated way. I will explain what I mean. Consider the following example. You have a polynomial of degree 3 and you would like to guess its roots.
Skip to 3 minutes and 14 seconds Well you decide that you are feeling lucky and you try x equals 1 minus 1, 2 minus 2, 3 minus 3– that is you plug them into P and see if you get 0, actually none of these give you 0. None of these values are roots. Well, you didn’t guess in an educated way because it turns out we could have known a priori that the only possible rational roots of this polynomial are plus and minus 1 and plus and minus 1/2. Why is that? It’s because of something called the rational root theorem which I now state. Here it is. Suppose we have a polynomial with integer coefficients whose constant term is non-zero. It might have rational roots.
Skip to 4 minutes and 0 seconds Any rational root can be expressed in the form p over q, where p and q are integers and the fraction is irreducible. In that case, the theorem says, the p must divide the constant term and the q must divide the leading coefficient. If you apply that statement to the example above, you will see that the q must divide 2 and the p must divide 1. That leads to 4 possibilities in all– plus and minus 1, plus and minus 1/2. When you try these four possibilities, it turns out that exactly one of them is a root, namely x equals a 1/2. Now note an important fact about this theorem.
Skip to 4 minutes and 45 seconds This theorem does not say there will be a rational root of the polynomial. It says if there is a root, then the p over q will be such that and so on. So it’s not a guarantee that there will be a rational root. It’s simply a good way to produce educated guesses. Here’s an example. We want to find– we have found, actually already, that x equals 1/2 is a root of this polynomial, how do we go on to go towards the root factorization? Well we divide by x minus 1/2. After the long division, we get a certain quadratic. Can we go on? Can we factor that quadratic? No, it’s irreducible. Why is it irreducible? Because its discriminant. It’s strictly negative.
Skip to 5 minutes and 33 seconds Therefore we have achieved the root factorization of p and p has exactly one root. Here is another example of a slightly different type. As usual, we want to begin by finding the first root that will give us our entry point to the procedure. What should be our guesses for that first root? By the rational root theorem, we know that any rational root p over q will be such that q has to be plus or minus 1 because the leading coefficient of our polynomial is 1. And the p has to be a divisor of 8. That produces, as you can see, 8 possibilities in all– plus and minus 1, 2, 4, 8. We try them all. Eight times we calculate P.
Skip to 6 minutes and 17 seconds And we discover that exactly one of these numbers is a root, namely 4. Given that 4 is a root, we divide P by x minus 4. We know it will divide evenly. And a certain quadratic polynomial then results. Is that quadratic polynomial susceptible to being factored further? Well, its discriminant is strictly positive. So, yes, it has two roots. And therefore, that quadratic polynomial will be the product of two linear terms, namely the two that you see. The root factorization of P is now complete. P has been factored down as far as it can be. A cubic polynomial can have no more than three roots.
# Finding roots
Root-finding strategy, the Rational Root Theorem
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# Measures of Central Tendency – Mean, Median, Mode
SSC CGL exam is conducted to hire personnel for government jobs in various government departments. SSC CGL exam is conducted in 4 phases or tiers. The SSC CGL Tier 2 exam consists of 4 papers: Quantitative Ability, English Comprehension, Statistics, General Studies ( Economics and Finance). Paper 3 of SSC CGL Tier 2 is Statistics and measures of central tendency is an important topic for statistics paper.
## Measures of Central Tendency
A measure of central tendency is a single value that tries to describe a complete data set by the process of identifying central positions within such data sets. In statistics mean, median and mode are the three most common measures of central tendency.
Let us now learn about the three measures of central tendency in detail below:
### Mean
Mean is the most common of the measures of central tendency. In statistics, the mean is usually referred to as the arithmetic mean. Mean can be used to calculate both discreet as well as continuous data. However, most often it is used for calculating continuous data sets.
In other words, the mean represents the sum of all values in a dataset divided by the number of values in the data set
It can be said that
Mean = Sum of all the values in a dataset / Total number of data values
#### Steps to calculate mean
Count the total number of data values and represent them as x1, x2 etc.)
Calculate the sum of all the data values (represented as )
Mean is denoted by the following formula
Or, it can be represented with the following formula
OR
In the case of data presented in a tabular form, the following formula is used to calculate the mean
Mean = ∑f x / ∑f
Where ∑f = N
N = Total number of observations
### Median
Median is defined as the middlemost value that is obtained after arranging the observations in ascending or descending order.
For example, consider this data 3,4,7,9,6
On arranging the data in ascending order we get,
3,4,6,7,9
In the above data set 6 is the middlemost value as it falls in the middle of other observations. Now, let’s see how we can determine the median in the case of grouped and ungrouped data.
#### Grouped Data
When the data presented is continuous and in the form of a frequency distribution, then the median can be calculated as follows:
Step 1: Find the median class
Now N = Total number of observations or sum of all the frequencies
Median is the middlemost value; therefore, a median class will be the class where the value of N/2 lies
Step 2: Now, in the case of grouped data, the median can be calculated using the following formula
Where,
l = Lower limit of the median class
c = Cumulative frequency of the class preceding or before the median class
h = Class size
f = Frequency of the median class
#### Ungrouped Data
For calculating the median of ungrouped data, the following steps are observed:
1. Arrange the data in ascending order
2. Assume the total number of observations as N
Now we have to see if the total number of observations(n) is odd or even
In case n is odd
Then median = (n + 1)/2th observation
In case n is even,
Then median = mean of (n/2)th observation and [(n/2)+1]th observation
### Mode
Mode is referred to as the observation that occurs most frequently in a data set or we can say mode is the observation with the highest frequency. The mode can be calculated for grouped and ungrouped data.
#### Grouped Data
Grouped data is said to be formed when the data is continuous. In such cases, the mode can be found out by first finding the modal class and then using the formula
Where
l = Lower limit of modal class
fm= frequency of modal class
f1= frequency of class preceding modal class,
f2= frequency of class succeeding modal class
h = class width
#### Ungrouped Data
For ungrouped data, the observation that occurs the maximum number of times is considered as mode.
Step 2: Create a free Oliveboard account or login using your existing Oliveboard account details
### Conclusion
SSC CGL Tier 2 exam is very important for the candidates and scoring well in this phase paves the way for progress to further phases. Statistics paper in SSC CGL Tier 2 is important for the JSO post. We hope that the blog on mean, median and mode will help candidates appearing for SSC CGL Tier 2 Statistics in their preparation.
## FAQs
What are three common measures of central tendency?
The three common measures of central tendency are mean, median and mode.
How do I calculate mode?
Mode is that value that occurs most frequently in a data set. In other words, the mode is the observation with the highest frequency in a data set.
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# Whats three consecutive even integers whose sum is 54?
Mar 13, 2018
16, 18, 20
#### Explanation:
To get to the next even number you have to 'jump over' an odd number. So every second number is even if you start from one.
Let the first even number be $n$ so we have:
$n , \textcolor{w h i t e}{\text{d")n+2, color(white)("d}} n + 4$
Adding these up (their sum) we have:
$\left(n\right) + \left(n + 2\right) + \left(n + 4\right) \leftarrow$The brackets just demonstrate the
$\textcolor{w h i t e}{\text{dddddddddddddddddddddd}}$grouping. They serve no other$\textcolor{w h i t e}{\text{dddddddddddddddddddddd}}$ purpose.
Their sum is $3 n + 6 = 54$
Subtract 6 from both sides
$\textcolor{w h i t e}{\text{dddddddddddd.}} 3 n = 48$
Divide both sides by 3
$\textcolor{w h i t e}{\text{ddddddddddddd.}} n = 16$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
First number $\textcolor{w h i t e}{\text{d..}} \to 16$
Second number $\to 18$
Third number $\textcolor{w h i t e}{\text{.d")->ul(20larr" Add as a check}}$
$\textcolor{w h i t e}{\text{ddddddddddddddd}} 54$
Mar 13, 2018
$16$, $18$, and $20$
#### Explanation:
Let $x =$ our first even integer.
We can write the problem as:
$x + \left(x + 2\right) + \left(x + 4\right) = 54$
$3 x + 6 = 54$
Rearrange and solve for $x$:
$3 x = 54 - 6$
$3 x = 48$
$x = 16$
Therefore our three consecutive even integers are $16$, $18$, and $20$.
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# How do you find the integral of 5/[sqrt(9x^2-16)] dx?
Apr 1, 2018
The answer is $= \frac{5}{3} \ln \left(| \frac{3}{4} x + \sqrt{{\left(\frac{3}{4} x\right)}^{2} - 1} |\right) + C$
#### Explanation:
Perform some simplification
$\frac{5}{\sqrt{9 {x}^{2} - 16}} = \frac{5}{4 \sqrt{\left({\left(\frac{3}{4} x\right)}^{2}\right) - 1}}$
Let $\frac{3}{4} x = \sec u$, $\implies$, $\frac{3}{4} \mathrm{dx} = \sec u \tan u \mathrm{du}$
$\sqrt{{\left(\frac{3}{4} x\right)}^{2} - 1} = \sqrt{{\sec}^{2} u - 1} = \tan u$
Therefore, the integral is
$I = \int \frac{5 \mathrm{dx}}{\sqrt{9 {x}^{2} - 16}} = \frac{5}{4} \int \frac{\frac{4}{3} \sec u \tan u \mathrm{du}}{\tan u}$
$= \frac{5}{3} \int \sec u \mathrm{du}$
$= \frac{5}{3} \int \frac{\sec u \left(\sec u + \tan u\right) \mathrm{du}}{\sec u + \tan u}$
Let $v = \sec u + \tan u$, $\implies$, $\mathrm{dv} = \left({\sec}^{2} u + \sec u \tan u\right) \mathrm{du}$
Therefore,
$I = \frac{5}{3} \int \frac{\mathrm{dv}}{v}$
$= \frac{5}{3} \ln \left(v\right)$
$= \frac{5}{3} \ln \left(\sec u + \tan u\right)$
$= \frac{5}{3} \ln \left(| \frac{3}{4} x + \sqrt{{\left(\frac{3}{4} x\right)}^{2} - 1} |\right) + C$
|
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# Elementary Algebra: Basic Operations with Polynomials
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Elementary Algebra: Basic Operations with Polynomials)
2. Which of the following expressions represents the product of 3 less than twice x and 2 more than the quantity 3 times x ?
A. -6x2 + 25x + 6
B. 6x2 + 5x + 6
C. 6x2 - 5x + 6
D. 6x2 - 5x - 6
E. 6x2 - 13x - 6
(Elementary Algebra: Substituting Values into Algebraic Expressions)
3. If x = -1 and y = 2, what is the value of the expression 2x3 - 3xy ?
A. 8
B. 4
C. -1
D. -4
E. -8
(Intermediate Algebra: Rational Expressions)
4. For all r 2, = ?
A.
B.
C.
D.
E.
(Coordinate Geometry: Linear Equations in Two Variables)
5. What is the equation of the line that contains the points with (x,y) coordinates(-3,7) and (5,-1) ?
A. y = 3x - 2
B. y = x + 10
C. y = - x + 8
D. y = - x +
E. y = -x + 4
https://brainmass.com/math/basic-algebra/elementary-algebra-basic-operations-polynomials-263499
#### Solution Preview
Elementary Algebra: Basic Operations with Polynomials)
2. Which of the following expressions represents the product of 3 less than twice x and 2 more than the quantity 3 times x ?
A. -6x2 + 25x + 6
B. 6x2 + 5x + 6
C. 6x2 - 5x + 6
D. 6x2 - 5x - 6
E. 6x2 - 13x - 6
Solution:
We can write 3 less than twice x as algebraic expression as (2x - 3)
And 2 more than the quantity 3 times x as algebraic expression as (3x + 2)
We have to find the product now
(2x - 3)(3x + 2) ...
#### Solution Summary
The expert examines the basic operations with polynomial elementary algebra.
\$2.19
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Synthetic Division of Polynomials
## Concise method of dividing polynomials when the divisor is x minus a constant.
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Practice Synthetic Division of Polynomials
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Synthetic Division of Polynomials
The volume of a rectangular prism is $2x^3 + 5x^2 - x - 6$ . Determine if $2x + 3$ is the length of one of the prism's sides.
### Guidance
Synthetic division is an alternative to long division from the previous concept. It can also be used to divide a polynomial by a possible factor, $x - k$ . However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.
#### Example A
Divide $2x^4-5x^3-14x^2-37x-30$ by $x - 2$ .
Solution: Using synthetic division, the setup is as follows:
Therefore, 2 is a solution, because the remainder is zero. The factored polynomial is $2x^3-x^2-16x+15$ . Notice that when we synthetically divide by $k$ , the “leftover” polynomial is one degree less than the original. We could also write $(x-2)(2x^3-x^2-16x+15)=2x^4-5x^3-14x^2+47x-30$ .
#### Example B
Determine if 4 is a solution to $f(x)=5x^3+6x^2-24x-16$ .
Using synthetic division, we have:
The remainder is 304, so 4 is not a solution. Notice if we substitute in $x = 4$ , also written $f(4)$ , we would have $f(4)=5(4)^3+6(4)^2-24(4)-16=304$ . This leads us to the Remainder Theorem.
Remainder Theorem: If $f(k) = r$ , then $r$ is also the remainder when dividing by $(x - k)$ .
This means that if you substitute in $x = k$ or divide by $k$ , what comes out of $f(x)$ is the same. $r$ is the remainder, but also is the corresponding $y-$ value. Therefore, the point $(k, r)$ would be on the graph of $f(x)$ .
#### Example C
Determine if $(2x - 5)$ is a factor of $4x^4-9x^2-100$ .
Solution: If you use synthetic division, the factor is not in the form $(x - k)$ . We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put $\frac{5}{2}$ up in the left-hand corner box. Also, not every term is represented in this polynomial. When this happens, you must put in zero placeholders. In this example, we need zeros for the $x^3-$ term and the $x-$ term.
This means that $\frac{5}{2}$ is a zero and its corresponding binomial, $(2x - 5)$ , is a factor.
Intro Problem Revisit
If $2x + 3$ divides evenly into $2x^3 + 5x^2 - x - 6$ then it is the length of one of the prism's sides.
If we use synthetic division, the factor is not in the form $(x - k)$ . We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put $\frac{-3}{2}$ up in the left-hand corner box.
When we perform the synthetic division, we get a remainder of 0. This means that $(2x + 3)$ is a factor of the volume. Therefore, it is also the length of one of the sides of the rectangular prism.
### Guided Practice
1. Divide $x^3-9x^2+12x-27$ by $(x + 3)$ . Write the resulting polynomial with the remainder (if there is one).
2. Divide $2x^4-11x^3+12x^2+9x-2$ by $(2x + 1)$ . Write the resulting polynomial with the remainder (if there is one).
3. Is 6 a solution for $f(x)=x^3-8x^2+72$ ? If so, find the real-number zeros (solutions) of the resulting polynomial.
1. Using synthetic division, divide by -3.
The answer is $x^2+6x-6-\frac{9}{x+3}$ .
2. Using synthetic division, divide by $-\frac{1}{2}$ .
The answer is $2x^3-12x^2+18x-\frac{2}{2x+1}$ .
3. Put a zero placeholder for the $x-$ term. Divide by 6.
The resulting polynomial is $x^2-2x-12$ . While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.
$x=\frac{2 \pm \sqrt{2^2-4(1)(-12)}}{2}=\frac{2 \pm \sqrt{4+48}}{2}=\frac{2 \pm 2 \sqrt{13}}{2}=1 \pm \sqrt{13}$
The solutions to this polynomial are 6, $1+\sqrt{13} \approx 4.61$ and $1-\sqrt{13} \approx -2.61$ .
### Vocabulary
Synthetic Division
An alternative to long division for dividing $f(x)$ by $k$ where only the coefficients of $f(x)$ are used.
Remainder Theorem
If $f(k) = r$ , then $r$ is also the remainder when dividing by $(x - k)$ .
### Practice
Use synthetic division to divide the following polynomials. Write out the remaining polynomial.
1. $(x^3+6x^2+7x+10) \div (x+2)$
2. $(4x^3-15x^2-120x-128) \div (x-8)$
3. $(4x^2-5) \div (2x+1)$
4. $(2x^4-15x^3-30x^2-20x+42) \div (x+9)$
5. $(x^3-3x^2-11x+5) \div (x-5)$
6. $(3x^5+4x^3-x-2) \div (x-1)$
7. Which of the division problems above generate no remainder? What does that mean?
8. What is the difference between a zero and a factor?
9. Find $f(-2)$ if $f(x)=2x^4-5x^3-10x^2+21x-4$ .
10. Now, divide $2x^4-5x^3-10x^2+21x-4$ by $(x + 2)$ synthetically. What do you notice?
Find all real zeros of the following polynomials, given one zero.
1. $12x^3+76x^2+107x-20; -4$
2. $x^3-5x^2-2x+10; -2$
3. $6x^3-17x^2+11x-2; 2$
Find all real zeros of the following polynomials, given two zeros.
1. $x^4+7x^3+6x^2-32x-32; -4, -1$
2. $6x^4+19x^3+11x^2-6x; 0, -2$
### Vocabulary Language: English
Oblique Asymptote
Oblique Asymptote
An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Oblique Asymptotes
Oblique Asymptotes
An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Remainder Theorem
Remainder Theorem
The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$.
Synthetic Division
Synthetic Division
Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.
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# Lesson 14
More Arithmetic with Complex Numbers
### Problem 1
Select all expressions that are equivalent to $$8+16i$$.
A:
$$2(4+8i)$$
B:
$$2i(8-4i)$$
C:
$$4(2i-4)$$
D:
$$4i(4-2i)$$
E:
$$\text-2i(\text-8-4i)$$
### Problem 2
Which expression is equivalent to $$(\text-4 + 3i)(2-7i)$$?
A:
$$\text-29 - 22i$$
B:
$$\text-29 + 34i$$
C:
$$13 - 22i$$
D:
$$13 + 34i$$
### Problem 3
Match the equivalent expressions.
### Problem 4
Write each expression in $$a+bi$$ form.
1. $$(\text-8 + 3i) - (2 +5i)$$
2. $$7i(4 - i)$$
3. $$(3i)^3$$
4. $$(3 + 5i)(4 + 3i)$$
5. $$(3i)(\text-2 i)(4i)$$
### Problem 5
Here is a method for solving the equation $$\sqrt{5+x}+10=6$$. Does the method produce the correct solution to the equation? Explain how you know.
\begin{align} \sqrt{5+x}+10 &= 6 \\ \sqrt{5+x} &= \text-4 &\text{ (after subtracting 10 from each side)} \\ 5+x &= 16 &\text{ (after squaring both sides)} \\ x &= 11 \\ \end{align}
### Solution
(From Unit 3, Lesson 7.)
### Problem 6
Write each expression in the form $$a+bi$$, where $$a$$ and $$b$$ are real numbers.
1. $$4(3-i)$$
2. $$(4+2i) + (8-2i)$$
3. $$(1+3i)(4+i)$$
4. $$i(3+5i)$$
5. $$2i \boldcdot 7i$$
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ISEE Middle Level Quantitative : Range
Example Questions
← Previous 1
Example Question #1 : Range
Which is the greater quantity?
(A) The midrange of the data set
(B) The midrange of the data set
(B) is greater
(A) is greater
It is impossible to determine which is greater from the information given
(A) and (B) are equal
(A) is greater
Explanation:
The midrange of a data set is the mean of its least and greatest elements. The midrange of the first data set is ; that of the second data set is . (A) is greater.
Example Question #2 : Range
Set is defined as:
Set is made by doubling the values in set .
What is the range of values in set ?
Explanation:
To find the range of a set of numbers, you do not even have to put them in order. You merely need to subtract the smallest value from the largest. Given the way of constructing set by doubling set 's values, the largest and smallest values in will directly correlate to the same in set :
Smallest:
Largest:
Therefore, the range is:
Example Question #3 : Range
The members of set are defined as the values for:
For values of between and .
What is the range of set ?
Explanation:
To find the range of a set of numbers, you do not even have to put them in order. You merely need to subtract the smallest value from the largest. Given the way that we construct set from the function , we merely need to use that function to find the smallest and largest values. Luckily, that is pretty easy for this question. The smallest will be and the largest will be
Smallest:
Largest:
Therefore, the range is:
Example Question #4 : Range
Set is defined as:
The members of set are defined by the function:
, where is a member of set .
So, for instance, set T contains because for , we get:
What is the range of set ?
Explanation:
We first need to determine the members of set . Using our function, we will get:
Our largest value is , and our smallest value is ; therefore, the range is
Example Question #5 : Range
Given the below set of numbers find the range:
Explanation:
Range for a set of data is defined as the difference between the biggest and smallest number.
First we find the biggest number which is 15, we then subtract the smallest number in this set which is negative 2 as shown below:
Remember, when subtracting a negative number we must add the numbers.
Example Question #6 : Range
Given the below set of numbers, find the range:
Explanation:
In order to find the range, we must subtract the smallest number from the biggest number.
We must convert the fractions to have a common denominator which is 10.
Therefore, the range of this set is
.
Example Question #7 : Range
Given the below set of numbers, find the range:
Explanation:
In order to find the range, we must subtract the smallest number from the biggest number.
We must convert the fractions to have a common denominator which is 10.
Therefore, the range of this set is
.
Example Question #9 : Find Range
Find the range of the data set provided:
Explanation:
In order to answer this question correctly, we need to recall the definition of range:
Range: The range of a data set is the difference between the highest value and the lowest value in the set.
In order to find the range, we need to first organize the data from least to greatest to find the lowest and highest values:
Next, we can solve for the difference between the highest value and the lowest value:
The range for this data set is
Example Question #11 : Data Analysis
Find the range of the data set provided:
Explanation:
In order to answer this question correctly, we need to recall the definition of range:
Range: The range of a data set is the difference between the highest value and the lowest value in the set.
In order to find the range, we need to first organize the data from least to greatest to find the lowest and highest values:
Next, we can solve for the difference between the highest value and the lowest value:
The range for this data set is
Example Question #12 : Data Analysis
Find the range of the data set provided:
Explanation:
In order to answer this question correctly, we need to recall the definition of range:
Range: The range of a data set is the difference between the highest value and the lowest value in the set.
In order to find the range, we need to first organize the data from least to greatest to find the lowest and highest values:
Next, we can solve for the difference between the highest value and the lowest value:
The range for this data set is
← Previous 1
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## Presentation on theme: "Chapter 16 Quadratic Equations."— Presentation transcript:
Solving Quadratic Equations by Completing the Square
§ 16.2 Solving Quadratic Equations by Completing the Square
Completing the Square Solving a Quadratic Equation by Completing a Square If the coefficient of x2 is NOT 1, divide both sides of the equation by the coefficient. Isolate all variable terms on one side of the equation. Complete the square (half the coefficient of the x term squared, added to both sides of the equation). Factor the resulting trinomial. Use the square root property.
Solving Equations Example Solve by completing the square. y2 + 6y = 8
y = 4 or 2
Solving Equations Example (y + ½)2 = Solve by completing the square.
y2 + y – 7 = 0 y2 + y = 7 y2 + y + ¼ = 7 + ¼ (y + ½)2 =
Solving Equations Example Solve by completing the square.
2x2 + 14x – 1 = 0 2x2 + 14x = 1 x2 + 7x = ½ x2 + 7x = ½ = (x + )2 =
The Quadratic Formula Another technique for solving quadratic equations is to use the quadratic formula. The formula is derived from completing the square of a general quadratic equation.
The Quadratic Formula A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions.
Solve 11n2 – 9n = 1 by the quadratic formula. 11n2 – 9n – 1 = 0, so a = 11, b = -9, c = -1
Solve x2 + x – = 0 by the quadratic formula. x2 + 8x – 20 = (multiply both sides by 8) a = 1, b = 8, c = 20
Solve x(x + 6) = 30 by the quadratic formula. x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution.
Example x y Graph y = 2x2 – 4. (–2, 4) (2, 4) x y 2 4 1 –2 (–1, – 2) (1, –2) –4 –1 –2 (0, –4) –2 4
Intercepts of the Parabola
Although we can simply plot points, it is helpful to know some information about the parabola we will be graphing prior to finding individual points. To find x-intercepts of the parabola, let y = 0 and solve for x. To find y-intercepts of the parabola, let x = 0 and solve for y.
Characteristics of the Parabola
If the quadratic equation is written in standard form, y = ax2 + bx + c, 1) the parabola opens up when a > 0 and opens down when a < 0. 2) the x-coordinate of the vertex is To find the corresponding y-coordinate, you substitute the x-coordinate into the equation and evaluate for y.
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# How do you evaluate the integral of x*(cos(x^2))^5?
Mar 23, 2016
${\sin}^{5} \frac{{x}^{2}}{10} - {\sin}^{3} \frac{{x}^{2}}{3} + \sin \frac{{x}^{2}}{2} + C$
#### Explanation:
First, let
$u = {x}^{2} \text{ "=>" } \mathrm{du} = 2 x \mathrm{dx}$
So, we have
$\int x {\cos}^{5} \left({x}^{2}\right) \mathrm{dx}$
Rearrange the $x$ and $\mathrm{dx}$ and add a $2$ to achieve $2 x \mathrm{dx}$. To balance the $2$ inside the integral, multiply by $1 / 2$ outside of the integral.
$= \frac{1}{2} \int {\underbrace{{\cos}^{5} \left({x}^{2}\right)}}_{{\cos}^{5} \left(u\right)} \cdot {\underbrace{2 x \mathrm{dx}}}_{\mathrm{du}}$
Substituting, this equals
$= \frac{1}{2} \int {\cos}^{5} \left(u\right) \mathrm{du}$
Before we integrate this, we want to modify this so that we can have a $v$ and $\mathrm{dv}$ term both in the integral. We can do this by writing ${\cos}^{5} \left(u\right)$ as
${\cos}^{5} \left(u\right) = \cos \left(u\right) {\cos}^{4} \left(u\right) = \cos \left(u\right) {\left({\cos}^{2} \left(u\right)\right)}^{2} = \cos \left(u\right) {\left(1 - {\sin}^{2} \left(u\right)\right)}^{2}$
Giving us the integral
$= \frac{1}{2} \int \cos \left(u\right) {\left(1 - {\sin}^{2} \left(u\right)\right)}^{2} \mathrm{du}$
Now, we can substitute again:
$v = \sin \left(u\right) \text{ "=>" } \mathrm{dv} = \cos \left(u\right) \mathrm{du}$
Rearrange the integral:
$= \frac{1}{2} \int {\underbrace{{\left(1 - {\sin}^{2} \left(u\right)\right)}^{2}}}_{{\left(1 - {v}^{2}\right)}^{2}} \cdot {\underbrace{\cos \left(u\right) \mathrm{du}}}_{\mathrm{dv}}$
Which equals:
$= \frac{1}{2} \int {\left(1 - {v}^{2}\right)}^{2} \mathrm{dv}$
Distribute ${\left(1 - {v}^{2}\right)}^{2}$.
$= \frac{1}{2} \int \left(1 - 2 {v}^{2} + {v}^{4}\right) \mathrm{dv}$
Splitting up the integral, we obtain:
$= \frac{1}{2} \int {v}^{4} \mathrm{dv} - 2 \left(\frac{1}{2}\right) \int {v}^{2} \mathrm{dv} + \frac{1}{2} \int 1 \mathrm{dv}$
Using the rule:
$\int {v}^{n} \mathrm{dv} = {v}^{n + 1} / \left(n + 1\right) + C$
Yields, after integrating term by term:
$= \frac{1}{2} \left({v}^{5} / 5\right) - {v}^{3} / 3 + \frac{1}{2} \left(v\right) + C$
$= {v}^{5} / 10 - {v}^{3} / 3 + \frac{v}{2} + C$
Using $v = \sin \left(u\right)$:
$= {\sin}^{5} \frac{u}{10} - {\sin}^{3} \frac{u}{3} + \sin \frac{u}{2} + C$
Now using $u = {x}^{2}$:
$= {\sin}^{5} \frac{{x}^{2}}{10} - {\sin}^{3} \frac{{x}^{2}}{3} + \sin \frac{{x}^{2}}{2} + C$
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Fun with numbers and triangles
# Fun with numbers and triangles
During my interview with Arthur Benjamin, he spoke briefly about the joy of discovering something in mathematics, even if one is relatively certain that the given relationship has been known for years, or even centuries. Along the same lines, I’ve been playing with right triangles of late.
They are called Pythagorean triples—three numbers, a, b, and c, where aSquared+bSquared=cSquared, and a b and c are all whole numbers. I wanted a way to generate them.
I started by noticing that the difference between two squares, such that one root is one less than the other, is always an odd number.
2*2=4
1*1=1
4-1=3
3*3=9
2*2=4
9-4=5
4*4=16
3*3=9
16-9=7
The result climbs up the odd number line, 3, 5, 7, 9, 11…
Another way to put this, is to say that the difference between consecutive whole number squares is on the line defined by 2x+1.
This can be shown algebraically:
(x+1)Squared-xSquared=2x+1
Expanding (x+1)Squared gives us (xSquared+2x+1)
So (xSquared+2x+1)-xSquared=2x+1
Add xSquared to each side and we get:
XSquared+2x+1=xSquared+2x+1
Take xSquared from each side and we get:
2x+1=2x+1
Take 2x from each side and we get
1=1
That’s about as true as numbers can get.
I’m not the first to have noticed that relation, but I did run across it independently.
Next, in order to confine the results to whole numbers, we substitute some whole number that is greater than or equal to 1 for x, which will give you side a.
2*2+1=3
Next we square a: 3*3=9
Subtract 1: 9-1=8
And divide by 2: 8/2=4
That gives us side b=4.
Next since we know the difference in length from b to c is 1, c=b+1: 4+1=5
3, 4, 5
3*3=9
4*4=16
9+16=25
5*5=25
Let’s try using 2 in place of x. 2x+1
2*2=4, 4+1=5, a=5
5*5=25, 25-1=24, 24/2=12, b=12
12+1=13 c=13
5, 12, 13
5*5=25
12*12=144
25+144=169
13*13=169
This is much easier to do in your head than it is to write it out.
3*2=6+1=7
7*7=49-1=48/2=24
24+1=25
7, 24, 25
4*2=8+1=9
9*9=81-1=80/2=40
40+1=41
9, 40, 41
When I looked it up, it turned out that Euclid used this trick to prove that the set of Pythagorean triples is infinite, since the set of odd numbers is infinite and you can use any odd number to generate sides a b and c.
Q.E.D.
Next, I got to wondering if you could generalize the method, so that you could generate triples when the difference between side b and c, c-b could equal any whole number.
Yes you can! If d=c-b than the line that your numbers need to sit on is (2*d)x+dSquared, where x is any whole number greater or equal to 0, unless d=1.
If d=1, the line is 2x+1. If x=0, you end up with 1, 0, 1. And while 1*1+0*0 does equal 1*1, that’s not really a triangle; it’s a line segment.
By the by, when you use d=1 you get 1*2=2 and 1*1=1, so the line is 2x+1, which we already knew.
Let’s try d=2.
2*2=4 and 2*2=4, so our line is given by 4x+4
Put 0 in place of x.
4*0=0+4=4, so a=4
4*4=16-4=12/4=3 so b=3
3+2=5 so c=5
4, 3, 5, which is our old friend 3, 4, 5
Using 4x+4 and replacing x with 1
4*1=4+4=8, a=8
8*8=64-4=60/4=15, so b=15
15+2=17
8, 15, 17
Using x=2, 2*4=8+4=12, a=12
12*12=144-4=140/4=35, b=35
35+2=37
12, 35, 37
Now let’s set d to 3
3*2=6 and 3*3=9, so the line is 6x+9
Starting with 0
6*0=0+9=9, a=9
9*9=81-9=72/6=12, b=12
12+3=15
9, 12, 15
X=1
6*1=6+9=15 a=15
15*15=225-9=216/6=36, b=36
36+3=39
15, 36, 39
X=2
6*2=12+9=21, a=21
21*21=441-9=432/6=72, b=72
72+3=75
21, 72, 75
In fact, there are simpler methods for making triples, but I found this all on my own, and I can’t even describe the joy of it, when the numbers click together and it all makes sense.
As any whole number greater than 0 can be substituted for d, and the set of whole numbers greater than 0 is infinite, this adds another layer of infinity on top of the one Euclid found, just in case one infinity wasn’t enough
Q.E.D.
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# Example Construction of Frequency Distribution
Construct a frequency distribution with suitable class interval size of marks obtained by 50 students of a class are given below:
23, 50, 38, 42, 63, 75, 12, 33, 26, 39, 35, 47, 43, 52, 56, 59, 64, 77, 15, 21, 51, 54, 72, 68, 36, 65, 52, 60, 27, 34, 47, 48, 55, 58, 59, 62, 51, 48, 50, 41, 57, 65, 54, 43, 56, 44, 30, 46, 67, 53
Solution:
Arrange the marks in ascending order as
12, 15, 21, 23, 26, 27, 30, 33, 34, 35, 36, 38, 39, 41, 42, 43, 43, 44, 46, 47, 47, 48, 48, 50, 50, 51, 51, 52, 52, 53, 54, 54, 55, 56, 56, 57, 58, 59, 59, 60, 62, 63, 64, 65, 65, 67, 68, 72, 75, 77
Minimum Value = $12$ Maximum = $77$
Range = Maximum Value – Minimum Value = $77 - 12$ = $65$
Number of Classes = $1 + 3.322\log N$
Number of Classes = $1 + 3.322\log 50$
Number of Classes = $1 + 3.322(1.69897)$
Number of Classes = $1 + 5.64$ = $6.64$ or $7$ approximate
Class Interval Size ($h$) = $\frac{{Range}}{{No.{\text{ }}of{\text{ }}Classes}}$ = $\frac{{65}}{7}$ = $9.3$ or $10$
Marks Class Limits C.L Number of Students $f$ Class Boundary C.B Class Marks $x$ $10 - 19$ $2$ $9.5 - 19.5$ $\frac{{10 + 19}}{2} = 14.5$ $20 - 29$ $4$ $19.5 - 29.5$ $\frac{{20 + 29}}{2} = 24.5$ $30 - 39$ $7$ $29.5 - 39.5$ $\frac{{30 + 39}}{2} = 34.5$ $40 - 49$ $10$ $39.5 - 49.5$ $\frac{{40 + 49}}{2} = 44.5$ $50 - 59$ $16$ $49.5 - 59.5$ $\frac{{50 + 59}}{2} = 54.5$ $60 - 69$ $8$ $59.5 - 69.5$ $\frac{{60 + 69}}{2} = 64.5$ $70 - 79$ $3$ $69.5 - 79.5$ $\frac{{70 + 79}}{2} = 74.5$ $50$
Note: For finding the class boundaries, we take half of the difference between lower class limit of the 2nd class and upper class limit of the 1st class$\frac{{20 - 19}}{2} = \frac{1}{2} = 0.5$. This value is subtracted from lower class limit and added in upper class limit to get the required class boundaries.
Frequency Distribution by Exclusive Method
Class Boundray C.B Frequency $f$ $10 - 19$ $2$ $20 - 29$ $4$ $30 - 39$ $7$ $40 - 49$ $10$ $50 - 59$ $16$ $60 - 69$ $8$ $70 - 79$ $3$ $50$
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Have you ever compared your random draws to find that they actually don’t seem that random at all? In fact, you might’ve noticed a pretty substantial overlap in the number of people listed in each selection. Well, it turns out that this is actually quite expected. And with a short lesson in probability, we’ll show you why.
For starters, let’s consider a consortium with 100 people in it. We do an initial draw of 20%. So, the probability of any one person being selected is 20 in 100 which can be simplified to 1 in 5. If we do another draw, how much overlap is expected between the first and second draws?
Probability tells us that, if you know the probability of 2 events happening independently, you can find the probability that they’ll happen together by multiplying their individual probabilities. In this case, our first event is Jim getting selected in the first draw. The probability of that occurring is 1 in 5 as noted above. The second event it Jim getting selected in another draw. We’re doing exactly the same thing as before so the probability is the same: 1 in 5. To get the probability that both of these events occur, we multiply the two probabilities (1/5 x 1/5) to get the probability of Jim getting selected for both draws: 1/25.
To summarize, if we do 2 random draws of 20% of the people from a 100-person consortium, we should actually expect the draws to have 100 x 1/25 = 4 people in common out of the 20 selected. It seems much higher than what our intuition would tell us but, like our math teachers told us, probability never lies.
So, let’s test this. To produce the image below I performed 20 real random selections in TestVault and, for each pair (e.g., selection 2 and selection 1, selection 3 and selection 1, selection 3 and selection 2, etc.), I found the number of overlaps. Then, I tallied the number of occurrences of each number. For example, you can see in the chart that there were 56 pairs that had 4 people in common. In fact, it is clear from the chart that 4 is the clear winner in number of overlaps and that the other frequencies circle around 4. However, while 4 is the winner, we saw overlaps of 0 people all the way up to 10 (half of the number of people selected).
As you can see, we can’t make any real assumptions on how much overlap to expect for any two selections. That number can vary tremendously. I hope, however, that this article helps to inform your intuition for future selections. Happy testing and good luck.
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# OBJECTIVE
After completing this unit, you should be able to:
• Understand the cartesian coordinate system.
• Understand the Cartesian coordinates of the plane.
• Understand the Cartesian coordinates of three-dimensional space.
• Explain the difference between polar and rectangular coordinated.
• Identify the programmable axes on a CNC machining.
# THE CARTESIAN COORDINATE SYSTEM
Cartesian coordinates allow one to specify the location of a point in the plane, or in three-dimensional space. The Cartesian coordinates or rectangular coordinates system of a point are a pair of numbers (in two-dimensions) or a triplet of numbers (in three-dimensions) that specified signed distances from the coordinate axis. First we must understand a coordinate system to define our directions and relative position. A system used to define points in space by establishing directions(axis) and a reference position(origin). A coordinate system can be rectangular or polar.
Just as points on the line can be placed in one to one correspondence with the real number line, so points in plane can be placed in one to one correspondence with pairs of real number line by using two coordinate lines. To do this, we construct two perpendicular coordinate line that intersect at their origins; for convenience. Assign a set of equally space graduations to the x and y axes starting at the origin and going in both directions, left and right (x axis) and up and down (y axis) point along each axis may be established. We make one of the number lines vertical with its positive direction upward and negative direction downward. The other number lines horizontal with its positive direction to the right and negative direction to the left. The two number lines are called coordinate axes; the horizontal line is the x axis, the vertical line is the y axis, and the coordinate axes together form the Cartesian coordinate system or a rectangular coordinate system. The point of intersection of the coordinate axes is denoted by O and is the origin of the coordinate system. See Figure 1.
Figure 1
It is basically, Two Real Number Lines Put Together, one going left-right, and the other going up-down. The horizontal line is called x-axis and the vertical line is called y-axis.
# The Origin
The point (0,0) is given the special name “The Origin”, and is sometimes given the letter “O”.
# Real Number Line
The basis of this system is the real number line marked at equal intervals. The axis is labeled (X, Y or Z). One point on the line is designated as the Origin. Numbers on one side of the line are marked as positive and those to the other side marked negative. See Figure 2.
Figure 2. X-axis number line
# Cartesian coordinates of the plane
A plane in which a rectangular coordinate system has been introduced is a coordinate plane or an x-y-plane. We will now show how to establish a one to one correspondence between points in a coordinate plane and pairs of real number. If A is a point in a coordinate plane, then we draw two lines through A, one perpendicular to the x-axis and one perpendicular to the y-axis. If the first line intersects the x-axis at the point with coordinate x and the second line intersects the y-axis at the point with coordinate y, then we associate the pair (x,y) with the A( See Figure 2). The number a is the x-coordinate or abscissa of P and the number b is the y-coordinate or ordinate of p; we say that A is the point with coordinate (x,y) and denote the point by A(x,y). The point (0,0) is given the special name “The Origin”, and is sometimes given the letter “O”.
Abscissa and Ordinate:
The words “Abscissa” and “Ordinate” … they are just the x and y values:
• Abscissa: the horizontal (“x”) value in a pair of coordinates: how far along the point is.
• Ordinate: the vertical (“y”) value in a pair of coordinates: how far up or down the point is.
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# 9th grade math skills: Find out what you need to know for your student
In ninth grade, your child will be learning a wide range of math skills in algebra, geometry and other areas.
## Algebra
Overview
For high school students, math skills and understandings are organized not by grade level but by concept. In algebra, students work with creating and reading expressions, rational numbers and polynomials, and the conventions of algebraic notations. They apply these skills and understandings to solve real-world problems.
Understanding equations
Understand an equation as a mathematical statement that uses letters to represent unknown numbers (such as 2x-6y+z=14) and is a statement of equality between two expressions (“this equals that”). Explain each step in solving a simple equation, and construct a practical argument to justify a solution method. Graph these equations on coordinate axes with labels and scales.
Rewriting expressions
Identify ways to rewrite the structure of an expression.
Equation solutions
Understand that some equations have no solutions in a given number system, but have a solution in a larger system. For example, the solution of x + 1 = 0 is an integer, not a whole number; the solution of 2x + 1 = 0 is a rational number, not an integer; the solutions of x² – 2 = 0 are real numbers, not rational numbers; and the solutions of x² + 2 = 0 are complex numbers, not real numbers.
Polynomials
Add, subtract, and multiply polynomials (expressions with multiple terms, such as 5xy² + 2xy - 7). Understand the relationship between the zeros and the factors of polynomials.
Polynomial identities
Use polynomial identities to solve real-world problems.
Example:
A rectangular garden has a length of x + 2 ft. and a width of x + 8 ft. What must x be in order for the garden to have an area of 91 sq. ft.?
One-variable equations
Create equations and inequalities in one variable, and use them to solve problems, including weighted averages, calculation of mortgage and interest rates, and rate of travel.
Example:
A plane takes off from Chicago O’Hare airport, heading east and traveling at 580 miles an hour. Another plane takes off from O’Hare at the same time, heading west and traveling at 530 miles an hour. The two planes will be 1,000 miles apart in how many hours?
##### Graphs
Represent, interpret, and solve equations and inequalities on graphs, plotted in the coordinate plane, and using technology to graph the functions and make tables of values.
## Geometry
Overview
For high school students, math skills and understandings are organized not by grade level but by concept. In geometry, students work primarily with plane, or Euclidean geometry (with and without coordinates). Students build on geometry concepts learned through 8th grade, using more precise definitions and develop careful proofs of theorems (statements that can be proved true).
Transformation
Understand geometric transformation (moving a shape so it is in a different position, but still has same size, area, angles, and lengths) – especially rigid motions: translations, rotations, reflections, and combinations of these – involving angles, circles, perpendicular lines, parallel lines, and line segments.
Geometric theorems
Understand and prove geometric theorems about lines and angles, triangles, parallelograms, and circles. For example, Pythagorean Theorem, Line Intersection Theorem, Exterior Angle Theorem.
Trigonometry
Understand trigonometry as a measurement of triangles (and circles, such as orbits). Apply trigonometry to general triangles. Define the sine, cosine, and tangent trigonometric ratios.
Algebraic reasoning
Understand and use algebraic reasoning to prove geometric theorems.
Volume formulas
Explain volume formulas and use them to solve problems.
Example:
What is the volume of a cylinder that is 10m high, and has a radius of 9m? (Use π = 3.14)
Real-life situations
Apply geometric concepts to model real-life situations.
• Use measures and properties of geometric shapes to describe objects – for example, model a tree trunk or a human torso as a cylinder.
• Apply concepts of density based on area and volume – for example, persons per square mile, BTUs per cubic foot.
• Design objects or structures to satisfy specific physical constraints or minimize cost.
## Numbers
Overview
For high school students, math skills and understandings are organized not by grade level but by concept. In number and quantity, students extend their understanding of number to imaginary numbers and complex numbers, and work with a variety of measurement units in modeling. Emphasis is on using numbers – in calculations, equations, and measurements – to solve real-world problems, including those that students themselves quantify and define.
Rational and irrational numbers
Understand and explain why:
• the sum of two rational numbers is rational (sum can be written as a fraction or decimal)
• the sum of a rational number and an irrational number is irrational (sum cannot be written as a fraction; written in decimal form, is non-repeating and unending)
Interpreting and converting units
Consistently choose and interpret units in formulas; scale drawings and figures in graphs, data displays and maps. Convert rates and measurements (grams to centigrams, inches to feet, meters to kilometers, miles to kilometers, square inches into square feet, etc.).
Real-world problems
Use measurement units in modeling to solve real-world problems – for example: acceleration, currency conversions, per capita income, safety statistics, disease incidence, batting averages, etc.)
Complex numbers
Understand that complex numbers are formed by real numbers and imaginary numbers – imaginary numbers that, when squared, give a negative result: i² = -1. Use the relation i² = -1 to add, subtract and multiply complex numbers.
Understanding vectors
Understand a vector as a quantity that has both magnitude (length) and direction. Add and subtract vectors.
Velocity
Solve problems involving velocity and other quantities represented by vectors.
Example:
• Drew leaves home for a morning walk. He goes 13.5 km south and 5.5 km west. What is his velocity relative to his brother, who is still asleep in bed at home?
• Jack is doing push-ups. Which requires smaller muscular force – if his hands are 0.25m apart, or his hands are 0.5m apart?
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# How do you write the expression 4^(4/3) in radical form?
Apr 18, 2017
$\sqrt[3]{{4}^{4}}$
#### Explanation:
${a}^{\frac{\textcolor{red}{n}}{\textcolor{g r e e n}{m}}} = \sqrt[\textcolor{g r e e n}{m}]{{a}^{\textcolor{red}{n}}}$
So ${4}^{\frac{\textcolor{red}{4}}{\textcolor{g r e e n}{3}}} = \sqrt[\textcolor{g r e e n}{3}]{{4}^{\textcolor{red}{4}}}$
Apr 18, 2017
See the entire solution process below:
#### Explanation:
First, we can use this rule of exponents to rewrite the expression:
${x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}} = {\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}}$
${4}^{\frac{4}{3}} = {4}^{\textcolor{red}{4} \times \textcolor{b l u e}{\frac{1}{3}}} = {\left({4}^{\textcolor{red}{4}}\right)}^{\textcolor{b l u e}{\frac{1}{3}}} = {256}^{\frac{1}{3}}$
We can now use this rule of exponents to write this expression in radical form:
${x}^{\frac{1}{\textcolor{red}{n}}} = \sqrt[\textcolor{red}{n}]{x}$
${256}^{\frac{1}{\textcolor{red}{3}}} = \sqrt[\textcolor{red}{3}]{256}$
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# Belarus MO 2004 Problem 7
Problem 7 of the Belarus Mathematical Olympiad 2004
Let there be given two similar triangles such that the altitudes of the first triangle are equal to the sides of the other. Find the largest possible value of the similarity ratio of the triangles.
Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Alex Bogomolny, Ph.D.)
Let the first triangle be ABC and the feet from A, B and C to the opposite sides be D, E and F, respectively.
Now let BC = a, AC = b, AB = c and AD = d, BE = e and CF = f.
Assume abc. Because twice the area of triangle ABC = ad = be = cf, our assumption makes fed.
To find the largest possible value of the similarity ratio of the triangles we need to find the largest possible ratio \frac{f}{a}
or the largest possible cos∠FCB.
The similarity of the triangles ADB and CFB gives us \frac{d}{c} = \frac{f}{a} (1)
and the similarity of triangles AEB and AFC, \frac{e}{c} = \frac{f}{b} (2)
And because the altitudes of the first triangle are equal to the sides of the second,
we also have \frac{f}{a} = \frac{e}{b} (3)
From (2), e = \frac{cf}{b} ; substitute it into (3), we then have = ac
Now the law of cosines gives us = + - 2ac·cos∠ABC
or ac = + - 2ac·cos∠ABC
or cos∠ABC = \frac{a² + c² - ac}{2ac} = \frac{a² + c²}{2ac} - \frac{1}{2}
But ∠ABC + ∠FCB = 90°; therefore, cos∠FCB = sin∠ABC = \sqrt{1 - cos²∠ABC}
Hence, cos∠FCB is largest when cos²∠ABC is smallest or when \frac{a² + c²}{2ac} - \frac{1}{2} is smallest, or when \frac{a² + c²}{2ac} is smallest.
Let x = \frac{a}{c}. Then \frac{a² + c²}{2ac} = \frac{f(x)}{2}. Function f(x) = x + \frac{1}{x} has no maximum, but has a minimum when f '(x) = 0.
f '(x) = 1 - x^{-2} = 0 gives x^{2} - 1 = 0 so that the only minimum is attained for x = 1, i.e., when a = c. Since, by our assumption, abc, all three sides are equal and both triangles are equilateral. Thus \frac{a}{f} = \frac{2\sqrt{3}}{3}. A pripori, it is not clear which of the two possible ratios is meant to be maximized, the answer is the largest of the two, \frac{a}{f} = \frac{2\sqrt{3}}{3} or \frac{f}{a} = \frac{\sqrt{3}}{2}, which is the former - \frac{2\sqrt{3}}{3} - and is attained for the equilateral triangles.
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Home » Multiples » Multiples of 567
# Multiples of 567
Here you can find the multiples of 567. The multiples of five hundred and sixty-seven are the numbers which contain 567 n times; the integer n is called coefficient or multiplier. Also, likewise important, we show you the submultiples of 567.
±567, ±1134, ±1701, ±2268, ±2835, ±3402, ±3969, ±4536, ±5103, ±5670
If you like to learn what is a multiple or how to find multiples of a number like 567 then read our article multiples which you can find in the header menu. On this page you can only learn the multiples of the number 567. We start with giving you the definition right below.
A multiple b of 567 is a number such that b = n * 567, n . In other words, a multiple of 567, divided by 567, produces an integer. When you have a certain number b, you can find out if b is multiple of 567 by dividing that number by 567. If the remainder, also known as modulo, is zero, then b is a multiple of 567. If the modulo is different from 0, that is if the division renders decimals, then b is not part of the multiples of 567.
In the same way you can test any number, not just the number five hundred and sixty-seven .
## What are the Multiples of 567
The numbers are 0, 567, 1134, 1701, … successively. As you can see, getting the list of multiples of 567 is really easy: b = n * 567 with n = 1, 2, 3, … and, although not used very often, 0, by definition, is also a multiple of five hundred and sixty-seven . Below we have compiled the first 101 numbers:
0, 567, 1134, 1701, 2268, 2835, 3402, 3969, 4536, 5103, 5670, 6237, 6804, 7371, 7938, 8505, 9072, 9639, 10206, 10773, 11340, 11907, 12474, 13041, 13608, 14175, 14742, 15309, 15876, 16443, 17010, 17577, 18144, 18711, 19278, 19845, 20412, 20979, 21546, 22113, 22680, 23247, 23814, 24381, 24948, 25515, 26082, 26649, 27216, 27783, 28350, 28917, 29484, 30051, 30618, 31185, 31752, 32319, 32886, 33453, 34020, 34587, 35154, 35721, 36288, 36855, 37422, 37989, 38556, 39123, 39690, 40257, 40824, 41391, 41958, 42525, 43092, 43659, 44226, 44793, 45360, 45927, 46494, 47061, 47628, 48195, 48762, 49329, 49896, 50463, 51030, 51597, 52164, 52731, 53298, 53865, 54432, 54999, 55566, 56133, 56700.
Now you already know what are the multiples of 567. Do you know how many multiples 567 has? The answer is that the number is unlimited. The list of all multiples of 567 is infinite.
BTW: Besides all the multiples of 567, other searched terms on our website include:
## The Submultiples of 567
An integer a is a submultiple of 567 only if 567 is a multiple of a. By coming to this page chances are high that you are in fact looking for the submultiples of 567. We will list the numbers right below. Just let us add that you can learn how to get the submultiples of 567 by reading the instructions on our page entitled multiples.
The submultiples of 567 are: 1, 3, 7, 9, 21, 27, 63, 81, 189, 567.
How many submultiples does 567 have? The quantity of submultiples of five hundred and sixty-seven is 10.
You can obtain the submultiples of any integer using the calculator below. Insert the number, our calculator then produces the result without the need to press a button.
This brings us to the end of our post. We are sure you can now answer questions including, but not limited to, first five multiples of 567?, and name 4 multiples of 567 for example.
If this article about the multiples and submultiples of 567 has been useful to you, then kindly hit the social buttons to let your friends know about our site. If you have any question or suggestion about what are multiples of 567 let us know by filling in the form below or sending us an email.
Thanks for visiting multiples of five hundred and sixty-seven on timestable.net.
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# The foot of a 20 ft. ladder is 12 ft. from the base of a house. How far up the side of the house does the ladder reach?
Mar 15, 2016
The ladder reaches $16$ feet up the side of the house.
#### Explanation:
Let $c$ denote the ladder.
$c = 20$ feet
$b$ denotes distance of the base of the ladder to the house.
$b = 12$ feet.
we are required to calculate the value of $a :$
Using the Pythagoras Theorem:
${c}^{2} = {b}^{2} + {a}^{2}$
${20}^{2} = {12}^{2} + {a}^{2}$
$400 = 144 + {a}^{2}$
$400 - 144 = {a}^{2}$
${a}^{2} = 256$
$a = \sqrt{256}$
$a = 16$ feet.
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1. Home
3. Operations with Decimal Numbers
Multiplying decimals
You went to the grocery store today to buy a couple of things 1 liter of milk, 24 eggs, a pound of tomatoes, 2 lbs of potatoes and 3 2 liters of Coca Cola. You checked the price tag for each item and you found out that 1 liter of whole fat milk costs C$3.11, 12 large eggs costs$4.28, 2 lbs of tomatoes costs $4.02, 2 lbs of potatoes costs C$ 2.57 and a 2 liter of Coca Cola costs C\$ 2.38. To come up with the total amount of money you need to pay, you will need to apply the different operations with the set of decimal numbers given above.
In this chapter, we will learn about the proper skills that will help you solve the problem above. This chapter has four parts. In the first chapter we will learn how to add and subtract decimal numbers. The rule is simple as you only need to align the given numbers according to the location of the decimal point. Then, depending on the instructions, we will need to round off the resulting number to the nearest tenth. Rounding off numbers is mostly used to simplify the results.
For the second part of this chapter, we will look at how to multiply decimals. In this operation, we will simply multiply the decimal numbers like a regular whole number by ignoring the decimal point. We will only put the decimal point in place after we get the product. By counting the combined number of decimal places of the two numbers we have multiplied we will determine where to put the decimal point.
Following this discussion is the third part of the chapter where we will study how to divide decimals numbers. If the divisor is is not a whole number, we will move the decimal point to the right until the number becomes a whole number. The dividend's decimal point should also be moved accordingly. After the adjustments are made, we can proceed as usual. The quotient's decimal point should be placed directly above that of the dividend. In some cases, we may encounter repeating numbers in the quotient like 1/3 which will give us the repeating decimal 0.3333….
Finally, in the last part of this chapter, we will discuss the Order of operations PEMDAS, which stands for P Parentheses first, E Exponents (ie Powers and Square Roots), MD Multiplication or Division (left-to-right) and AS Addition or Subtraction (left-to-right).
Multiplying decimals
Previously, we learned how to add and subtract decimal numbers. In this section, we will learn how to multiply decimal numbers. As learned in previous section, when adding and subtracting decimal numbers, the decimal points must be lined up. In contrast, when multiplying decimal numbers, it is not important that the decimal points be lined up. Instead, it is important to line up the digits in the lowest place values of both numbers. In order to figure out where to place the decimal point in the answer, we must count how many digits total, between the two numbers being multiplied together, are behind the decimal points.
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 14.E: Partial Differentiation (Exercises)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
These are homework exercises to accompany David Guichard's "General Calculus" Textmap. Complementary General calculus exercises can be found for other Textmaps and can be accessed here.
## 14.1: Functions of Several Variables
Q14.1.1 Let $$f(x,y)=(x-y)^2$$. Determine the equations and shapes of the cross-sections when $$x=0$$, $$y=0$$, $$x=y$$, and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer)
Q14.1.2 Let $$f(x,y)=|x|+|y|$$. Determine the equations and shapes of the cross-sections when $$x=0$$, $$y=0$$, $$x=y$$, and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer)
Q14.1.3 Let $$f(x,y)=e^{-(x^2+y^2)}\sin(x^2+y^2)$$. Determine the equations and shapes of the cross-sections when $$x=0$$, $$y=0$$, $$x=y$$, and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer)
Q14.1.4 Let $$f(x,y)=\sin(x-y)$$. Determine the equations and shapes of the cross-sections when $$x=0$$, $$y=0$$, $$x=y$$, and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer)
Q14.1.5 Let $$f(x,y)=(x^2-y^2)^2$$. Determine the equations and shapes of the cross-sections when $$x=0$$, $$y=0$$, $$x=y$$, and describe the level curves. Use a three-dimensional graphing tool to graph the surface. (answer)
Q14.1.6 Find the domain of each of the following functions of two variables:
1. $$\ds\sqrt{9-x^2}+\sqrt{y^2-4}$$
2. $$\arcsin(x^2+y^2-2)$$
3. $$\ds\sqrt{16-x^2-4y^2}$$ (answer)
Q14.1.7 Below are two sets of level curves. One is for a cone, one is for a paraboloid. Which is which? Explain.
## 14.2: Limits and Continuity
Determine whether each limit exists. If it does, find the limit and prove that it is the limit; if it does not, explain how you know.
Q14.2.1 $$\lim_{(x,y)\to(0,0)}{x^2\over x^2+y^2}$$ (answer)
Q14.2.2 $$\lim_{(x,y)\to(0,0)}{xy\over x^2+y^2}$$ (answer)
Q14.2.3 $$\lim_{(x,y)\to(0,0)}{xy\over 2x^2+y^2}$$ (answer)
Q14.2.4 $$\lim_{(x,y)\to(0,0)}{x^4-y^4\over x^2+y^2}$$ (answer)
Q14.2.5 $$\lim_{(x,y)\to(0,0)}{\sin(x^2+y^2)\over x^2+y^2}$$ (answer)
Q14.2.6 $$\lim_{(x,y)\to(0,0)}{xy\over \sqrt{2x^2+y^2}}$$ (answer)
Q14.2.7 $$\lim_{(x,y)\to(0,0)} {e^{-x^2-y^2}-1\over x^2+y^2}$$ (answer)
Q14.2.8 $$\lim_{(x,y)\to(0,0)}{x^3+y^3\over x^2+y^2}$$ (answer)
Q14.2.9 $$\lim_{(x,y)\to(0,0)}{x^2 + \sin^2 y\over 2x^2+y^2}$$ (answer)
Q14.2.10 $$\lim_{(x,y)\to(1,0)}{(x-1)^2\ln x\over(x-1)^2+y^2}$$ (answer)
Q14.2.11 $$\lim_{(x,y)\to(1,-1)}{3x+4y}$$ (answer)
Q14.2.12 $$\lim_{(x,y)\to(0,0)}{4x^2y\over x^2+y^2}$$ (answer)
Q14.2.13 Does the function $$f(x,y)={x-y\over 1+x+y}$$ have any discontinuities? What about $$f(x,y)={x-y\over 1+x^2+y^2}$$? Explain.
## 14.3: Partial Differentiation
Q14.3.1 Find $$f_x$$ and $$f_y$$ where $$f(x,y)=\cos(x^2y)+y^3$$. (answer)
Q14.3.2 Find $$f_x$$ and $$f_y$$ where $$f(x,y)={xy\over x^2+y}$$. (answer)
Q14.3.3 Find $$f_x$$ and $$f_y$$ where $$f(x,y)=e^{x^2+y^2}$$. (answer)
Q14.3.4 Find $$f_x$$ and $$f_y$$ where $$f(x,y)=xy\ln(xy)$$. (answer)
Q14.3.5 Find $$f_x$$ and $$f_y$$ where $$f(x,y)=\sqrt{1-x^2-y^2}$$. (answer)
Q14.3.6 Find $$f_x$$ and $$f_y$$ where $$f(x,y)=x\tan(y)$$. (answer)
Q14.3.7 Find $$f_x$$ and $$f_y$$ where $$f(x,y)={1\over xy}$$. (answer)
Q14.3.8 Find an equation for the plane tangent to $$2x^2+3y^2-z^2=4$$ at $$(1,1,-1)$$. (answer)
Q14.3.9 Find an equation for the plane tangent to $$f(x,y)=\sin(xy)$$ at $$(\pi,1/2,1)$$. (answer)
Q14.3.10 Find an equation for the plane tangent to $$f(x,y)=x^2+y^3$$ at $$(3,1,10)$$. (answer)
Q14.3.11 Find an equation for the plane tangent to $$f(x,y)=x\ln(xy)$$ at $$(2,1/2,0)$$. (answer)
Q14.3.12 Find an equation for the line normal to $$x^2+4y^2=2z$$ at $$(2,1,4)$$. (answer)
Q14.3.13 Explain in your own words why, when taking a partial derivative of a function of multiple variables, we can treat the variables not being differentiated as constants.
Q14.3.14 Consider a differentiable function, $$f(x,y)$$. Give physical interpretations of the meanings of $$f_x(a,b)$$ and $$f_y(a,b)$$ as they relate to the graph of $$f$$.
Q14.3.15 In much the same way that we used the tangent line to approximate the value of a function from single variable calculus, we can use the tangent plane to approximate a function from multivariable calculus. Consider the tangent plane found in Exercise 11. Use this plane to approximate $$f(1.98, 0.4)$$.
Q14.3.16 Suppose that one of your colleagues has calculated the partial derivatives of a given function, and reported to you that $$f_x(x,y)=2x+3y$$ and that $$f_y(x,y)=4x+6y$$. Do you believe them? Why or why not? If not, what answer might you have accepted for $$f_y$$?
Q14.3.17 Suppose $$f(t)$$ and $$g(t)$$ are single variable differentiable functions. Find $$\partial z/\partial x$$ and $$\partial z/\partial y$$ for each of the following two variable functions.
1. $$z=f(x)g(y)$$
2. $$z=f(xy)$$
3. $$z=f(x/y)$$
## 14.4: The Chain Rule
Q14.4.1 Use the chain rule to compute $$dz/dt$$ for $$z=\sin(x^2+y^2)$$, $$x=t^2+3$$, $$y=t^3$$. (answer)
Q14.4.2 Use the chain rule to compute $$dz/dt$$ for $$z=x^2y$$, $$x=\sin(t)$$, $$y=t^2+1$$. (answer)
Q14.4.3 Use the chain rule to compute $$\partial z/\partial s$$ and $$\partial z/\partial t$$ for $$z=x^2y$$, $$x=\sin(st)$$, $$y=t^2+s^2$$. (answer)
Q14.4.4 Use the chain rule to compute $$\partial z/\partial s$$ and $$\partial z/\partial t$$ for $$z=x^2y^2$$, $$x=st$$, $$y=t^2-s^2$$. (answer)
Q14.4.5 Use the chain rule to compute $$\partial z/\partial x$$ and $$\partial z/\partial y$$ for $$2x^2+3y^2-2z^2=9$$. (answer)
Q14.4.6 Use the chain rule to compute $$\partial z/\partial x$$ and $$\partial z/\partial y$$ for $$2x^2+y^2+z^2=9$$. (answer)
Q14.4.7 Chemistry students will recognize the ideal gas law, given by $$PV=nRT$$ which relates the Pressure, Volume, and Temperature of $$n$$ moles of gas. (R is the ideal gas constant). Thus, we can view pressure, volume, and temperature as variables, each one dependent on the other two.
• a. If pressure of a gas is increasing at a rate of $$0.2 Pa/\hbox{min}$$ and temperature is increasing at a rate of $$1 K/\hbox{min}$$, how fast is the volume changing?
• b. If the volume of a gas is decreasing at a rate of $$0.3 L/\hbox{min}$$ and temperature is increasing at a rate of $$.5 K/\hbox{min}$$, how fast is the pressure changing?
• c. If the pressure of a gas is decreasing at a rate of $$0.4 Pa/\hbox{min}$$ and the volume is increasing at a rate of $$3 L/\hbox{min}$$, how fast is the temperature changing? (answer)
Q14.4.8 Verify the following identity in the case of the ideal gas law: ${\partial P\over \partial V} {\partial V\over \partial T} {\partial T\over \partial P}=-1$
Q14.4.9 The previous exercise was a special case of the following fact, which you are to verify here: If $$F(x,y,z)$$ is a function of 3 variables, and the relation $$F(x,y,z)=0$$ defines each of the variables in terms of the other two, namely $$x=f(y,z)$$, $$y=g(x,z)$$ and $$z=h(x,y)$$, then ${\partial x\over \partial y} {\partial y\over \partial z} {\partial z\over \partial x}=-1$
## 14.5: Directional Derivatives
Q14.5.1 Find $$D_{\bf u} f$$ for $$f=x^2+xy+y^2$$ in the direction of $${\bf u}=\langle 2,1\rangle$$ at the point $$(1,1)$$. (answer)
Q14.5.2 Find $$D_{\bf u} f$$ for $$f=\sin(xy)$$ in the direction of $${\bf u}=\langle -1,1\rangle$$ at the point $$(3,1)$$. (answer)
Q14.5.3 Find $$D_{\bf u} f$$ for $$f=e^x\cos(y)$$ in the direction 30 degrees from the positive $$x$$ axis at the point $$(1,\pi/4)$$. (answer)
Q14.5.4 The temperature of a thin plate in the $$x$$-\)y\) plane is $$T=x^2+y^2$$. How fast does temperature change at the point $$(1,5)$$ moving in a direction 30 degrees from the positive $$x$$ axis? (answer)
Q14.5.5 Suppose the density of a thin plate at $$(x,y)$$ is $$1/\sqrt{x^2+y^2+1}$$. Find the rate of change of the density at $$(2,1)$$ in a direction $$\pi/3$$ radians from the positive $$x$$ axis. (answer)
Q14.5.6 Suppose the electric potential at $$(x,y)$$ is $$\ln\sqrt{x^2+y^2}$$. Find the rate of change of the potential at $$(3,4)$$ toward the origin and also in a direction at a right angle to the direction toward the origin. (answer)
Q14.5.7 A plane perpendicular to the $$x$$-\)y\) plane contains the point $$(2,1,8)$$ on the paraboloid $$z=x^2+4y^2$$. The cross-section of the paraboloid created by this plane has slope 0 at this point. Find an equation of the plane. (answer)
Q14.5.8 A plane perpendicular to the $$x$$-\)y\) plane contains the point $$(3,2,2)$$ on the paraboloid $$36z=4x^2+9y^2$$. The cross-section of the paraboloid created by this plane has slope 0 at this point. Find an equation of the plane. (answer)
Q14.5.9 Suppose the temperature at $$(x,y,z)$$ is given by $$T=xy+\sin(yz)$$. In what direction should you go from the point $$(1,1,1)$$ to decrease the temperature as quickly as possible? What is the rate of change of temperature in this direction? (answer)
Q14.5.10 Suppose the temperature at $$(x,y,z)$$ is given by $$T=xyz$$. In what direction can you go from the point $$(1,1,1)$$ to maintain the same temperature? (answer)
Q14.5.11 Find an equation for the plane tangent to $$x^2-3y^2+z^2=7$$ at $$(1,1,3)$$. (answer)
Q14.5.12 Find an equation for the plane tangent to $$xyz=6$$ at $$(1,2,3)$$. (answer)
Q14.5.13 Find an equation for the line normal to $$x^2+2y^2+4z^2=26$$ at $$(2,-3,-1)$$. (answer)
Q14.5.14 Find an equation for the line normal to $$x^2+y^2+9z^2=56$$ at $$(4,2,-2)$$. (answer)
Q14.5.15 Find an equation for the line normal to $$x^2+5y^2-z^2=0$$ at $$(4,2,6)$$. (answer)
Q14.5.16 Find the directions in which the directional derivative of $$f(x,y)=x^2+\sin(xy)$$ at the point $$(1,0)$$ has the value 1. (answer)
Q14.5.17 Show that the curve $${\bf r}(t) = \langle\ln(t),t\ln(t),t\rangle$$ is tangent to the surface $$xz^2-yz+\cos(xy) = 1$$ at the point $$(0,0,1)$$.
Q14.5.18 A bug is crawling on the surface of a hot plate, the temperature of which at the point $$x$$ units to the right of the lower left corner and $$y$$ units up from the lower left corner is given by $$T(x,y)=100-x^2-3y^3$$.
1. If the bug is at the point $$(2,1)$$, in what direction should it move to cool off the fastest? How fast will the temperature drop in this direction?
2. If the bug is at the point $$(1,3)$$, in what direction should it move in order to maintain its temperature? (answer)
Q14.5.19 The elevation on a portion of a hill is given by $$f(x,y) = 100 -4x^2 - 2y$$. From the location above $$(2,1)$$, in which direction will water run? (answer)
Q14.5.20 Suppose that $$g(x,y)=y-x^2$$. Find the gradient at the point $$(-1, 3)$$. Sketch the level curve to the graph of $$g$$ when $$g(x,y)=2$$, and plot both the tangent line and the gradient vector at the point $$(-1,3)$$. (Make your sketch large). What do you notice, geometrically? (answer)
Q14.5.21 The gradient $$\nabla f$$ is a vector valued function of two variables. Prove the following gradient rules. Assume $$f(x,y)$$ and $$g(x,y)$$ are differentiable functions.
1. $$\nabla(fg)=f\nabla(g)+g\nabla(f)$$
2. $$\nabla(f/g)=(g\nabla f - f \nabla g)/g^2$$
3. $$\nabla((f(x,y))^n)=nf(x,y)^{n-1}\nabla f$$
## 14.6: Higher order Derivatives
Q14.6.1 Let $$f=xy/(x^2+y^2)$$; compute $$f_{xx}$$, $$f_{yx}$$, and $$f_{yy}$$. (answer)
Q14.6.2 Find all first and second partial derivatives of $$x^3y^2+y^5$$. (answer)
Q14.6.3 Find all first and second partial derivatives of $$4x^3+xy^2+10$$. (answer)
Q14.6.4 Find all first and second partial derivatives of $$x\sin y$$. (answer)
Q14.6.5 Find all first and second partial derivatives of $$\sin(3x)\cos(2y)$$. (answer)
Q14.6.6 Find all first and second partial derivatives of $$e^{x+y^2}$$. (answer)
Q14.6.7 Find all first and second partial derivatives of $$\ln\sqrt{x^3+y^4}$$. (answer)
Q14.6.8 Find all first and second partial derivatives of $$z$$ with respect to $$x$$ and $$y$$ if $$x^2+4y^2+16z^2-64=0$$. (answer)
Q14.6.9 Find all first and second partial derivatives of $$z$$ with respect to $$x$$ and $$y$$ if $$xy+yz+xz=1$$. (answer)
Q14.6.10 Let $$\alpha$$ and $$k$$ be constants. Prove that the function $$u(x,t)=e^{-\alpha^2k^2t}\sin(kx)$$ is a solution to the heat equation $$u_t=\alpha^2u_{xx}$$
Q14.6.11 Let $$a$$ be a constant. Prove that $$u=\sin(x-at)+\ln(x+at)$$ is a solution to the wave equation $$u_{tt}=a^2u_{xx}$$.
Q14.6.12 How many third-order derivatives does a function of 2 variables have? How many of these are distinct?
Q14.6.13 How many $$n$$th order derivatives does a function of 2 variables have? How many of these are distinct?
## 14.7: Maxima and minima
Q14.7.1 Find all local maximum and minimum points of $$f=x^2+4y^2-2x+8y-1$$. (answer)
Q14.7.2 Find all local maximum and minimum points of $$f=x^2-y^2+6x-10y+2$$. (answer)
Q14.7.3 Find all local maximum and minimum points of $$f=xy$$. (answer)
Q14.7.4 Find all local maximum and minimum points of $$f=9+4x-y-2x^2-3y^2$$. (answer)
Q14.7.5 Find all local maximum and minimum points of $$f=x^2+4xy+y^2-6y+1$$. (answer)
Q14.7.6 Find all local maximum and minimum points of $$f=x^2-xy+2y^2-5x+6y-9$$. (answer)
Q14.7.7 Find the absolute maximum and minimum points of $$f=x^2+3y-3xy$$ over the region bounded by $$y=x$$, $$y=0$$, and $$x=2$$. (answer)
Q14.7.8 A six-sided rectangular box is to hold $$1/2$$ cubic meter; what shape should the box be to minimize surface area? (answer)
Q14.7.9 The post office will accept packages whose combined length and girth is at most 130 inches. (Girth is the maximum distance around the package perpendicular to the length; for a rectangular box, the length is the largest of the three dimensions.) What is the largest volume that can be sent in a rectangular box? (answer)
Q14.7.10 The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost. (answer)
Q14.7.11 Using the methods of this section, find the shortest distance from the origin to the plane $$x+y+z=10$$. (answer)
Q14.7.12 Using the methods of this section, find the shortest distance from the point $$(x_0,y_0,z_0)$$ to the plane $$ax+by+cz=d$$. You may assume that $$c\not=0$$; use of Sage or similar software is recommended. (answer)
Q14.7.13 A trough is to be formed by bending up two sides of a long metal rectangle so that the cross-section of the trough is an isosceles trapezoid, as in figure 6.2.6. If the width of the metal sheet is 2 meters, how should it be bent to maximize the volume of the trough? (answer)
Q14.7.14 Given the three points $$(1,4)$$, $$(5,2)$$, and $$(3,-2)$$, $$\ds(x-1)^2+(y-4)^2+(x-5)^2+(y-2)^2+(x-3)^2+(y+2)^2$$ is the sum of the squares of the distances from point $$(x,y)$$ to the three points. Find $$x$$ and $$y$$ so that this quantity is minimized. (answer)
Q14.7.15 Suppose that $$f(x,y)=x^2+y^2+kxy$$. Find and classify the critical points, and discuss how they change when $$k$$ takes on different values.
Q14.7.16 Find the shortest distance from the point $$(0,b)$$ to the parabola $$y=x^2$$. (answer)
Q14.7.17 Find the shortest distance from the point $$(0,0,b)$$ to the paraboloid $$z=x^2+y^2$$. (answer)
Q14.7.18 Consider the function $$f(x,y)=x^3-3x^2y+y^3$$.
1. Show that $$(0,0)$$ is the only critical point of $$f$$.
2. Show that the discriminant test is inconclusive for $$f$$.
3. Determine the cross-sections of $$f$$ obtained by setting $$y=kx$$ for various values of $$k$$.
4. What kind of critical point is $$(0,0)$$?
Q14.7.19 Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid $$2x^2+72y^2+18z^2=288$$. (answer)
## 14.8: Lagrange Multipliers
Q14.8.1 A six-sided rectangular box is to hold $$1/2$$ cubic meter; what shape should the box be to minimize surface area? (answer)
Q14.8.2 The post office will accept packages whose combined length and girth are at most 130 inches (girth is the maximum distance around the package perpendicular to the length). What is the largest volume that can be sent in a rectangular box? (answer)
Q14.8.3 The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost. (answer)
Q14.8.4 Using Lagrange multipliers, find the shortest distance from the point $$(x_0,y_0,z_0)$$ to the plane $$ax+by+cz=d$$. (answer)
Q14.8.5 Find all points on the surface $$xy-z^2+1=0$$ that are closest to the origin. (answer)
Q14.8.6 The material for the bottom of an aquarium costs half as much as the high strength glass for the four sides. Find the shape of the cheapest aquarium that holds a given volume $$V$$. (answer)
Q14.8.7 The plane $$x-y+z=2$$ intersects the cylinder $$x^2+y^2=4$$ in an ellipse. Find the points on the ellipse closest to and farthest from the origin. (answer)
Q14.8.8 Find three positive numbers whose sum is 48 and whose product is as large as possible. (answer)
Q14.8.9 Find all points on the plane $$x+y+z = 5$$ in the first octant at which $$f(x,y,z) = xy^2z^2$$ has a maximum value. (answer)
Q14.8.10 Find the points on the surface $$x^2 -yz = 5$$ that are closest to the origin. (answer)
Q14.8.11 A manufacturer makes two models of an item, standard and deluxe. It costs \\)40 to manufacture the standard model and \\)60 for the deluxe. A market research firm estimates that if the standard model is priced at $$x$$ dollars and the deluxe at $$y$$ dollars, then the manufacturer will sell $$500(y-x)$$ of the standard items and $$45,000+500(x-2y)$$ of the deluxe each year. How should the items be priced to maximize profit? (answer)
Q14.8.12 A length of sheet metal is to be made into a water trough by bending up two sides as shown in figure 14.8.3. Find $$x$$ and $$\phi$$ so that the trapezoid--shaped cross section has maximum area, when the width of the metal sheet is 27 inches (that is, $$2x+y=27$$). (answer)
Figure 14.8.3. Cross-section of a trough.
Q14.8.13 Find the maximum and minimum values of $$f(x,y,z)=6x+3y+2z$$ subject to the constraint $$g(x,y,z) = 4x^2+2y^2 + z^2 - 70 = 0$$. (answer)
Q14.8.14 Find the maximum and minimum values of $$f(x,y)=e^{xy}$$ subject to the constraint $$g(x,y) = x^3+y^3 - 16 = 0$$. (answer)
Q14.8.15 Find the maximum and minimum values of $$f(x,y) = xy + \sqrt{9-x^2-y^2}$$ when $$x^2+y^2 \leq 9$$. (answer)
Q14.8.16 Find three real numbers whose sum is 9 and the sum of whose squares is a small as possible. (answer)
Q14.8.17 Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere. (answer)
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### is reflected in the line y = -x, its image is (-b,-a).
```
Reflection: a transformation that uses a line to reflect
an image.
A reflection is an isometry, but its orientation
changes from the preimage to the image
Line of reflection is the line that acts like a mirror
A reflection in a line (m) maps every point (P) in the
plane to a point (P΄) so that for each point, one of
the following is true:
If P is on m, then P=P΄
P
P΄
m
Or, if P is not on m, then m is the ⊥ bisector
PP΄
P
P΄
Reflect AB:
1. across the x-axis
2. across the y-axis
3. across the line y=x
4. across the line y=-x
If (a,b) is reflected in the x-axis, its image is (a,-b).
If (a,b) is reflected in the y-axis, its image is (-a,b).
If (a,b) is reflected in the line y = x,
its image is (b,a).
If (a,b) is reflected in the line y = -x,
its image is (-b,-a).
Reflection Matrices
Across x-axis:
1 0
0 -1
Reflection
matrix
X
Across y-axis:
D E F
1 3 4
–1 0
2 3 0
0 1
Polygon
matrix
Reflection
matrix
X
D E
1 3
F
4
2 3
0
Polygon
matrix
Use matrix multiplication to reflect a polygon
The vertices of DEF are D(1, 2), E(3, 3), and F(4, 0).
Find the reflection of DEF in the y-axis using matrix
multiplication. Graph DEF and its image.
SOLUTION
STEP 1
–1 0
0 1
Multiply the polygon matrix by the matrix
for a reflection in the y-axis.
D E F
1 3 4
X
2 3 0
Reflection
matrix
Polygon
matrix
EXAMPLE 5
=
Use matrix multiplication to reflect a polygon
–1(1) + 0(2)
–1(3) + 0(3)
–1(4) + 0(0)
0(1) + 1(2)
0(3) + 1(3)
0(4) + 1(0)
D′ E′ F′
–1 –3 –4
=
2
3 0
Image
matrix
Graph reflections in horizontal and vertical lines
The vertices of ABC are A(1, 3), B(5, 2), and C(2, 1).
Graph the reflection of ABC described.
a. In the line n : x = 3
SOLUTION
Point A is 2 units left of n,
so its reflection A′ is 2 units
right of n at (5, 3). Also, B′ is
2 units left of n at (1, 2), and
C′ is 1 unit right of n at (4,
1).
Graph reflections in horizontal and vertical lines
The vertices of ABC are A(1, 3), B(5, 2), and C(2, 1).
Graph the reflection of ABC described.
b. In the line m : y = 1
SOLUTION
Point A is 2 units above m,
so A′ is 2 units below m at
(1, –1). Also, B′ is 1 unit
below m at (5, 0). Because
point C is on line m, you
know that C = C′.
Real World: Find a minimum distance
You are going to meet a friend on the beach shoreline.
Where should you meet in order to minimize the
distances you both have to walk.?
house is at (9,6). At what point on the
shoreline (x-axis) should you meet?
6
4
2
Shoreline
-10
-5
5
-2
-4
-6
10
```
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Refer to Go Math Grade 8 Answer Key Chapter 13 Volume for answers and explanations. It is very easy to get good marks in the exam by practicing with the help of Go Math Grade 8 Solution Key. Top-ranked student’s first choice is Go Math Grade 8 Answer Key Chapter 13 Volume. Learn and practice the maths in the best way using Go Math Grade 8 Answer Key.
Lesson 1: Volume of Cylinders
Lesson 2: Volume of Cones
Lesson 3: Volume of Spheres
Model Quiz :
Review :
### Guided Practice – Volume of Cylinders – Page No. 402
Question 1.
Vocabulary Describe the bases of a cylinder.
Type below:
____________
The ends of a cylinder are the bases of the cylinder of the two flat surfaces.
Question 2.
Figure 1 shows a view from above of inch cubes on the bottom of a cylinder. Figure 2 shows the highest stack of cubes that will fit inside the cylinder. Estimate the volume of the cylinder. Explain your reasoning.
________ in3
427in³
Explanation:
Number of inch cubes that fit in the base of the cylinder = 61
Number of inch cubes that fit in the length of the cylinder = 7
Volume of cylinder = basearea x height
V= 61 x 7
V = 427 cubic units.
Volume of each cube = 1 in³
Volume of cylinder = 427in³
Question 3.
Find the volume of the cylinder to the nearest tenth. Use 3.14 for π.
The volume of the cylinder is approximately _____ m3.
________ m3
The volume of the cylinder is approximately 1695.6 m3.
1695.6m³
Explanation:
V = πr²h
V = π . 6² . 15
V = 3.14 × 36 × 15
V = 1695.6m³
Question 4.
A Japanese odaiko is a very large drum that is made by hollowing out a section of a tree trunk. A museum in Takayama City has three odaikos of similar size carved from a single tree trunk. The largest measures about 2.7 meters in both diameter and length, and weighs about 4.5 metric tons. Using the volume formula for a cylinder, approximate the volume of the drum to the nearest tenth.
The volume of the drum is about _____ m3.
The volume of the drum is about ___________ m3
The volume of the drum is about 15.5 m3.
Explanation:
Diameter of base of drum = 2.7 m
The radius of the base of the drum = 2.7/2
R = 1.35 m
The volume of cylinder = πr²h
Height (h) = 2.7 m
Volume = 3.14 × (1.35) × 2.7
V = 15.4511 m³
V = 15.5 m³
ESSENTIAL QUESTION CHECK-IN
Question 5.
How do you find the volume of a cylinder? Describe which measurements of a cylinder you need to know.
Type below:
____________
The volume of the cylinder is = πr²h
Explanation:
We need to find the radius of the base, r, and the height of the cylinder, h.
The volume of the cylinder is = πr²h
### 13.1 Independent Practice – Volume of Cylinders – Page No. 403
Find the volume of each figure. Round your answers to the nearest tenth if necessary. Use 3.14 for π.
Question 6.
_________ cm3
569.9 cm³
Explanation:
Radius of base = 11 cm
Height of cylinder = 1.5 cm
The volume of cylinder = πr²h
V = 3.14 × (11)² × 1.5
V = 569.91
V = 569.9 cm³
Question 7.
_________ in3
1205.8 in³
Explanation:
Radius of base = 4 in
Height of cylinder = 24 in
The volume of cylinder = πr²h
V = 3.14 × (4)² × 24
V = 1205.76
V = 1205.8 in³
Question 8.
_________ m3
1256 m³
Explanation:
Radius of base = 5 m
Height of cylinder = 16 m
The volume of cylinder = πr²h
V = 3.14 × (5)² × 16
V = 1256
V = 1256 m³
Question 9.
_________ in3
942 in³
Explanation:
Diameter of base = 10 in
Radius of base = 5 in
Height of cylinder = 12 in
The volume of cylinder = πr²h
V = 3.14 × (5)² × 12
V = 942 in³
Question 10.
A cylinder has a radius of 4 centimeters and a height of 40 centimeters.
_________ cm3
2009.6 cm³
Explanation:
Radius of base = 4 cm
Height of cylinder = 40 cm
The volume of cylinder = πr²h
V = 3.14 × (4)² × 40
V = 2009.6 cm³
Question 11.
A cylinder has a radius of 8 meters and a height of 4 meters.
_________ m3
803.8 m³
Explanation:
The radius of base = 8 m
Height of cylinder = 4 m
The volume of cylinder = πr²h
V = 3.14 × (8)² × 4
V = 803.84 m³
V = 803.8 m³
Round your answer to the nearest tenth, if necessary. Use 3.14 for π.
Question 12.
The cylindrical Giant Ocean Tank at the New England Aquarium in Boston is 24 feet deep and has a radius of 18.8 feet. Find the volume of the tank.
_________ ft3
26635.2 ft³
Explanation:
Base radius of the tank = 18.8 ft
Depth of the tank in the ocean = 24 ft
The volume of cylinder = πr²h
V = 3.14 × (18.8)² × 24
V = 3.14 × 354.44 × 24
V = 26635.2384 ft³
V = 26635.2 ft³
Question 13.
A standard-size bass drum has a diameter of 22 inches and is 18 inches deep. Find the volume of this drum.
_________ in3
6838.9 in³
Explanation:
Base diameter of drum = 22 in
Base radius of the drum = 22/2 = 11 in
Depth of the bass drum = 18 in
Volume of the bass drum = πr²h
V = 3.14 × (11)² × 18
V = 3.14 × 121 × 18
V = 6838.92 in³
V = 6838.9 in³
Question 14.
Grain is stored in cylindrical structures called silos. Find the volume of a silo with a diameter of 11.1 feet and a height of 20 feet.
_________ ft3
1934.4 ft³
Explanation:
Base diameter of silo = 11.1 ft
Base radius of the silo = 11.1/2 = 5.55 ft
Depth of the silo = 20 ft
Volume of the silo = πr²h
V = 3.14 × (5.55)² × 18
V = 3.14 × 30.8025 × 18
V = 1934.397 ft³
V = 1934.4 ft³
Question 15.
The Frank Erwin Center, or “The Drum,” at the University of Texas in Austin can be approximated by a cylinder that is 120 meters in diameter and 30 meters in height. Find its volume.
_________ m3
339120 m³
Explanation:
Base diameter of the drum = 120 m
Base radius of the drum = 120/2 = 60 m
Height of the drum = 30 m
Volume of the drum = πr²h
V = 3.14 × (60)² × 30
V = 3.14 × 3600 × 30
V = 339120 m³
### Volume of Cylinders – Page No. 404
Question 16.
A barrel of crude oil contains about 5.61 cubic feet of oil. How many barrels of oil are contained in 1 mile (5280 feet) of a pipeline that has an inside diameter of 6 inches and is completely filled with oil? How much is “1 mile” of oil in this pipeline worth at a price of $100 per barrel? __________ barrels$ __________
184.7 barrels
$18470 Explanation: Volume of barrel = 5.61 cubic feet Length of the pipe = 1 mile = 5280 feet Diameter of the pipe = 6 inches = 0.5 feet Radius of the pipe = 6/2 inches = 3 inches = 0.25 feet Volume of oil in the pipe = πr²h = 3.14 × (0.25)² × 5280 = 1036.2 cubic feet Number of barrels in the pipe = 1036.2/5.61 = 184.7 barrels Cost of one barrel =$100
Cost of 184.7 barrels =184.7 × $100 =$18470
Question 17.
A pan for baking French bread is shaped like half a cylinder. It is 12 inches long and 3.5 inches in diameter. What is the volume of uncooked dough that would fill this pan?
_________ in3
57.697 in³
Explanation:
The length of the pan = 12 in
The diameter of the pan = 3.5 in
Radius = 3.5/2 = 1.75 in
The volume of uncooked dough = Half the volume of the full cylinder of the above dimensions.
= (πr²h)/2 = (3.14 × (1.75)² × 12)/2 = 115.395/2 = 57.697 in³
FOCUS ON HIGHER ORDER THINKING
Question 18.
Explain the Error A student said the volume of a cylinder with a 3-inch diameter is two times the volume of a cylinder with the same height and a 1.5-inch radius. What is the error?
Type below:
_______________
The volume of the cylinder of 3 in is four times the volume of the new cylinder of radius 1.5 in
Explanation:
The volume of a cylinder is directly proportional to the square of the radius of the cylinder. The volume does not depend on the radius linearly.
Volume = πr²h
V1 = π(3)²h
V2 = π(1.5)²h
V1/V2 = (π(3)²h)/(π(1.5)²h)
V1/V2 = 4
V1 = 4V2
Question 19.
Communicate Mathematical Ideas Explain how you can find the height of a cylinder if you know the diameter and the volume. Include an example with your explanation.
Type below:
_______________
Let the diameter be D.
Volume = πr²h
Volume = π(D/2)²h
V = π((D)²/4)h
h = 4V/π(D)²
To find the height of a cylinder with diameter D = 2 m
Let the volume be 10 m³
h = 4V/π(D)²
h = (4 × 10)/(3.14 × 2²)
h = 3.18 m³
Question 20.
Analyze Relationships Cylinder A has a radius of 6 centimeters. Cylinder B has the same height and a radius half as long as cylinder A. What fraction of the volume of cylinder A is the volume of cylinder B? Explain.
Fraction: $$\frac{□}{□}$$
$$\frac{VA}{4}$$
Explanation:
rA = 6 cm
rB = half of the radius of cylinder A = 3 cm
hA = hB
VA = πrA²h
VB = πrB²h
VA/VB = (πrA²h)/(πrB²h)
VA/VB = 6²/3² = 36/9 = 4
Thus VB = VA/4
### Guided Practice – Volume of Cones – Page No. 408
Question 1.
The area of the base of a cylinder is 45 square inches and its height is 10 inches. A cone has the same area for its base and the same height. What is the volume of the cone?
The volume of the cone is _____ in3.
_________ in3
150 in³
Explanation:
In the question, the area of the base of the cylinder, B = 45 in²
Height of the cylinder, h = 10 inch
Volume of the cylinder, V cylinder = B × h = 45 × 10 = 450 inch³
Volume of the cone, V Cone = 1/3 V cylinder
=1/3(450 inch) = 150 inch³
So, the volume of the cone is
Vcone = 150 in³
Question 2.
A cone and a cylinder have congruent height and bases. The volume of the cone is 18 m3.What is the volume of the cylinder? Explain.
_________ m3
54 m3
Explanation:
The volume of the cone is 18 m3.
Vcone = 1/3 Vcylinder
Vcylinder = 3Vcone
Vcylinder = 3.18
Vcylinder = 54 m3
Find the volume of each cone. Round your answer to the nearest tenth if necessary. Use 3.14 for π.
Question 3.
_________ ft3
65.94 ft³
Explanation:
the diameter of the cone is 6ft.
so, the radius of the cone is 3ft.
the height of the cone is 7ft.
the volume of the cone = 1/3 × πr²h = 1/3 × 3.14 × 3² × 7 = 65.94 ft³
Question 4.
_________ in3
113982in³
Explanation:
The radius is 33inch and the height is 100 inch
Volume of the cone = 1/3 × πr²h = 1/3 × π(33)²100 = 113982in³
Question 5.
Gretchen made a paper cone to hold a gift for a friend. The paper cone was 15 inches high and had a radius of 3 inches. Find the volume of the paper cone to the nearest tenth. Use 3.14 for π.
_________ in3
141.3in³
Explanation:
the radius of the cone is 3inch and the height of the cone is 15inch.
Volume of the cone = 1/3 × πr²h = 1/3 × π(3)² × 15 = 141.3in³
Question 6.
A cone-shaped building is commonly used to store sand. What would be the volume of a cone-shaped building with a diameter of 50 meters and a height of 20 meters? Round your answer to the nearest tenth. Use 3.14 for π.
_________ m3
13083.33 m³
Explanation:
The diameter of the cone is 50 meters. So, the radius of the cone is 25 meters. The height of the cone is 20 meters.
Volume of the cone = 1/3 × πr²h = 1/3 × π(25)² × 20 = 13083.33 m³
ESSENTIAL QUESTION CHECK-IN
Question 7.
How do you find the volume of a cone?
Type below:
____________
V cone = 1/3 V cylinder
V cone = 1/3 πr²h
### 13.2 Independent Practice – Volume of Cones – Page No. 409
Find the volume of each cone. Round your answers to the nearest tenth if necessary. Use 3.14 for π.
Question 8.
_________ mm3
410.3 mm³
Explanation:
height = 8 mm
Volume of cone = 1/3 πr²h
Volume = 1/3(3.14)(7)²(8)
Volume = 410.29 mm³
Volume = 410.3 mm³
Question 9.
_________ in3
25.1 in3
Explanation:
Height = 6 in
Volume of cone = 1/3 πr²h
Volume = 1/3(3.14)(2)²(6)
Volume = 25.12 in3
Volume = 25.1 in3
Question 10.
A cone has a diameter of 6 centimeters and a height of 11.5 centimeters.
_________ cm3
108.3 cm3
Explanation:
Diameter of base = 6 cm
Radius = 6/2 cm = 3 cm
Height = 11.5 cm
Volume of cone = 1/3 πr²h
Volume = 1/3(3.14)(3)² (11.5)
Volume = 108.33 cm3
Volume = 108.3 cm3
Question 11.
A cone has a radius of 3 meters and a height of 10 meters.
_________ m3
94.2 m3
Explanation:
Height = 10 m
Volume of cone = 1/3 πr²h
Volume = 1/3(3.14)(3)²(10)
Volume = 94.2 m3
Round your answers to the nearest tenth if necessary. Use 3.14 for π.
Question 12.
Antonio is making mini waffle cones. Each waffle cone is 3 inches high and has a radius of $$\frac{3}{4}$$ inch. What is the volume of a waffle cone?
_________ in3
1.8 in3
Explanation:
Height = 3 in
Volume of each waffle cone = 1/3 πr²h
Volume = 1/3 (3.14) (0.75)² (3)
Volume = 1.76625 in3
Volume = 1.8 in3
Question 13.
A snack bar sells popcorn in cone-shaped containers. One container has a diameter of 8 inches and a height of 10 inches. How many cubic inches of popcorn does the container hold?
_________ in3
167.5 in3
Explanation:
Diameter of base = 8 in
Radius = 8/2 in = 4 in
Height = 10 in
Volume of cone = 1/3 πr²h
Volume = 1/3 (3.14) (4)² (10)
Volume = 167.466 in3
Volume = 167.5 in3
Question 14.
A volcanic cone has a diameter of 300 meters and a height of 150 meters. What is the volume of the cone?
_________ m3
3534291.7 m3
Explanation:
Diameter of base = 300 m
Radius = 300/2 m = 150 m
Height = 150 m
Volume of cone = 1/3 πr²h
Volume = 1/3 (3.14) (150)² (150)
Volume = 3534291.735 m3
Volume = 3534291.7 m3
Question 15.
Multistep Orange traffic cones come in a variety of sizes. Approximate the volume, in cubic inches, of a traffic cone that has a height of 2 feet and a diameter of 10 inches. Use 3.14 for π.
_________ in3
628 in³
Explanation:
The radius of the cone is Diameter/2 = 10/2 = 5
The height of the cone is 2 ft = 2 . 12 = 24 in
Vcone = 1/3 πr²h
Vcone = 1/3 (3.14) (5)² (24)
Vcone = 628 in³
Find the missing measure for each cone. Round your answers to the nearest tenth if necessary. Use 3.14 for π.
Question 16.
height = 6 in.
volume = 100.48 in3
_________ in.
4 in.
Explanation:
Height = 6 in
Volume = 100.4 in
Volume of cone = 1/3 πr²h
√(3v/hπ) = R
√((3 × 100.48)/(18.84)) = R
√(301.44/18.84) = R
R = √(16)
R = 4 in
Question 17.
diameter = 6 cm
height = _______
volume = 56.52 cm3
_______ cm
height = 6 cm
h = 6 cm
Explanation:
Let height be h
Diameter = 6 cm
Radius = 6/2 = 3 cm
Volume = 56.52 cm
Volume of cone = 1/3 πr²h
(3V/r²h) = h
(3 × 56.52)/(3² × 3.14) = h
169.56/28.26 = h
h = 6 cm
Question 18.
The diameter of a cone-shaped container is 4 inches, and its height is 6 inches. How much greater is the volume of a cylinder-shaped container with the same diameter and height? Round your answer to the nearest hundredth. Use 3.14 for π.
Type below:
____________
The volume of the cylinder is 50.24 in³ greater than the volume of the cone.
Explanation:
The diameter of a cone, d = 4 inch
radius of a cone, r = d/2 = 4/2 = 2 inches
height of a cone, h = 6 inches.
So, volume of a cone, V cone = 1/3 πr²h
= 1/3 (3.14) (2)² (6)
= 25.12 in³
And volume of a cylinder with same diameter and height,
V cylinder = πr²h = (3.14) (2)² (6) = 75.36 in³
The volume of the cylinder is 50.24 in³ greater than the volume of the cone.
### FOCUS ON HIGHER ORDER THINKING – Volume of Cones – Page No. 410
Question 19.
Alex wants to know the volume of sand in an hourglass. When all the sand is in the bottom, he stands a ruler up beside the hourglass and estimates the height of the cone of sand.
a. What else does he need to measure to find the volume of sand?
____________
To find the volume of the sand, he needs to measure the radius of the base of the hourglass.
Question 19.
b. Make a Conjecture If the volume of sand is increasing at a constant rate, is the height increasing at a constant rate? Explain.
____________
The volume of the cone is linearly proportional to the height of the cone. Therefore, if the volume is increasing at a constant rate, the height is also increasing at a constant rate.
Question 20.
Problem Solving The diameter of a cone is x cm, the height is 18 cm, and the volume is 301.44 cm3. What is x? Use 3.14 for π.
________ cm
8 cm
Explanation:
V cone = 1/3 πr²h
301.44 = 1/3 . 3.14 . r² . 18
r² = 904.32/56.52
r² = 16
r = 4 cm
The diameter of the circle is twice its radius, therefore
x = 2 . r
x = 2 . 4
x = 8 cm
Question 21.
Analyze Relationships A cone has a radius of 1 foot and a height of 2 feet. How many cones of liquid would it take to fill a cylinder with a diameter of 2 feet and a height of 2 feet? Explain.
________ cones
3 cones
Explanation:
The diameter of the base of the cylinder is 2 feet, which means that its radius is 1 foot. Its height is 2 feet. The volume of this cylinder is
V cylinder = πr²h
V cylinder = (3.14) (1)² (2)
V cylinder = 6.28
The radius of the cone is 1 foot and the height of the cone is 2 feet. The volume of the cone is:
V cone = 1/3 πr²h
V cone = 1/3 (3.14) (1)² (2)
V cone = 1/3 × 6.28
V cone = 1/3 . V cylinder
V cone = 2.09
It would take 3 cones of liquid to fill the cylinder.
Question 22.
Critique Reasoning Herb knows that the volume of a cone is one third that of a cylinder with the same base and height. He reasons that a cone with the same height as a given cylinder but 3 times the radius should therefore have the same volume as the cylinder, since $$\frac{1}{3}$$ ⋅ 3 = 1. Is Herb correct? Explain.
____________
The volume of the given cylinder is V cylinder = πr²h
The volume of the cone with the same height h as a given cylinder but 3 times the radius r is
V cone = 1/3 π(3r)²h
V cone = 3 πr²h
V cone = 3 V cylinder
As we can see, Herb is not correct. The volume of the cone is not equal to the volume of the cylinder. But it is three times the volume of the cylinder.
### Guided Practice – Volume of Spheres – Page No. 414
Question 1.
Vocabulary A sphere is a three-dimensional figure with all points _____ from the center.
Type below:
____________
A sphere is a three-dimensional figure with all points at equal distance from the center.
Question 2.
Vocabulary The _____ is the distance from the center of a sphere to a point on the sphere.
Type below:
____________
Explanation:
The radius is the distance from the center f the sphere to a point on the sphere
Find the volume of each sphere. Round your answers to the nearest tenth if necessary. Use 3.14 for π.
Question 3.
_______ in3
4.12 in3
Explanation:
V = 4/3πr³
V = 4/3 (3.14) (1)³
V = 4.12 in3
Question 4.
_______ cm3
4186.7 cm³
Explanation:
Diameter = 20 cm
Radius r = 20/2 = 10 cm
Volume of sphere = 4/3πr³
Volume = 4/3 (3.14) (10)³
Volume = 4186.66 cm³
Volume = 4186.7 cm³
Question 5.
A sphere has a radius of 1.5 feet.
_______ ft3
14.1 ft³
Explanation:
The volume of sphere = 4/3πr³
Volume = 4/3 (3.14) (1.5)³
Volume = 14.13 ft³
Volume = 14.1 ft³
Question 6.
A sphere has a diameter of 2 yards.
_______ yd3
4.2 yd³
Explanation:
Diameter = 2 yards
Volume of sphere = 4/3πr³
Volume = 4/3 (3.14) (1)³
Volume = 4.1866 yd³
Volume = 4.2 yd³
Question 7.
A baseball has a diameter of 2.9 inches. Find the volume of the baseball. Round your answer to the nearest tenth if necessary. Use 3.14 for π.
_______ in3
12.8 in³
Explanation:
Diameter of baseball = 2.9 in
Radius of baseball = 1.45 in
The volume of sphere = 4/3πr³
Volume = 4/3 (3.14) (1.45)³
Volume = 12.763 in³
Volume = 12.8 in³
Question 8.
A basketball has a radius of 4.7 inches. What is its volume to the nearest cubic inch. Use 3.14 for π.
_______ in3
1304 in³
Explanation:
Radius of baseball = 4.7 in
The volume of sphere = 4/3πr³
Volume = 4/3 (3.14) (4.7)³
Volume = 1304.0168 in³
Volume = 1304 in³
Question 9.
A company is deciding whether to package a ball in a cubic box or a cylindrical box. In either case, the ball will touch the bottom, top, and sides.
a. What portion of the space inside the cylindrical box is empty? Explain.
Type below:
_______________
The volume of the cylinder is V cylinder = πr²h
Since the ball touches the bottom, top, and sides, then the height of the cylinder will be equal to 2r.
V cylinder = πr²(2r) = 2πr³
On the other hand, the volume of the sphere is
V sphere = 4/3 πr³
The volume of the empty space inside the cylindrical box is found by subtracting the volume of the sphere from the volume of the cylinder
V cylinder – V sphere = 2πr³ – 4/3 πr³
= (2 – 4/3)πr³
= 2/3πr³
Question 9.
b. Find an expression for the volume of the cubic box.
Type below:
_______________
The volume of a cube with side a is V cube = a³
Since the ball touches the bottom, top, and sides, then the side of the cube will be equal to 2r.
V cube = (2r)³
V cube = 8r³
Question 9.
c. About what portion of the space inside the cubic box is empty? Explain
Type below:
_______________
The volume of the empty space inside the cubical box is found by subtracting the volume of the sphere from the volume of the cube:
V cube – V sphere = 8r³ – 4/3 πr³
= (8 – 4/3π)r³
= (8 – 4.2)r³
= 3.8r³
ESSENTIAL QUESTION CHECK-IN
Question 10.
Explain the steps you use to find the volume of a sphere.
Type below:
_______________
Step 1: The radius of the sphere is found out.
Step 2: The volume of the sphere is 4/3 πr³; where R is the radius.
Step 3: Put the value of radius in the equation of volume.
Step 4: Calculate the volume.
### 13.3 Independent Practice – Volume of Spheres – Page No. 415
Find the volume of each sphere. Round your answers to the nearest tenth if necessary. Use 3.14 for π.
Question 11.
_______ m3
124.7 m³
Explanation:
The volume of the sphere with a radius of 3.1 meter is 4/3 πr³
V = 4/3 . (3.14) . (3.1)³
V = 124.7 m³
Question 12.
diameter of 18 inches
_______ in3
3052.1 in³
Explanation:
The diameter of the sphere is 18 inches, which means that its radius is 9 inches. The volume of the sphere is
V = 4/3 πr³
V = 4/3 . (3.14) . (9)³
V = 3052.08 in³
V = 3052.1 in³
Question 13.
r = 6 in.
_______ in3
904.3 in³
Explanation:
The volume of the sphere with a radius of 6 inches is
V = 4/3 πr³
V = 4/3 (3.14) (6)³
V = 904.32
V = 904.3 in³
Question 14.
d = 36 m
_______ m3
24416.6 m³
Explanation:
d = 36 m
radius r = 36/2 = 18 m
Volume = 4/3 πr³
= 4/3 (3.14) (18)³
= 24416.64
Volume = 24416.6 m³
Question 15.
_______ cm3
5572.5 cm³
Explanation:
The volume of the sphere with a radius of 11 centimeters is
V = 4/3 πr³
V = 4/3 (3.14) (11)³
V = 5572.5 cm³
Question 16.
_______ ft3
8.2 feet³
Explanation:
The diameter of the sphere is 2.5 feet, which means that its radius is 1.25 feet. The volume of the sphere is
V = 4/3 πr³
V = 4/3 . (3.14) . (1.25)³
V = 8.2 feet³
The eggs of birds and other animals come in many different shapes and sizes. Eggs often have a shape that is nearly spherical. When this is true, you can use the formula for a sphere to find their volume.
Question 17.
The green turtle lays eggs that are approximately spherical with an average diameter of 4.5 centimeters. Each turtle lays an average of 113 eggs at one time. Find the total volume of these eggs, to the nearest cubic centimeter.
_______ cm3
5389 cm³
Explanation:
The diameter of an egg (sphere) is 4.5 centimeters, which means that its radius is 2.25 centimeters. The volume of a single egg is
V = 4/3 πr³
V = 4/3 (3.14) (2.25)³
V = 47.68875 cm³
Therefore, the total volume of 113 eggs is
113 . V = 113(47.68875)
= 5388.82875
= 5389 cm³
Question 18.
Hummingbirds lay eggs that are nearly spherical and about 1 centimeter in diameter. Find the volume of an egg. Round your answer to the nearest tenth.
_______ cm3
0.5 cm³
Explanation:
The diameter of an egg (sphere) is 1 centimeter, which means that its radius is 0.5 centimeters. The volume of a single egg is
V = 4/3 πr³
V = 4/3 (3.14) (0.5)³
V = 0.5 cm³
Question 19.
Fossilized spherical eggs of dinosaurs called titanosaurid sauropods were found in Patagonia. These eggs were 15 centimeters in diameter. Find the volume of an egg. Round your answer to the nearest tenth.
_______ cm3
1766.25 cm³
Explanation:
Diameter of an egg = 15 cm
Its radius = 15/2 = 7.5 cm
Volume = 4/3 πr³
V = 4/3 (3.14) (7.5)³
V = 1766.25 cm³
Question 20.
Persevere in Problem Solving An ostrich egg has about the same volume as a sphere with a diameter of 5 inches. If the eggshell is about $$\frac{1}{12}$$ inch thick, find the volume of just the shell, not including the interior of the egg. Round your answer to the nearest tenth.
_______ in3
6.8 in³
Explanation:
Diameter including the eggshell
= 5 + (2 . 1/2)
= (5 + 1/6) in
= 5.166 in
Radius including egg shell = 5.166/2 = 2.5833 in
Volume = 4/3 πr³
Volume = 4/3 (3.14) (2.5833)³
=72.176 in³
Volume with shell = 72.2 in³
Radius excluding shell = 5/2 = 2.5 in
Volume = 4/3 (3.14) (2.5)³
= 65.416 in³
Volume (without shell) = 65.4 in³
Volume of shell = Total volume – Inner Volume
= 72.2 – 65.4
= 6.8 in³
Question 21.
Multistep Write the steps you would use to find a formula for the volume of the figure at right. Then write the formula.
Type below:
_____________
5/3πr³
Explanation:
The radius of hemisphere = r
Height of cylinder = r
Step 1: Find the formula for the volume of a hemisphere
The volume of hemisphere = 4/3 π/2 r³
= 2/3πr³
Step 2: Find the formula for the volume of a cylinder
The volume of cylinder = πr²h
=πr³
Step 3: Add both the volume expressions:
Total volume = 2/3πr³ + πr³
= 5/3πr³
### Volume of Spheres – Page No. 416
Question 22.
Critical Thinking Explain what happens to the volume of a sphere if you double the radius.
Type below:
_____________
Volume V1 = 4/3πr³
Volume V2 = 4/3π(2r)³
= 8 . 4/3πr³
= 8 V1
= 8(initial volume)
By doubling the radius of sphere we make the voulme 8 times the intial value.
Question 23.
Multistep A cylindrical can of tennis balls holds a stack of three balls so that they touch the can at the top, bottom, and sides. The radius of each ball is 1.25 inches. Find the volume inside the can that is not taken up by the three tennis balls.
_______ in3
12.3 in³
Explanation:
Radius of the ball = 1.25 inch
Height of the cylinder = (2 × 1.25) × 3
= (2.5) × 3
= 7.5 in
radius of base of cylinder = 1.25 in.
Volume of cylinder = πr²h
= (3.14) (1.25)² (7.5)
= 36.7968
= 36.8 in³
Volume of a ball (all three) = 3 × 4/3πr³
= 4 (3.14) (1.25)³
= 24.53125 in³
= 24.5 in³
Volume of empty space = Volume of cylinder – Volume of ball
= 36.8 – 24.5 = 12.3 in³
FOCUS ON HIGHER ORDER THINKING
Question 24.
Critique Reasoning A sphere has a radius of 4 inches, and a cube-shaped box has an edge length of 7.5 inches. J.D. says the box has a greater volume, so the sphere will fit in the box. Is he correct? Explain.
_____________
The volume of sphere = 4/3πr³
= 4/3 (3.14) (4)³
= 267.9466
= 268
The volume of cube = (7.5)³
= 421.875
=421.9
The volume of cube > Volume of a sphere
But the base of the cube has an area of (7.5 × 7.5) = 56.25 while the cross-action area of the sphere.
πr² = (3.14) (4)² = 50.24
The cross-section area of the cube is less than that of a sphere. thus J.D. is wrong and the ball (sphere) will not fit in the cube.
Question 25.
Critical Thinking Which would hold the most water: a bowl in the shape of a hemisphere with radius r, a cylindrical glass with radius r and height r, or a cone-shaped drinking cup with radius r and height r? Explain.
_____________
The volume of a sphere with radius r is
V sphere = 4/3πr³
Therefore, the volume of a hemisphere is
V hemisphere = V sphere/2
V hemisphere = 2/3πr³
The volume of a cylinder with radius r and height r is
V cylinder = πr²h
V cylinder = πr³
The volume of a cone with radius r and height r is
V cone = 1/3πr²h
V cone = 1/3πr³
V cone < V hemisphere < V cylinder
Therefore, the cylindrical glass with radius r and height r will hold the most water.
Question 26.
Analyze Relationships Hari has models of a sphere, a cylinder, and a cone. The sphere’s diameter and the cylinder’s height are the same, 2r. The cylinder has radius r. The cone has diameter 2r and height 2r. Compare the volumes of the cone and the sphere to the volume of the cylinder.
Type below:
_____________
Radius of sphere = 2r/2 = r
Volume of sphere = 4/3πr³
Height of cylinder = 2r
volume of cylinder = πr²(2r)
V cylinder = 2πr³
Radius of cone = 2r/2 = r
Height of cone = 2r
Volume of cone = 1/3 πr²(2r)
V cone = 2/3πr³
Volume of cylinder > Volume of sphere > Volume of cone
2πr³ > 4/3πr³ > 2/3πr³
Question 27.
A spherical helium balloon that is 8 feet in diameter can lift about 17 pounds. What does the diameter of a balloon need to be to lift a person who weighs 136 pounds? Explain.
_______ feet
Diameter of ballon = 8 ft
Weight it could lift = 17 pound
Volume = 4/3 π(8/2)³
= 4³(4π/3)
4³/x(4π/3) = 17/36
1/x = 1/8 × 3/4π × 1/48
x = 4π/3 . 4³ . 2³
x = 4/3. π . 8³
The volume of ballon which can lift 136 pounds is equal to 4/3. π . 8³
The radius of that ballon = 8ft
Diameter = 8 . 2 = 16 ft
### Ready to Go On ? – Model Quiz – Page No. 417
13.1 Volume of Cylinders
Find the volume of each cylinder. Round your answers to the nearest tenth if necessary. Use 3.14 for π.
Question 1.
_______ ft3
904.8 ft³
Explanation:
Radius of base = 6 ft
Height of cylinder = 8 ft
The volume of cylinder = πr²h
Volume = (3.14) (6)² (8)
Volume = 904.77 ft³
Volume = 904.8 ft³
Question 2.
A can of juice has a radius of 4 inches and a height of 7 inches. What is the volume of the can?
_______ in3
351.7 in³
Explanation:
Radius if cylindrical can = 4 in
Height of cylindrical can = 7 in
The volume of cylinder = πr²h
Volume = (3.14) (4)² (7)
Volume = 351.68 in³
Volume = 351.7 in³
13.2 Volume of Cones
Find the volume of each cone. Round your answers to the nearest tenth if necessary. Use 3.14 for π.
Question 3.
_______ cm3
565.2 cm³
Explanation:
Radius of base of cone = 6 cm
Height of cone = 15 cm
Volume of cone = 1/3πr²h
Volume = 1/3 (3.14) (4)² (7)
Volume = 565.2 cm³
Question 4.
_______ in3
3014.4 in³
Explanation:
The radius of the base of cone = 12 in
Height of cone = 20 in
The volume of cone = 1/3πr²h
Volume = 1/3 (3.14) (12)² (20)
Volume = 3014.4 in³
13.3 Volume of Spheres
Find the volume of each sphere. Round your answers to the nearest tenth if necessary. Use 3.14 for π.
Question 5.
_______ in3
113 in³
Explanation:
Radius of sphere = 3 ft
Volume of sphere = 4/3πr³
Volume = 4/3 (3.14) (3)³
Volume = 113.04 ft³
Volume = 113 in³
Question 6.
_______ cm3
1149.8 cm³
Explanation:
Diameter = 13 cm
Radius = 13/2 cm = 6.5 cm
Volume of sphere = 4/3πr³
Volume = 4/3 (3.14) (6.5)³
Volume = 1149.7633 cm³
Volume = 1149.8 cm³
ESSENTIAL QUESTION
Question 7.
What measurements do you need to know to find the volume of a cylinder? a cone? a sphere?
Type below:
___________
Sphere: To find the volume of the sphere, the radius is to be measured.
Cylinder: To measure the volume of the cylinder, we need to find out the base radius of the base of the cylinder along with the height of the cylinder.
Cone: To calculate the volume of the cone, we need to calculate the base radius of the base of the cone along with the height of the cone.
Explanation:
The volume of sphere = 4/3 πr³
Sphere: For finding the volume of the sphere, the radius is to be measured
The volume of Cylinder = πr²h
Cylinder: To calculate the volume of the cylinder, we need to find out the base radius of the base of the cylinder along with the height of the cylinder
The volume of Cone = 1/3 πr²h
Cone: To calculate the volume of the cone, we need to measure the base radius of the base of the cone along with the height of the cone
### Selected Response – Mixed Review – Page No. 418
Question 1.
The bed of a pickup truck measures 4 feet by 8 feet. To the nearest inch, what is the length of the longest thin metal bar that will lie flat in the bed?
Options:
a. 11 ft 3 in.
b. 10 ft 0 in.
c. 8 ft 11 in.
d. 8 ft 9 in.
d. 8 ft 9 in.
Explanation:
The length of the longest thin metal bar that will lie flat in the bed’s equal to the length of the bed’s hypotenuse. Let a = 4 and b = 8. Using the Pythagorean Theorem
a² + b² = c²
4² + 8² = c²
16 + 64 = c²
80 = c²
Rounding the length of the hypotenuse to the nearest tenth of a foot
c = 8.9 ft
Therefore, the length of the longest thin metal bar that will lie flat in the bed is 8 ft. 9 in.
Question 2.
Using 3.14 for π, what is the volume of the cylinder below to the nearest tenth?
OPtions:
a. 102 cubic yards
b. 347.6 cubic yards
c. 1,091.6 cubic yards
d. 4,366.4 cubic yards
c. 1,091.6 cubic yards
Explanation:
Diameter of the base of the cylinder = 11.4 yd
Radius = 11.4/2 yd = 5.7 yd
Height = 10.7 ys
Volume of cylinder = πr²h
Volume = (3.14) (5.7)² (10.7)
Volume = 1091.599 yd³
Volume = 1091.6 yd³
Question 3.
Rhett made mini waffle cones for a birthday party. Each waffle cone was 3.5 inches high and had a radius of 0.8 inches. What is the volume of each cone to the nearest hundredth?
Options:
a. 1.70 cubic inches
b. 2.24 cubic inches
c. 2.34 cubic inches
d. 8.79 cubic inches
c. 2.34 cubic inches
Explanation:
Height of each waffle cone = 3.5 in
Radius of base = 0.8 in
Volume of cone = 1/3 πr²h
Volume = 1/3 (3.14) (0.8)² (3.5)
Volume = 2.344533 in³
Volume = 2.34 in³
Question 4.
What is the volume of a cone that has a height of 17 meters and a base with a radius of 6 meters? Use 3.14 for π and round to the nearest tenth.
Options:
a. 204 cubic meters
b. 640.6 cubic meters
c. 2,562.2 cubic meters
d. 10,249 cubic meters
b. 640.6 cubic meters
Explanation:
Height of the cone = 17 m
Radius of base = 6 m
Volume of cone = 1/3 πr²h
Volume = 1/3 (3.14) (6)² (17)
Volume = 640.56 m³
Volume = 640.6 m³
Question 5.
Using 3.14 for π, what is the volume of the sphere to the nearest tenth?
Options:
a. 4,180 cubic centimeters
b. 5,572.5 cubic centimeters
c. 33,434.7 cubic centimeters
d. 44,579.6 cubic centimeters
b. 5,572.5 cubic centimeters
Explanation:
Diameter of the base of the sphere = 22 cm
Radius = 22/2 yd = 11 cm
Volume of sphere = 4/3 πr³
Volume = 4/3 (3.14) (11)³
Volume = 5572.4533 cm³
Volume = 5572.5 cm³
Question 6.
A diagram of a deodorant container is shown. It is made up of a cylinder and half of a sphere.
Use 3.14 for π and round answers to the nearest tenth.
a. What is the volume of the half sphere?
_______ cm3
8.574 cm³
Explanation:
The radius of the cylinder as well as the hemisphere = 1.6 cm
Height = 6.2 cm
the volume of the hemisphere = 2/3 πr³
the volume of the hemisphere = 2/3 (3.14) (1.6)³
the volume of the hemisphere = 8.574 cm³
Question 6.
b. What is the volume of the cylinder?
_______ cm3
49.838 cm³
Explanation:
The volume of cylinder = πr²h
= (3.14) (1.6)² (6.2)
= 49.838 cm³
Question 6.
c. What is the volume of the whole figure?
_______ cm3
|
# NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.4
Go back to 'Introduction to Trigonometry'
## Chapter 8 Ex.8.4 Question 1
Express the trigonometric ratios $$\text{sin} \;A$$, $$\text{sec} \;A$$ and $$\text{ tan} \;A$$ in terms of $$\text{cot} \;A$$.
#### Reasoning:
\begin{align}{ \text{cosec} ^{2} A=1+\cot ^{2} A} \\ {\sec ^{2} A=1+\tan ^{2} A}\end{align}
#### Steps:
Consider a \begin{align}\Delta {ABC}\end{align} with \begin{align}\angle {B }=\text{ }90{}^\circ \end{align}
Using the Trigonometric Identity,
$$\text{cose}{{\text{c}}^{2}}\,{A}=1+{{\cot }^{2}}{A}$$
\begin{align} \frac{1}{\text{cose}{{\text{c}}^{2}}\,{A}}&=\frac{\text{1}}{\text{1+co}{{\text{t}}^{\text{2}}}{A}} \end{align} (By taking reciprocal both the sides)
\begin{align} &{{\text{sin}}^{2}}{A}=\frac{1}{1+{{\cot }^{2}}{A}} \\ & \left( \text{As }\frac{1}{\text{cose}{{\text{c}}^{2}}\,\text{A}}={{\sin }^{2}}\text{A} \right) \\ \end{align}
Therefore,
\begin{align}\text{sin}\,{A}\,=+\frac{1}{\sqrt{1+{{\cot }^{2}}{A}}} \end{align}
For any sine value with respect to an angle in a triangle, sine value will never be negative. Since, sine value will be negative for all angles greater than 180°.
Therefore, \begin{align}\text{sin}\,{A}\,=\frac{\text{1}}{\sqrt{\text{1+co}{{\text{t}}^{\text{2}}}{A}}} \end{align}
We know that,
\begin{align}\text{tan} A=\frac{\text{sin}} A{\cos \,{A}}\end{align}
However, Trigonometric Function,
\begin{align}\text{cot}\,{A=}\frac{\cos \,{A}}{\text{Sin}\,{A}}\end{align}
Therefore, Trigonometric Function,
\begin{align}\text{tan}\,A=\frac{1}{\cot \,{A}}\end{align}
Also,
\begin{align}\text{se}{{\text{c}}^{\text{2}}} A = 1 + {{\text{tan}}^{\text{2}}}{A}\end{align} (Trigonometric Identity)
\begin{align}=1+\frac{1}{{{\cos }^{2}}{A}} \\ =\frac{{{\cot }^{2}}\,{A+1}}{{{\cot }^{2}}{A}} \\ \end{align}
\begin{align}\text{sec}\,{A}=\frac{\sqrt{{{\cot }^{2}}{A+1}}}{\cot \,A}\end{align}
## Chapter 8 Ex.8.4 Question 2
Write all the other trigonometric ratios of$$\angle A$$ in terms of $$\text{sec} \;A$$.
#### Reasoning:
\begin{align}{\sin ^{2} A+\cos ^{2} A=1} \\ {\text{cosec} ^{2} A=1+\cot ^{2} A} \\ {\sec ^{2} A=1+\tan ^{2} A}\end{align}
#### Steps:
We know that, Trigonometric Function,
\begin{align}\text{cos}\,{A = }\frac{1}{\sec {A}}\ldots \text{Equation }\left( \text{1} \right)\end{align}
Also,
\begin{align} & \text{si}{{\text{n}}^{\text{2}}}A + \text{co}{{\text{s}}^{\text{2}}}{A} = {1 } \\ & \left( \text{Trigonometric identity} \right) \\ \\& \text{si}{{\text{n}}^{\text{2}}}{A} = {1 }-\text{ co}{{\text{s}}^{\text{2}}}{A } \\ & \left( \text{By transposing} \right) \\ \end{align}
Using value of $$\text{cos} \;A$$ from Equation $$(1)$$ and simplifying further,
\begin{align}\sin {{A} }&=\sqrt {1 - {{\left( {\frac{1}{{\sec {{A}}}}} \right)}^2}} \\&= \sqrt {\frac{{{{\sec }^2}\,{{A}} - 1}}{{{{\sec }^2}\,{{A}}}}} \\ &= \frac{{\sqrt {{{\sec }^2}{{A}} - 1} }}{{\sec {{A}}}}\ldots (2)\end{align}
\begin{align} & {\text{ta}}{{\text{n}}^{\text{2}}}{{A + 1 }}={{\text{sec}}^{\text{2}}}{{A }} \\ & \left( {{\text{Trigonometric identity}}} \right)\\\\ & {\text{ta}}{{\text{n}}^{\text{2}}}{{A}}={{\text{sec}}^{\text{2}}}{{A - 1 }} \\ & \left( {{\text{By transposing}}} \right) \\\end{align}
Trigonometric Function,
\begin{align} {\text{tan}}\,{\text{A}}&=\sqrt {{\text{se}}{{\text{c}}^2}{\text{A}} - {\text{1}}}\,\,\dots\left( {\text{3}} \right)\\ {\text{cot}}\,{\text{A}}\,& =\,\frac{{{\text{cosA}}}}{{{\text{sinA}}}} \\&= \frac{{\frac{1}{{\sec \,{\text{A}}}}}}{{\frac{{\sqrt {{{\sec }^2}} {\text{A}} - 1}}{{\sec \,{\text{A}}}}}}\\& \begin{bmatrix}\text{ (By substituting}\\ \text{equations (1) and (2)}\end{bmatrix}\\\\&= \frac{1}{{\sqrt {{{\sec }^2}{\text{A}} - {\text{1}}} }} \\{\rm{cosec}}\,{\text{A}}&=\frac{1}{{\sin {\text{A}}}}\\&= \frac{{\sec {\text{A}}}}{{\sqrt {{{\sec }^2}\,{\text{A}} - 1} }}\\\\&\begin{bmatrix}\text{(By substituting}\\\text{ Equation (2) and}\\\text{simplifying)}\end{bmatrix}\end{align}
## Chapter 8 Ex.8.4 Question 3
Evaluate
(i) $$\,\,\frac{{{\sin }^{2}}{{63}^{{}^\circ }}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\cos }^{2}}{{17}^{{}^\circ }}+{{\cos }^{2}}{{73}^{{}^\circ }}}$$
(ii) $$\,\,\,\text{sin}{{25}^{^\circ }}\text{cos}{{65}^{^\circ }}+\text{cos}{{25}^{^\circ }}\text{ sin}{{65}^{^\circ }}$$
#### Reasoning:
\begin{align}{\sin ^{2} A+\cos ^{2} A=1} \\ {\sin \left(90^{\circ}-\theta\right)=\cos \theta} \\ {\cos \left(90^{\circ}-\theta\right)=\sin \theta}\end{align}
#### Steps:
(i)\begin{align}\frac{{{\sin }^{2}}{{63}^{{}^\circ }}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\cos }^{2}}{{17}^{{}^\circ }}+{{\cos }^{2}}{{73}^{{}^\circ }}}\end{align}
\begin{align} & =\frac{{{\left[ \sin \left( {{90}^{{}^\circ }}-27 \right) \right]}^{2}}+{{\sin }^{2}}27}{{{\left[ \cos \left( {{90}^{{}^\circ }}-{{73}^{{}^\circ }} \right) \right]}^{2}}+{{\cos }^{2}}{{73}^{{}^\circ }}} \\ & =\frac{{{[\cos 27]}^{2}}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\left[ \sin {{73}^{{}^\circ }} \right]}^{2}}+{{\cos }^{2}}{{73}^{{}^\circ }}} \\ & \begin{bmatrix} \sin \left( {{90}^{{}^\circ }}-\text{ }\!\!\theta\!\!\text{ } \right)=\cos \text{ }\!\!\theta\!\! \\ \And \\ \cos \left( {{90}^{{}^\circ }}-\theta \right)=\sin \theta \end{bmatrix} \\ & =\frac{1}{1} \; \begin{bmatrix} \text{By identity} \\ \sin ^{2} A+\cos ^{2} A=1 \end{bmatrix} \\& =1\end{align}
(ii) $$\sin {{25}^{{}^\circ }}\cos {{65}^{{}^\circ }}+\cos {{25}^{{}^\circ }}\sin {{65}^{{}^\circ }}$$
\begin{align} &= \left( \sin {{25}^{{}^\circ }} \right)\left\{ \cos \left( {{90}^{{}^\circ }}-{{25}^{{}^\circ }} \right) \right\} +\cos {{25}^{{}^\circ }} \left\{ \sin \left( {{90}^{{}^\circ }}-{{25}^{{}^\circ }} \right) \right. \\ & \begin{bmatrix} \sin \left( {{90}^{{}^\circ }}-\text{ }\!\!\theta\!\!\text{ } \right)=\cos \text{ }\!\!\theta\!\! \\ \And \cos \left( {{90}^{{}^\circ }}-\theta \right)=\sin \theta \end{bmatrix} \\ & = \begin{pmatrix} \left( \sin {{25}^{{}^\circ }} \right)\left( \sin {{25}^{{}^\circ }} \right) +\cos {{25}^{{}^\circ }}\left( \cos {{25}^{{}^\circ }} \right) \end{pmatrix}\\ & ={{\sin }^{2}}{{25}^{{}^\circ }}+{{\cos }^{2}}{{25}^{{}^\circ }} \\ & =1 \\ \\& \begin{bmatrix} \text{By identity} \\ \sin ^{2} A+\cos ^{2} A=1 \end{bmatrix} \end{align}
## Chapter 8 Ex.8.4 Question 4
Choose the correct option. Justify your choice.
(i) $$\text{ 9 sec 2A - 9 tan 2A} = \_\_\_\_\_\_\_\_\_\_\_\_\_$$
(A) $$1$$
(B) $$9$$
(C) $$8$$
(D) $$0$$
(ii) $$\left( \text{1 + tan } \theta\text{ + sec }\theta \right) \left( \text{1 + cot } \theta -\text{cosec }\theta \right) = \_\_\_\_\_\_\_\_\_\_\_\_\_$$
(A) $$0$$
(B) $$1$$
(C) $$2$$
(D) $$-1$$
(iii) $$\left( \text{sec A + tan A} \right) \left( \text{1 -sin A} \right) =\_\_\_\_\_\_\_\_\_$$
(A) $$\text{ sec A }$$
(B) $$\text{ sin A }$$
(C) $$\text{ cosec A }$$
(D) $$\text{ cos A}$$
(iv) $$\frac{\text{1+ta}{{\text{n}}^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}} = \_\_\_\_\_$$
(A) $$\text{ sec2A}$$
(B) $$1$$
(C) $$\text{cot2A}$$
(D) $$\text{tan2A}$$
#### Steps:
(i)
\begin{align} & \text{ 9 se}{{\text{c}}^{\text{2}}}\text{A }-\text{ 9 ta}{{\text{n}}^{\text{2}}}\text{A} \\ \\ &=9\text{ (se}{{\text{c}}^{\text{2}}}\text{A}-\text{ta}{{\text{n}}^{\text{2}}}\text{A) } \\& =\text{9 }\left( \text{1} \right) \\ & \begin{bmatrix} \text{By the identity,} \\ 1+ \sec ^2 A = \tan ^2 A \\ \Rightarrow \sec ^2 A - \tan^2A=1 \end{bmatrix} \\ & = 9\end{align}
(ii)
\begin{align} \left( \text{1 + tan }\theta\text{ + sec } \theta \right) \left( \text{1 + cot }\theta\text{ - cosec }\theta \right) \ldots (1) \end{align}
We know that the trigonometric functions,
\begin{align} & \tan (x)=\frac{\sin (x)}{\cos (x)} \\ & \cot (x)=\frac{\cos (x)}{\sin (x)}=\frac{1}{\tan (x)} \end{align}
And
\begin{align} \sec (x)&=\frac{1}{\cos (x)} \\ {cosec}(x)&=\frac{1}{\sin (x)} \end{align}
By substituting the above function in Equation $$(1)$$,
\begin{align}& = \begin{bmatrix} \left( 1+ \frac{\text{sin }\theta }{\text{cos }\theta } + \frac{1}{\text{cos }\theta } \right) \left( 1+ \frac{\text{cos }\theta }{\text{sin }\theta } - \frac{1}{\text{sin }\theta} \right) \end{bmatrix} \\ & = \begin{bmatrix} \left( \frac{\text{cos }\theta\text{ +sin }\theta+1}{\text{cos }\theta} \right) \left( \frac{\text{sin }\theta\text{ +cos }\theta -1}{\text{sin }\theta} \right)\end{bmatrix}\\& \text{(By taking LCM and multiplying)}\\\\& = \frac{{{\text{(sin }\theta +\text{cos} \theta )}^2}-(1)^2}{\text{sin }\theta \text{cos }\theta}\\& \text{(Using )}\\ \\&= \frac{ \begin{Bmatrix}\text{sin}^2 \theta \text{ +co}{{\text{s}}^2} \theta \\ \text{ +2sin }\theta\text{ cos }\theta -1 \end{Bmatrix} }{\text{sin }\theta\text{ cos }\theta} \\ & = \frac{\text{1+2sin }\theta\text{ cos }\theta -1 }{\text{sin }\theta\text{ cos}\theta}\\&\text{(Using identify)}\\\\&=\frac{\text{2sin }\theta\text{ cos }\theta\text{ }}{\text{sin }\theta\text{ cos }\theta\text{ }}\text{= 2}\end{align}
Hence, option (C) is correct.
(iii)
\begin{align}\left( \text{sec A + tan A} \right) \left( 1 - \text{sin A} \right) \ldots \left( 1 \right)\end{align}
We know that the trigonometric functions,
\begin{align}\tan (x)=\frac{\sin (x)}{\cos (x)}\end{align}
And
\begin{align}\sec (x)=\frac{1}{\cos (x)}\end{align}
By substituting the above function in Equation $$(1)$$,
\begin{align} &= \begin{bmatrix} \left( \frac{1}{\text{cosA}} + \frac{\text{sinA}}{\text{cosA}} \right) (1-\text{sinA)} \end{bmatrix} \\ & =\left( \frac{\text{1+sinA}}{\text{cosA}} \right) (1-\text{sinA)} \\ & = \frac{1-\text{si}{{\text{n}}^{2}}\text{A}}{\text{cosA}} \\ & = \frac{\text{co}{{\text{s}}^{2}}\text{A}}{\text{cosA}} \\ & \begin{bmatrix} \text{By identity}\\\sin^2 \theta+ \cos^{2} \theta=1 \\ \Rightarrow 1- {\rm{sin^2}} θ = {\rm{cos^2}} θ \end{bmatrix} \\ &= \cos A\end{align}
Hence, option (D) is correct.
(iv) \begin{align}\frac{1+{{\tan }^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}}\end{align}
We know that the trigonometric functions,
\begin{align} \tan (x)&=\frac{\sin (x)}{\cos (x)} \\ \cot (x)&=\frac{\cos (x)}{\sin (x)} \\ & =\frac{1}{\tan (x)} \end{align}
By substituting the above function in Equation ($$1$$),
\begin{align}\frac{{{\rm{1 + ta}}{{\rm{n}}^{{2}}}{{A}}}}{{{\rm{1 + co}}{{\rm{t}}^{{2}}}{{A}}}}&= \frac{{{{1 + }}\frac{{{\rm{sin}}^2{{A}}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}}}{{{{1 + }}\frac{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}{{{\rm{si}}{{\rm{n}}^{{2}}}{\rm{A}}}}}}\\ &= \frac{{\frac{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}} + {\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}}}{{\frac{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}} + {\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}}}\\&=\frac{{\frac{{{1}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}}}{{\frac{1}{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}}}\\&=\frac{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}\\&= \rm{tan}^2 A\end{align}
Hence, option ($$D$$) is correct
## Chapter 8 Ex.8.4 Question 5
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) \begin{align} {{\text{(cosec}\,\text{ }\!\!\theta\!\!\text{ }-\text{cot}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}\text{=}\frac{\text{1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ }}{\text{1+cos}\,\text{ }\!\!\theta\!\!\text{ }} \end{align}
(ii) \begin{align} \frac{\text{cos}\,A}{\text{1+sin}\,A}\,\,\text{+}\,\frac{\text{1+sin}\,A}{\text{cos}\,A}\text{=2sec}A \end{align}
(iii) \begin{align} \begin{bmatrix} \frac{\text{tan} \theta }{1-\text{cot} \theta} +\frac{\text{cot} \theta}{1-\text{tan} \theta}\\ =1+\text{sec} \theta \text{cosec} \theta \end{bmatrix} \end{align}
(iv) \begin{align} \frac{\text{1+sec}\; A}{\text{sec}\,A}\text{=}\frac{\text{si}{{\text{n}}^{\text{2}}}A}{\text{1}-\text{cos}\,A} \end{align}
(v) \begin{align} \begin{bmatrix} \frac{\text{cos}\,\text{A}-\text{sin}\,\text{A+1}}{\text{cos}\,\text{A+sin}\,\text{A+1}} \\ =\text{cosec}\,\text{A}\,\text{+}\,\text{cotA} \end{bmatrix} \end{align}
(vi) \begin{align} \sqrt{\frac{\text{1+sinA}}{\text{1-sinA}}}\,\text{=}\,\text{sec}\,\text{A}\,\text{+}\,\text{tan}\,\text{A} \end{align}
(vii) \begin{align} \frac{\text{sin}\,\text{ }\!\!\theta\!\!\text{ }-\text{2si}{{\text{n}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2cos}\,\text{ }\!\!\theta\!\!\text{ }-\text{cos}\,\text{ }\!\!\theta\!\!\text{ }}\,\text{=}\,\text{tan}\,\text{ }\!\!\theta\!\!\text{ } \end{align}
(viii) \begin{align} \begin{bmatrix} {{\text{(sin} A\,\text{+}\,\text{cosec}\,A)}^{\text{2}}}\\+(cos\,A\,\text{+}\,\text{sec}\,A{{\text{)}}^{\text{2}}} \\ =\text{7+ta}{{\text{n}}^{\text{2}}}A\,\text{+}\,\text{co}{{\text{t}}^{\text{2}}}A \end{bmatrix} \end{align}
(ix) \begin{align} \begin{bmatrix} \text{(cosecA}-\text{sinA)}\\\text{(secA}-\text{cosA)} \\ =\frac{\text{1}}{\,\text{tan}\,\text{A+cot}\,\text{A}} \end{bmatrix} \end{align}
(x) \begin{align} \begin{bmatrix} \left( \frac{1+{{\tan }^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}} \right) \\ ={{\left( \frac{1-\tan \text{A}}{1-\cot \text{A}} \right)}^{2}}\\ = \rm tan^2 A \end{bmatrix} \end{align}
#### Reasoning:
\begin{align}{\sin ^{2} A+\cos ^{2} A=1} \\ {\text{cosec} ^{2} A=1+\cot ^{2} A} \\ {\sec ^{2} A=1+\tan ^{2} A}\end{align}
#### Steps:
(i)\begin{align} \;\; {(\text{cosec}\; \theta-\cot \theta)^{2}}&={\frac{1-\cos \theta}{1+\cos \theta}} \end{align}
\begin{align} {\text { L.H.S }}&={(\text{cosec}\; \theta-\cot \theta)^{2}} \\ &\qquad \quad \ldots (1) \end{align}
We know that the trigonometric functions,
\begin{align} \cot \,(x)&=\frac{\cos \,(x)}{\sin \,(x)} =\frac{1}{\tan \,(x)} \\ \operatorname{cosec}\,(x) &=\frac{1}{\sin \,(x)} \end{align}
By substituting the above function in Equation $$(1)$$
\begin{align} \rm (cosec \theta - cot \theta)^2 &\!\!= \!\!{{\left( \frac{1}{\text{sin } \theta }-\frac{\text{cos} \theta }{\text{sin} \theta} \right)}^{2}}\end{align}
\begin{align} \text{=}\,\frac{{{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}{{{\text{(sin}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}\,\,\,\,\,\,\end{align}
\begin{align} \text{=}\,\frac{{{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}{\text{si}{{\text{n}}^{\text{2}}}\,\text{ }\!\!\theta\!\!\text{ }} \end{align}
\begin{align} \text{=}\,\frac{{{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}{\text{1}-\text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}\,\,\,\,\,\,\,\, \end{align} (By identity $$\sin^2 A$$ $$+$$ $$\cos^2 A =$$ $$1$$ Hence, $$1-\cos^2 A$$$$= \sin^2 A$$)
\begin{align} \text{=}\,\frac{{{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )(1+cos}\,\text{ }\!\!\theta\!\!\text{ )}} \end{align} [Using $$a^2 - b^2$$$$=$$$$(a+b)(a-b)$$]
\begin{align} \text{=}\,\frac{\text{1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ }}{\text{1+cos}\,\text{ }\!\!\theta\!\!\text{ }} \end{align}
=R.H.S.
(ii) \begin{align} \frac{\text{cos}A}{\text{1+sin}A}&+\frac{\text{1+sin}A}{\text{cos}A}\,\,=\,\text{2sec}A \end{align}
LHS\begin{align} =\,\frac{\text{cosA}}{\text{1+sinA}}\text{+}\frac{\text{1+sinA}}{\text{cosA}} \end{align}
\begin{align} \text{=}\,\frac{\text{co}{{\text{s}}^{\text{2}}}\text{A+(1+sinA}{{\text{)}}^{\text{2}}}}{\text{(1+sinA)}\,\text{(cosA)}} \end{align}
\begin{align} \text{=}\,\,\frac{\text{co}{{\text{s}}^{\text{2}}}A\text{+1+si}{{\text{n}}^{\text{2}}}A\text{+2sin}A}{\text{(1+sin}A)\,\text{(cos}A)} \end{align}
\begin{align} \text{=}\frac{\text{si}{{\text{n}}^{\text{2}}}A\text{+co}{{\text{s}}^{\text{2}}}A\text{+1+2sin}A}{\text{(1+sin}A)\text{(cos}A)} \end{align}
\begin{align} \text{=}\frac{\text{1+1+2sin}A}{\text{(1+sin}A)\text{(cos}A)} \end{align} (By identify $$\sin^2 A + \cos ^2 A=1$$)
\begin{align} \text{=}\frac{\text{2+2sin}A}{\text{(1+sin}A)\text{(cos}A)} \end{align}
\begin{align} \text{=}\frac{\text{2(1+sin}A)}{\text{(1+sin}A)\text{(cos}A)}\,\,\,\,\,\,\,\,\,\, \end{align} (By taking $$2$$ common and simplifying)
\begin{align} \text{=}\frac{\text{2}}{\text{cos}A} \end{align}
\begin{align} \text{=2sec}A \end{align}
=R.H.S.
(iii) \begin{align} \begin{bmatrix} \frac{\text{tan} \theta }{1-\text{cot} \theta} +\frac{\text{cot} \theta}{1-\text{tan} \theta}\\ =1+\text{sec} \theta \text{cosec} \theta \end{bmatrix} \end{align}
LHS\begin{align} {=\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}} \dots(1) \end{align}
We know that the trigonometric functions,
\begin{align} & \tan \,(x)=\frac{\sin (x)}{\cos (x)} \\ & \cot \,(x)=\frac{\cos (x)}{\sin (x)}=\frac{1}{\tan (x)} \end{align}
By substituting the above relations in Equation $$(1)$$,
\begin{align} &=\frac{\frac{\sin \theta }{\cos \theta }}{1-\frac{\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{1-\frac{\cos \theta }{\sin \theta }} \\ \\&=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{\sin \theta -\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{\frac{\cos \theta -\sin \theta }{\cos \theta }} \\ \\ & = \begin{bmatrix} \frac{{{\sin }^{2}}\theta }{\cos \theta (\sin \theta -\cos \theta )} \\ + \frac{{{\cos }^{2}}\theta }{\sin \theta (\sin \theta -\cos \theta )} \end{bmatrix}\end{align}
Taking \begin{align} \,\,\frac{\text{1}}{\text{(sin }\!\!\theta\!\!\text{ }-\text{cos }\!\!\theta\!\!\text{ )}} \end{align} as common
\begin{align} \text{=}\,\,\frac{\text{1}}{\text{(sin }\!\!\theta\!\!\text{ }-\text{cos }\!\!\theta\!\!\text{ )}}\left[ \frac{\text{si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{cos }\!\!\theta\!\!\text{ }}-\frac{\text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{sin }\!\!\theta\!\!\text{ }} \right] \\ \text{=}\,\,\frac{\text{1}}{\text{(sin }\!\!\theta\!\!\text{ }-\text{cos }\!\!\theta\!\!\text{ )}}\left[ \frac{\text{si}{{\text{n}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }-\text{co}{{\text{s}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }}{\text{sin}\,\text{ }\!\!\theta\!\!\text{ }\,\text{cos }\!\!\theta\!\!\text{ }} \right] \end{align}
Using
$$a^3 -b^3 =(a-b) \left( a^2 + \, ab \,+ \, b^2 \right)$$
\begin{align} =\frac{1}{\text{(sin}\theta-\text{cos} \theta)} \end{align} \begin{align} \begin{bmatrix} \frac{ \begin{pmatrix} \text{(sin} \theta -\text{cos} \theta) \\ \begin{pmatrix} \text{si}{{\text{n}}^{2}} \theta + \text{co}{{\text{s}}^2} \theta \\+\text{sin} \theta \text{cos } \theta \end{pmatrix} \end{pmatrix} }{\text{sin } \theta \text{cos } \theta } \end{bmatrix} \end{align}
\begin{align}\text{=}\frac{\text{(1}\,\text{+}\,\text{sin}\,\text{ }\!\!\theta\!\!\text{ }\,\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}{\text{(sin}\,\text{ }\!\!\theta\!\!\text{ cos}\,\text{ }\!\!\theta\!\!\text{ )}}\quad\end{align} (By identity $$\sin^2 A$$ $$+$$ $$\cos^2 A$$ $$= 1$$ )
\begin{align} \text{=1+}\,\text{sec}\,\text{ }\!\!\theta\!\!\text{ }\,\text{cosec}\,\text{ }\!\!\theta\!\!\text{ } \\ \end{align}
=R.H.S.
(iv) \begin{align} \frac{1+\sec A}{\sec A}&=\frac{\sin ^{2} A}{1-\cos A} \\ \end{align}
LHS \begin{align}=\frac{1+\sec A}{\sec A} \dots \dots (1) \end{align}
We know that the trigonometric functions,
\begin{align}\sec \,(x)=\frac{1}{\cos \,(x)}\end{align}
By substituting the above function in Equation $$(1)$$,
\begin{align} \frac{{{\rm{1}}\,{\rm{ + }}\,{\rm{sec}}\,{\rm{A}}}}{{{\rm{sec}}\,{\rm{A}}}}{\rm{ = }}\,\frac{{{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{cosA}}}}}}{{\frac{{\rm{1}}}{{{\rm{cosA}}}}}}\\ {\rm{ = }}\,\frac{{\frac{{{\rm{cosA + 1}}}}{{{\rm{cosA}}}}}}{{\frac{{\rm{1}}}{{{\rm{cosA}}}}}}\\ = \frac{{{\rm{cosA + 1}}}}{{{\rm{cosA}}}} \times \frac{{{\rm{cosA}}}}{1}\\ {\rm{ = }}\,{\rm{(1 + cosA)}} \end{align}
By multiplying $$\text{( 1 - cos A)}$$ in both denominator and numerator
\begin{align} &\Rightarrow\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)} \\ &=\frac{1-\cos ^{2} A}{1-\cos A} \\ &=\frac{\sin ^{2} A}{1-\cos A} \\ & \quad \begin{bmatrix} \text { By Identity } \\ \sin ^{2} A+\cos ^{2} A =1\end{bmatrix} \\ & = \text{R.H.S.} \end{align}
(v)\begin{align}\begin{bmatrix}\frac{\text{cos}\,\text{A}-\text{sin}\,\text{A+1}}{\text{cos}\,\text{A+sin}\,\text{A+1}}\\ =\text{cosec}\,\text{A}\,\text{+}\,\text{cotA} \end{bmatrix} \end{align}
$${\rm{L}}.{\rm{H}}.{\rm{S}} = \frac{{\cos {\rm{A}} - \sin {\rm{A}} + 1}}{{\cos {\rm{A}} + \sin {\rm{A}} - 1}}$$
Diving both numerator and denominator by $$\text{sin A}$$
\begin{align}\text{=}\frac{\frac{\text{cos}\,\text{A}}{\text{sin}\,\text{A}}-\frac{\text{sin}\,\text{A}}{\text{sin}\,\text{A}}\text{+}\frac{\text{1}}{\text{sinA}}}{\frac{\text{cos}\,\text{A}}{\text{sin}\,\text{A}}\text{+}\frac{\text{sin}\,\text{A}}{\text{sin}\,\text{A}}\text{+}\frac{\text{1}}{\text{sin}\,\text{A}}}\end{align}
We know that the trigonometric functions,
\,\begin{align} \text{cot}\,(x)= \frac{\text{cos}\,(x)}{\text{sin}\,(x)}\,\,\,&=\frac{1}{\text{tan}\,(x)} \\ \text{cosec}\,(x) &=\frac{\text{1}}{\text{sin}\,(x)} \end{align}
\,\begin{align} \text{We}\,\text{get} \\ \,\,\,&\Rightarrow =\frac{\text{cot}\,\text{A - 1+}\,\,\text{cosec}\,\text{A}}{\text{cot}\,\text{A}\,\text{+ 1- cosec}\,\text{A}} \\ &\Rightarrow =\frac{\text{cot}\,\text{A - (1-}\,\,\text{cosec}\,\text{A)}}{\text{cot}\,\text{A}\,\text{+ (1- cosec}\,\text{A)}}\end{align}
We know that,
\,\begin{align}\text{1 + co}{{\text{t}}^{\text{2}}}\text{A = Cose}{{\text{c}}^{\text{2}}}\text{A}\end{align}
Hence multiplying $$[\cot A – (1 – \rm{cosec}\; A)]$$ in numerator and denominator
\begin{align} & = \frac{ \begin{bmatrix} \{\text{ (cot}\,\text{A)}-(1-\text{cosec}\,\text{A) }\} \\ \{\text{ (cot}\,\text{A)}-\text{(1-cosec}\,\text{A) }\} \end{bmatrix} }{ \begin{bmatrix} \{\text{ (cot}\,\text{A)}\,+ \,(1-\text{cosec}\,\text{A) }\} \\ \{\text{ (cot}\,\text{A)}-(1-\text{cosec}\,\text{A) }\} \end{bmatrix} } \\ & = \frac{{{\text{(cot}\,\text{A}-1+ \text{cosec}\,\text{A)}}^{2}}}{{{\text{(cot}\,\text{A)}}^{2}}-{{(1-\text{cosecA)}}^{2}}} \\ & =\frac{ \begin{bmatrix} \text{co}{{\text{t}}^{2}}\text{A}\,+ 1+ \\ \text{cose}{{\text{c}}^{2}}\text{A}-2 \,\text{cot}\,\text{A} \\ -\text{2cosec}\,\text{A}\,\text{+} \\ \text{2}\,\text{cot}\,\text{A}\,\text{cosecA} \end{bmatrix} }{ \begin{bmatrix} \text{co}{{\text{t}}^{2}}\text{A} \\ -\begin{pmatrix} \text{1+cose}{{\text{c}}^{2}}\text{A} \\ -\text{2cosec}\,\text{A} \end{pmatrix} \end{bmatrix} } \\ & = \frac{ \begin{bmatrix}2\,\text{cose}{{\text{c}}^{2}}\text{A}\,+\,\text{2cot}\,\text{A}\,\,\text{cosecA} \\ - \text{2cot}\,\text{A}-\text{2cosecA} \end{bmatrix} }{ \begin{bmatrix} \text{co}{{\text{t}}^{2}}\text{A}-1+ \,\text{cose}{{\text{c}}^{\text{2}}}\text{A}\, \\ +\,\text{2}\,\text{cosecA} \end{bmatrix} }\,\, \\ & = \frac{ \begin{bmatrix} \text{2cosec}\,\text{A}\,\text{(cosec}\,\text{A}\,+ \,\text{cot}\,\text{A)}\,\\ -\text{2(cot}\,\text{A}-\text{cosec}\,\text{A)} \end{bmatrix} }{ \begin{bmatrix} \text{co}{{\text{t}}^{\text{2}}}\text{A}-\,\text{cose}{{\text{c}}^{2}}\text{A} \\ -1+2 \text{cosec}\,\text{A} \end{bmatrix} } \\ & = \frac{ \begin{bmatrix} \text{(cosec}\,\text{A} + \text{cotA)} \\ (2 \text{cosecA}-2) \end{bmatrix} }{ 1-1+ \,\text{2cosec}\,\text{A} } \\ & = \frac{ \begin{bmatrix} \text{(cosecA}\,+\,\text{cotA)} \\ (2\,\text{cosecA}\,-2) \end{bmatrix} }{(2 \,\text{cosec}\,\text{A}-2)} \\ & =\text{cosec A + cot A} \\ & =\text{R.H.S} \\ \end{align}
(vi) \begin{align} \sqrt{\frac{1+\sin A}{1-\sin A}}&=\sec A+\tan A \\\end{align}
LHS \begin{align} =\sqrt{\frac{1+\sin A}{1-\sin A}} \dots \dots (1) \end{align}
Multiplying and dividing by \begin{align}\sqrt{\left( \text{1+}\,\text{sin}\,\text{A} \right)}\end{align}
$\begin{array}{l} \Rightarrow \sqrt {\frac{{{\rm{(1 + }}\,{\rm{sin}}\,{\rm{A)(1 + }}\,{\rm{sin}}\,{\rm{A)}}}}{{{\rm{(1}} - {\rm{sin}}\,{\rm{A)(1 + }}\,{\rm{sin}}\,{\rm{A)}}}}} \\ = \sqrt {\frac{{{{{\rm{(1 + }}\,{\rm{sin}}\,{\rm{A)}}}^2}}}{{{\rm{(1}} - {\rm{si}}{{\rm{n}}^2}{\rm{A)}}}}} \\ \begin{bmatrix} a^2 - b^2 = \\ \left( {a-b} \right)\left( {a + b} \right), \end{bmatrix} \\ \,\begin{array}{*{20}{l}} \begin{array}{l} {\rm{ = }}\frac{{{\rm{(1 + }}\,{\rm{sinA)}}}}{{\sqrt {{\rm{1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{A}}} }}\,\\ {\rm{ = }}\,\frac{{{\rm{1 + sin}}\,{\rm{A}}}}{{\sqrt {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{A}}} }} \end{array}\\ \begin{array}{l} {\rm{ = }}\frac{{{\rm{1 + }}\,{\rm{sin}}\,{\rm{A}}}}{{{\rm{cos}}\,{\rm{A}}}}\\ = \frac{1}{{{\rm{cos}}\,{\rm{A}}}}\,\, + \frac{{{\rm{sin}}\,{\rm{A}}}}{{{\rm{cos}}\,{\rm{A}}}}\\ {\rm{ = }}\,\,{\rm{sec}}\,{\rm{A}}\,{\rm{ + }}\,{\rm{tan}}\,{\rm{A}} \end{array}\\ {{\rm{ = R}}{\rm{.H}}{\rm{.S}}} \end{array} \end{array}$
(vii) \begin{align} { \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos \theta-\cos \theta}}&={\tan \theta} \end{align}
LHS \begin{align} &={\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}}\end{align}
Taking $$\text{Sin }\!\!\;\theta\!\!\;\text{ and Cos }\!\!\;\theta\!\!\text{ }$$ common in both numerator and denominator respectively.
\begin{align}=\Rightarrow \frac{\sin \text{ }\!\!\theta\!\!\text{ }\left( 1-2{{\sin }^{2}}\text{ }\!\!\theta\!\!\text{ } \right)}{\cos \text{ }\!\!\theta\!\!\text{ }\left( 2{{\cos }^{2}}\text{ }\!\!\theta\!\!\text{ }-1 \right)}\end{align}
By Identity $$\text{si}{{\text{n}}^{\text{2}}}\text{A + co}{{\text{s}}^{\text{2}}}\text{A = 1}$$ hence, $$\text{co}{{\text{s}}^{\text{2}}}\text{A = 1 - si}{{\text{n}}^{\text{2}}}\text{A}$$ and substituting this in the above equation,
$\begin{array}{l} \Rightarrow \frac{{{\rm{sin}}\,{\rm{\theta }}\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)}}{{{\rm{cos}}\,{\rm{\theta }}\left\{ {{\rm{2}}\left( {{\rm{1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right) - {\rm{1}}} \right\}}}\\ = \frac{{{\rm{sin}}\,{\rm{\theta }}\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)}}{{{\rm{cos}}\,{\rm{\theta }}\left( {{\rm{2}} - 2{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }} - {\rm{1}}} \right)}}\\ {\rm{ = }}\,\frac{{{\rm{sin}}\,{\rm{\theta }}\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)}}{{{\rm{cos}}\,{\rm{\theta }}\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)}}\\ {\rm{ = }}\,\frac{{{\rm{sin}}\,{\rm{\theta }}}}{{{\rm{cos}}\,{\rm{\theta }}}}\,\\ {\rm{ = }}\,{\rm{tan}}\,{\rm{\theta }}\\ {\rm{ = RHS}} \end{array}$
(viii) \begin{align} \begin{bmatrix} {{\text{(sin} A\,\text{+}\,\text{cosec}\,A)}^{\text{2}}}\\+(cos\,A\,\text{+}\,\text{sec}\,A{{\text{)}}^{\text{2}}} \\ =\text{7+ta}{{\text{n}}^{\text{2}}}A\,\text{+}\,\text{co}{{\text{t}}^{\text{2}}}A \end{bmatrix} \end{align}
LHS \begin{align} = \begin{bmatrix} (\sin \mathrm{A}+\text{cosec} \mathrm{A})^{2}\\+(\cos \mathrm{A}+\sec \mathrm{A})^{2} \end{bmatrix} \end{align}
By using\begin{align}{{\left( \text{a + b } \right)}^{\text{2}}}&\text{ = }{{\text{a}}^{\text{2}}}\text{ + 2ab +}\,{{\text{b}}^{\text{2}}}\end{align}
\begin{align} \Rightarrow \quad & = \begin{bmatrix} \text{si}{{\text{n}}^{\text{2}}}\text{A}\,+ \,\text{cose}{{\text{c}}^{2}}\text{A}\,\\+ \,2 \,\text{sin}\,\text{A}\,\,\text{cosec}\,\text{A}\, \\ + \,\text{co}{{\text{s}}^{\text{2}}}\text{A}\,+ \,\text{se}{{\text{c}}^{2}}\text{A}\,\\+ \,2 \,\text{cosA}\,\,\,\text{secA} \end{bmatrix} \end{align}
By rearranging and using
\begin{align}\text{sec A } & = \frac{1}{\text{cos A}} \\ \\ & \qquad \rm and \\\\ \rm cosec A &= \frac{{1}}{\sin \rm A}\end{align}
\begin{align} \Rightarrow \begin{bmatrix} \left( \text{sin }^2\text{A}\,+ \,\text{co}{{\text{s}}^{2}}\text{A} \right) \\ + \left( \text{cose}{{\text{c}}^{2}}\text{A}\,+ \,\text{se}{{\text{c}}^{2}}\text{A} \right)\, \\ + \,2 \,\text{sin}\,\text{A}\left( \frac{1}{\text{sin}\,\text{A}} \right)\,\\ + \,\text{2cos}\,\text{A}\left( \frac{1}{\text{cos}\,\text{A}} \right) \end{bmatrix} \end{align}
Hence
\begin{align} & \left( {{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{A}}\,+ \,{\rm{co}}{{\rm{s}}^2}{\rm{A}}} \right)\,= \,1, \\ & {\rm{cose}}{{\rm{c}}^2}{\rm{A}}\, = \left( {{\rm{1 + co}}{{\rm{t}}^2}{\rm{A}}} \right)\\ & \qquad {\rm{ and }} \\ & \left( {{\rm{se}}{{\rm{c}}^2}{\rm{A}}\, - {\rm{ta}}{{\rm{n}}^2}{\rm{A}}} \right)\, = 1 \end{align}
\begin{align} & = \begin{bmatrix} (1)+\begin{pmatrix} 1+{{\cot }^{2}}\text{A}\\ +1+{{\tan }^{2}}\text{A} \end{pmatrix} \\ +(2)+(2) \end{bmatrix} \\ & =7+{{\tan }^{2}}\text{A}+{{\cot }^{2}}\text{A} \\ & = \text{R.H.S.} \end{align}
(ix) \begin{align} \begin{bmatrix} \text{(cosecA}-\text{sinA)}\\\text{(secA}-\text{cosA)} \\ =\frac{\text{1}}{\,\text{tan}\,\text{A+cot}\,\text{A}} \end{bmatrix} \end{align}
LHS \begin{align} = \begin{bmatrix} (\operatorname{cosec}\text{A}-\sin \text{A}) \\ (\sec \text{A}-\cos \text{A}) \end{bmatrix} \ldots (1) \\ \end{align}
We know that the trigonometric functions,
\begin{align} & \sec \,(x)=\frac{1}{\cos \,(x)} \\ & \operatorname{cosec}\,(x)=\frac{1}{\sin \,(x)} \end{align}
By substituting the above relations in Equation $$(1)$$
\begin{align} \Rightarrow \, & \begin{bmatrix} \left( {\frac{1}{{{\rm{sin}}\,{\rm{A}}}} - {\rm{sin}}\,{\rm{A}}} \right) \\ \left( {\frac{1}{{{\rm{cosA}}}} - {\rm{cos}}\,{\rm{A}}} \right) \end{bmatrix} \\ \\ &= \left( {\frac{{1 - {\rm{si}}{{\rm{n}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A}}}}} \right)\left( {\frac{{1 - {\rm{co}}{{\rm{s}}^2}{\rm{A}}}}{{{\rm{cos}}\,{\rm{A}}}}} \right)\\ & = \,\frac{{{\rm{co}}{{\rm{s}}^2}{\rm{A si}}{{\rm{n}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}\\ &= \frac{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}{1}\\ &= \frac{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{A + co}}{{\rm{s}}^{\rm{2}}}{\rm{A}}}} \\ & \begin{bmatrix} {\left( {{\rm{si}}{{\rm{n}}^2 }{\rm{A}}\, + \,{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{A}}} \right)\, = 1} \end{bmatrix} \\ &= \frac{1}{{\frac{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{A + co}}{{\rm{s}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}}}\\ \\ & \begin{bmatrix} \text{Dividing numerator}\\ \text{ and denominator by } \\ \left( {\rm{sin}}\,{\rm{A cos}}\,{\rm{A}} \right) \end{bmatrix}\\ \\ &= \frac{1}{{\frac{{{\rm{si}}{{\rm{n}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}} + \frac{{{\rm{co}}{{\rm{s}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}}}\\ &= \frac{1}{{\frac{{{\rm{sin A}}}}{{{\rm{cos}}\,{\rm{A}}}} + \frac{{{\rm{cos A}}}}{{{\rm{sin}}\,{\rm{A}}}}}}\\ &= \frac{1}{{\tan {\rm{A}} + \cot {\rm{A}}}}\\ &= {\rm{RHS}} \end{align}
(x) \begin{align}\,\,\left( \frac{1+{{\tan }^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}} \right)={{\left( \frac{1-\tan \text{A}}{1-\cot \text{A}} \right)}^{2}}\end{align}
Taking LHS, \begin{align}\left( \frac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)\end{align}
\begin{align} & \text{=}\,\frac{\text{se}{{\text{c}}^{\text{2}}}\text{A}}{\text{cose}{{\text{c}}^{\text{2}}}\text{A}} \\ & \text{=}\,\frac{\text{se}{{\text{c}}^{\text{2}}}\text{A}}{\text{cose}{{\text{c}}^{\text{2}}}\text{A}} \\ & \text{=}\,\frac{\frac{\text{1}}{\text{co}{{\text{s}}^{\text{2}}}\text{A}}}{\frac{\text{1}}{\text{si}{{\text{n}}^{\text{2}}}\text{A}}} \\ & \text{=}\,\frac{\text{1}}{\text{co}{{\text{s}}^{\text{2}}}\text{A}}\,\text{ }\!\!\times\!\!\text{ }\,\text{si}{{\text{n}}^{\text{2}}}\text{A} \\ & \text{=}\,\text{ta}{{\text{n}}^{\text{2}}}\text{A} \\&= \text{RHS} \\ \end{align}
Taking , \begin{align}{{\left( \frac{1-\tan \text{A}}{1-\cot \text{A}} \right)}^{2}}\end{align}
$$\begin{array}{c} {\rm{ = }}\,{\left( {\frac{{{\rm{1}} - {\rm{tanA}}}}{{{\rm{1}} - \frac{{\rm{1}}}{{{\rm{tanA}}}}}}} \right)^{\rm{2}}}\\ {\rm{ = }}\,{\left( {\frac{{{\rm{1}} - {\rm{tanA}}}}{{\frac{{{\rm{tanA}} - 1}}{{\tan {\rm{A}}}}}}} \right)^{\rm{2}}}\\ = {\left( {\left( {{\rm{1}} - {\rm{tanA}}} \right) \times \frac{{{\rm{tanA}}}}{{{\rm{tanA}} - 1}}} \right)^{\rm{2}}}\\ {\rm{ = }}\,{{\rm{(}} - {\rm{tanA)}}^{\rm{2}}}\\ {\rm{ = }}\,{\rm{ta}}{{\rm{n}}^{\rm{2}}}{\rm{A}}\\ {\rm{ = RHS}} \end{array}$$
Hence, L.H.S = R.H.S.
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Number System: Finding Unit and Last two digits of an expression
To find the last digit of an expression which is in the format of ab
Cyclicity of unit digit of an expression:
From the table it is clearly visible that for all the numbers whose unit digit in the format of 2b, the unit digits are respectively 21 = 2, 22 = 4, 23 = 8, 24 = 6, 25 = 2
Similarly we can find unit digits for the remaining numbers easily.
Please observe, The cyclicity of the numbers 2, 3, 7, 8 is 4, and for 4, 9 is 2 as the pattern is repeating after power 4. The cyclicity of 0, 1, 5, 6 is 1.
Problem 1:
What is the unit digit of the expression 317171
Solution:
Here we can concentrate only on the unit digit of the base and the power. Unit digit of the base is 7 so from the table its cyclicity is 4.
Let us find the remainder when 171 is divided by 4. For the divisibility rule for the 4 is to find the remainder of the last two digits of 171, so 71 when divided by 4 gives a remainder 3. So from the table unit digit of 73 is 3.
Problem 2:
Find the unit digit of the expression ${1}^{781}+{2}^{781}$ + ${3}^{781}+{4}^{781}$ + . . . . . . . + ${9}^{781}$
Solution:
we know that 781 when divided by 4 gives a remainder 1. As is visible clearly from the table that for every unit digit after the power 4 the same unit digit repeats.
So unit digit = 1 + 2 + 3 .........9 = 45 so unit digit is 5
Last two digits of an expression:
If we need to find the last two digits of an expression we need to consider the last two digits of the base. We need to consider two cases separately.
Case 1: Numbers which base end with 1.
These numbers are in the format of $...abc{1}^{...xyz}$.
Unit digit of this expression is always 1 as the base ends with 1. For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit
Example: the last two digits of ${2341}^{369}$ = (4 x 9),1 = 61
Case 2: Numbers which end with 5 as unit digit
The last two digits are always 25 or 75. Let the give number is $..ab{5}^{xyz}$. If the product of units digit of the power (i.e., z) and digit left to the 5 in the base (i.e.,b), is even then last two digits of the expression is 25, If the power is odd then it is 75.
Example: last two digits of ${2345}^{369}$ are 25 as the product 4*9 = 36 which is even.
Case 3: Numbers which end with 3, 7, 9.
we need to change the unit digits 3, 7, 9 to 1 by little modification. From the unit digit table we can find that 3, 7, 9 may give unit digit 1 for the powers of 4, 4, 2 respectively.
In finding the last two digits of an expression ${2343}^{4747}$ we can re-write the expression ${43}^{4747}$ as we are concerned with only last two digits only.
Now we consider ${\left({43}^{4}\right)}^{1186}{.43}^{3}$ $⇒{\left({43}^{2}×{43}^{2}\right)}^{1186}{.43}^{3}$
${43}^{2}=1849$ so
We have to consider the last two digits of 1849 $⇒{\left(49×49\right)}^{1186}{.43}^{3}$.
Now49 x 49 will give unit digits as 01, $⇒{\left(01\right)}^{1186}{.43}^{3}$
$⇒01×07⇒07$.
Case 4: Numbers which end with 2, 4, 6.
Firstly we should by-heart these two rules: ${2}^{10}$ raised to the even power always give last two digits as 76, and when raised to the odd power give the last two digits as 24.
Then we should take 2's separately from the given number and we need to apply the above rule.
Example: Find the last two digits of ${48}^{199}$
By re-writing ${48}^{199}$ = ${\left({2}^{4}×3\right)}^{199}$ = ${2}^{796}×{3}^{199}$ = ${2}^{10×79+6}×{3}^{199}$ = ${\left({2}^{10}\right)}^{79}{.2}^{6}{.3}^{199}$
Now ${2}^{10}$ raised to an odd power gives 24 as the last two digits. So
$\left(24\right){.2}^{6}{.3}^{199}⇒24×64×\left({3}^{4×49+3}\right)⇒24×64×{81}^{49}×{3}^{3}$ $⇒24×64×21×27$ = 12.
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# $\cos{(A+B)}$ formula Proof
## Formula
$\cos{(A+B)}$ $\,=\,$ $\cos{A}\cos{B}$ $-$ $\sin{A}\sin{B}$
### Construction of sum of angles triangle
The expansion of $\cos{(A+B)}$ formula is originally derived in geometric method by constructing a right triangle whose angle is sum of two angles.
1. $\small \Delta FDE$ is a right triangle. Draw a straight line from point $\small D$ and it divides the angle as two angles $\small A$ and $\small B$ and intersects the side $\small \overline{FE}$ at point $\small G$.
2. Later, draw a line from point $\small G$ to side $\small \overline{DF}$ but it should be perpendicular to the side $\small \overline{DG}$ and it interests the side $\small \overline{DF}$ at point $\small H$.
3. Draw a perpendicular line to side $\small \overline{DE}$ from point $\small H$. It intersects the side $\small \overline{DG}$ at point $\small I$ firstly and then side $\small \overline{DE}$ at point $\small J$.
4. Lastly, draw a perpendicular line from point $\small G$ to side $\small \overline{HJ}$ and also draw a perpendicular line from point $\small I$ to side $\small \overline{FE}$.
### Express Cos of sum of angles in its ratio form
The angle of $\Delta HDJ$ is a sum of the angles $A$ and $B$. So, write cos of angle $A+B$ in its ratio form for deriving expansion of cosine of sum of two angles.
1. $\small \overline{DJ}$ is Adjacent side (or) Base
2. $\small \overline{DH}$ is hypotenuse
Express, cosine of sum of the angles in ratio form of the sides by considering right triangle $HDJ$.
$\cos{(A+B)} \,=\, \dfrac{DJ}{DH}$
### Express length of the side in alternative form
The side $\overline{HJ}$ perpendicularly divides the side $\overline{DE}$ at point $J$. So, the sum of the lengths of the sides $\overline{DJ}$ and $\overline{JE}$ is equal to length of the side $\overline{DE}$.
$DE = DJ+JE$
$\implies DJ = DE-JE$
The length of the side $\overline{DJ}$ in expansion of $\cos{(A+B)}$ can be replaced by its equal value.
$\implies$ $\cos{(A+B)}$ $\,=\,$ $\dfrac{DE-JE}{DH}$
The perpendicular intersections of sides $\overline{DE}$ and $\overline{KG}$ with the sides $\overline{HJ}$ and $\overline{EF}$ form a rectangle. It is the main reason for the equality of the sides $\overline{JE}$ and $\overline{KG}$.
$JE = KG$
Replace the length of the side $\overline{JE}$ in the expansion of $\cos{(A+B)}$ identity.
$\implies$ $\cos{(A+B)}$ $\,=\,$ $\dfrac{DE-KG}{DH}$
$\implies$ $\cos{(A+B)}$ $\,=\,$ $\dfrac{DE}{DH}-\dfrac{KG}{DH}$
### Write each side in a trigonometric function form
The length of the side $\overline{DE}$ is part of the right triangle $EDG$ and its angle is $A$. Now, write length of the side $\overline{DE}$ in another form by considering $\Delta EDG$.
In $\Delta EDG$, $A$ is angle and $\overline{DE}$ is adjacent side. So, The relation between them is represented by cosine. So, express length of side ${DE}$ in terms of cos of angle $A$.
$\cos{A} = \dfrac{DE}{DG}$
$\implies DE = DG\cos{A}$
Substitute the length of the side $\overline{DE}$ in the expansion of $\cos{(A+B)}$ rule by its new equal value.
$\implies$ $\cos{(A+B)}$ $\,=\,$ $\dfrac{DG\cos{A}}{DH}$ $-$ $\dfrac{KG}{DH}$
### Find Angle of the right triangle
The expansion of cos of sum of angles is derived in the previous step as follows.
$\cos{(A+B)} \,=\, \dfrac{DG\cos{A}}{DH}-\dfrac{KG}{DH}$
$KG$ is length of the opposite side ($\overline{KG}$) of right triangle $GHK$ and the angle of this triangle is unknown. So, it is not possible to express the length of side $\overline{KG}$ in terms of a trigonometric function but it is possible by finding the angle of this triangle.
The $\angle GHK$ is about to find in three steps geometrically.
#### Find the angle of triangle by similarity
$\Delta EDG$ is a right triangle and its angle is $A$. $\Delta LIG$ is another right triangle and its angle is unknown.
Geometrically, the triangles $\Delta EDG$ and $\Delta LIG$ are similar.
$\Delta EDG \, \sim \, \Delta LIG$
Therefore, the corresponding angles $\angle LIG$ and $\angle EDG$ should be equal.
$\implies \angle LIG = \angle EDG$
$\,\,\, \therefore \,\,\,\,\,\, \angle LIG = A$
Therefore, the $\angle LIG$ is equal to $A$.
#### Equality of Interior alternate angles
$\overline{DG}$ is a transversal of parallel lines $\overline{IL}$ and $\overline{KG}$. The $\angle LIG$ and $\angle KGI$ are known as interior alternate angles.
It is proved that the alternate interior angles formed by the parallel lines and their transversal are equal geometrically.
$\implies \angle KGI \,=\, \angle LIG$
$\,\,\, \therefore \,\,\,\,\,\, \angle KGI = A$
Therefore, the $\angle KGI$ is equal to $A$.
#### Find Complementary angle of triangle
The side $\overline{HG}$ is a perpendicular line to side $\overline{DG}$. So, $\angle DGH = 90^°$.
The intersection of side $\overline{KG}$ divides the $\angle DGH$ as two angles and they are $\angle DGK$ and $\angle KGH$. Therefore, the sum of them is equal to $\angle DGH$.
$\angle DGK + \angle KGH = \angle DGH$
$\implies \angle DGK + \angle KGH = 90^°$
Geometrically, $\angle DGK = \angle KGI$ but $\angle KGI = A$ as per the above step.
$\angle KGI + \angle KGH = 90^°$
$\implies A + \angle KGH = 90^°$
$\,\,\, \therefore \,\,\,\,\,\, \angle KGH = 90^°-A$
#### Find unknown angle of triangle
$\Delta KHG$ is a right triangle and its angle is unknown but the other two angles are known. So, the angle this triangle can be evaluated by using sum of interior angles in a triangle rule.
$\angle KHG + \angle HGK + \angle GKH \,=\, 180^°$
$\implies \angle KHG + \angle KGH + \angle GKH \,=\, 180^°$
$\angle GKH$ is a right angle and it is derived in the previous step that $\angle KGH = 90^°-A$. Substitute them and get the value of $\angle KHG$ geometrically.
$\implies \angle KHG + 90^°-A + 90^° \,=\, 180^°$
$\implies \angle KHG + 180^°-A \,=\, 180^°$
$\implies \angle KHG \,=\, 180^°-180^°+A$
$\implies \angle KHG \,=\, \require{cancel} \cancel{180^°}-\cancel{180^°}+A$
$\,\,\, \therefore \,\,\,\,\,\, \angle KHG \,=\, A$
Therefore, the angle of $\Delta KHG$ is $A$.
### Continue transforming lengths of sides as trigonometric functions
Return to deriving the expansion of $\cos{(A+B)}$ formula.
$\cos{(A+B)}$ $\,=\,$ $\dfrac{DG\cos{A}}{DH}-\dfrac{KG}{DH}$
The angle of $\Delta KHG$ is determined in the above step. So, the length of the side can be expressed in the form a trigonometric function.
$KG$ is opposite site of the $\Delta KHG$ and the angle of this triangle is $A$. So, the relationship between them can be represented by sine function.
$\sin{A} \,=\, \dfrac{KG}{HG}$
$\implies KG = HG \times \sin{A}$
Now, replace the length of the side KG in the expansion of cos of sum of two angles.
$\implies \cos{(A+B)}$ $\,=\,$ $\dfrac{DG\cos{A}}{DH}-\dfrac{HG\sin{A}}{DH}$
$\implies \cos{(A+B)}$ $\,=\,$ $\cos{A} \times \dfrac{DG}{DH}$ $-$ $\sin{A} \times \dfrac{HG}{DH}$
### Express each ratio in trigonometric function form
$DG$, $HG$ and $DH$ are lengths of three sides of the right angled triangle ($\Delta HDG$) whose angle is $B$. Now, express value of each ratio of the sides in the form of corresponding trigonometric function.
$(1) \,\,\,\,\,\,$ $\sin{B} \,=\, \dfrac{HG}{DH}$
$(2) \,\,\,\,\,\,$ $\cos{B} \,=\, \dfrac{DG}{DH}$
Now, substitute the ratios of lengths of the sides by their associated trigonometric functions in the expansion of $\cos{(A+B)}$ identity.
$\implies \cos{(A+B)}$ $\,=\,$ $\cos{A} \times \cos{B}$ $-$ $\sin{A} \times \sin{B}$
$\,\,\, \therefore \,\,\,\,\,\, \cos{(A+B)}$ $\,=\,$ $\cos{A}\cos{B}$ $-$ $\sin{A}\sin{B}$
Therefore, it is proved that cos of sum of two angles is expanded as the subtraction of product of sin of angles from product of cos of angles.
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#### Project Euler 16 Problem Statement
The sum of the squares of the first ten natural numbers is 1² + 2² + … + 10² = 385.
The square of the sum of the first ten natural numbers is (1 + 2 + … + 10)² = 55² = 3025.
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
#### Solution
Solving this problem is helped by knowing two established equations:
1. The sum of the first n numbers (triangular numbers, used in Project Euler Problem 1):
$\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$
For a visual example of triangular numbers, imagine setting bowling pins:
Let’s apply this equation to our example:
The square of the sum of the first ten natural numbers is 10(11)/2 = 55, and 552 = 3025✓.
2. The sum of the first n square numbers (square pyramidal numbers):
$\sum\limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$
For a visual example of square pyramidal numbers, imagine stacking jaw breakers:
Now, let’s apply this equation to our example:
The sum of the squares of the first ten natural numbers is 10(11)(21)/6 = 385✓.
Finally, calculate the difference, 3025 – 385 = 2640✓. The answer to the example above, all done without loops. These equations will be useful for solving other problems.
Now, I guess for the purists one could perform the subtraction of these two summations and derive the following equation:
$\frac{n(n-1)(n+1)(3n+2)}{12}$
And giving it a try with our example: 10(9)(11)(32)/12 = 2640✓.
#### HackerRank version
Avoiding loops to solve these types of problems in favor of equations and closed-form calculations becomes apparent with the more demanding hackerRank version. It requires you to solve up to 10,000 test cases with a higher limit of n ≤ 10,000 in a fixed amount of time, typically less than a tenth of a second.
#### Last Word
Reference: The On-Line Encyclopedia of Integer Sequences (OEIS) A000217: Triangular numbers: a(n) = binomial(n+1,2) = n(n+1)/2 = 0 + 1 + 2 + ... + n.
Reference: The On-Line Encyclopedia of Integer Sequences (OEIS) A000330: Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.
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# 33 Mathematical Ideas: Basic Concepts of Set Theory
Jenna Lehmann
##### Symbols and Terminology
A set is a collection of objects of values that are in this case called elements or members. They can be described using words, lists, or set-builder notation.
• Words: a set of odd numbers less than 6
• Listing: {1,3,5}
• Set-builder notation: {x|x is an odd counting number less than 6}
If a set has no elements, it’s called an empty or null set and its symbol is Ø. Make sure not to write this as {Ø}, because that is technically incorrect.
It is important to make sure that a set is well-defined, meaning that there’s no room for subjective interpretation about whether something belongs in a set or not. An example of a well-defined set is a set of all numbers between 1 and 10. We can say for sure that 5 belongs and 13 doesn’t. A set that is not well defined is a set of all numbers that are aesthetically pleasing. It’s not clear what would define aesthetically pleasing so we’re unsure about whether 5 or 13 would fit.
The symbols and are used to describe whether something is or isn’t an element of a set. Going back to our example of all numbers between 1-10 (we’ll name this set A) we can say that 5 ∈ A while 13 ∉ A.
##### Important Definitions for Sets
Here are some important definitions before moving forward:
• Natural numbers or Counting numbers are all integers starting at 1: {1, 2, 3, 4,…}
• Whole numbers are all integers starting at 0: {0, 1, 2, 3, 4…}
• Integers are all whole numbers from -∞ to ∞: {… -3, -2, -1, 0, 1, 2, 3…}
• Rational numbers are numbers that can be created by dividing two integers (like 1/2 or 9/10 or 4/1).
• Real Numbers are any number that isn’t imaginary, so the typical integer, fraction, or decimal we’re used to. An imaginary number is when a negative number is square rooted (√-1)
• Irrational Numbers are decimals that can’t be expressed as the fraction of two integers. The square root of 2 would be an example of this because the decimals are ongoing and there is no discernible pattern to the decimals.
• A cardinal number is the number of elements in a set and it is written as n(A) and spoken as “n of A.” So if I was given set Z = {1, 5, 7, 2, 9, 10}, I would say n(Z) is 6.
• A finite set is a set that has a whole number as its cardinal number. So we could technically count how many elements are in a set.
• An infinite set is a set where the number of elements is infinite and we couldn’t possibly count them.
• Two sets are equal if two conditions are met: (1) every element of the first set is an element of the second set and (2) if every element of the second set is an element of the first one. That being said, it does not matter if the elements are written in a different order ({a, b, c, d} = {a, c, d, b}) and repeating elements doesn’t add a new element ({a, b, a, c, d, d} = {a, b, c, d}).
##### Venn Diagrams and Subsets
The universe in which we are working, or the area that we’re concerned within a set, is called the universal set. This consists of everything in the wider set. Venn diagrams are often used to discuss commonalities and differences between sets in the universal set. The universal set is everything within the rectangle encompassing the Venn diagram including the Venn diagram itself. The Venn diagram is made up of sets within U and can overlap. Set A is everything in circle A, set B is everything in circle B, and where they overlap are all the elements that they have in common.
Complements of a set are everything that a set is not. The complement of A is A’ (spoken as A prime) and it includes everything in U except what is included in A.
##### Subsets of a Set
A subset is any set that is also part of another set. For example, if U = {1, 2, 3, 4, 5} and A = {1, 4, 5}, then we would say that A is a subset of U. This is denoted like this: A ⊆ U. If B = {1, 2, 4, 7}, because 7 is not part of U, we would say B is not a subset of U, also denoted like this: B ⊄ U.
There are different kinds of subsets. Any subset can be called a subset, but some can be described as a proper subset. A proper subset is a subset that has elements of a set but not exactly all of the elements in that set. For example, if set Y = {1, 2, 3} and Z = {1, 2, 3, 4}, then we could say that Y is a subset of Z and we could also say that Y is a proper subset of Z because it does not include all of the elements of Z. This is written as such: Y ⊂ Z.
Sometimes you will be asked to calculate how many subsets exist within a set. This can be calculated using powers of 2. For example, if I have the set {1, 4, 6, 2, 7}, I can see that there are 5 elements in the set. I make that 5 an exponent of 2 (2^5) to calculate how many subsets are possible: 32. To calculate the number of proper subsets, the equation is (2^n) – 1.
##### Set Operations
An intersection of sets is the elements of two sets that they have in common. For example, {1, 4, 5} ∩ {6, 9, 5} = {5}. Put in other words, A ∩ B = {x|x ∈ A and x ∈ B}. If two sets have no elements in common, they are called disjointed sets and written as such: A ∩ B = ø. A union of sets is the set of all elements belonging to either set one or set two, written as such: A ∪ B = {x|x ∈ A or x ∈ B}. A difference of sets is the set of all elements of the first set and not the second. For example, if set A is {1, 2, 3, 4, 5, 6} and B is {1, 3, 5}, then the difference would be {2, 4,6}. In other words, A – B = {x|x ∈ A and x ∉ B}.
When elements are placed in {braces}, it doesn’t matter in what order they are listed. When elements are placed in (parentheses), it’s called an ordered pair and it does matter what order they are listed in. In other words (a,b) ≠ (b,a). In the ordered pair (a,b), a is the first component and b is the second component.
A cartesian product of sets is a way of creating a set of ordered pairs. It’s written like A X B and when presented with a problem asking you to find cartesian products, you have to create ordered pairs with each number in each set in the order that the notation dictates. For example, if A = {1, 2, 3} and B = {8, 9}, and you were asked to find A X B, then the answer would be {(1,8), (1,9), (2,8), (2,9), (3,8), (3,9)}. If you were asked to find B X A, the answer would be {(8,1), (8,2), (8,3), (9,1), (9,2), (9,3)}. The cardinal number of a cartesian product is going to be the cardinal number of set 1 times the cardinal number of set 2, or n(A) x n(B).
This chapter was originally posted to the Math Support Center blog at the University of Baltimore on November 11, 2019.
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## Working Systematically
Working Systematically is part of our Thinking Mathematically collection.
Mathematicians often talk about the importance of working systematically. This means that rather than working in a haphazard and random way, there is a methodical, organised and logical approach. The problems below will challenge you to work systematically, and will help you appreciate the benefits of working in this way.
Scroll down to see our complete collection of problems, or explore the two sub-collections.
## Noticing Patterns
The key to solving these problems is to notice patterns or properties. Organising your work systematically allows you to notice what might not otherwise be obvious.
## Finding All Solutions
These problems challenge you to find all possible solutions. One of the best answers to "How do you know you have found them all" is to be able to say "I worked systematically!"
## Two and Two
How many solutions can you find to this sum? Each of the different letters stands for a different number.
## Summing Consecutive Numbers
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?
## Isosceles Triangles
Draw some isosceles triangles with an area of \$9\$cm\$^2\$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
## American Billions
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
## 1 Step 2 Step
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
## Can They Be Equal?
Can you find rectangles where the value of the area is the same as the value of the perimeter?
## Pick's Theorem
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
## Sticky Numbers
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
How many different symmetrical shapes can you make by shading triangles or squares?
## Shifting Times Tables
Can you find a way to identify times tables after they have been shifted up or down?
## Peaches Today, Peaches Tomorrow...
A monkey with peaches, keeps a fraction of them each day, gives the rest away, and then eats one. How long can his peaches last?
## Charlie's Delightful Machine
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
## Nine Colours
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
## What's Possible?
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
## Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
## Triangles to Tetrahedra
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
## Making a Difference
How many different differences can you make?
## Where Can We Visit?
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
## Cinema Problem
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
By selecting digits for an addition grid, what targets can you make?
## Squares in Rectangles
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
## Gabriel's Problem
Gabriel multiplied together some numbers and then erased them. Can you figure out where each number was?
## Multiples Sudoku
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
## Number Daisy
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
## Special Numbers
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
## Consecutive Negative Numbers
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
Can you find a cuboid that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
Different combinations of the weights available allow you to make different totals. Which totals can you make?
## Sociable Cards
Move your counters through this snake of cards and see how far you can go. Are you surprised by where you end up?
## Product Sudoku
The clues for this Sudoku are the product of the numbers in adjacent squares.
Ben, Jack and Emma passed counters to each other and ended with the same number of counters. How many did they start with?
The items in the shopping basket add and multiply to give the same amount. What could their prices be?
Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
## Warmsnug Double Glazing
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
## Difference Sudoku
Use the differences to find the solution to this Sudoku.
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## Learning Path
Exploring problem solving with 5th and 6th class learning path.
This learning path gives a brief overview of how to develop a classroom culture of sharing knowledge, valuing mistakes and providing cognitively challenging tasks. Problem solving strategies can be explicitly taught using this learning path with an introduction to the concept of Low Threshold High Ceiling Tasks also. 11 resources in this Learning Path
## Developing Maths problem solving
a pre assessment to see how children are able to respond and complete problem
## How it maps to the curriculum
Strand: Number
Suggestions for use: Assessment of student learning
## Same But Different
The images on this website can be used for developing classroom discussion, critical thinking and collaboration. Key components for problem solving.
Strand: Useful Websites
Suggestions for use: This can website can be used by teachers as a mini problem-solving lessons (10 mins approx.) to develop a classroom climate conducive to problem-solving. Teachers should encourage all pupils to ‘have a go’ and ‘value all contributions’. Promoting higher order skills of reasoning and discussion. Success is based on effort and skills rather than answers.
## Low Threshold High Ceiling
A low threshold high ceiling task is one which is designed to be mathematically accessible, and to have built-in extension opportunities. In other words, everyone can get started and everyone can get stuck. In this updated feature, NRich brings together their favourite low threshold high ceiling tasks, as well as two articles which will support you in creating a low threshold high ceiling classroom.
Suggestions for use: Developing a positive classroom climate in maths through mathematical discussion and collaboration
## Problem Solving with 5th and 6th Class - Sandwiches (Trial and Improvement)
The problem is particularly valuable as it gives students an opportunity to work on a proof to explain why something is impossible.
Strand: Algebra
Suggestions for use: Using number to predict, generalise and verify
## Problem Solving with 5th and 6th Class - Tea Cups (Working Systematically)
The problem lends itself for small group work, so that the learners have an opportunity to decide on approaches. Learners can collaborate, share and discuss the different solutions, and each method's strengths and weaknesses.
Suggestions for use: Allowing learners to read the accompanying story with the problem can support student’s comprehension and decision making about which pieces of information are relevant. Development of higher order maths skills: Applying and problem-solving, Communicating and expressing, Integrating and connecting, Reasoning
## Problem Solving with 5th and 6th Class - Tables without tens (Pattern spotting)
This problem provides an interesting way of revising multiplication tables. It is also very useful for getting learners to predict what they think they will find out and spot pattern between times tables.
Strand unit: Number Theory
Content objective: This resource should enable a child to:
• identify common factors and multiples
• identify factors and multiples
• identify simple prime and composite numbers
Suggestions for use: times tables
## Problem Solving with 5th and 6th Class - Counting Cards (Working Backwards)
While this problem provides the “trick” element to it, it is firmly rooted in mathematical concepts and problem solving strategies.
## Problem Solving with 5th and 6th Class - Baravelle (Visualising)
The aim of this problem is to encourage discussion about the different ways of seeing, and to pose questions that can form the focus of further investigation.
Strand: Shape & Space
Strand unit: 2-D Shapes
• make informal deductions about 2-D shapes and their properties
• plot simple co-ordinates and apply where appropriate
• tessellate combinations of 2-D shapes
• use 2-D shapes and properties to solve problems
Suggestions for use: Allows for creative pattern design. A good follow up problem can be found here https://nrich.maths.org/2132/index (Inside 7 squares)
## Problem Solving with 5th and 6th Class - Planning a school trip (Reasoning)
This problem will encourage learners to organise information, identify redundant information and to check their work.
Suggestions for use: The activity lends itself to collaborative working, both for children who are inexperienced at working in a group and children who are used to working in this way. By working together on this problem, the task is shared and therefore becomes more manageable than if working alone. A number of cross curricular links can be used to extend this lesson.
## Problem Solving with 5th and 6th Class - Magic V's (Conjecture)
Task card involving placing numbers 1 to 5 in the V shape so that the two arms of the V have the same total. Short task.
Strand unit: Operations: Addition & Subtraction
Suggestions for use: Print and display on maths station or use separately in a teacher led problem solving activity. Magic V gives opportunities for children to make conjectures, prove these conjectures and make generalisations. They will be practising addition and subtraction, and applying their knowledge of odd/even numbers. Supporting video clip on the problem can be found at this link https://www.youtube.com/watch?v=-JrZcMbsNdA
Strand unit: Operations
• add and subtract whole numbers and decimals (to three decimal places) without and with a calculator
## Problem Solving with 5th and 6th Class - Money Bags (Conjecture)
This problem is a good example of a challenge which does not require high-level mathematics, but does need a systematic approach. It also lends itself to a focus on different ways of recording and learners can discuss on the merits of the different ways of recording findings.
Strand: Measures
Strand unit: Money - Euro
• compare 'value for money' using unitary method
• explore value for money
## Registering for a Scoilnet Account – your first step to contributing and sharing
What you need....
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## Roughwood Primary School
Where a love of learning grows.
## Working systematically with NRICH maths challenges in Class 6
As part of our reasoning and applying our knowledge within mastery maths sessions, the children in Year 5 have completed another investigative NRICH reasoning investigation using their systematic thinking and recording.
The investigation generated lots of discussion and at the end, children wanted to investigate further links between divisors, like 3, 6 and 9, to see if they could see a similar pattern there too.
Using our school learning characters, we’ve needed to be like professor thought shower and captain team work to apply our prior knowledge and understanding.
Look at how the children recorded their answers systematically.
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• Problem Solving and Mastery Ideas
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• MASTERY An excellent resource with ideas to support teaching 'mastery' you will need to set up Oxford Owls, which is free to do.
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• Topmarks KS1 Links to lesson support materials and activities
• Investigating number bonds
• Developing tasks to promote problem solving skills with your class NRICH With good links to tasks to encourage working systematically and other skills
• Conjecturing and Generalising in KS1
• Button.docx
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## Year Four's Blog
Beech, cedar and hawthorne show off, problem-solving, a puzzling problem.
## Cedar solve the dreaded Coded Hundred Square!
Today we have been solving the riddle of the Coded Hundred Square. Well done to all the children who persevered and tried to find a solution. Here are two examples from William and Alex and Hala and Nabil.
## Live Problem: Which Scripts?
Which Scripts?
If you feel like a challenge, why not try solving this problem? Remember, if you’re finding it hard, keep searching for a solution; draw the problem; discuss it with someone, but don’t give up straight away! On the left-hand side of the page there are some tips for getting started on the problem.
When you have found a solution, you can also submit it to the Nrich website. You never know; the people at Nrich may publish your solution on the website where it will remain forevermore. Good luck!
## Tea Cup Challenge!
‘Mathematics is not about numbers, but about life. It is about the world in which we live. It is about ideas. And far from being dull and sterile, as it is so often portrayed, it is full of creativity.’
Some light reading for any interested parents out there: The Elephant in the Classroom: Helping Children Learn & Love Maths by Jo Boaler
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## Problem solving - Working Systematically Bundle - Reasoning and fluency (Yr 3 &4)
Subject: Mathematics
Age range: 7-11
Resource type: Worksheet/Activity
Last updated
28 January 2018
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## Not quite what you were looking for? Search by keyword to find the right resource:
#### IMAGES
1. Systematic Problem Solving PowerPoint Template
2. PPT
3. A Systematic Approach to Problem Solving
4. systematic approach to problem solving
5. Systematic Problem Solving PowerPoint Template
6. Systematic Problem Solving PowerPoint Template
#### VIDEO
1. How to Solve a Problem in Four Steps: The IDEA Model
2. 36. AQA GCSE (8525) SLR7
3. Why is the Systematic Approach Important?
4. Problem Solving Approach
5. Problem Solving Strategies & Polya's 4-step Process
6. 5 ทักษะการแก้ปัญหาสำหรับคนทำงาน
1. Troubleshooting Kohler Engines
A systematic approach to solving the problem is often the best way to track down and then correct engine problem, according to Kohler Power. The first step of troubleshooting to figure out what’s wrong with a Kohler engine is identifying th...
2. What Are the Six Steps of Problem Solving?
The six steps of problem solving involve problem definition, problem analysis, developing possible solutions, selecting a solution, implementing the solution and evaluating the outcome. Problem solving models are used to address issues that...
3. How to Solve Common Maytag Washer Problems
Maytag washers are reliable and durable machines, but like any appliance, they can experience problems from time to time. Fortunately, many of the most common issues can be solved quickly and easily. Here’s a look at how to troubleshoot som...
4. Working Systematically at KS2
Scroll down to see our complete collection of KS2 problems that require children to work systematically, or explore the two sub-collections focusing on
5. Working Systematically
This means that rather than working in a haphazard and random way, there is a methodical, organised and logical approach. The problems below will challenge you
6. Working systematically
Work with objects, numbers and pictures in a systematic way to solve finding all possibilities problems. Use simple lists and practical resources to organize
7. Problem-solving activities: ideas for the classroom
Many of the activities featured in this document have been sourced from the. University of Cambridge's NRICH. Roadshow. You can access them free of charge at
8. Exploring Problem Solving with 5th and 6th Class Learning Path
Problem Solving with 5th and 6th Class - Tea Cups (Working Systematically) · https://nrich.maths.org/32 Added: 27 Mar 2020 Contributor: PDST Resource type
9. Year 3
solve the problem; present and interpret the solution in the context of the problem.
10. Progression in Reasoning
• Use a systematic way to solve a. Example learning outcomes: Example learning
11. Working systematically with NRICH maths challenges in Class 6
... NRICH reasoning investigation using their systematic thinking ... The problems we had to solve were around Ben having five coins in his pocket.
12. Problem Solving and Mastery Ideas
... NRICH With good links to tasks to encourage working systematically and other skills. Conjecturing and Generalising in KS1. Useful Problem Solving Tasks. Button
13. problem-solving
Beech class tried to piece together a multiplication grid puzzle this morning. Reasoning skills and systematic thinking was the order of day
14. Working Systematically Bundle
Inspired by Nrich & White Rose hub materials this is a bundle of 11 different problem solving activities, to practice working systematically
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Period and Frequency
## Horizontal distance traveled before y values repeat; number of complete waves in 2pi.
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Changes in the Period of a Sine and Cosine Function
Your mission, should you choose to accept it, as Agent Trigonometry is to find the period of the cosine function y=cos[π(2x+4)]\begin{align*}y=\cos [\pi (2x + 4)]\end{align*}.
### Period
The last thing that we can manipulate on the sine and cosine curve is the period.
The normal period of a sine or cosine curve is 2π\begin{align*}2 \pi\end{align*}. To stretch out the curve, then the period would have to be longer than 2π\begin{align*}2 \pi\end{align*}. Below we have sine curves with a period of 4π\begin{align*}4 \pi\end{align*} and then the second has a period of π\begin{align*}\pi\end{align*}.
To determine the period from an equation, we introduce b\begin{align*}b\end{align*} into the general equation. So, the equations are y=asinb(xh)+k\begin{align*}y=a\sin b(x-h)+k\end{align*} and y=acosb(xh)+k\begin{align*}y=a\cos b(x-h)+k\end{align*}, where a\begin{align*}a\end{align*} is the amplitude, b\begin{align*}b\end{align*} is the frequency, h\begin{align*}h\end{align*} is the phase shift, and k\begin{align*}k\end{align*} is the vertical shift. The frequency is the number of times the sine or cosine curve repeats within 2π\begin{align*}2 \pi\end{align*}. Therefore, the frequency and the period are indirectly related. For the first sine curve, there is half of a sine curve in 2π\begin{align*}2 \pi\end{align*}. Therefore the equation would be y=sin12x\begin{align*}y=\sin \frac{1}{2}x\end{align*}. The second sine curve has two curves within 2π\begin{align*}2 \pi\end{align*}, making the equation y=sin2x\begin{align*}y=\sin 2x\end{align*}. To find the period of any sine or cosine function, use 2π|b|\begin{align*}\frac{2 \pi}{|b|}\end{align*}, where b\begin{align*}b\end{align*} is the frequency. Using the first graph above, this is a valid formula: 2π12=2π2=4π\begin{align*}\frac{2 \pi}{\frac{1}{2}}=2 \pi \cdot 2=4 \pi\end{align*}.
Let's determine the period of the following sine and cosine functions.
1. y=3cos6x\begin{align*}y=-3 \cos 6x\end{align*}
The 6 in the equation tells us that there are 6 repetitions within 2π\begin{align*}2 \pi\end{align*}. So, the period is 2π6=π3\begin{align*}\frac{2 \pi}{6}=\frac{\pi}{3}\end{align*}.
1. y=2sin14x\begin{align*}y=2 \sin \frac{1}{4}x\end{align*}
The 14\begin{align*}\frac{1}{4}\end{align*} in the equation tells us the frequency. The period is 2π14=2π4=8π\begin{align*}\frac{2 \pi}{\frac{1}{4}}=2 \pi \cdot 4=8 \pi\end{align*}.
1. y=sinπx7\begin{align*}y=\sin \pi x -7\end{align*}
The π\begin{align*}\pi\end{align*} is the frequency. The period is 2ππ=2\begin{align*}\frac{2 \pi}{\pi}=2\end{align*}.
Now, let's graph y=3cos6x\begin{align*}y=-3 \cos 6x\end{align*} from [0,2π]\begin{align*}[0, 2 \pi]\end{align*}, determine where the maximum and minimum values occur, and state the domain and range.
The amplitude is -3, so it will be stretched and flipped. The period is π3\begin{align*}\frac{\pi}{3}\end{align*} (from above) and the curve should repeat itself 6 times from 0 to 2π\begin{align*}2 \pi\end{align*}. The first maximum value is 3 and occurs at half the period, or x=π6\begin{align*}x=\frac{\pi}{6}\end{align*} and then repeats at x=π2,5π6,7π6,3π2,\begin{align*}x=\frac{\pi}{2}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \ldots\end{align*} Writing this as a formula we start at π6\begin{align*}\frac{\pi}{6}\end{align*} and add π3\begin{align*}\frac{\pi}{3}\end{align*} to get the next maximum, so each point would be (π6±π3n,3)\begin{align*}\left(\frac{\pi}{6} \pm \frac{\pi}{3}n,3\right)\end{align*} where n\begin{align*}n\end{align*} is any integer.
The minimums occur at -3 and the x\begin{align*}x\end{align*}-values are multiples of π3\begin{align*}\frac{\pi}{3}\end{align*}. The points would be (±π3n,3)\begin{align*}\left(\pm \frac{\pi}{3}n, -3\right)\end{align*}, again n\begin{align*}n\end{align*} is any integer. The domain is all real numbers and the range is y[3,3]\begin{align*}y \in [-3,3]\end{align*}.
Finally, let's find all the solutions from the function y=2sin14x\begin{align*}y=2 \sin \frac{1}{4}x\end{align*} from [0,2π]\begin{align*}[0, 2 \pi]\end{align*}.
Now that the period can be different, we can have a different number of zeros within [0,2π]\begin{align*}[0, 2\pi]\end{align*}. In this case, we will have 6 times the number of zeros that the parent function. To solve this function, set y=0\begin{align*}y = 0\end{align*} and solve for x\begin{align*}x\end{align*}.
00=3cos6x=cos6x\begin{align*}0 &=-3 \cos 6x \\ 0 &=\cos 6x\end{align*}
Now, use the inverse cosine function to determine when the cosine is zero. This occurs at the multiples of \begin{align*}\frac{\pi}{2}\end{align*}.
\begin{align*}6x=\cos^{-1}0=\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2},\frac{9\pi}{2}, \frac{11\pi}{2}, \frac{13\pi}{2}, \frac{15 \pi}{2}, \frac{17\pi}{2}, \frac{19\pi}{2}, \frac{21\pi}{2}, \frac{23\pi}{2}\end{align*}
We went much past \begin{align*}2 \pi\end{align*} because when we divide by 6, to get \begin{align*}x\end{align*} by itself, all of these answers are going to also be divided by 6 and smaller.
\begin{align*}x=\frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{5\pi}{4}, \frac{17 \pi}{12}, \frac{19\pi}{12}, \frac{21\pi}{2}, \frac{23\pi}{12}\end{align*}
\begin{align*}\frac{23 \pi}{12}<2\pi\end{align*} so we have found all the zeros in the range.
### Examples
#### Example 1
Earlier, you were asked to find the period of \begin{align*}y=\cos [\pi (2x + 4)]\end{align*}.
First, we need to get the function in the form \begin{align*}y=a\cos b(x-h)+k\end{align*}. Therefore we need to factor out the 2.
\begin{align*}y=\cos [\pi (2x + 4)]\\ y = \cos [2\pi(x + 2)]\end{align*}
The \begin{align*}2\pi\end{align*} is the frequency. The period is therefore \begin{align*}\frac{2 \pi}{2\pi}=1\end{align*}.
#### Example 2
Determine the period of the function \begin{align*}y=\frac{2}{3}\cos\frac{3}{4}x\end{align*}.
The period is \begin{align*}\frac{2 \pi}{\frac{3}{4}}=2 \pi \cdot \frac{4}{3}=\frac{8 \pi}{3}\end{align*}.
#### Example 3
Find the zeros of the function from Example 2 from \begin{align*}[0, 2\pi]\end{align*}.
The zeros would be when \begin{align*}y\end{align*} is zero.
\begin{align*}0 &=\frac{2}{3} \cos \frac{3}{4}x \\ 0 &=\cos \frac{3}{4}x \\ \frac{3}{4}x &=\cos^{-1}0=\frac{\pi}{2},\frac{3 \pi}{2} \\ x &=\frac{4}{3}\left(\frac{\pi}{2},\frac{3 \pi}{2}\right) \\ x &=\frac{2\pi}{3},2\pi\end{align*}
#### Example 4
Determine the equation of the sine function with an amplitude of -3 and a period of \begin{align*}8\pi\end{align*}.
The general equation of a sine curve is \begin{align*}y=a\sin bx\end{align*}. We know that \begin{align*}a = -3\end{align*} and that the period is \begin{align*}8 \pi\end{align*}. Let’s use this to find the frequency, or \begin{align*}b\end{align*}.
\begin{align*}\frac{2\pi}{b} &=8\pi \\ \frac{2\pi}{8\pi} &=b \\ \frac{1}{4} &=b\end{align*}
The equation of the curve is \begin{align*}y=-3\sin \frac{1}{4}x\end{align*}.
### Review
Find the period of the following sine and cosine functions.
1. \begin{align*}y=5\sin 3x\end{align*}
2. \begin{align*}y=-2\cos 4x\end{align*}
3. \begin{align*}y=-3\sin 2x\end{align*}
4. \begin{align*}y=\cos \frac{3}{4}x\end{align*}
5. \begin{align*}y=\frac{1}{2}\cos 2.5x\end{align*}
6. \begin{align*}y=4\sin 3x\end{align*}
Use the equation \begin{align*}y=5\sin 3x\end{align*} to answer the following questions.
1. Graph the function from \begin{align*}[0, 2\pi]\end{align*} and find the domain and range.
2. Determine the coordinates of the maximum and minimum values.
3. Find all the zeros from \begin{align*}[0, 2\pi]\end{align*}.
Use the equation \begin{align*}y=\cos \frac{3}{4}x\end{align*} to answer the following questions.
1. Graph the function from \begin{align*}[0, 4\pi]\end{align*} and find the domain and range.
2. Determine the coordinates of the maximum and minimum values.
3. Find all the zeros from \begin{align*}[0, 2\pi]\end{align*}.
Use the equation \begin{align*}y=-3\sin 2x\end{align*} to answer the following questions.
1. Graph the function from \begin{align*}[0, 2\pi]\end{align*} and find the domain and range.
2. Determine the coordinates of the maximum and minimum values.
3. Find all the zeros from \begin{align*}[0, 2\pi]\end{align*}.
4. What is the domain of every sine and cosine function? Can you make a general rule for the range? If so, state it.
Write the equation of the sine function, in the form \begin{align*}y=a\sin bx\end{align*}, with the given amplitude and period.
1. Amplitude: -2 Period: \begin{align*}\frac{3 \pi}{4}\end{align*}
2. Amplitude: \begin{align*}\frac{3}{5}\end{align*} Period: \begin{align*}5 \pi\end{align*}
3. Amplitude: 9 Period: 6
4. Challenge Find all the zeros from \begin{align*}[0, 2\pi]\end{align*} of \begin{align*}y=\frac{1}{2}\sin 3\left(x-\frac{\pi}{3}\right)\end{align*}.
To see the Review answers, open this PDF file and look for section 14.4.
### Notes/Highlights Having trouble? Report an issue.
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# Group Theory
Group Theory is the study of objects in mathematics called groups. The concept of the group crops up in many areas of mathematics and science, and so this entry aims to provide a guide to what groups are, along with some of their properties and results that relate to them.
### What is a Group?
A group consists of two parts; a set and a mathematical operation. The set is essentially a collection of elements, such as numbers, that are part of the group. The operation1 must be a binary operation, meaning that it should take two elements as inputs and output another element. For example, we can regard multiplication as a binary operation, as we can multiply two elements to get an output of another element. The presence of an operator therefore allows us to produce another element from two of the elements in the group, thereby linking different elements together.
We denote the operation using a symbol, common examples of which are the '+' used in addition and the '×' used in multiplication. When describing rules that apply to all binary operations, we will use the generic symbol '*' (a subscript asterisk) to denote an operation. We denote the result of applying an operation '*', to elements a and b, to be a*b. For example, if we add a to b we get a+b as a result.
If we denote our set as G, and the operation *, then the group can be written in shorthand as (G,*). In this Entry we will simply call the group G, unless it is necessary to clarify the binary operation we are referring to. Also, it will be convenient to write a*b as ab, as these two expressions have the same meaning for any particular operation.
### Conditions
In fact, not every binary operation will turn a set into a group. There are certain conditions that must be met, to turn a set G and operation *, into a group (G,*). These are as follows:
#### 1 - Closure
The first condition is that if a and b are in our set G, then a*b is also in G. This simply means that our set is self-contained, and we do not need to consider anything outside the set G when applying our operation.
#### 2 - Identity
This condition states that there is at least one element e of G, such that x*e = e*x = x, for any x in G. Such an element is called an identity element of G, and doesn't affect itself or other members of the group.
For example, if G is the set of whole numbers, and * is the addition (+) operation, then 0 is an identity, as adding 0 to any number doesn't change it (x+ 0 = 0+x = x). Also, if G is the set of rational numbers2 except for 0, and * is multiplication (×), then 1 is an identity (x ×1 = 1×x = x). Sometimes 0 and 1 are used to denote identities in other situations where operations are defined that share properties of addition and multiplication respectively.
Note that we haven't said that only one identity element exists for any operation, although this is true, as we shall see later.
#### 3 - Inverse
This condition states that for each x in G, the group must contain at least one element y, such that x*y = y*x = e. This makes y the 'inverse' of x, as it cancels out x. For example, if G is the set of whole numbers, and * is addition, then the inverse of x is -x. Each element has only one inverse, as we shall see later, and we write the inverse of x as x-1.
Note that if y is the inverse of x, then x is the inverse of y. We can pair off all members of G with their inverses, with some exceptions. Some elements of G will be self-inverses, meaning that they are their own inverses. The identity element must be its own inverse, for example.
#### 4 - Associativity
This condition states that x*(y*z) = (x*y)*z for any choice of x, y and z in G. The brackets here indicate that we evaluate the operation in the bracket first. For example, to perform 3 + (4 + 5), we add the 4 and 5 to get 9, leaving 3 + 9 = 12. This condition allows us to ignore brackets when evaluating an expression, as all that matters is the order the elements are written in. In the expression 3 + 4 + 5, for example, it does not matter whether we add 3 to 4 to get 7, and then add 5, or add 4 to 5 to get 9, then add the 3. Note that this doesn't necessarily mean that the order in which the elements are written doesn't matter.
#### Abelian Groups
If x*y = y*x, for any x and y in G, then the operation * is commutative. If G is group, then if its operation * is commutative, it is an Abelian group, or just Abelian3. Many common operations in mathematics are commutative. For example, addition and multiplication are commutative, on the set of whole numbers. However, there are some non-commutative operations, such as subtraction. For example, 5 - 2 = 3, whereas 2 - 5 = -3. The operations of subtraction and division are neither associative nor commutative, and therefore cannot be used as the main operations in groups.
### The Result
These laws give us a set of elements which includes an identity, an inverse of every element in the group, and every element that can possibly be generated from these elements using the operation. The main operation used must be associative, and so the same result is achieved whenever the operation is applied to three elements written in a particular order.
### Some Inherent Properties
From these initial conditions, we can prove fairly simply some elementary properties of groups.
• Each Group Has Only One Identity Element
• First we will show that there is only one identity element. Suppose that there are two identity elements e and f. Then e*f = e, as f is an identity element. But e*f = f, as e is an identity element, thus e = f.
• Each Element Has Only One Inverse
• Secondly, we can show that inverses are unique. Consider an element x, and suppose we have two inverses y and z. x*y = x*z = e, as both y and z are inverse of x. But then x*y = x*z, and by multiplying by y on the left we get y*(x*y) = y*(x*z), and by associativity (y*x)*y = (y*x)*z. Finally, y*x = e, thus y = z.
• Cancellation Laws
• We can also show that if k*a = k*b or a*k = b*k, then a = b. If k*a = k*b, then multiply both sides by k-1 to get k-1k*a = k-1k*b. But k cancels with k-1, so a = b. A very similar argument tells us that a*k = b*k implies that a = b. These results are called cancellation laws.
### Cayley Tables
A Cayley table4 is a way of summarising the properties of a finite group5.
The Cayley table for a group consists of a table with n + 1 rows and columns, where n is the number of elements in the group. The top left box will contain the symbol for the operation defined on the group, eg, *. The other boxes in the top row contain the elements of the group, and similarly the left-hand column, both in the same order.
The other boxes contain the result of applying the operation with the element in the left-hand column in the same row on the left, and the element in the top row in the same column on the right. This is equivalent to multiplication tables, where the operation is multiplication. However, the order in which we perform the operation may matter, if the group is non-Abelian.
Below is an example of a Cayley table, for a group K, with 3 elements:
+3 0 1 2 0 0 1 2 1 1 2 0 2 2 0 1
This is the Cayley table for the group consisting of the set {0,1,2}6, with the operation +3. This operation is called 'addition modulo 3'. It is similar to ordinary addition, but after addition, both numbers are replaced by their remainders when divided by 3. For example, 1+32. 1 + 2 = 3, but 3 divides by 3, so has remainder 0, thus 1+32 = 0. This is also called clock arithmetic, as it resembles the way that two hours after 11 o'clock is 1 o'clock, the result of adding 11 and 2 modulo 12.
The Cayley table has various properties, corresponding to the definition of the group, and some other properties of groups:
• We can see the closure property, as the only elements present in the table are those listed on the top row and left-hand column.
• We can see that 0 is an identity element, as the 0 row and 0 column are identical to the top row and left-hand column respectively. This shows that it doesn't affect any of the elements of the group.
• We can observe that two elements x and y are inverses of each other, by the fact that the element in the x row and y column (or vice versa), is the identity. For example, we can see that 1 and 2 are inverses of each other, as the element in the 1 row, and 2 column (and the 1 column and 2 row), is 0, the identity.
• The table, apart from the left column and top row, is a Latin square. That is, it contains each element once on each row, and once on each column. All groups will give this property. This is due to the cancellation laws mentioned, and the closure property. One of the laws states that k*a = k*b only if a = b. In other words, if a and b are different elements, then ka and kb are different. On each row we have ka, kb, kc.., where a, b, c are different, so k*a, k*b, k*c, the elements of the boxes, will be different. Each of a, b, c must appear due to the closure property. Thus each element must appear once, and only once, in each row. Similarly, the other cancellation law means that each element must occur once and only once in each column.
• The fact that this group is Abelian can be seen as the group is symmetric about a line from top-left to bottom-right. A reflection through such a line would swap the rows and columns. The only distinction between rows and columns is the order that the operation is carried out in, and order doesn't matter for Abelian groups.
• Unfortunately, the Associativity property cannot be verified merely by looking at the Cayley table, as the Cayley table only looks at pairs of elements, and the associativity property concerns trios of elements.
### Cyclic subgroups
A subgroup H, of a group G, is a group where all of its elements are elements of G. A cyclic subgroup is a particular type of subgroup, being the smallest one containing a particular element p, with all the other elements in the subgroup being generated by p. This is written as <p>, and called the subgroup of G generated by p.
Consider a group G, and one element, p. We wish to find <p>, the smallest subgroup that contains p. Thus, we want <p> to contain only those elements of G that it must to be a group.
We start with the identity element e, and p. But if p is in <p> then p*p must be, by the closure property. Similarly p*p*p = (p*p)*p must be in <p>. This can continue on to give us that p*p*p*p, p*p*p*p*p, etc, must all be in <p>. We can write n lots of p as pn7, so p3 = p*p*p, and we can define p0 = e
Then, we need each pn to have an inverse p-n. This will constitute n lots of p-1, the inverse of p, as n lots of p will cancel with n lots of the inverse, to get e.
So, we have established that our group <p> must contain pn for all whole numbers n (positive, negative or zero). We can also see that the set <p> only contains these elements, as the set H, containing these elements, fulfils all the rules for being a group:
• <p> is closed, as pn*pm = pn+m, which is also a member of H8.
• It includes the identity element e = p0.
• Each element pn has an inverse p-n.
• Finally the associativity property applies, as it does in our larger set G, so the set H is a group, and thus <p> = H
We can also see that <p> will be Abelian, as pn*pm = pm*pn = pn+m.
If <p> is the whole of the group G, then G itself is called a cyclic group. It is clear that a cyclic subgroup will be a cyclic group by this definition, as it is the cyclic subgroup of itself generated by our original p. By this definition, we don't need to consider a cyclic group as being part of some larger group, we can check that a group is cyclic by checking that one element of the group generates the whole group.
There are essentially three types of cyclic group. The first simply consists of only an identity element. It can only be generated by an identity element, as any other element will generate a group containing itself and the identity, so must contain at least two elements.
The second consists of a single element p, and its 'powers', p2..,pn = e, where n is some positive integer and e is the identity. Here the other 'powers', such as pn+1, or p-3, are equal to one of the pi with i between 1 and n. In other words, after a certain stage, no new elements are being produced and the group is complete.
For an example of this, consider the group K considered earlier, and a particular element 1. The group <1> here will consist of 0, 1, 1+31 = 2,1+31+31 = 2+31 = 0, etc, and their inverses. But here, we cover all the elements of the group K in the subgroup. This means that K itself is a cyclic group. In fact, if we consider the group G consisting of the set {0, ..., n-1}, and the operation of addition modulo n, then G is a finite cyclic group, and in fact all cyclic groups are in some way equivalent to one such group. We call such groups Zn.
We can observe that the subsequent rows of its Cayley table are copies of the row above, shifted one place to the right. This is true for all finite Cyclic groups.
The final category is infinite cyclic groups. Here we need all the 'powers' listed, as they all produce different elements. Consider for example (Z,+), the group consisting of the set of whole numbers under the operation of addition. <1>. The element 1 generates the whole of (Z,+) as any whole number n can be written as n lots of 1 added together, or -n lots of -1 added together, or 0. Thus (Z,+) = <1>, and so it is cyclic. In a sense, which we will explore later, all infinite cyclic groups are essentially the same as (Z,+).
### Lagrange's Theorem
Lagrange's Theorem is one of the first 'big' results that most mathematicians see in group theory. It is named after Italian mathematician and astronomer Joseph Louis Lagrange (1736 - 1813), responsible for a number of contributions to physics and mathematics.
One of the consequences of the theorem is that the order of (number of elements in) a group G is cleanly divisible by the order of any subgroup H. This result allows us to prove a result about cyclic groups, that in fact all groups with prime numbers as orders are cyclic.
To show this, consider a group G with n elements, where n is a prime number. Then consider an element p other than the identity, and consider the cyclic subgroup H generated by it. This is a subgroup, so its order must be either 1 or n, as those are the only numbers that divide n. But H cannot be order 1, as it contains both the identity and p, thus it has at least 2 elements. So H must have order n, the same as G, thus as H is contained within G, H must be G. Thus G has a cyclic subgroup H that contains all of G, and so G is cyclic.
### Isomorphism
Isomorphism is a concept that formalises what it means for two groups to be essentially the same. It is an important concept not only in group theory, but in many other areas of mathematics. First we will see a sketch of the general concept of isomorphism, then a more concrete explanation.
To get the general idea of what isomorphisms are, consider stories, and what it means for two stories to be essentially the same. One idea, which we will consider here, is that two stories are the same if the relationships between the characters in both stories are the same. To show this, we would need to pair off the characters in the stories, so that the relationship between each pair of characters in one story is the same as that between the corresponding characters in the other story.
For example, if we have two love stories, one where Alice and Barry plan to marry, but Barry meets another woman Carol, and a second where Amelia and Bob's planned marriage is threatened by Bob's meeting of Cathy. Here we can pair off Albert with Arthur, Barry with Bob, and Carol and Cathy. Then the relationship between Alice and Barry (planned marriage) is the same as that between Amelia and Bob, and relationships hold for each other pair of characters. Thus we can say, according to this idea, that the two stories are essentially the same. You may wish to consider more details of each story, such as the ages of the people involved, the setting, etc, but most comparisons will involve pairing the elements of one story with those of the other. If we have such a pairing off of elements, then we can say informally that the two stories are isomorphic, or that there is an isomorphism between them. In particular we can call such a pairing an isomorphism.
In groups, the idea of isomorphism is very similar, although more precise. Suppose we have two groups G and H with associated operations *1 and *2 respectively. Then a pairing that associates x in G with x' in H is an isomorphism, if a*1b is associated with a'*2b', for any a and b in G. G and H are isomorphic if some isomorphism exists between them.
If two groups G and H are isomorphic, then their Cayley table will be equivalent, in the sense that if the elements of G are replaced with the associated elements of H, then its Cayley tables will be identical to that of H. Given that the Cayley table gives all the information about the structure of a group, then this means that two isomorphic groups are essentially the same group with different labels for their elements.
### Implications of Isomorphism
Two Isomorphic groups have essentially the same structure, with the only difference being how they are labelled. This means that they share the same properties. For example, if we have two Isomorphic groups, if one of them is Abelian, then so is the other. This means that if we prove one of these properties for one group, then it is proven for any other isomorphic group.
In fact, the four conditions for a set being a group are shared in a similar way. If we have two sets, each with an operation defined on them, then if one of them has (for example), the closure property, then so does the other, if they are isomorphic.
Given the idea of isomorphism, we can count the number of different groups, if we consider isomorphic groups the same. While there are infinitely many different labellings we can use there are only a finite number of groups of each order, or size, although the actual number can vary quite a lot. For example, there are only 2 groups of order 50, but 1543 of order 192.
Two further consequence refers back to the cyclic groups. Firstly it turns out that cyclic groups of the same order are isomorphic. To see this, consider two cyclic groups G and H, of order n, with operations * and + respectively. They are generated by two elements, p and q, and so consist of elements p,p*p,p*p*p,...,pn, and q,q+q,q+q+q,...,qn, respectively with the other powers being duplicates of these. Then there is an isomorphism between pm and qm. This is an isomorphism as pa*pb = pa+b, and so is associated with qa+b. But qa+b = qa+qb, so pa*pb is associated with qa+qb.
Thus all cyclic groups, of order n, are isomorphic, and isomorphic to cyclic group Zn.
This leads to the conclusion that for any prime number p, there is only one group with order p, as all the groups with order p will be cyclic, and there is essentially only one group order p, again Zp.
Further, all infinite cyclic groups are isomorphic. Again, consider two infinite cyclic groups G and H, which are generated by p and q. The groups then consist of powers of p and q respectively, and an equivalent isomorphism can be used associating pm and qm. This means that all infinite cyclic groups are isomorphic to each other. They are also therefore isomorphic to (Z,+).
1Examples of operations are addition, subtraction, multiplication and division.2All numbers that can be written as a fraction.3Named after Norwegian mathematician Niels Henrik Abel (1802 - 1829). Before his death after contracting tuberculosis, at the age of 27, he made a considerable contribution to mathematics.4Named after British mathematician Arthur Cayley (1821 - 1895), who first introduced this abstract concept of the group, and devised these tables to give their properties.5A group with a limited number of elements, as opposed to an infinite number.6The notation {a,b,c,d}, for example, indicates a set containing the elements a, b, c and d.7Note that this does not necessarily mean that p has been multiplied by itself n times, as if the operator is '+' it means that p has been added to itself n times, and so on.8In other words, as the group contains pn for all values of n, you can combine any two elements in the group and just get one with another value of n.
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# Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.5
## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.5
Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
The value of $$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{DA}}+\overrightarrow{\mathrm{CD}}$$ is ………………
Solution:
(c) $$\overrightarrow{0}$$
Question 2.
If $$\vec{a}+2 \vec{b}$$ and $$3 \vec{a}+m \vec{b}$$ are parallel, then the value of m is ………………
(a) 3
(b) $$\frac{1}{3}$$
(c) 6
(d) $$\frac{1}{6}$$
Solution:
(c) 6
Question 3.
The unit vector parallel to the resultant of the vectors $$\hat{i}+\hat{j}-\hat{k}$$ and $$\hat{i}-2 \hat{j}+\hat{k}$$ is ………………
Solution:
Question 4.
A vector $$\overrightarrow{O P}$$ makes 60° and 45° with the positive direction of the x and y axes respectively. Then the angle between $$\overrightarrow{O P}$$ and the z-axis is …………….
(a) 45°
(b) 60°
(c) 90°
(d) 30°
Solution:
(b) 60°
α = 60°, β = 45°
We know cos2α + cos2β + cos2γ = 1
(i.e.,) $$\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}$$ + cos2γ = 1
cos2γ = 1 – $$\frac{1}{4}-\frac{1}{2}=\frac{1}{4}$$
cos γ = $$\frac{1}{2}$$ ⇒ y = π/3 = 60°
Question 5.
If $$\overrightarrow{B A}=3 \hat{i}+2 \hat{j}+\hat{k}$$ and the position vector of B is $$\hat{i}+3 \hat{j}-\hat{k}$$, then the position vector A is …………………
Solution:
Question 6.
A vector makes equal angle with the positive direction of the coordinate axes. Then each angle is equal to …………..
Solution:
Question 7.
The vectors $$\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}$$ are ……………
(a) parallel to each other
(b) unit vectors
(c) mutually perpendicular vectors
(d) coplanar vectors
Solution:
(d) coplanar vectors
Question 8.
If ABCD is a parallelogram, then $$\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{C B}+\overrightarrow{C D}$$ is equal to ……………
Solution:
Question 9.
One of the diagonals of parallelogram ABCD with $$\vec{a}$$ and $$\vec{b}$$ as adjacent sides is $$\vec{a}+\vec{b}$$. The other diagonal $$\overrightarrow{\mathrm{BD}}$$ is ……………
Solution:
Question 10.
If $$\vec{a}$$, $$\vec{b}$$ are the position vectors A and B, then which one of the following points whose position vector lies on AB, is ………….
Solution:
(c) $$\frac{2 \vec{a}+\vec{b}}{3}$$
Question 11.
If $$\vec{a}, \vec{b}, \vec{c}$$ are the position vectors of three collinear points, then which of the following is true?
Solution:
(b) $$2 \vec{a}=\vec{b}+\vec{c}$$
Question 12.
If $$\vec{r}=\frac{9 \vec{a}+7 \vec{b}}{16}$$, then the point P whose position vector $$\vec{r}$$ divides the line joining the points with position vectors $$\vec{a}$$ and $$\vec{b}$$ in the ratio ………………
(a) 7 : 9 internally
(b) 9 : 7 internally
(c) 9 : 7 externally
(d) 7 : 9 externally
Solution:
Question 13.
If $$\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}$$ is a unit vector, then the value of λ is ……………..
(a) $$\frac{1}{3}$$
(b) $$\frac{1}{4}$$
(c) $$\frac{1}{9}$$
(d) $$\frac{1}{2}$$
Solution:
Question 14.
Two vertices of a triangle have position vectors $$3 \hat{i}+4 \hat{j}-4 \hat{k}$$ and $$2 \hat{i}+3 \hat{j}+4 \hat{k}$$ .If the position vector of the centroid is $$\hat{i}+2 \hat{j}+3 \hat{k}$$, then the position vector of the third vertex is ………………….
Solution:
Question 15.
(a) 42
(b) 12
(c) 22
(d) 32
Solution:
(c) 22
Question 16.
If $$\vec{a}$$ and $$\vec{b}$$ having same magnitude and angle between them is 60° and their scalar product is $$\frac{1}{2}$$ then $$|\vec{a}|$$ is ……………
(a) 2
(b) 3
(c) 7
(d) 1
Solution:
(d) 1
Question 17.
The value of θ ∈ (0, $$\frac{\pi}{2}$$) for which the vectors are perpendicular, is equal to …………………
Solution:
Question 18.
(a) 15
(b) 35
(c) 45
(d) 25
Solution:
(d) 25
Question 19.
Vectors $$\vec{a}$$ and $$\vec{b}$$ are inclined at an angle θ = 120°
is equal to ………….
(a) 225
(b) 275
(c) 325
(d) 300
Solution:
(d) 300
Question 20.
If $$\vec{a}$$ and $$\vec{b}$$ are two vectors of magnitude 2 and inclined at an angle 60°, then the angle between $$\vec{a}$$ and $$\vec{a}+\vec{b}$$ is ………………
(a) 30°
(b)60°
(c) 45 °
(d) 90°
Solution:
(a) 30°
Question 21.
If the projection of $$5\hat{i} -\hat{j}-3 \hat{k}$$ on the vector $$\hat{i}+3 \hat{j}+\lambda \hat{k}$$ is same as the projection of $$\hat{i}+3 \hat{j}+\lambda \hat{k}$$ on $$5\hat{i}- \hat{j}-3 \hat{k}$$ then λ is equal to ………………
(a) ±4
(b) ±3
(c) ±5
(d) ±1
Solution:
(c) ±5
Question 22.
If (1, 2, 4) and (2, – 3λ, – 3) are the initial and terminal points of the vector $$\hat{i}+5 \hat{j}-7 \hat{k}$$, then the value of λ is equal to ……………..
(a) $$\frac{7}{3}$$
(b) $$-\frac{7}{3}$$
(c) $$-\frac{5}{3}$$
(d) $$\frac{5}{3}$$
Solution:
Question 23.
If the points whose position vector $$10 \hat{i}+3 \hat{j}, 12 \hat{i}-5 \hat{j}$$ and $$\vec{a} \hat{i}+11 \hat{j}$$ are collinear then a is equal to ………………
(a) 6
(b) 2
(c) 5
(d) 8
Solution:
(d) 8
equating $$\hat{j}$$ components
⇒ -8 = 8t ⇒ t = -1
equation $$\hat{i}$$ components
t(a – 10) = 2
(i.e.,) (-1) (a – 10) = 2
a – 10 = -2
a = – 2 + 10 = -8
Question 24.
If then x is equal to …………..
(a) 5
(b) 7
(c) 26
(d) 10
Solution:
(c) 26
Question 25.
If $$\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k},|\vec{b}|=5$$ and the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\frac{\pi}{6}$$, then the area of the triangle formed by these two vectors as two sides, is …………….
(a) $$\frac{7}{4}$$
(b) $$\frac{15}{4}$$
(c) $$\frac{3}{4}$$
(d) $$\frac{17}{4}$$
Solution:
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# Mathematics word problem solver
This Mathematics word problem solver provides step-by-step instructions for solving all math problems. We will also look at some example problems and how to approach them.
## The Best Mathematics word problem solver
logarithm is the natural logarithm to the base e. It is used to solve equations with a base of e. The logarithm solve for x is: When solving logarithms, it is important to remember that the answer in this case is the base e raised to an integer power (i.e., 1 + 2 = 3). Logarithms are most useful when solving exponential equations, and they are especially useful when you are solving problems with large exponents. For example, if you have an equation that looks like this: y = 4x² + 9x - 14 Then using a logarithm solve for x, you would solve y = log10(4) + log10(9) + log10(14) = 5log10(4) + log10(3.4) = 5log2(4) = 2.06 Example 1: If you want to find out how many hours it takes for water to boil on a stove top, then solve for x: y = 4x² + 9x - 14 Here's what the math looks like: fp = 4 * x^2 + 9 * x - 14 yp = 4 * x^2 + 9 * x - 14 Here's what it means: First, find out how much water there is in the pot.
The main drawback with Wolfram is that it doesn’t always have all of the answers. For example, it might not know that 4x^3 + 2x^2 + y^3 = 0 because it doesn’t know what “+ y^3” means. You can also get stuck in the Wolfram Alpha sandbox if you accidentally click on something. The only solution is to close the window and start again from scratch.
However, a better way is to subtract or add terms. This can be done using one of three strategies: If you have two numbers and one is bigger than the other, you can ignore the smaller one and just add or subtract that one’s value from both sides of the inequality. For example: 3x > 4 5 + x In this case, you would subtract 4 from both sides, leaving 3 > 5 6 – 4 , which is true because 6 > 5. This method can also be used to turn an inequality into a statement about addition or subtraction, as in “I am more than \$100 poorer than my friend.” If you have two numbers and one is less than the other, you can ignore the bigger one and just add or subtract that one’s value from both sides of the inequality. For example: 6 10 12 + 8 = ? = 15 20 In this case, you would add 8 to both sides, leaving 6 10 12 – 8 , which is true because 12 20 . This method can also be
This works really well for my son it not only gives the answer but it shows the steps and you can also push the back button and it goes back bit by bit which is really useful and he said he is able to learn at a pace that makes him feel comfortable instead of being left pressured and mystified.
### Jayleen Harris
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In how many ways can $7$ identical erasers be distributed among $4$ kids in such a way that each kid gets at least one eraser but nobody gets more than $3$ erasers?
1. $16$
2. $20$
3. $14$
4. $15$
Given that, $7$ identical erasers and $4$ kids (distinct).
We have to distribute $7$ erasers in such a way that each kid gets at least one eraser and not more than $3$.
Now, We can give them all one eraser each. So the remaining erasers will be $7-4=3$.
Since no student can get more than $3$ erasers, we can’t give the remaining $3$ erasers to one student only.
Therefore, $2$ cases are possible.
$\textbf{Case 1:}$ Give $1$ eraser each to $3$ students
• $a,b,c,d \quad \text{(Let the name of the kids)}$
• $1,1,1,1 \quad \text{(Already given)}$
• $1,1,1,0$
• $\vdots \;\;\; \vdots \;\; \vdots \;\; \vdots$
Number of possible ways $= \frac{4!}{3!} = 4$ ways
$\textbf{Case 2:}$ Give $1$ eraser to $1$ student and $2$ erasers to other student.
• $a,b,c,d$
• $1,1,1,1 \quad \text{(Already given)}$
• $1,2,0,0$
• $\vdots \;\;\; \vdots \;\; \vdots \;\; \vdots$
Number of possible ways $= \frac{4!}{2!} = \frac{4\times3\times2!}{2!} =12$ ways
$\therefore$ Total number of ways $=4+12=16$ ways.
Correct Answer $:\text{A}$
10.3k points
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Discover x's worth in the displayed triangle.
Welcome to Warren Institute! In this article, we will explore the exciting world of Mathematics education by tackling a common problem: finding the value of x in a triangle. By utilizing our critical thinking skills and applying various geometric principles, we will unravel the mystery behind this mathematical conundrum. Get ready to unleash your inner problem solver as we dive into the depths of triangle calculations. So, *find the value of x* and join us on this journey of discovery!
Understanding the Triangle Angle Sum Theorem
The Triangle Angle Sum Theorem states that the sum of the interior angles of any triangle is always equal to 180 degrees. This theorem forms the basis for solving problems involving finding the value of x in a given triangle. By applying the theorem, we can determine the relationship between the known angles and x, allowing us to find its value.
Applying the Exterior Angle Theorem
The Exterior Angle Theorem states that the measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles. By using this theorem, we can derive equations involving x and the known angles of the triangle. Solving these equations will lead us to the value of x.
Utilizing Congruent Triangles
Congruent triangles have corresponding sides and angles that are equal. By identifying congruent triangles within the given triangle, we can establish congruence relationships and use them to solve for x. This approach relies on recognizing patterns and applying the properties of congruent triangles.
Employing Trigonometric Ratios
Trigonometric ratios such as sine, cosine, and tangent can be used to find missing side lengths or angles in a triangle. By setting up and solving trigonometric equations involving x and the known values, we can determine the value of x and complete the triangle. This approach is particularly useful when dealing with right triangles.
How can we use the given information to find the value of x in the triangle shown below?
We can use the given information, such as the measurements of the angles and sides of the triangle, along with relevant mathematical principles and formulas, to find the value of x.
What techniques or formulas can be applied to determine the value of x in the given triangle?
In the context of Mathematics education, to determine the value of x in a given triangle, various techniques and formulas can be applied depending on the specific information provided. One commonly used technique is applying the properties of similar triangles, where the ratios of corresponding sides are equal. The Pythagorean theorem can also be employed in right triangles to find missing side lengths and subsequently solve for x. Additionally, trigonometric functions such as sine, cosine, and tangent can be utilized when angle measures are known.
Are there any specific rules or theorems that can help us solve for the value of x in this triangle?
Yes, the Pythagorean theorem is a specific rule that can help us solve for the value of x in a right triangle. According to this theorem, in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. By applying this theorem, we can find the value of x in a right triangle.
Can we use similar triangles or trigonometric ratios to find the value of x in the triangle provided?
Yes, we can use similar triangles or trigonometric ratios to find the value of x in the given triangle.
Is there a specific method or step-by-step process that should be followed to find the value of x in this triangle?
Yes, there is a specific method or step-by-step process that should be followed to find the value of x in this triangle. The process typically involves using geometric principles and properties such as the angles in a triangle, congruent triangles, and the Pythagorean theorem. By applying these concepts, one can set up and solve equations to determine the value of x.
In conclusion, the process of finding the value of x in the triangle shown above is a great exercise for students to enhance their problem-solving skills and deepen their understanding of geometric concepts. By applying the principles of angle relationships and algebraic reasoning, students can confidently determine the value of x and unravel the mysteries of this triangular puzzle. It is crucial for educators to provide ample opportunities for students to engage in such problem-solving tasks, fostering their mathematical thinking and encouraging them to persevere through challenges. With strong support and guidance, students can unlock the beauty and elegance of mathematics, empowering them to become critical thinkers and lifelong learners in the field of mathematics education.
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How do you simplify Square root of 63 - Square root of 28?
Mar 3, 2018
$\setminus \sqrt{63} - \setminus \sqrt{28} = \setminus \sqrt{{3}^{2} \cdot 7} - \setminus \sqrt{{2}^{2} \cdot 7} = 3 \setminus \sqrt{7} - 2 \setminus \sqrt{7} = \setminus \sqrt{7}$
Mar 3, 2018
$\sqrt{7}$
Explanation:
We must use our surd knowledge:
$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$
$\implies \sqrt{63} = \sqrt{9 \cdot 7} = \sqrt{9} \cdot \sqrt{7} = 3 \sqrt{7}$
$\implies \sqrt{28} = \sqrt{4 \cdot 7} = \sqrt{4} \cdot \sqrt{7} = 2 \sqrt{7}$
$\therefore \sqrt{63} - \sqrt{28} \equiv 3 \sqrt{7} - 2 \sqrt{7}$
$= \sqrt{7}$
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## Tuesday, 7 February 2017
### The Sine Rule, the Cosine Rule and the Area of a Triangle
To understand the sine and cosine rules, we need to view a triangle like this:
In these triangles, the small case letters are the sides and the capital equivalents to those letters represent the angles directly opposite them.
### The Sine Rule
If, in a question, you are asked to find an angle in a triangle, and you are given two sides and another angle, you need to use the sine rule for angles.
Let's imagine that you are given a question like this and you have to find the angle:
Use the first diagram and label the sides and their corresponding angles. Here the missing angle would be A and 7cm would be a, as that is the side opposite the missing angle. 81° would be B, meaning that 18cm would be b.
Putting these values in the formula shows us that:
If you are asked to find a missing side in a triangle with two angles and one side, you have to use the sine rule for sides, as illustrated below.
The method used here is similar to the method illustrated above.
### The Cosine Rule
If you are given a question where you have to find the missing side given three sides, you would have to use the cosine rule for angles.
Let's imagine you are given a question like this where you had to find the angle:
If the missing angle is A, then 9cm would be a. 10cm would then be b and 11cm would then be c.
Putting these values in the formula shows us that:
If you are asked to find a missing side with two sides and an angle, you would have to use the cosine rule of sides.
### The Area of a Triangle
If you have an angle and two sides in a triangle and you are asked to find the area, you have to use this formula:
Let us consider the second example on this page. A is 50.5°, b is 10cm and c is 11cm.
Putting these values in the formula gives us:
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# Areas of two similar triangles are $36 \mathrm{~cm}^{2}$ and $100 \mathrm{~cm}^{2}$. If the length of a side of the larger triangle is $20 \mathrm{~cm}$, find the length of the corresponding side of the smaller triangle.
Given:
Areas of two similar triangles are $36\ cm^2$ and $100\ cm^2$. If the length of a side of the larger triangle is $20\ cm$.
To do:
We have to find the length of the corresponding side of the smaller triangle.
Solution:
As given, area of smaller triangle $=36\ cm^2$
Area of larger triangle $=100\ cm^2$
Length of the side of the larger triangle $=20\ cm$
Let $x\ cm$ be the length of the corresponding side of the smaller triangle.
We know that,
$\frac{\text { area( larger triangle) }}{\text { area( smaller triangle) }}=\frac{\text { (side of larger triangle })^2}{\text { (side of smaller triangle })^2}$
$\Rightarrow \frac{36}{100}=(\frac{x}{20})^2$
$\Rightarrow \frac{6^2}{10^2}=(\frac{x}{20})^2$
$\Rightarrow (\frac{6}{10})^2=(\frac{x}{20})^2$
$\Rightarrow \frac{6}{10}=\frac{x}{20}$
$\Rightarrow x=2\times6$
$\Rightarrow x=12$
Hence, the corresponding side of the smaller triangle is $12\ cm$.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
20 Views
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# 4.8 L’hôpital’s rule
Page 1 / 7
• Recognize when to apply L’Hôpital’s rule.
• Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L’Hôpital’s rule in each case.
• Describe the relative growth rates of functions.
In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule , uses derivatives to calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to determine its exact value.
## Applying l’hôpital’s rule
L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider
$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}.$
If $\underset{x\to a}{\text{lim}}f\left(x\right)={L}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to a}{\text{lim}}g\left(x\right)={L}_{2}\ne 0,$ then
$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\frac{{L}_{1}}{{L}_{2}}.$
However, what happens if $\underset{x\to a}{\text{lim}}f\left(x\right)=0$ and $\underset{x\to a}{\text{lim}}g\left(x\right)=0?$ We call this one of the indeterminate forms , of type $\frac{0}{0}.$ This is considered an indeterminate form because we cannot determine the exact behavior of $\frac{f\left(x\right)}{g\left(x\right)}$ as $x\to a$ without further analysis. We have seen examples of this earlier in the text. For example, consider
$\underset{x\to 2}{\text{lim}}\frac{{x}^{2}-4}{x-2}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}.$
For the first of these examples, we can evaluate the limit by factoring the numerator and writing
$\underset{x\to 2}{\text{lim}}\frac{{x}^{2}-4}{x-2}=\underset{x\to 2}{\text{lim}}\frac{\left(x+2\right)\left(x-2\right)}{x-2}=\underset{x\to 2}{\text{lim}}\left(x+2\right)=2+2=4.$
For $\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}$ we were able to show, using a geometric argument, that
$\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}=1.$
Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.
The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions $f$ and $g$ such that $\underset{x\to a}{\text{lim}}f\left(x\right)=0=\underset{x\to a}{\text{lim}}g\left(x\right)$ and such that ${g}^{\prime }\left(a\right)\ne 0$ For $x$ near $a,$ we can write
$f\left(x\right)\approx f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)$
and
$g\left(x\right)\approx g\left(a\right)+{g}^{\prime }\left(a\right)\left(x-a\right).$
Therefore,
$\frac{f\left(x\right)}{g\left(x\right)}\approx \frac{f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)}{g\left(a\right)+{g}^{\prime }\left(a\right)\left(x-a\right)}.$
Since $f$ is differentiable at $a,$ then $f$ is continuous at $a,$ and therefore $f\left(a\right)=\underset{x\to a}{\text{lim}}f\left(x\right)=0.$ Similarly, $g\left(a\right)=\underset{x\to a}{\text{lim}}g\left(x\right)=0.$ If we also assume that ${f}^{\prime }$ and ${g}^{\prime }$ are continuous at $x=a,$ then ${f}^{\prime }\left(a\right)=\underset{x\to a}{\text{lim}}{f}^{\prime }\left(x\right)$ and ${g}^{\prime }\left(a\right)=\underset{x\to a}{\text{lim}}{g}^{\prime }\left(x\right).$ Using these ideas, we conclude that
$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }\left(x\right)\left(x-a\right)}{{g}^{\prime }\left(x\right)\left(x-a\right)}=\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}.$
Note that the assumption that ${f}^{\prime }$ and ${g}^{\prime }$ are continuous at $a$ and ${g}^{\prime }\left(a\right)\ne 0$ can be loosened. We state L’Hôpital’s rule formally for the indeterminate form $\frac{0}{0}.$ Also note that the notation $\frac{0}{0}$ does not mean we are actually dividing zero by zero. Rather, we are using the notation $\frac{0}{0}$ to represent a quotient of limits, each of which is zero.
## L’hôpital’s rule (0/0 case)
Suppose $f$ and $g$ are differentiable functions over an open interval containing $a,$ except possibly at $a.$ If $\underset{x\to a}{\text{lim}}f\left(x\right)=0$ and $\underset{x\to a}{\text{lim}}g\left(x\right)=0,$ then
$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)},$
assuming the limit on the right exists or is $\infty$ or $\text{−}\infty .$ This result also holds if we are considering one-sided limits, or if $a=\infty \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-\infty .$
Differentiation and integration
yes
Damien
proper definition of derivative
the maximum rate of change of one variable with respect to another variable
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
what is calculus?
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
what is x and how x=9.1 take?
what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
by using integration product formula
Roha
find derivative f(x)=1/x
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
Roha
please try to guide me how?
Meher
what do u want to ask
Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
👍
Roha
Roha are u A Student
Lutf
yes
Roha
What is the first fundermental theory of Calculus?
do u mean fundamental theorem ?
Roha
I want simple integral
for MSc chemistry... simple formulas of integration
aparna
hello?
funny
how are you
funny
I don't understand integration
aparna
r u insane
aparna
integration is so simple not typical..
funny
tell me any questions about integration then i will solve.
funny
we use integration for whole values or for sum of values any there are some basic rule for integration..
funny
I just formulas
aparna
I just want formulas of integration
aparna
value of log ax cot-x cos-x
aparna
there are many formulas about integration
funny
more then one formula are exist about integration..
funny
so I want simple formulas Because I'm studying MSc chem...Nd have done bsc from bio...
aparna
I am M.sc physics now i am studying in m.phil
funny
so what can i do
aparna
I will send you basic formula for integration after two mint first of all i write then i will send you.
funny
send me your messenger id where i can send you formulas about integration because there is no option for image sending..
funny
integration f(X) dx this is basic formula of integration sign is not there you can look integration sign in methematics form... and f(X) my be any function any values
funny
you send me your any ID where i can send you information about integration
funny
funny
Hi
RIZWAN
I don't understand the formula
who's formula
funny
which formula?
Roha
what is the advantages of mathematical economics
Mubarak
What is a independent variable
a variable that does not depend on another.
Andrew
which can be any no... does not need to find its value by any other variable.. often x is independent and y is dependent
Roha
solve number one step by step
x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
how to prove 1-sinx/cos x= cos x/-1+sin x?
1-sin x/cos x= cos x/-1+sin x
Rochel
how to prove 1-sun x/cos x= cos x / -1+sin x?
Rochel
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# Ex.2.2 Q2 Polynomials Solution - NCERT Maths Class 10
Go back to 'Ex.2.2'
## Question
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \begin{align}\frac{1}{4},\, - 1\end{align} (ii) \begin{align}\sqrt 2 ,\;\frac{1}{3}\end{align} (iii) $${0,\;5}$$ (iv) $$1, \;1$$ (v) \begin{align}\frac{1}{4},\;\frac{1}{4} \end{align} (vi) $$4,\;1$$
Video Solution
Polynomials
Ex 2.2 | Question 2
## Text Solution
What is known?
The sum and product of zeroes of quadratic polynomials.
What is unknown?
A quadratic polynomial with the given numbers as the sum and product of its zeroes respectively.
Reasoning:
This question is straight forward - the value of sum of roots and product of roots is given. You have to form a quadratic polynomial. Put the values in the general equation of the quadratic polynomial i.e.
$k\left( {{x^2}-{\text{ }}\left( {{\text{sum of roots}}} \right)x{\text{ }} + {\text{ product of roots}}} \right)$
Steps:
(i) \begin{align}\,\,\frac{1}{4},\; - 1\end{align}
We know that the general equation of a quadratic polynomial is:
\begin{align}&{k\left( {{x^2} - \left( {{\text{sum of roots}}} \right){\text{ }}x + {\text{ product of roots }}} \right)}\\&{k\left\{ {{x^2} - \frac{1}{4}x + \frac{1}{4} \times - 1} \right\}}\\&{k\left\{ {{x^2} - \frac{1}{4}x - \frac{1}{4}} \right\}}\end{align}
(ii) \begin{align}\sqrt 2,\;\frac{1}{3}\end{align}
We know that the general equation of a quadratic polynomial is:
\begin{align}&k\left( {{x^2} - \left( {{\text{sum}}\,\,{\text{of}}\,\,{\text{roots}}} \right)x + {\text{product}}\,{\text{of}}\,{\text{roots}}} \right)\\&k\left\{ {{x^2} - \sqrt 2 x + \frac{1}{3}} \right\}\end{align}
(iii) $$\,0,\;\sqrt 5$$
We know that the general equation of a quadratic polynomial is
\begin{align}&k\left( {{x^2}-{\text{ }}\left( {{\text{sum of roots}}} \right)x{\text{ }} + {\text{ product of roots}}} \right)\\&k\left\{ {{x^2}-{\text{ }}0.x{\text{ }} + {\text{ }}\sqrt 5 } \right\}\\ &k\left\{ {{x^2}+ {\text{ }}\sqrt 5 } \right\}\end{align}
(iv) $$\,1,\;1$$
We know that the general equation of a quadratic polynomial is
\begin{align}&k\left( {{x^2}-{\text{ }}\left( {{\text{sum of roots}}} \right)x{\text{ }} + {\text{ product of roots}}} \right)\\&k\left\{ {{x^2}-{\text{ }}1x{\text{ }} + {\text{ }}1} \right\}\\&k\left\{ {{x^2}-{\text{ }}x{\text{ }} + {\text{ }}1} \right\}\end{align}
(v) \begin{align} - \frac{1}{4},\;\frac{1}{4}\end{align}
We know that the general equation of a quadratic polynomial is:
\begin{align}& k\left( {{x^2} - \left( {{\text{sum}}\,\,{\text{of}}\,{\text{roots}}} \right)\,x + {\text{product}}\,{\text{of}}\,{\text{roots}}} \right)\\&k\left( {{x^2} - \left( { - \frac{1}{4}x} \right) + \frac{1}{4}} \right) \\& k\left( {{x^2} + \frac{1}{4}x + \frac{1}{4}} \right) \\ \end{align}
(vi) $$4,\;1$$
We know that the general equation of a quadratic polynomial is:
\begin{align}&k\left( {{x^2}-\left( {{\text{sum of roots}}} \right)x + {\text{ product of roots}}} \right)\\&k\left\{ {{x^2}-{\text{ }}4x + {\text{ }}1} \right\}\end{align}
Learn from the best math teachers and top your exams
• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school
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# What is 3/4 - 1/8?
getcalc.com's two fractions subtraction calculator is an online basic math function tool to find what's the equivalent fraction for difference between two fractional numbers with different, unequal or unlike denominators 3/4 and 1/8.
3/4 - 1/8 = 20/32 as simplified as 5/8 in fraction form
3/4 - 1/8 = 0.625 in decimal form
This calculator, formula, step by step calculation and associated information to find the difference between 3/4 & 1/8 may help students, teachers, parents or professionals to learn, teach, practice or verify such two fractions subtraction calculations efficiently.
## How to Find Equivalent Fraction for 3/4-1/8?
The below workout with step by step calculation shows how to find what is the difference between two fractions by subtracting a fraction 1/8 from 3/4. For different, unequal or unlike denominators, use any one of the following methods to find the equivalent fraction for 3/4 minus 1/8.
1. LCM method
2. Cross multiplication method.
Problem & Workout - LCM Method
What is 3/4 minus 1/8 as a fraction by subtraction using LCM method?
step 1 Address formula, input parameters & values.
Input parameters & values:
3/4 & 1/8
3/4-1/8 = ?
step 2 For unequal denominators, find the LCM (least common multiple) for both denominators
The LCM for 4 and 8 is 8.
step 3 To have the common denominator, multiply the LCM with both numerator & denominator
=(3 x 8)/(4 x 8)-(1 x 8)/(8 x 8)
=6/8-1/8
=(6 - 1)/8
step 4 Find difference between numerators and rewrite it in a single form.
=5/8
3/4-1/8=5/8
By using LCM method, 5/8 is the equivalent fraction by subtracting 1/8 from 3/4.
Problem & Workout - Cross Multiplication Method
step 1 Address formula, input parameters & values.
Input parameters & values:
3/4 & 1/8
3/4 - 1/8 = ?
step 2 Cross multiply both numerator & denominator, multiply both denominators and rewrite as below
= (3 x 8) - (1 x 4)/(4 x 8)
step 3 Simplify & rewrite the fraction
= (24 - 4)/32
= 20/32
= 5/8
3/4-1/8 = 5/8
By using cross multiplication method, 5/8 is the difference between two fractions 3/4 and 1/8.
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2.10
# Conversion between different units
Whenever we have two or more systems of measuring, conversion questions come up. These give us a rich natural family of examples for direct proportionalities. But in fact there are two fundamentally different kinds of conversions: between different units in the same system, and between different systems.
In this lesson, you will
• get an overview of some useful conversions
• learn how to mathematically think about the problem of converting from one unit to another.
## Different units within the same system
Within a given system, there are often many different kinds of units, designed for convenience at different scales. Also there is often quite a high degree of historical whimsiness involved.
The British Imperial system of lengths in based on the foot, which traces back to the ancient Egyptians, and was passed on to the Greeks and Romans. The introduction of the yard as a unit of length came later. Some believe the origin was the double cubit, from the Egyptian, but was in any case standardized as the distance from the tip of the nose to the end of the thumb of King Henry I (reigned 1100–1135). Here are some common equivalences:
Imperial Length Proportionalities
12 inches = 1 foot
3 feet = 1 yard
1760 yards = 1 mile
There are also links, chains, furlong and poles but these are more specialized and not in common usage these days.
In the metrical system, everything is based on powers of ten, with
Metric Length Proportionalities
10 millimetres = 1 centimetre
100 centimetres = 1 metre
1000 metres = 1 kilometre
If we want to convert say $\normalsize{100}$ yards to inches, what would we do? We employ basic proportionalities: $1\,\mbox{ yard} = 3\,\mbox{ feet}$ and $1 \,\mbox{ foot} = 12 \,\mbox{ inches}$. These equalities can be combined as:
So $1 \, \mbox{ yard} = 36\,\mbox{ inches}$. Multiplying this by $100$ gives the desired conversion:
Q1 (E): Convert $\normalsize{200}$ yards to miles.
Q2 (E): Convert $\normalsize{300}$ kilometres to centimetres.
Another way of performing this conversion is to use direct proportionality. If we let $x = \mbox{yards}$ and $y = \mbox{inches}$, then a change of one yard corresponds to a change of 36 inches and
and the equation of this line is $y=36 x$. When $x=100$, then $y = 3600$.
## Converting from one system to another
If we want to convert from British Imperial to metrical units, we need to know how to convert. These are usually decided by international agreement and are not as neat as the conversions between units in the same system.
International Agreement Date of Agreement
1 inch = 2.54 centimetres 1896
1 mile = 1.609344 kilometres 1959
So if we want to convert $\normalsize{30}$ inches to metres, we could proceed as follows: let $y=\mbox{centimetres}$ and $x=\mbox{inches}$. A change in one inch corresponds to a change in 2.54 centimetres:
So our direct proportionality between centimetres and inches is $y = 2.54\, x$. If we let $z = \mbox{ metres}$ the direct proportionality between metres ($z$) and centimetres ($y$) can be expressed as:
Combining these equations gives:
Q3 (E): Substitute $x=30$ into the above equation to convert 30 inches to metres.
Q4 (M): Convert $\normalsize{14,000}$ feet to kilometers.
Q5 (C): Convert $\normalsize{46}$ miles/hour to metres/sec.
A1. 200 yards = 0.114 miles
A2. 300 kilometres = 30 million (30,000,000) centimetres
A3. 30 inches = 0.762 metres
A4. 14 000 feet = 4.27 kilometres
A5. 46 miles per hour = 20.56 metres per second
## Get a taste of this course
Find out what this course is like by previewing some of the course steps before you join:
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# Family of Parabolas y=(x-p)²+k (combination of horizontal and vertical shifts)
🏆Practice parabola of the form y=(x-p)²+k
### Family of Parabolas$y=(x-p)²+k$
Combination of Horizontal and Vertical Shift
$K$ determines the amount of steps and the vertical direction in which the function will shift - upwards or downwards.
$P$ determines the amount of steps and the horizontal direction in which the function will shift - to the right or to the left.
## Test yourself on parabola of the form y=(x-p)²+k!
Which equation represents the function:
$$y=x^2$$
moved 2 spaces to the right
and 5 spaces upwards.
## Let's look at an example of combining both displacements together.
For example, in the function:
$y=(x-4)^2+3$
The changes will be:
according to $P=4$ -The parabola will shift $4$ steps to the right.
According to $K=3$ -The parabola will shift $3$ steps upwards.
Let's see it in the illustration:
We can see that the vertex of the parabola is:
$(4,3)$
### Zero Point or Root of the Function - Graphical and Algebraic Solution when$Y=0$
The zeros of a function are the intersections with the $X$ axis.
## Algebraic Solution
when $K$ positive - This equation has no solution except in the case where $K$ equals$0$ and also $P$ equals$X$.
when $K$ negative- Generally, this equation will have $2$ solutions.
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## Graphical Solution
The graphical solution are the points of intersection of the parabola with the $X$ axis
that is, the zeros of the function.
Do you know what the answer is?
## examples with solutions for parabola of the form y=(x-p)²+k
### Exercise #1
Which equation represents the function:
$y=x^2$
moved 2 spaces to the right
and 5 spaces upwards.
### Video Solution
$y=(x-2)^2+5$
### Exercise #2
Find the corresponding algebraic representation of the drawing:
### Video Solution
$y=(x+2)^2+7$
### Exercise #3
Find the corresponding algebraic representation of the drawing:
### Video Solution
$y=x^2-4$
### Exercise #4
Choose the equation that represents the function
$y=-x^2$
moved 3 spaces to the left
and 4 spaces up.
### Video Solution
$y=-(x+3)^2+4$
### Exercise #5
Find the corresponding algebraic representation of the drawing:
### Video Solution
$y=(x-5)^2+4$
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# GCSE Ratio jfrosttiffin kingston sch uk Dr Frost
• Slides: 28
GCSE : : Ratio jfrost@tiffin. kingston. sch. uk @Dr. Frost. Maths Objectives: Understanding what is meant by ratio. Solve problems where a value and a ratio is given. Combine ratios. Solve problems involving changing ratios. Last modified: 29 th December 2018
www. drfrostmaths. com Everything is completely free. Why not register? Register now to interactively practise questions on this topic, including past paper questions and extension questions (including UKMT). Teachers: you can create student accounts (or students can register themselves), to set work, monitor progress and even create worksheets. With questions by: Dashboard with points, trophies, notifications and student progress. Teaching videos with topic tests to check understanding. Questions organised by topic, difficulty and past paper.
Overview Ratio allows us to express the ‘relative size’ of different quantities. Lesson 1: Simplifying ratios and solving problems where one amount or total is given. Lesson 2: Combining or splitting ratios. Lesson 3: Further Ratio Problem Solving (Advanced) “Javier and Carmen share some money in the ratio 5 : 6. When each are given £ 4 more, the ratio of the money they have is now 6 : 7. How much money did each initially have? ”
Stand up if… You have two colours to pick from only. Red Blue Stand up if you prefer (a) Red (b) Blue Suppose the results were the following: Red 18 Blue 12 Suppose also we were unconcerned by exact numbers who voted for each but wanted to express the relative size of how many voted for each. What ways could we do this? Percentages: Ratio: Red 60%, Blue 40% ? Percentages are good for indicating proportion of some total. Ratio of red to blue: ? (We’ll explore this on 3: 2 the next slide…)
What is ratio? For each three people who chose What specifically does this mean? red, two chose blue. It could be that we just had 3 red people and 2 blue. The above statement would be true! And similarly we could have 9 red and 6 blue. This means we can times (or divide) each part in the ratio by the same number, and end up with an equivalent ratio.
Simplifying ratios This suggests we can ‘simplify’ ratios in the same way as fractions. Just like with fractions, we prefer the numbers to be as small as possible. a ? b ? ? c Quickfire Simplifying: a b c d e f ? ? ? g ?
Splitting an amount into a given ratio Method 1: Identify what ‘ 1 part’ is worth In the ratio 3: 1, we have “ 3 parts muffin” to “ 1 part cupcake”. The given value in the question is the 20 cakes. How many ‘parts’ does this represent? ? 4 parts = 20 1 part = 5 ? 3 parts = 15 ? There are 15 cupcakes. Method 2: The ‘scaling’ way ?
Test Your Understanding Method 1: Identify what ‘ 1 part’ is worth (The one value given is the € 56, which is the total amount Jack and Kate share. So how many parts does this represent? ) 14 parts = € 56 1 part = € 4 ? 6 parts = € 24 Lila gets € 24 Method 2: The ‘scaling’ way ?
When one of the quantities is given In a death match of monkeys vs squirrels, the monkeys outnumber the squirrels by a ratio of 5: 2. There’s 35 monkeys. How many squirrels are there? Method 1: Identify what ‘ 1 part’ is worth The one value given is the 35 monkeys. How many parts in the ratio does this represent? 5 parts = 35 1 part = 7 ? 2 parts = 14 Therefore 14 squirrels. Method 2: The ‘scaling’ way ?
When the difference is given Alice and Bob share some money in the ratio 7: 4. Alice received £ 12 more than Bob. What did they receive in total? Method 1: Identify what ‘ 1 part’ is worth 3 parts = £ 12 1 part = £ 4 11 parts = £ 44 This time £ 12 represents the DIFFERENCE in the parts. ? We want the total amount, so need to find the value of the total parts. Method 2: The ‘scaling’ way ?
Test Your Understanding 1 The ratio of cats to dogs at Battersea Dogs & Cats home is 2: 3. There are 45 animals in total. How many cats are there? 2 The ratio of Tiffin Boys students to Tiffin Girls students is 8: 9. There’s 2700 students at Tiffin Girls. How many at Tiffin Boys? 2400 ? 18 ? 3 Tom, Dick and Harry share some money in the ratio 4: 7: 2. Dick receives £ 35 more than Harry. How much did Tom and Dick receive in total? £ 77 ? 4 ?
Ratios to/from equations or worded descriptions ? ? The number of yellow balls is twice the number of red balls and the number of green balls is 4 times the number of yellow balls. Determine the ratio of red to yellow to green balls. ?
Exercise 1 1 ? a c ? (on provided worksheet) ? ? b d 2 a b What is the ratio of shaded to non-shaded (small) triangles? 2: 6=1: 3 ? If £ 30 is split between two people in the ratio 3: 2, what does each get? £ 18, £ 12 ? 5 ? 7 3 4 6 To make orange squash I mix concentrate and water in the ratio 2: 7. If I use 150 ml of concentrate, how much water do I use? 525 ml ? ? 8 ? 9 ? 10 ?
Exercise 1 (on provided worksheet) 11 14 ? 15 ? 12 ? 16 ? 13 ? 17 ? ?
Combining and Subdividing Ratios One new type of question to the 2017+ GCSE syllabus is combining two ratios into one:
Further Example ?
Test Your Understanding 24 15? 20
Subdividing a Ratio ? ? ?
Exercise 2 (on provided worksheet) 4 1 ? ? ? a b c d e 2 ? 5 ? ? 3 6 ? ?
Exercise 2 (on provided worksheet) 9 7 ? ? 8 10 ? ? ?
Updated Ratios after Changing Amounts ?
A few new ratio skills… Before we move on to harder problems, here are two new key skills that will help us tackle them: The ratio of the width to height of a TV is 16 : 9. How could we represent all possible widths and heights of the TV, that ensures this ratio is maintained? ? ?
Updated Ratios after Changing Amounts (harder) Jamie and Alastair share some sweets in the ratio 7 : 5. Jamie gives 2 sweets to Alastair. The ratio of sweets is now 13 : 11. How many sweets did each initially have? Possible numbers of sweets before: (multiples of 7 : 5) Jamie 7 14 21 28 35 … Alastair 5 10 15 20 25 Trial and Error-ey but easiest method: We could just scale each of the ratios until we see a pair of scalings that work… Possible numbers of sweets after: (using 13 : 11) Jamie 13 26 ? Alastair 11 22 At this stage we can see that we could have started with 28 and 20 sweets, and after the exchange, each would have 26 and 22, as above.
Updated Ratios after Changing Amounts (harder) Jamie and Alastair share some sweets in the ratio 7 : 5. Jamie gives 2 sweets to Alastair. The ratio of sweets is now 13 : 11. How many sweets did each initially have? But what if we wanted to use an algebraic method? Let’s try and use the two new skills we have just learnt. ? Cross-multiply.
Test Your Understanding Javier and Carmen share some money in the ratio 5 : 6. When each are given £ 4 more, the ratio of the money they have is now 6 : 7. How much money did each initially have? ?
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# One-Sided Limits and Continuity
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#### Recommended Lessons and Courses for You
Lesson Transcript
Instructor: Robert Egan
Over the river and through the woods is only fun on a continuous path. What happens when the path has a discontinuity? In this lesson, learn about the relationship between continuity and limits as we walk up and down this wildlife path.
## Continuous and Discontinuous Paths
Consider for a minute a sidewalk that goes over the river and through the woods and perhaps over some hills. If I take a look at the elevation of the sidewalk as a function of location, then its function might look something like this. Now this is a continuous path; I can trace this path without lifting up my finger, so what about the limits along this continuous path? Well, If I look at the elevation as I approach the treeline, I might find that the elevation is 100 feet. Let's say there was a gigantic earthquake! And the earthquake split the ground at the treeline. Now, if I approach the treeline from the river, then the limit might be 100 feet. But if I approach the treeline from the woods, then the limit might be 120 feet. If I want to trace this path, it's now discontinuous; I have to lift my finger up from the paper to continue tracing it because of this discontinuity at the treeline.
## One-Sided Limits
What can we learn from our treeline? First, limits can be different when you approach a point from the left- or right-hand side. These are called one-sided limits. A mathematical example of this might be the function f(x) where it equals x for x<1 and it equals x + 1 for x is greater than or equal to 1. This is a lot like our earthquake example. For values less than 1, f(x)=x. At 1, this line jumps because f(x)=x + 1. At this point here, we have a limit approaching 1 on the left-hand side that's different from the limit approaching 1 from the right-hand side. So let's look at the limit from the left-hand side. We're going to differentiate this limit from the limit that's approaching 1 from the right-hand side by putting a minus sign by the number that we're approaching. The limit as x approaches 1 from the left side is 1, and the limit as x approaches 1 from the right side - which is designated by a plus sign - is 2.
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9.7: Triangle Classification by Angles
Difficulty Level: At Grade Created by: CK-12
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Practice Triangle Classification by Angles
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Do you know about different types of triangles? Well, Cassie is learning all about them.
"As if protractors weren't bad enough," Cassie complained sitting at the kitchen table.
"What's the matter?" her brother Kyle asked.
"Well, look at this," Cassie said showing him the book. "I have to identify these triangles."
"That's not so bad if you know what to look for," Kyle explained.
But Kyle is right. There are things to look for when classifying triangles. Pay attention to this Concept and you will know what Kyle is talking about by the end of it.
Guidance
This next Concept is all about triangles; the prefix “tri” means three-triangle means three angles.
When we classify a triangle according to its angles, we look at the angles inside the triangle. We will be using the number of degrees in these angles to classify the triangle. Let’s look at a picture of a triangle to explain.
Here is a triangle. We can look at the measure of each angle inside the triangle to figure out what kind of triangle it is. There are four types of triangles based on angle measures.
What are the four kinds of triangles?
The first type of triangle is a right triangle. A right triangle is a triangle that has one right angle and two acute angles. One of the angles in the triangle measures and the other two angles are less than 90. Here is a picture of a right triangle.
Can you figure out which angle is the one just by looking at it?
Sure, you can see that the 90 degree angle is the one in the bottom left corner. You can even draw in the small box to identify it as a 90 degree angle. If you look at the other two angles you cans see that those angles are less than 90 degrees and are acute.
Here we have one angle and two angles. We can find the sum of the three angles.
The sum of the three angles of a triangle is equal to .
The second type of triangle is an equiangular triangle. If you look at the word “equiangular” you will see that the word “equal” is right in the word. This means that all three of the angles in a equiangular triangle are equal.
The three angles of this triangle are equal. This is an equiangular triangle.
In an equiangular triangle, all of the angle measures are the same. We know that the sum of the three angles is equal to , therefore, for all three angles to be equal, each angle must be equal to .
The sum of the angles is equal to .
The next type of triangle is an acute triangle. The definition of an acute triangle is in the name “acute.” All three angles of the triangle are less than 90 degrees. Here is an example of an acute triangle.
All three of these angles measure less than 90 degrees.
The sum of the angles is equal to .
The last type of triangle that we are going to learn about is called an obtuse triangle. An obtuse triangle has one angle that is obtuse or greater than 90 and two angles that are less than 90 or are acute.
The sum of the angles is equal to .
Now it is time to practice. Identify each type of triangle according to its angles.
Example A
A triangle with angles that are all 60 degrees is called _________________.
Solution: An Equiangular Triangle
Example B
A triangle with one angle that is 90 degrees is called _________________.
Solution: A Right Triangle
Example C
A triangle with one angle that is 120 degrees is called _______________.
Solution: An Obtuse Triangle
Now back to Cassie and the triangles. Here is the original problem once again.
"As if protractors weren't bad enough," Cassie complained sitting at the kitchen table.
"What's the matter?" her brother Kyle asked.
"Well, look at this," Cassie said showing him the book. "I have to identify these triangles."
"That's not so bad if you know what to look for," Kyle explained.
To identify each triangle by angles, Kyle knows that Cassie needs to look at the interior angles of each triangle. Let's use the information that you just learned in this Concept to classify each triangle.
The first one has three angles less than 90, so this is an acute triangle.
The second one has one right angle, therefore it is a right triangle.
The third triangle has one angle greater than 90, so it is an obtuse triangle.
The last triangle has one angle greater than 90, so it is also an obtuse triangle.
Vocabulary
Here are the vocabulary words in this Concept.
Triangle
a three sided figure with three angles. The prefix “tri”means three.
Acute Triangle
all three angles are less than 90 degrees.
Right Triangle
One angle is equal to 90 degrees and the other two are acute angles.
Obtuse Triangle
One angle is greater than 90 degrees and the other two are acute angles.
Equiangular Triangle
all three angles are equal
Guided Practice
Here is one for you to try on your own.
True or false. An acute triangle can also be an equiangular triangle.
This is true. Because all of the angles in an acute triangle are less than 90 and all of the angles in an equiangular triangle are 60 degrees, an acute triangle can also be an equiangular triangle.
Practice
Directions: Classify each triangle according to its angles.
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5.
Directions: Classify the following triangle by looking at the sum of the angle measures.
6.
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Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English
TermDefinition
Acute Triangle An acute triangle has three angles that each measure less than 90 degrees.
Equilateral Triangle An equilateral triangle is a triangle in which all three sides are the same length.
Obtuse Triangle An obtuse triangle is a triangle with one angle that is greater than 90 degrees.
Right Triangle A right triangle is a triangle with one 90 degree angle.
Triangle A triangle is a polygon with three sides and three angles.
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Properties of a Parallelogram
Definition: A parallelogram is a quadrilateral with parallel opposite sides. Rectangles, rhombi and squares are all parallelograms.
A parallelogram has the following properties:
– Opposite sides are congruent and parallel
– Opposite angles are congruent
– Diagonals bisect each other
– Consecutive angles are supplementary
Example: Find the lengths of side BC and side DC
Solution:
Use the given information to determine the side lengths of sides BA and AD.
Step 1: The figure indicates that this quadrilateral has two pairs of parallel lines, so we can conclude that it is a parallelogram
Step 2: Since parallelograms have two pairs of congruent sides, we know side BA is congruent to side CD,
and side BC is congruent to side AD.
Step 3: Since side AB = 5 mm, then CD must also have a length of 5 mm
Since side AD = 3 + 4 = 7 mm, then side BC must also equal 7mm.
Based on the properties of a parallelogram, we can conclude that side BC = 7mm, side DC = 5mm.
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# PROPORTIONS SSS SIMILARITY PARALLEL TRANSVERSALS PROBLEM 1 PROBLEM 2
## Presentation on theme: "PROPORTIONS SSS SIMILARITY PARALLEL TRANSVERSALS PROBLEM 1 PROBLEM 2"— Presentation transcript:
PROPORTIONS SSS SIMILARITY PARALLEL TRANSVERSALS PROBLEM 1 PROBLEM 2
STANDARDS 4 and 5 PROPORTIONS SSS SIMILARITY PARALLEL TRANSVERSALS PROBLEM 1 PROBLEM 2 PROBLEM 5 AA SIMILARITY JOINING MIDPOINTS IN A TRIANGLE SAS SIMILARITY PROBLEM 6 ARE POLYGONS SIMILAR? PROBLEM 3 PARALLEL TO SIDE OF TRIANGLE PROBLEM 4 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Students prove basic theorems involving congruence and similarity.
Standard 4: Students prove basic theorems involving congruence and similarity. Los estudiantes prueban teoremas básicos que involucran congruencia y semejanza. Standard 5: Students prove triangles are congruent or similar and are able to use the concept of corresponding parts of congruent triangles. Los estudiantes prueban que triángulos son congruentes o semejantes y son capaces de usar el concepto de partes correspondientes de triángulos congruentes. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
In a classroom there are 3 boys and 6 girls
In a classroom there are 3 boys and 6 girls. What is the ratio of boys to girls? . 3 3 6 1 2 = 3 to 6 or 3:6 or 5 2 1.0 (0.5) (100%) = 50% =0.5 - 1 0 STANDARD 1.3 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
A A to B A : B B C C to D C : D D A C = B D
How do you express the ratio of A to B? A B A to B A : B How do you express the ratio of C to D? C D C to D C : D Now when we equal to ratios, we get a PROPORTION: A B D C = PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
The product of the MEANS is equal to the product of the EXTREMES
= B D B C B and C are the MEANS A D A and D are the EXTREMES = A D B C = Cross-multiplying: (A)(D)=(C)(B) The product of the MEANS is equal to the product of the EXTREMES PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
STANDARDS 4 and 5 If A K and B L and C M then Triangles are SIMILAR when the corresponding sides are proportional: SSS similarity A CA AB BC B C LM MK KL KL AB MK CA LM BC OR K M CAB MKL Both triangles are similar ( ) L PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
The triangles below are similar, find CA=? and TS=?
6 4 A C T X+6 S 2X+3 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
The triangles below are similar, find CA=? and TS=?
RT B BC 6 4 A C S T CA X+6 TS 2X+3 4 6 X+6 2X+3 CA = X + 6 = 12 4(2X+3) = 6(X+6) = 18 8X +12 = 6X + 36 TS = 2X + 3 8X = 6X + 24 = 2( ) + 3 12 -6X -6X = 2X = 24 = 27 X = 12 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
What is the value of X and JL if JKL RST. JK= X+2, KL=10, RS=X+6,
ST= 20, and JL = 5X + 2 JK X+2 RS X+6 KL 10 ST 20 J R K S L T 5X+2 = JL = 5X + 2 = 5( ) + 2 2 20(X+2) = 10(X+6) = 20X +40 = 10X + 60 = 12 20X = 10X + 20 -10X -10X 10X = 20 X = 2 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
X A Y Z C B if A X and C Z then ABC XYZ By AA similarity
X XY A AB AC Y Z C B if and A X then ABC XYZ By SAS similarity XZ
First Prove that all triangles in the figure are similar among them:
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. E S F R G T U H First Prove that all triangles in the figure are similar among them: 1. Two lines cut by a common perpendicular transversal are parallel. 2. A line perpendicular to one line is perpendicular to any line parallel to it. 3. Two perpendicular lines form 4 right angles. 4. Alternate interior angles are congruent. 5. Corresponding angles are congruent STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
All highlighted triangles in figure are similar by AA SIMILARITY! H
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. E S F R G T U All highlighted triangles in figure are similar by AA SIMILARITY! H STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
All highlighted triangles in figure are similar by AA SIMILARITY! H
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. 25 15 3 E S F R G T U All highlighted triangles in figure are similar by AA SIMILARITY! H RF UR = EG GF = 3 15 RF 25 15RF = (3)(25) 15RF = 75 RF = 5 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Applying the Pythagorean Theorem EG GF RF = UF + UR
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. 3 E S F R G T U 5 H RF UR = Applying the Pythagorean Theorem EG GF RF = UF + UR 2 = 3 15 RF 25 5 = UF + 3 2 25 = UF + 9 2 15RF = (3)(25) 15RF = 75 UF = 16 2 UF = 4 RF = 5 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Applying the Pythagorean Theorem EG GF RF = UF + UR 25RS = (15)(15) =
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. 25 15 10 E S F R G T U 5 H RS RT = GF EG RF UR = 5+10 25 RS 15 = Applying the Pythagorean Theorem EG GF RF = UF + UR 2 25RS = (15)(15) = 3 15 RF 25 5 = UF + 3 2 25RS = 225 25 = UF + 9 2 15RF = (3)(25) RS = 9 15RF = 75 UF = 16 2 UF = 4 RF = 5 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Applying the Pythagorean Theorem EG GF RF = UF + UR 25RS = (15)(15) =
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. E S F R G T U H RS RT = GF EG RF UR = 5+10 25 RS 15 = Applying the Pythagorean Theorem EG GF RF = UF + UR 2 25RS = (15)(15) = 3 15 RF 25 5 = UF + 3 2 25RS = 225 25 = UF + 9 2 15RF = (3)(25) RS = 9 15RF = 75 UF = 16 2 UF = 4 RF = 5 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
A AD AE E D DB EC B C If DE BC then STANDARDS 4 and 5
A AD AE E D DB EC B C If then DE BC STANDARDS 4 and 5
Find the value for X B 6 S TA SB 18 = CT CS C X 6 (24) (24) = 40 24 24
(24) (24) = (6)(24) 18 X = 144 18 X X = 8 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
C BC B AB A EF DE E F D STANDARDS 4 and 5
Find the values for Y and Z: Y Y+5 = 6 7
Find the values for Y and Z: Y Y+5 = 6 7
30 +5 Y = 30 Z Z+1 = Y Y+5 30 Z 30 Z+1 30+5 = Z 30 Z+1 35 = 35Z = 30(Z+1) 35Z = 30Z +30 -30Z -30Z 5Z = 30 Z = 6 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
K S R L M and RS= LM 1 2 If KR RL KS SM and then RS LM
K S R M L Find in the problem below the value for RS: 10 1 7 LM If
120 RS = LM 1 2 If then RS = (120) 1 2 RS = 60 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
The two irregular polygons are similar find values for X and Y: 30 98
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# ML Aggarwal Circles ISC Class-11 Maths Understanding
ML Aggarwal Circles ISC Class-11 Maths Understanding Solutions Chapter-12. Step by step Solutions of ML Aggarwal ISC Class-11 Mathematics with Exe-1, Exe-2, Exe-3 Exe-4, Exe-5, Exe-6, and Chapter Test Questions. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
## ML Aggarwal Circles ISC Class-11 Maths Understanding Ch-12
Board ISC Class 11 Subject Mathematics Chapter-12 Circles Session 2024-25 Topics Solutions of ML Aggarwal
### Exe-12.2
ML Aggarwal Circles ISC Class-11 Maths Understanding Ch-12
Exe-12.3
### Exe-12.4
ML Aggarwal Circles Class-11 Maths Understanding Ch-12
#### How to Find the Equation of the Circle?
Here, some solved problems are given to find the equation of a circle on both cases such as when the centre of a circle is origin and centre is not an origin is given below.
Example 1:
Consider a circle whose centre is at the origin and radius is equal to 8 units.
Solution:
Given: Centre is (0, 0), radius is 8 units.
We know that the equation of a circle when the centre is origin:
x2y= a2
For the given condition, the equation of a circle is given as
xy= 82
xy2= 64, which is the equation of a circle
Example 2:
Find the equation of the circle whose centre is (3,5) and the radius is 4 units.
Solution:
Here, the centre of the circle is not an origin.
Therefore, the general equation of the circle is,
(x-3)+ (y-5)2 = 42
x– 6x + 9 + y-10y +25 = 16
x+y-6x -10y + 18 =0
Exe-12.5
### Exe-12.6
ML Aggarwal Circles ISC Class-11 Maths Understanding Ch-12
Chapter Test
### ML Aggarwal Circles ISC Class-11 Maths Understanding Ch-12
-: End of Circles ISC Class-11 ML Aggarwal Maths Understanding Chapter-12 Solution :-
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# Rounding Numbers
Top
Sub Topics The study of rounding decimals is important specifically in decimal topics of mathematics. Rounding decimals also covered fraction independently. The main purpose of using rounding number method is just to simplify the number keeping its closes value. Rounding number method is used to short the number basically.This is applicable on the right on the numbers only or after the decimal point also involves a number which can increase or decrease the next digit . These numbers helps make values easier or avoid confusion of numbers.Find more detailed information of rounding numbers below;
## Rounding Decimals
Round numbers are nothing but the nearest tenths, hundredths etc integer. These numbers are necessary to calculate when we no need to keep many numbers after decimal point.
To rounding decimal step by step method is as follows;
1) Find the positive value of the number.
2) Drop the digits after the decimal if digits are less then five.
3) If the digits after the decimal are equal or grater than five then add 1 to rounding digit.
## Rounding Numbers Rules
There are basically 3 rules of rounding numbers. Rounding numbers rules are as follows;
Rule 1: Verify, if the right side decimals are 1,2,3,4 of the given number then simply slum all digits in the right side.
Rule 2: If the right side decimals are 5,6,7,8,9, then add 1 and delete others digits to get the rounding digit.
Rule 3: If above rule are not applicable Then use Banker's rule. Suppose we have some numbers like 3.315; here 5 is dropping as first digit and other decimals are preceding so this will be 3.325 or 3.32.This is also applicable when there is no digit after 5 or digits are following zeros.
The last rule works approximately partially. Half of the time its rounded up and in another half its rounded down.
## Rounding off Numbers
To round off whole numbers: Find the place value you want (the "rounding digit") and look to the digit just to the right of it. If that digit is less than 5, do not change the "rounding digit" but change all digits to the right of the "rounding digit" to zero.
## Rounding Numbers to the Nearest 10
There are some basic rules to find the nearest 10 rounding number, The rules are as follows;
Rule 1: First find the given round number and the tenth decimal's place of the number. The tenth decimal place is after one place in right side. Let digit after tenth decimal places is less than five, delete all digits after tenth decimal place.
Rule 2: First find the given round number and the tenth decimal's place of the number. The tenth decimal place is after one place in right side. Let digit after tenth decimal places is equal or grater than five, Add 1 in tenth digit to round the hundredth decimal place.
Example:
Let a number is 15.367721 round the number in hundredth decimal places.
Solution:
Our ‘cut-off point’ comes after the ‘6’ 15.3 | 67721
This number is greater than 15.3, but lower than 15.4. Look at which is it closer to?
See the 2nd number from the decimal point, that is the number after that ‘cut-off point’.
If the number is 0,1,2,3, or 4 round DOWN to 15.3
If the number is 5,6,7,8, or 9 round UP to the next number 15.4.
Correct answer is 15.4 to 1 decimal places
## Rounding Numbers to the Nearest 100
There are some basic rules to find the nearest 100 rounding number, The rules are as follows;
Rule 1: First find the given round number and the tenth and hundredth decimal's place of the number. The hundredth decimal place is after two place in right side. Let digit after hundredth decimal places is less than five, delete all digits after tenth decimal place.
Rule 2: First find the given round number and the tenth and hundredth decimal's place of the number. The hundredth decimal place is after two place in right side. Let digit after hundredth decimal places is equal or grater than five, Add 1 in hundredth digit to round the thousand decimal place. Repeat the same for tenth decimal place.
## Rounding Numbers Worksheets
Some of the work problems are given below with solution;
Problem 1:
Let a number is 19.367721; round decimals to the nearest hundredth .
Solution:
Given 19.367721
Cut off
19.36 | 7721
19.36<19.367721<19.37
Here the cut-off point is 3rd number after the decimal point
Verify if, 0,1,2,3, or 4 round DOWN to 19.36
Verify if, 5,6,7,8, or 9 round UP to the next number 19.37
So the answer = 19.37 to 2 decimal places
Problem 2:
Round the number 546 to nearest 10
Solution:
The given number is 546
The number is ten places is 4 and the number one place is 6
Since number one place is 6 we make it zero
So the number 546 grounded to nearest 10 becomes 550.
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This post follows on from this post where the logic for the below is discussed. I am not going to define here what easy means!
Here is the strategy/guiding principle:
Fundamental Principle of Solving ‘Easy’ Equations
Identify what is difficult or troublesome about the equation and get rid of it. As long as you do the same thing to both numbers (the “Lhs” and the “Rhs”), the equation will be replaced by a simpler equation with the same solution.
## Inverses
For the moment think of a function as an object $f$ that associates to an input number $x$ an output number $f(x)$:
$x\mapsto f(x)$,
which could be read as $x$ is sent to $f(x)$.
Examples of functions include:
• the add-three function, e.g. $4\mapsto 7$.
• the subtract-seven function, e.g. $26\mapsto 19$.
• the multiply-by-eight function, e.g. $7\mapsto 56$.
• the divide-by-four function, e.g. $44\mapsto 11$.
• the squaring function, e.g. $5\mapsto 25$.
• the square root function, e.g. $100\mapsto 10$.
• a power function, e.g. $2\mapsto 2^4=16$.
• a root function, e.g. $27\mapsto \sqrt[3]{27}=3$.
Some functions have inverses $f^{-1}$ that undo the action of $f$.
$x\mapsto f(x)\mapsto f^{-1}(f(x))=x$.
Call $f(x)$ by $y$:
$y=f(x)\mapsto f^{-1}(f(x))=x\mapsto f(x)=y$,
that is if $f^{-1}$ is an inverse of $f$ then, usually, $f$ is an inverse of $f^{-1}$.
For example, we have inverses of the above:
• the inverse of the plus three function is the subtract three function, e.g. $4\mapsto 7\mapsto 4$.
• the inverse of the subtract seven function is the add seven function, e.g $26\mapsto 19\mapsto 26$.
• the inverse of the times eight function function is the divide-by-eight function, e.g. $7\mapsto 56\mapsto 7$
• the inverse of the divide-by-four is the times four function, e.g. $44\mapsto 11\mapsto 44$.
So… this “get rid of it” mentioned in the above principle is “apply the appropriate inverse”. In the below when I write $\underset{f^{-1}}{\Rightarrow}$ it means I applied the inverse function $f^{-1}$ to both of the numbers “Lhs” and “Rhs”.
## Examples
Solve each of the following for $x$. As per the previous post the first step is the implicit assumption that $x$ makes the equation true.
• $x-3=-10$.
What is troublesome here is we want to conclude that $x=$ some number but here we have the subtract 3 function:
$x\mapsto x-3$,
messing things up. Apply the inverse function:
$\displaystyle \underset{+3}{\Rightarrow}x=-7$.
• $\frac{x}{5}=1$.
What is troublesome here is we want to conclude that $x=$ some number but here we have the divide-by-five function:
$\displaystyle x\mapsto \frac{x}{5}$.
Apply the inverse function:
$\displaystyle \underset{\times 5}{\Rightarrow}x=5$.
• $\sqrt{x}=2$.
Going a bit faster now. The square root is in the way: apply the inverse:
$\displaystyle \underset{x^2}{\Rightarrow}x=4$.
• $-3x=12$.
The multiply-by-(-3) is in the way: apply the inverse:
$\displaystyle \underset{\div(-3)}{\Rightarrow} x=-4$.
It might us take a number of implications before we can conclude the value of $x$. As long as you follow the principle of doing the same thing to both numbers as explained in the previous post, there can be a number of ways to get to the correct conclusion about the value of $x$. For example, consider the equation:
• $2x+3=11$.
Solution 1: The plus three is in the way: apply the inverse:
$\displaystyle \underset{-3}{\Rightarrow}2x=8$.
The multiply-by-two is in the way: apply the inverse:
$\displaystyle \underset{\div 2}{\Rightarrow}x=4$.
Solution 2: The multiply-by-two is in the way: apply the inverse:
$\displaystyle \underset{\div 2}{\Rightarrow}x+\frac{3}{2}=\frac{11}{2}$.
Here we had to use the definition of division, multiplying out and commutativity to properly divide a sum:
$\displaystyle(a+b)\div c=(a+b)\times\frac{1}{c}=\frac{1}{c}(a+b)$
$\displaystyle \frac{1}{c}\times a+\frac{1}{c}\times b$
$=a\div c+b\div c$.
The plus $3/2$ is in the way: apply the inverse:
$\displaystyle \underset{-3/2}{\Rightarrow}x=4$.
Experience will tell us it is easier to deal with the $+3$ before the times two.
• $6x-7=2x+13$
Here there is a big problem. You want to conclude that $x$ must equal some number but here $x$ appears in both numbers. This is bad. The $2x$ shouldn’t be there: apply the inverse. The subtract seven is also in the way: apply that inverse too:
$\displaystyle \underset{-2x+7}{\Rightarrow} 4x=20$.
Can you finish this off?
• $\displaystyle \frac{2x}{3}+\frac{x}{4}=\frac{11}{6}$.
Here the difficulty is that we have fractions. We don’t really want the division by three, four and six: if we apply the inverses of the first two of these — multiply-by-three and multiply-by-four — we end up multiplying by twelve. This will also deal with the six we don’t want:
$\displaystyle \underset{\times 12}{\Rightarrow} 12\left(\frac{2x}{3}+\frac{x}{4}\right)=12\frac{11}{6}$
$\displaystyle \Rightarrow 4(2x)+3(x)=2(11)$
$\Rightarrow 8x+3x=22$,
where we used ‘multiplying out’. Can you finish this off?
• $\displaystyle 4=\frac{5}{x}$.
Here the problem is we want $x=$ some number but instead we have division-by-$x$: apply the inverse:
$\underset{\times x}{\Rightarrow}4x=5.$
Can you finish this off?
In the rest of this mini-series of posts we will talk about some subtleties to do with inverses and also solving equations with a number of variables.
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# ANGLE BETWEEN TWO VECTORS USING CROSS PRODUCT
## About "Angle Between Two Vectors Using Cross Product"
Angle Between Two Vectors Using Cross Product :
Here we are going to see how to find angle between two vectors using cross product.
## Angle between two vectors using cross product - Examples
Question 1 :
Find the angle between the vectors 2i vector + j vector − k vector and i vector+ 2j vector + k vector using vector product.
Solution :
Angle between two vectors using vector product
θ = sin-1 (|a vector x b vector|/|a vector||b vector|)
= i[1+2]-j[2+1]+k[4-1]
a vector x b vector= 3i vector + 3j vector + 3k vector
|a vector x b vector| = √32 + 32 + 32 = 3√3
|a vector| = √22 + 1+ 1= √6
|b vector| = √12 + 2+ 12 = √6
θ = sin-1 (3√3/√6√6)
θ = sin-1 (√3/2)
θ = π/3
Question 2 :
Let a vector, b vector, c vector be unit vectors such that a ⋅ b = a ⋅ c = 0 and the angle between b vector and c vector is π/3. Prove that a vector = ± (2/3) (b × c)
Solution :
From given information, we have a ⋅ b = a ⋅ c = 0
From this we may decide that a vector is perpendicular to b vector and a vector is perpendicular to c vector.
a vector is perpendicular to both b vector and c vector.So, a vector is proportional to (b x c) vector
a vector = ± λ (b vector x c vector)
|a vector| = ± λ |(b vector x c vector)| ----(1)
|a vector| = ± λ |b||c| sin θ
1 = ± λ sin π/3
λ = 2/
By applying the value λ = 2/3 in (1), we get
a vector = ± (2/3) (b × c)
Hence it is proved.
Question 3 :
For any vector a vector prove that
|a vector × i vector |2+|a vector × j vector|2+|a vector × k vector|2= 2 |a vector|2 .
Solution :
|a vector × i vector |2+|a vector × j vector|2+|a vector × k vector|2= 2 |a vector|2 .
Let a vector = xi vector + yj vector + zk vector, then
a vector x i vector =
= i[0-0] -j[0-z] + k[0-y]
a vector x i vector = zj vector - yk vector
|a vector x i vector| = √z2 + y2
|a vector x i vector|2 = (√z2 + y2)2
|a vector x i vector|2 = z2 + y -----(1)
||| ly
a vector x j vector =
= i[0-z] -j[0-0] + k[x-0]
a vector x j vector = -zi vector + xk vector
|a vector x j vector| = √z2 + x2
|a vector x j vector|2 = (√z2 + x2)2
|a vector x j vector|2 = z2 + x2 -----(2)
||| ly
a vector x k vector = i[y-0] -j[x-0] + k[0-0]
a vector x k vector = yi vector - xj vector
|a vector x k vector| = √y2 + x2
|a vector x k vector|2 = (√y2 + x2)2
|a vector x k vector|2 = y2 + x2 -----(3)
(1) + (2) + (3) ==>
|a vector × i vector |2+|a vector × j vector|2+|a vector × k vector|= 2x2 + 2y2 + 2z2
= 2(x2 + y2 + z2)
= 2|z|2
Hence proved.
After having gone through the stuff given above, we hope that the students would have understood, "Angle Between Two Vectors Using Cross Product"
Apart from the stuff given in "Angle Between Two Vectors Using Cross Product" if you need any other stuff in math, please use our google custom search here.
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# How do you find the product (4y^2-3)(4y^2+7y+2)?
Mar 12, 2017
See the entire solution process below:
#### Explanation:
To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.
$\left(\textcolor{red}{4 {y}^{2}} - \textcolor{red}{3}\right) \left(\textcolor{b l u e}{4 {y}^{2}} + \textcolor{b l u e}{7 y} + \textcolor{b l u e}{2}\right)$ becomes:
$\left(\textcolor{red}{4 {y}^{2}} \times \textcolor{b l u e}{4 {y}^{2}}\right) + \left(\textcolor{red}{4 {y}^{2}} \times \textcolor{b l u e}{7 y}\right) + \left(\textcolor{red}{4 {y}^{2}} \times \textcolor{b l u e}{2}\right) - \left(\textcolor{red}{3} \times \textcolor{b l u e}{4 {y}^{2}}\right) - \left(\textcolor{red}{3} \times \textcolor{b l u e}{7 y}\right) - \left(\textcolor{red}{3} \times \textcolor{b l u e}{2}\right)$
$16 {y}^{4} + 28 {y}^{3} + 8 {y}^{2} - 12 {y}^{2} - 21 y - 6$
We can now combine like terms:
$16 {y}^{4} + 28 {y}^{3} + \left(8 - 12\right) {y}^{2} - 21 y - 6$
$16 {y}^{4} + 28 {y}^{3} - 4 {y}^{2} - 21 y - 6$
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# Orthogonality in Statistics
Contents
## Orthogonality in mathematics
The word Orthogonality originates from a combination of two words in ancient Greek: orthos (upright), and gonia (angle). It has a geometrical meaning. It means two lines create a 90 degrees angle between them. So one line is perpendicular to the other line. Like so:
Even though Orthogonality is a geometrical term, it appears very often in statistics. You probably know that in a statistical context orthogonality means uncorrelated, or linearly independent. But why?
Why use a geometrical term to describe a statistical relation between random variables? By extension, why does the word angle appears in the incredibly common regression method least-angle regression (LARS)? Enough losing sleep over it (as you undoubtedly do), an extensive answer below.
## Orthogonality
Assume you have those two vectors x and y:
You are searching for the orthogonal projection of on . Why would you that? For now let’s say you are strange like that. You are looking to “cast” the vector such that it sits on top of the vector . Because it has to sit on top of the vector , there must be a scalar (call it ) which we could multiply the vector with, and get the vector we are looking for. Let’s colorfully appoint the name to that vector we are searching for. Another way to look at it: instead of searching for a scalar to multiply with, we can look for couple of numbers to add to such that we get the required vector . We can call those two numbers .
There is a simple formula to find that new vector:
Both and variables are vectors, so it is a vector multiplication (not element-wise).
Notice that we also found the scalar :
Because and are vectors we use the %*% operator for vector multiplication in the code, as follows:
And the scalar is
Now that we have our we can back out the vector :
Those are the numbers which must be added to the vector in order to “cast” it on the vector .
Let’s make sure that the angle between and is 90 degrees as it supposed to be, which would mean that and are indeed orthogonal. Of course and are also orthogonal because the angle between and is zero. We can use the cosine similarity function to check that.
The cosine of an angle between two vectors and say, is:
We know that the cosine of an angle which is 0 equals 1, and the cosine of an angle which is 90 degrees equals 0. So we can check our results are indeed correct.
Before we move on to discuss the relation of all this only semi-fascinating geometry with statistics, this is what we have so far:
## Orthogonality and Linear Regression
What is special about being, we now know, (2.74, 4.92)?
That point (2.74, 4.92), from all those (infinite number of) points sitting on the line between (0,0) and (5,9), is the closest point possible to the point (8,2). Move a bit further on the line towards (5,9) or a bit back towards the origin, and you are moving away from (8,2); Euclidean distance speaking.
In a linear regression context, we try to create a linear combination from our explanatory variables such that they best approximate the target variable . Basically we want to create the closest possible fit for using those variables. That scalar can then be seen as the coefficient which makes the linear combination of the ‘s to be as close as possible to the . In that sense, the coefficients of a linear regression
are trying to do the same thing. So linear regression is also, and simply, an orthogonal projection:
## Relation with independence
Recall the definition of covariance:
This is the theoretical definition, the empirical counterpart is
So if the variables are centered; then the nominator of the cosine function above and the nominator of the covariance are the same. So covariance = 0 means that the angle between the two vectors is 90 degrees. This means that two variables which are orthogonal are (linearly) independent. The reverse also holds of course; if they are independent they are orthogonal.
## Relation with Least Angle Regression (LARS)
Like in forward selection we start with all coefficients set equal to zero, and add variables one by one. We choose each “next” variable in a smart way. In forward selection, the next variable to enter is the one which has the highest correlation with the target . It is the same as checking all the pairwise angles between each variable and the target and add that variable with the least angle (which would imply highest correlation). In that sense, forward selection can also be called LARS. However, LARS is more elegant. It is more elegant in that we progress with the chosen variable, but don’t add it in full. Instead of adding the new chosen variable we only add a [portion]*[variable]. The [portion] is small, and we keep on adding small portions of the same variable until another variable “catches up”, meaning it now has the same correlation with the residuals as the previous variable.
You can see that there is a lot of geometry popping up in statistics. To be honest, you can get by without any geometrical intuition, most do in fact. But I think it is nice to know, especially if you like statistics.
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# Modular arithmetic for arbitrary number
The problem describes as:
When the even integer $n$ is divided by $7$，the remainder is $3$. What's is The remainder when $n$ is divided by $14$.
My simple solution is:
$n=7x+3$ where $x$ is odd, so, we can define $x = 2m+1$, then $n = 7(2m+1) + 3 = 14m + 7 +3= 14m + 10$. So reminder is: $10$
As I am learning the mod system and was trying to solve this problem with modulo arithmetic. But got stuck with "how to think this prob in modulo system" ie. if $$n \equiv 3 \pmod{7}$$ then $$n \equiv \ ? \pmod {14}$$
Any help regarding solving steps and learning reference would be appreciated.
• "how to think this prob in modulo system" well it depends of the meaning of those words, but IMHO I think you did it nicely that way indeed. You converted a problem in $\pmod{7}$ to a problem in $\pmod{14}$. Maybe is just a question of formatting your answer as a "modulo system problem": $x = 2m+1$, then $n = 7(2m+1) + 3 \pmod{14} \equiv 14m + 7 +3 \pmod{14} \equiv 14m + 10 \pmod {14} \equiv 14m \pmod{14} + 10 \pmod {14} \equiv 10 \pmod{14}$. – iadvd Apr 26 '17 at 5:22
• Edited the question a bit for better understandability of my points that is, I was thinking the format: if n mod 7 = 3 then n mod 14 = ?. That's how I was thinking. Thanks for replying! – neo-nant Apr 26 '17 at 5:46
• ok I have added a solution going backwards... I think that you meant that possibility. But the way you did it is easier and indeed my solution is based on your substitutions (but backwards, it is kind of tricky). – iadvd Apr 26 '17 at 7:03
As I said in the comments I think is fine the way you did it. I would suggest you to use "congruent to" instead of "equal to" as in my comment from the beginning, or add it at the end of the conversions you did as a last step.
Said that, maybe your question is more related with this point: how you could make a solution starting backwards? from $14m+r$.
$$n=14m+r$$
Then, let us suppose three possibilities:
1. $r \lt 7$
2. $r = 7$
3. $7 \lt r \lt 14$ so we can define $r=7+r'$, where $r' \lt 7$
• For the first case:
$$n=14m+r \pmod{7} \equiv 2\cdot 7m\pmod{7} + r \pmod{7} \equiv 0+r \equiv 3$$
Thus:
$$r=3$$
But that is not possible because it means that $n = 14m + 3$, but $14m+3$ is not even, and we know that $n$ is even, so that solution is not possible.
• For the second case:
$$n=14m+7 \pmod{7} \equiv 2\cdot 7\pmod{7} + 7 \pmod{7} \equiv 0 \equiv 3$$
So it is impossible for the residue to be $r=7$ because $14m+7 \not \equiv 3 \pmod {7}$, the residue does not comply the premise, being $\equiv {3} \pmod{7}$, it is indeed $\equiv {0} \pmod {7}$.
• For the third case:
$$n=14m+r \pmod{7} \equiv 2\cdot 7\pmod{7} + 7 + r' \pmod{7} \equiv 0+0+r' \equiv 3$$
Thus finally the unique remaining valid option is $7 \lt r = 10 \lt 14$:
$$r'=3, r=7+r'=7+3=10$$
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# 10 Most Common 7th Grade Georgia Milestones Assessment System Math Questions
Passing the 7th Grade Georgia Milestones Assessment System Math test and getting a high score on this test is an acquired skill that requires practice. It is best to remember that practice is one of the most important steps of the study. Therefore, to strengthen 7th Grade students’ math skills, we have seen the best tactics in preparing the best 7th Grade Georgia Milestones Assessment System Math practice questions. We have collected 10 common 7th Grade Georgia Milestones Assessment System Math practice questions in this article. We hope that by using these questions, students will achieve their goal of getting the best score on the 7th Grade Georgia Milestones Assessment System Math test.
Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.
## The Absolute Best Book to Ace 7th Grade Georgia Milestones Assessment System Math Test
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## 10 Sample 7th Grade Georgia Milestones Assessment System Math Practice Questions
1- What is the median of these numbers? $$2, 28, 28, 19, 67, 44, 35$$
A. 19
B. 28
C. 44
D. 35
2- Last week 24,000 fans attended a football match. This week three times as many bought tickets, but one-sixth of them canceled their tickets. How many are attending this week?
A. 48000
B. 54000
C. 60000
D. 72000
3- The following trapezoids are similar. What is the value of $$x$$?
A. 7
B. 8
C. 18
D. 45
4- If $$x=- 8$$, which equation is true?
A. $$x(2x-4)=120$$
B. $$8 (4-x)=96$$
C.$$2 (4x+6)=79$$
D. $$6x-2=-46$$
5- In a bag of small balls $$\frac{1}{3}$$ are black, $$\frac{1}{6}$$ are white, $$\frac{1}{4}$$ are red and the remaining 12 blue. How many balls are white?
A. 8
B. 12
C. 16
D. 24
6- A boat sails 40 miles south and then 30 miles east. How far is the boat from its start point?
A. 45
B. 50
C. 60
D. 70
7- Sophia purchased a sofa for $530.40. The sofa is regularly priced at$624. What was the percent discount Sophia received on the sofa?
A. $$12\%$$
B. $$15\%$$
C. $$20\%$$
D. $$25\%$$
8- The score of Emma was half as that of Ava and the score of Mia was twice that of Ava. If the score of Mia was 60, what is the score of Emma?
A. 12
B. 15
C. 20
D. 30
9- A bag contains 18 balls: two green, five black, eight blue, a brown, a red, and one white. If 17 balls are removed from the bag at random, what is the probability that a brown ball has been removed?
A. $$\frac{1}{9}$$
B. $$\frac{1}{6}$$
C. $$\frac{16}{17}$$
D. $$\frac{17}{18}$$
10- A rope weighs 600 grams per meter of length. What is the weight in kilograms of 12.2 meters of this rope? (1 kilograms = 1000 grams)
A. 0.0732
B. 0.732
C. 7.32
D. 7320
## Best 7th Grade Georgia Milestones Assessment System Math Prep Resource for 2022
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1- B
Write the numbers in order:
$$2, 19, 28, 28, 35, 44, 67$$
Since we have 7 numbers (7 is odd), then the median is the number in the middle, which is 28.
2- C
Three times 24,000 is 72,000. One-sixth of them canceled their tickets.
One sixth of 72,000 equals 12,000 $$(\frac{1}{6}) × 72000 = 12000$$.
60,000 $$72000 – 12000 = 60000$$ fans are attending this week
3- A
It’s needed to have a ratio to find value of $$x$$.
$$\frac{45}{40}=\frac{2x+4}{16}⇒ 40(2x+4)=45×16 ⇒ x=7$$
4- C
$$8 (4-(-8))=96$$
5- A
$$\frac{1}{3}x + \frac{1}{6}x + \frac{1}{4}x + 12= x$$
$$(\frac{1}{3} + \frac{1}{6} + \frac{1}{4}) x+ 12= x$$
$$(\frac{9}{12})x+ 12 = x$$
$$x = 48$$
In a bag of small balls $$\frac{1}{6}$$ are white then: $$\frac{48}{6} = 8$$
6- B
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: $$a^2 + b^2 = c^2$$
$$40^2 + 30^2 = c^2 ⇒ 1600 + 900 = c^2 ⇒ 2500 = c^2 ⇒ c = 50$$
7- B
The question is this: 530.40 is what percent of 624?
Use percent formula:
$$part = \frac{percent}{100}× whole$$
$$530.40= \frac{percent}{100}× 624 ⇒ 530.40 = \frac{percent ×624}{100}⇒53040 = percent ×624$$
$$⇒percent = \frac{53040}{624}= 85$$
530.40 is $$85 \%$$ of 624. Therefore, the discount is: $$100\% – 85\% = 15\%$$
8- B
If the score of Mia was 60, therefore the score of Ava is 30. Since, the score of Emma was half as that of Ava, therefore, the score of Emma is 15.
9- D
If 17 balls are removed from the bag at random, there will be one ball in the bag.
The probability of choosing a brown ball is 1 out of 18. Therefore, the probability of not choosing a brown ball is 17 out of 18 and the probability of having not a brown ball after removing 17 balls is the same.
10- C
The weight of 12.2 meters of this rope is: $$12.2 × 600 \space g = 7320 \space g$$
$$1\space kg = 1000 \space g$$ therefore,
$$7320 \space g ÷ 1000 = 7.32 \space kg$$
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## College Algebra 7th Edition
slope = $-\dfrac{4}{5}$ y-intercept = $2$ Refer to the image below for the graph.
RECALL: (1) The slope is equal to $\dfrac{rise}{run}$ and may be used to graph a line when one point on the line is known. (2) The slope-intercept form of a line's equation is $y=mx+b$ where $m$ = slope and $b$ = y-intercept. (3) The y-intercept is y-coordinate of the the point on the y-axis where the line passes through. Transform the given equation to slope-intercept form to obtain: $4x+5y=10 \\4x+5y-4x=-4x+10 \\5y=-4x+10 \\\dfrac{5y}{5} = \dfrac{-4x+10}{5} \\y = -\dfrac{4}{5}+2$ This equation has: slope (m) $=-\dfrac{4}{5}$ y-intercept (b) $=2$. To draw the graph of this line using the slope and y-intercept, perform the following steps: (1) Plot the y-intercept point$(0, 2)$. (2) From point $(0, 2)$, use the slope to locate/plot another point. The slope is $-\dfrac{4}{5}$, so down four units (the rise) and move to the right 5 units (the run) to end up at the point $(5, -2)$. (3) Connect the two points using a straight line to complete the graph. (refer to the attached image in the answer part above for the graph)
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# NCERT Solutions class 12 Maths Exercise 6.2 (Ex 6.2) Chapter 6 Application of Derivatives
## NCERT Solutions for Class 12 Maths Exercise 6.2 Chapter 6 Application of Derivatives – FREE PDF Download
Free PDF download of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 (Ex 6.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks.
# NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.2) Exercise 6.2
1. Show that the function given by is strictly increasing on R.
Ans. Given:
i.e., positive for all R
Therefore, is strictly increasing on R.
### 2. Show that the function given by is strictly increasing on R.
Ans. Given:
= > 0 i.e., positive for all R
Therefore, is strictly increasing on R.
### 3. Show that the function given by is (a) strictly increasing (b) strictly decreasing in (c) neither increasing nor decreasing in
Ans. Given:
(a) Since, > 0, i.e., positive in first quadrant, i.e., in
Therefore, is strictly increasing in
(b) Since, < 0, i.e., negative in second quadrant, i.e., in
Therefore, is strictly decreasing in
(c) Since > 0, i.e., positive in first quadrant, i.e., in and < 0, i.e., negative in second quadrant, i.e., in and .
Therefore, is neither increasing nor decreasing in
### 4. Find the intervals in which the function given by is (a) strictly increasing, (b) strictly decreasing.
Ans. Given:
……….(i)
Now
Therefore, we have two disjoint sub intervals and
(a) For interval taking (say), then from eq. (i), > 0.
Therefore, is strictly increasing in
(b) For interval taking (say), then from eq. (i), < 0.
Therefore, is strictly decreasing in
### 5. Find the intervals in which the function given by is (a) strictly increasing, (b) strictly decreasing.
Ans. (a) Given:
=
……….(i)
Now
or
or
Therefore, we have three disjoint sub-intervals and
For interval taking (say), from eq. (i),
> 0
Therefore, is strictly increasing in
For interval taking (say), from eq. (i),
< 0
Therefore, is strictly decreasing in
For interval taking (say), from eq. (i),
> 0
Therefore, is strictly increasing in
Hence, (a) is strictly increasing in and
(b) is strictly decreasing in
### 6. Find the intervals in which the following functions are strictly increasing or decreasing:
(a)
(b)
(c)
(d)
(e)
Ans. (a) Given:
……….(i)
Now
Therefore, we have two sub-intervals and
For interval taking (say), from eq. (i), < 0
Therefore, is strictly decreasing.
For interval taking (say), from eq. (i), > 0
Therefore, is strictly increasing.
(b) Given:
= ……….(i)
Now
Therefore, we have two sub-intervals and
For interval taking (say), from eq. (i),
> 0
Therefore, is strictly increasing.
For interval taking (say), from eq. (i),
< 0
Therefore, is strictly decreasing.
(c) Given:
……….(i)
Now = 0
or
Therefore, we have three disjoint intervals and
For interval , from eq. (i),
= < 0
Therefore, is strictly decreasing.
For interval , from eq. (i),
= > 0
Therefore, is strictly increasing.
For interval , from eq. (i),
= < 0
Therefore, is strictly decreasing.
(d) Given:
……….(i)
Now
Therefore, we have two disjoint intervals and
For interval , taking x=-6 (say)
from eq. (i), > 0
Therefore, is strictly increasing.
For interval , taking x= 0 (say)
from eq. (i), < 0
Therefore, is strictly decreasing.
(e) Given:
Here, factors and are non-negative for all x
Therefore, is strictly increasing if
And is strictly decreasing if
Hence, is strictly increasing in and is strictly decreasing in
### 7. Show that is an increasing function of throughout its domain.
Ans. Given:
= 11+x(4+2x2x)(2+x)211+x−(4+2x−2x)(2+x)2
……….(i)
Domain of the given function is given to be
Also and
From eq. (i), for all in domain and is an increasing function.
### 8. Find the value of for which is an increasing function.
Ans. Given:
[Applying Product Rule]
……….(i)
Therefore, we have four disjoint subintervals
For taking (say),
is decreasing.
For taking (say),
is increasing.
For taking (say),
is decreasing.
For taking (say),
is increasing.
thus is increasing in and
is decreasing in and
### 9. Prove that is an increasing function of in
Ans. Given:
Since and we have , therefore
for
Hence, is an increasing function of in
### 10. Prove that the logarithmic function is strictly increasing on
Ans. Given:
for all in
Therefore, is strictly increasing on
### 11. Prove that the function given by is neither strictly increasing nor strictly decreasing on
Ans. Given:
is strictly increasing if
i.e., increasing on the interval
is strictly decreasing if
i.e., decreasing on the interval
hence, is neither strictly increasing nor decreasing on the interval
### 12. Which of the following functions are strictly decreasing on
Ans. (A)
Since
thus for x in sinx is positive in first quadrent
So,
Therefore, is strictly decreasing on
(B)
Since
therefore
So,
Therefore, is strictly decreasing on
(C)
Since
thus two cases and
For
So,
Therefore, is strictly decreasing on
For
So ,
Therefore, is strictly increasing on
Hence, is neither strictly increasing not strictly decreasing on
(D)
> 0
Therefore, is strictly increasing on
### 13. On which of the following intervals is the function given by is strictly decreasing:
(A) (0, 1)
(B)
(C)
(D) None of these
Ans. Given:
(A) On (0, 1), therefore
And for
(0, 1 radian) = > 0
Therefore, is strictly increasing on (0, 1).
(B) For
= (1.5, 3.1) > 1 and hence > 100
For is in second quadrant and hence is negative and between and 0.
Therefore, is strictly increasing on .
(C) On = (0, 1.5) both terms of given function are positive.
Therefore, is strictly increasing on .
(D) Option (D) is the correct answer.
### 14. Find the least value of such that the function given by strictly increasing on (1, 2).
Ans.
Since is strictly increasing on (1, 2), therefore > 0 for all in (1, 2)
On (1, 2)
Minimum value of is and maximum value is
Since > 0 for all in (1, 2)
and
and
Therefore least value of is
### 15. Let I be any interval disjoint from Prove that the function given by is strictly increasing on I.
Ans. Given:
……….(i)
Here for every either or
for (say),
> 0
And for (say),
> 0
> 0 for all xIx∈I, hence is strictly increasing on II
### 16. Prove that the function given by is strictly increasing on and strictly decreasing on
Ans. Given:
On the interval i.e., in first quadrant,
> 0
Therefore, is strictly increasing on .
On the interval i.e., in second quadrant,
< 0
Therefore, is strictly decreasing on .
### 17. Prove that the function given by is strictly decreasing on and strictly increasing on
Ans. Given:
On the interval i.e., in first quadrant, is positive, thus < 0
Therefore, is strictly decreasing on .
On the interval i.e., in second quadrant, is negative thus > 0
Therefore, is strictly increasing on .
### 18. Prove that the function given by is increasing in R.
Ans. Given:
for all in R.
Therefore, is increasing on R.
### 19. The interval in which is increasing in:
(A)
(B)
(C)
(D) (0, 2)
Ans. Given:
In option (D), for all in the interval (0, 2).
Therefore, option (D) is correct.
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# If 40 is equal to x/30, then solve for x.
Title: Unraveling the Mystery: Solving Equations in Mathematics Education
Introduction:
Welcome to Warren Institute, where we explore the fascinating world of Mathematics education. In this article, we delve into the intriguing equation that involves the fraction x/30 and its equivalence to the number 40. Join us as we unravel the mystery behind this equation and discover the principles and techniques used to solve such mathematical puzzles. Through our in-depth analysis, we aim to shed light on the fundamental concepts of equations and empower educators and learners alike to tackle similar challenges with confidence. So, let's embark on this mathematical journey together!
## Simplifying the Fraction
When solving the equation "40 = x/30" in the context of Mathematics education, one important step is to simplify the fraction x/30. This involves reducing it to its simplest form by finding the greatest common divisor (GCD) of the numerator and denominator. By simplifying the fraction, we can gain a clearer understanding of its value and make further calculations easier.
## Finding the Value of x
In Mathematics education, the equation "40 = x/30" requires us to find the value of x. To do this, we can multiply both sides of the equation by 30, cancel out the denominators, and solve for x. By using algebraic techniques such as isolating the variable and performing necessary operations, we can determine the specific value of x that makes the equation true.
## Applying Proportional Reasoning
Proportional reasoning is a fundamental concept in Mathematics education. When dealing with the equation "40 = x/30," we can apply proportional reasoning to establish a relationship between the different quantities involved. By understanding that the ratio of x to 30 is equal to the ratio of 40 to 1, we can make connections and solve the equation using proportional relationships.
## Interpreting the Solution
Once we have found the value of x in the equation "40 = x/30," it is crucial to interpret the solution in the context of Mathematics education. We need to analyze what the value of x represents and how it relates to the problem being solved. By interpreting the solution, we can gain insight into the mathematical concepts and principles underlying the equation and its solution.
### How can I solve for the value of x if 40 is equal to the fraction x/30?
To solve for the value of x, we can set up an equation. Since 40 is equal to the fraction x/30, we can cross multiply to get 30 * 40 = x. Solving this equation, we find that x = 1200.
### What methods or strategies can I use to find the value of x in the equation 40 = x/30?
To find the value of x in the equation 40 = x/30, you can use the method of cross multiplication. Multiply both sides of the equation by 30 to get rid of the denominator: 30 * 40 = x. Simplify the equation to get 1200 = x. Therefore, x = 1200.
### Are there any specific steps or formulas that I should follow to determine the value of x when given the equation 40 = x/30?
To determine the value of x in the equation 40 = x/30, you can follow the steps of isolating the variable. Multiply both sides of the equation by 30 to get rid of the denominator: 30 * 40 = 30 * (x/30). Simplifying this gives 1200 = x. Therefore, the value of x is 1200.
### Can you explain the concept of fractions and how it relates to solving for x in the equation 40 = x/30 in mathematics education?
In mathematics education, fractions represent parts of a whole or a ratio between two quantities. In the equation 40 = x/30, we can solve for x by using the concept of fractions. We can multiply both sides of the equation by 30 to eliminate the denominator, resulting in 1200 = x. Therefore, x has a value of 1200.
### Is there a general approach or technique that can be applied to solve equations involving fractions, such as the equation 40 = x/30 in mathematics education?
Yes, there is a general approach or technique that can be applied to solve equations involving fractions in mathematics education. One common method is to eliminate the fractions by multiplying both sides of the equation by the least common multiple (LCM) of the denominators. In the given equation 40 = x/30, we can multiply both sides by 30 to get rid of the fraction. This simplifies the equation to 1200 = x. Therefore, x = 1200 is the solution.
In conclusion, understanding the concept of variables is crucial in Mathematics education. By carefully analyzing the given equation "40 = x/30", we can determine that x represents an unknown value. This equation highlights the importance of solving for x, which requires students to apply problem-solving strategies and mathematical operations. Through practice and guidance, students can enhance their critical thinking skills and develop a deeper understanding of algebraic concepts. Variables serve as powerful tools in problem-solving, allowing students to represent unknown quantities and solve equations efficiently. By mastering the concept of variables, students can navigate through complex mathematical problems with confidence and precision.
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# Evaluating Definite Integrals
## What is Integration in Maths?
• According to integration definition in Mathematics to find the whole, we generally add or sum up many parts to find the whole.
• We know that Integration is basically a reverse process of differentiation, which is defined as a process where we reduce the functions into smaller parts.
• To find the summation under a very large scale the process of integration is used.
• We can use calculators for the calculation of small addition problems which is a very easy task to do. We use integration methods to sum up many parts in problems where the limits reach infinity.
If $\frac{d}{dx}$ (F(x)) = f(x), then ∫f(x)dx = F(x) + c The function F(x) is called anti-derivative or integral or primitive of the given function f(x) and c is known as the constant of integration or the arbitrary constant. The function f(x) is called the integrand and f(x) dx is known as the element of integration.
## Some Elementary Standard Integrals in Integration
$\int x^{n}dx$ $\frac{x^{n+1}}{n+1} + C$, where n $\neq -1$ $\int sin x dx$ – cos x + C $\int cos x dx$ sin x + C $\int sec^{2} x dx$ tan x + C $\int cosec^{2} x dx$ – cot x + C $\int {\text{secx tanx dx}}$ sec x + C $\int {\text{cosecx cotx dx}}$ – cosec x + C
### Different Types of Integrals in Mathematics
Till now we have learned what Integration is. There are two types of Integrations or integrals in Mathematics
• Definite Integral
• Indefinite Integral
### What is Definite Integral?
• A Definite Integral has start and end values.
• In simpler words there is an interval [a, b].
• A definite integral is an integral that contains both the upper and the lower limits.
• Definite Integral is also known as Riemann Integral.
• Representation of a Definite Integral –
$\int_{a}^{b}f(x)dx$
• The variables a and b (called limits, bounds or boundaries) are put at the bottom and top of the S, like this:
(image will be uploaded soon) (image will be uploaded soon) Definite Integral (Limit goes from a to b where a is the lower limit and b is the upper limit) Indefinite Integral (Here there are no specific values, means there is no limit)
In this article we are going to discuss what definite integral is, properties of definite integrals which will help you solve definite integral problems and how to evaluate definite integral examples.
## The Quantity
$\int_{a}^{b}f(x)dx$ = F(b) – F(a)
It is known as the definite integral of f(x) from limit a to b. In the above given formula, F(a) is known to be the lower limit value of the integral and F(b) is known to be the upper limit value of any integral.
There is also a little bit of terminology that we can get out of the way. The number a at the bottom of the integral sign is called the lower limit and the number b at the top of the integral sign is called the upper limit. Although variable a and variable b were given as an interval the lower limit does not always need to be smaller than the upper limit that is b here. The variables a and b are often known as the interval of integration. Let’s understand the concept in a better way by solving definite integral problems.
1. ### Area Above – Area Below
The integral adds the area above the axis but the integral subtracts the area below, to obtain a net value.
$\int_{a}^{b}f(x)dx$ is equal to Area that lies above the x axis – Area that lies below the y axis.
The integral of the functions f and g (f+g) generally equals the integral of function f plus the integral of the function g:
$\int_{a}^{b}f(x) + g(x) dx$ is equal to $\int_{a}^{b}f(x)dx$ + $\int_{a}^{b}g(x)dx$
1. ### Reversing the Interval
When we reverse the direction of the interval it gives the negative of the original direction.
$\int_{a}^{b}f(x)dx$ is equal to – $\int_{b}^{a}f(x)dx$
1. ### Interval of Zero Length
When the interval of the integral starts and ends at the same place, in simpler words if the limit is same then the result is zero:
$\int_{a}^{a}f(x)dx$ = 0
$\int_{a}^{b}f(x)dx$ = $\int_{a}^{c}f(x)dx$ + $\int_{b}^{c}f(x)dx$
These properties will help you solve definite integral problems and how to evaluate definite integral examples.
Let’s evaluate definite integral examples and solve definite integral problems.
### Questions to be Solved
Question 1) Solve the following definite integral.
$\int_{-2}^{3} x^{3} dx$
Solution)$\int_{-2}^{3} x^{3} dx$
$\int_{-2}^{3} x^{3} dx = [\frac{x^{4}}{4}]_{-2}^{3}$
= $\frac{81}{4} – \frac{16}{4}$
= $\frac{65}{4}$
= 16.25
Question 2) Evaluate the integral given below.
$\int_{0}^{\frac{\pi}{2}} cosx dx$
Solution) Given, $\int_{0}^{\frac{\pi}{2}} cosx dx$
= $\int_{0}^{\frac{\pi}{2}} cosx dx$
On evaluating the given question,
= $sin(\frac{\pi}{2}) – sin(0)$
We know that the value of sin 0 is equal to zero and the value of sin ($\frac{\pi}{2}$) is equal to 1.
Therefore , putting the values ,
= 1- 0
= 1
Question 1) What does a Definite Integral Actually Calculate?
Answer) Evaluating a definite integral generally means to find the area enclosed by the graph of the function and the x-axis, over the given interval that is [a,b].
In the graph given below, the shaded area denotes the integral of f(x) on the given interval [a,b]. Now ,finding this area means that we are taking the integral of the function f(x), plugging the upper limit that is b into the result, and then subtracting from that whatever you get when you plug in the lower limit that is a.
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# Greatest Common Factor|Definition & Meaning
## Definition
The greatest common factor (or GCF) of two numbers is the largest factor shared by both numbers. Dividing the two numbers by the GCF always produces an integer for each one. Both share a few factors once all the numbers’ components have been identified.
The largest common factor is the one with the largest number among the common factors. HCF (Highest Common Factor) is another name for the GCF.
A factor is a quantity that produces the required quantities when multiplied by several other quantities. Factors are another name for the total that results. The numbers may often be factored into a variety of combinations.
Figure 1 below shows the GCF of 18.
Figure 1 – Representation of the greatest common factor of 18.
## How To Find the Greatest Common Factor?
We must first list each pair of digits’ prime factors to obtain the GCF for that pair. Multiplying the common features between the two numbers yields GCF. If there are no additional shared prime factors, the largest common denominator is 1. Numerous techniques may be used to figure out the GCF of two digits.
One of the simplest ways to determine the greatest common factor of two or more numbers is to use the prime factoring method.
The determination of the GCF of the given number set can be straightforward. But a few steps need to be followed to get the proper GCF. Before you can figure out the greatest component that any given number has in common, you must first figure out all of its components.
Figure 2 shows the GCF of 21.
Figure 2 – Representation of GCF of 21.
The GCF of two numbers can be calculated in several ways. By applying the prime factorization method, you can quickly get the GCF of two or more integers.
Finding the provided numbers’ highest common factor can be done through various methods. No of the approach, the GCF of the numbers will always give the same result. The GCF of two numbers can be calculated in one of three ways:
• GCF, through listing factors
• GCF by prime factorization
• GCF division approach
### Listing Factors Method for Greatest Common Factor
This approach involves listing each number’s elements and then identifying their shared factors. The most significant common factor among all the factors is then identified.
### Prime Factorization With Greatest Common Factor
We employ the below techniques to determine the GCF of integers using the prime factorization method. Let’s use the example below to understand this procedure.
• Step 1 is to identify the numbers that have common prime factors.
• Step 2: To find the GCF of those integers, multiply these frequent prime factors.
### Method of Greatest Common Factor by Division
The division technique may be used to get the GCF of two integers. Let’s break this down using the steps below and the example.
• Step 1: In this process, divide the bigger by, the smaller number, then determine the remaining.
• Step 2: Next, we split the remaining amount from the previous step into a new number, now the new dividend. We then divide again using long division.
• Step 3: We perform long division until the remainder equals 0. It should be remembered that the GCF of those two values will serve as the final divisor.
Figure 3 below shows GCF of 162 by division method.
Figure 3 – Representation of GCF by division.
## Multiple Numbers of a Greatest Common Factor
We may use either the “listing factors method” or the “prime factorization technique” to determine the GCF of several integers. However, there is little difference when dealing with several integers when utilizing the division approach.
## Greatest Common Factor of Three Numbers
We follow the steps below to determine the three-digit GCF. Let’s use the steps and example below to understand this.
• Step 1: Determine the GCFs of the two largest and smallest numbers provided.
• Step 2: After determining the GCF of the first two digits in Step 1, find the GCF of the third number.
• Step 3: This displays the three digits GCF.
## Greatest Common Factor of Prime Numbers
As far as we know, a prime quantity only contains two elements: 1 and the absolute amount. Let’s examine the prime elements 11 & 13, whose GCF may be determined by listing their constituent parts.
The components of 11 are 1, 11, & 13, respectively, are 1 and 13. The only thing that 11 & 13 have in common is 1. Consequently, the prime value’s GCF is always 1.
## Qualities of Greatest Common Factors
The GCF of x & y is traditionally understood to be the GCF between the digits a and b. Let’s look at some of the essential GCF attributes:
• Each number in a GCF greater than two is divided equally and without surplus.
• The GCF of one or more numbers is the component of each number.
• In most cases, the GCF of two integers or more is less than or equal to every one of the integers.
• The GCF is always 1, even though there are two or more prime factors.
## Examples of Greatest Common Factors
Here are a few examples of the greatest common factors that will further help clarify the concept.
### Example 1
A college student needs to find the GCF of 6, 72, and 120 using the method of listing the factors.
### Solution
The numbers 6, 72, and 120 are given. We will find the parts of each of these integers, and then we will see the GCF.
The number 6 is the only link between the numbers 72 and 120. Thus, the GCF for 6, 72, and 120 is 6.
### Example 2
While working on an assignment, a high school must find the GCF of the prime numbers 17 & 19. Using the methods above, find the GCF of the prime numbers.
### Solution
First, let’s list the components of 17 = 1, then we can determine the GCF of 17 & 19. Let’s now identify the components that make 19 equal 1. We can observe that between 17 and 19, there is just one component in common. The GCF of 17 & 19 is thus 1.
All Images are made using GeoGebra.
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# Differential Equations and Linear Algebra, 2.7: Laplace Transform: First Order Equation
From the series: Differential Equations and Linear Algebra
OK. This is the start of Laplace transforms. And that's going to take more than one short video. But I'll devote this video to first order equations, where the steps are easy and pretty quick. Then will come second order equations. So Laplace transforms starting now.
So let me tell you what-- I use a capital letter for the Laplace transform of little f, a function of t. The transform is capital F, a function of s. And you'll see where s comes in. Or if it's the solution I'm looking at, y of t, its transform is naturally called capital Y of s. So that's what we want-- we want to find y, and we know f. OK.
So can I do an example? Well, first tell you what the Laplace transform is. Suppose the function is f of t. Here is the transform. I multiply by e to the minus st, and I integrate from 0 to infinity. 0 to infinity. Very important. The function doesn't start until t equals 0, but it goes on to t equal infinity.
I integrate, and when I integrate, t disappears but s is still there. So I have a function of s. Well, I have to do an example. So to find the Laplace transform is to do an integration. And you won't be surprised that the good functions we know are the ones where we can do the integration and discover the transform, and make a little table of nice transforms.
And the number one function we know is the exponential. So can I find-- for that function, I'll compute its transform. So what do I have to do? I have to integrate from 0 to infinity-- you might say 0 to infinity is hard, but it's actually the best-- of my function, which is e to the at. So that's my function times e to the minus st dt.
OK. I can do that integral, because those combine into e to the a minus st. I can put those together into e to the a minus st. I integrate so I get e to the a minus st divided by a minus s. That's the integral of that. Because what I have in here is just that. To integrate the exponential, I just divide by the exponent there. And I have just substitute t equal infinity and t equal 0. So t equal infinity, starting at 0 to infinity. OK.
Infinity is the nice one. It's the easy one. I will look only at s's that are bigger than a. s larger than a means that this exponential is decreasing to 0. It gets to 0 at t equal infinity. So at t equal infinity, that upper limit of the integral ends up with a 0. So I just have to subtract the lower limit. And look how nice. Now I put in t equal 0. Well, then that becomes 1.
And it's a lower limit, so it comes with a minus sign. So it's just the 1 over, the minus sign will flip that s minus a. The most important Laplace transform in the world. Remember, the function was in to the at. The transform is a function of s. The original function depended on t and a parameter a. The result depends on s and a parameter a.
And an engineer would say, here we have the exponent. The growth rate is a. And over in the transform-- so this is the transform, remember. This is the transform f of x. In the transform, I see blow up-- a pole, that's called a pole-- at s equal a. 1/0 is a pole.
And I'm not surprised. So the answer is blowing up at s equal a. Well, of course. If s equals a, then this is the integral of 1 from 0 to infinity, and it's infinite. So I'm not surprised to see the pole showing up. The blow up showing up exactly at the exponent a. But this is a nice transform. OK.
I need to do one other-- oh, no. I could already solve the equation. So let me start with the equation dy dt minus ay equal 0. Oh, well, I can take the Laplace transform of 0 is 0, safe enough. The Laplace transform of y is capital Y. But what's the transform of this? Oh, I have to do one more transform for you.
I'm hoping that the transform of the derivative, dy dt, connects to the transform of y. So the transform of this guy is the integral from 0 to infinity of that function, whatever it is, times e to the minus st dt. This is the transform. So this Laplace transform.
Now what can I do with that integral? This is a step that goes back to the beginning of calculus. But it's easy to forget. When you see a derivative there inside that integral, you think, I could integrate by parts. I could integrate that term and take the derivative of that term. That's what integration by parts does. It moves the derivative away from that and onto that where it's no problem.
And do you remember that a minus sign comes in when I do this? So I have the integral from 0 to infinity of-- now the derivative is coming off of that, so that's just y of t. And the derivative is going onto that, so that's minus se to the minus st dt. Good. And then do you remember in integration by parts, there's also another term that comes from y times e to the minus st? This is ye to the minus st at 0 and infinity. OK.
I've integrated by parts. A very useful, powerful thing, not just a trick. OK. Now, can I recognize some of this? That is minus minus, no problem. I bring out-- that s is a constant. Bring it out, s. Now, what do I have left when I bring out that s? I have the integral of ye to the minus st dt. That is exactly the Laplace transform of y. It's exactly capital Y.
Put the equal sign here. I'll make that 0 a little smaller, get it out of the way. OK. sY of s. So that whole term has a nice form. When you take the derivative of a function, you multiply its Laplace transform by s. That's the rule. Take the derivative of the function, multiply the Laplace transform by s. If we have two derivatives, we'll multiply by s twice. Easy.
That's why the Laplace transform works. But now, here is a final term. y at infinity-- well, and e to the minus st at t equal infinity, 0. Forget it. So I just have to subtract off y at 0 times e to the minus st at 0, which is 1. e to the 0 is 1. So do you see that the initial condition comes into the transform? It's like, great.
We have the transform of Y. Now, all this is the transform. This is the transform of dy dt that we found. Now, why did I want that? Because I plan to take the transform of every term in my equation.
So like there are two steps to using the Laplace transform. One is to compute some transforms like this one, and some rules like this one. That's the preparation step. That comes from just looking at these integrals.
And then to use them, I'm going to take the Laplace transform of every term. So I have an equation. I take the Laplace transform of every term. I've got another equation. So the Laplace transform of this is sY of s minus y of 0. That was a Laplace transform of this part.
Now the Laplace transform of this is minus a, a constant, Y of x. And the Laplace transform of 0 is 0. Do you realize what we've done? I've taken a differential equation and I've produced an algebra equation. That's the point of the Laplace transform, to turn differential equations-- derivatives turn into multiplications, algebra.
So all the terms turn into that one. And now comes-- so that's big step one. Transform every term. Get an algebra problem for each s. We've changed from t, time in the differential equation, to s in Laplace transform.
Now solve this. So how am I going to solve that? I'm going to put y of 0 on to the right-hand side. And then I have Y of s times s minus a. So I will divide by s minus a. And that gives me Y of s. So that was easy to do. The algebra problem was easy to solve. The differential equation more serious.
OK. The algebra problem is easy. Are we finished? Got the answer, but we're in the s variable, the s domain. I've got to get back to-- so now this is going to be an inverse Laplace transform. That's the inverse transform. To give me back y of t equals what?
How am I going to do the inverse transform? So now I have the transform of the answer, and I want the answer. I have to invert that transform and get out of s and back to t. Well, y of 0 is a constant. Laplace transform is linear, no problem. So have y of 0 from that. And now I have 1 over s minus a.
So I'm asking myself, what is the function whose transform is 1 over s minus a? Then it's that function that I want to put in there. And what is the function whose transform is 1 over s minus a? It's the one we did. It's this one up here. 1 over s minus a came from the function e to the at.
So that 1 over s minus a, when I transform back, is the e to the at. And I'm golden. And that you recognize, of course, as the correct answer, correct solution to this differential equation. The initial value y of 0 takes off with exponential e to the at. No problem. OK.
Can I do one more example of a first order equation? Now I'm going to put it in an f of t. I'm going to put in a source term . So I'll do all the same stuff, but I'm going to have an f of t. And what shall I take for-- I'll take an exponential again, e to the ct. So that's my right-hand side.
Can I do the same idea, the central idea? Take my differential equation, transform every term. I've started with a time equation and I'm going to get an s equation. So again, dy dt minus ay, that transformed to-- what did that transform to? sY of s minus y of 0. Came from there. Minus aY of 0-- minus aY of s. Minus aY of s.
And on the right-hand side, I have the transform of e to the ct. We're getting good at this transform. 1 over s minus c, instead of a at c. OK. That's our equation transform. Now algebra. I just pull Y of s out of that. How am I going to pull Y of s out of this equation?
Well, I'll move y of 0 to the other side. And I'll divide by s minus a. Look at it. Y of s is-- OK. I have 1 over s minus c. And I have an s minus a that I'm dividing by. S minus a. And then I have the y of 0 over s minus a.
I've transformed the differential equation to an s equation. I've just done simple algebra to solve that equation. And I've got two terms. Two terms. You see that term? That's what I had before. That's what I had just there. The inverse transform was this. No problem. That's the null solution that's coming out of the initial value .
The new term that's coming from the e to the ct, coming from the force, is this one. And I have to do its inverse transform. I have to figure out what function has that transform. And you may say, that's completely new. But we can connect it to the one we know.
OK. So that will give me the same inverse transform, the growing exponential. But what does this one give? That's a key question. We have to be able to do-- invert, figure out what function has that transform? The function will involve a and c and t, the time. And s, the transform variable, will become t, the time period.
So that's the big question. What do I do with this? And notice, it has two poles. It blows up at s equal a, and it blows up at s equal c. And I have to figure out-- well, actually, by good luck, I want to separate those two poles. Because if I separate the two poles, I know what to do with a blow up at s equal a and a blow up at s equal c.
The problem is, right now I have both blow ups at once. So I'm going to separate that. And that's called partial fractions. So I will have to say more about partial fractions. Right now, let me just do it. That expression there, I'll take this guy away. Because it gives that term that we know.
It's this one. Is that one. It's this two poles thing that I want to separate those two poles. So this is algebra again. Partial fractions is just algebra. No calculus. No derivatives are in here. I just want to write that as 1 over s minus c. It turns out-- look. There's a way to remember the answer.
s minus c times c minus a and 1 over s minus a. And now a minus c. Do you see that that has only one pole at s equals c? This is just a number. This has one pole at s equals a. That's just a number. In fact, those numbers are the opposite.
So now, are we golden? I can take the inverse transform with just one pole. So now that gives me the solution y from-- so this is just a constant, 1 over c minus a. And what is the inverse transform of this? That's the simple pole at c. It came from a pure exponential, e to the ct. Right?
And now this guy, this one. OK. Well, this has a minus c, which is the opposite of c minus a. So if I put a minus sign, I can put them all over c minus a. Look at that. Look at this. C minus a is in both of these. Here it is with a plus sign. And that transform came from that function. Here it is with a minus sign. So I want a minus there.
And what function gave me that transform? e to the at, right? That's the one we know. The unforgettable transform of a simple exponential, e to the at. That is the particular solution. So the Laplace transform, we transformed the differential equation. We got an algebra equation. We solved that algebra equation, and then we had to go backwards to find what function had this transform y.
And to see that, clearly we had to use this partial fraction idea which separated these two poles into one pole there, when s is c, and another pole, when s is a. We've got two easy fractions. The easy fractions each gave me an exponential. And the final result was this one. And I don't know if you remember that. That is the correct solution to the first order linear constant coefficient equation, the simple equation there, when the right-hand side is e to the ct.
So our final solution then is the null solution with the initial value in it. And that function comes from the right-hand side, comes from the force, e to the ct. And so that's how Laplace transforms work. Take the Laplace transform of every term. Solve for y of s, and try your best to invert that transform. OK. More of that coming in the next lecture on Laplace transforms. Thank you.
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# polynomials
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Category: Education
## Presentation Description
Intro. to Polynomials
By: RAMPRIY (128 month(s) ago)
## Presentation Transcript
### z :
z MDVM (Parle) School Session : 2011-2012 Submitted By : - Aditya Rathore
### Slide 2:
Contents INTRODUCTION GEOMETRICAL MEANING OF ZEROS OF POLYNOMIAL RELATION BETWEEN ZEROS AND COEFFICIENTS OF A POLYNOMIAL DIVISION ALGORITHM FOR POLYNOMIALS SUMMARY QUESTION AND EXERCISE
### Slide 3:
Polynomials 3x2+2x=9 X3-3x2+9x-1=5 2x3-3x2+7x-5=0 3x5+5x4+x-2=3 9x2+3y=4
### Slide 4:
A polynomial is an expression of finite length constructed from variables and constants, using only the operations of addition, subtraction, multiplication, and non-negative, whole number exponents. Polynomials appear in a wide variety. Introduction :
### Slide 5:
Cont… Let x be a variable n, be a positive integer and as, a1 , a2 ,..….an be constants (real nos.) Then f(x)= anxn + an-1xn-1+…….a1x+x0 anxn , an-1xn-1 …….a1x and a0 are known as the terms of polynomial. an, an-1, an-2……a1 and a0 are their coefficients. For example : p(x) = 3x – 2 is a polynomial in variable x. q(x) = 3y 2 – 2y + 4 is a polynomial in variable y. f(u) = 1/2u 3 – 3u 2 + 2u – 4 is a polynomial in variable u. NOTE : 2x 2 – 3√x + 5, 1/x 2 – 2x +5 , 2x 3 – 3/x +4 are not polynomials .
### Slide 6:
Degree of polynomial The exponent of the highest degree term in a polynomial is known as its degree . For example : f(x) = 3x + ½ is a polynomial in the variable x of degree 1. g(y) = 2y2 – 3/2y + 7 is a polynomial in the variable y of degree 2. p(x) = 5x3 – 3x2 + x – 1/√2 is a polynomial in the variable x of degree 3. q(u) = 9u 5 – 2/3u 4 + u 2 – ½ is a polynomial in the variable u of degree 5.
### Slide 7:
Constant Polynomial: A polynomial of degree zero is called a constant polynomial. For example: f(x) = 7, g(x) = -3/2, h(x) = 2 are constant polynomials. The degree of constant polynomials is not defined
### Slide 8:
Linear polynomial: A polynomial of degree one is called a linear polynomial. For example: p(x) = 4x – 3, q(x) = 3y are linear polynomials. Any linear polynomial is in the form ax + b, where a, b are real nos. and a ≠ 0. It may be a monomial or a binomial. F(x) = 2x – 3 is binomial whereas g(x) = 7x is monomial.
### Slide 9:
Types of polynomial: QUADRATIC POLYNOMIALS A polynomial of degree two is called a quadratic polynomial. f(x) = √3x 2 – 4/3x + ½, q(w) = 2/3w 2 + 4 are quadratic polynomials with real coefficients. Any quadratic is always in the form f(x) = ax 2 + bx +c where a, b, c are real nos . and a ≠ 0. CUBIC POLYNOMIALS A polynomial of degree three is called a cubic polynomial. f(x) = 9/5x 3 – 2x 2 + 7/3x +1/5 is a cubic polynomial in variable x. Any cubic polynomial is always in the form f(x = ax3 + bx2 +cx + d where a, b, c, d are real nos.
### Slide 10:
Value’s & zero’s of Polynomial : If f(x) is a polynomial and y is any real no. then real no. obtained by replacing x by y in f(x) is called the value of f(x) at x = y and is denoted by f(x). Value of f(x) at x = 1 f(x) = 2x 2 – 3x – 2 f(1) = 2(1) 2 – 3 x 1 – 2 = 2 – 3 – 2 = –3 A real no. x is a zero of the polynomial f(x), is f(x) = 0. Finding a zero of the polynomial means solving polynomial equation f(x) = 0. Zero of the polynomial f(x) = x 2 + 7x + 12 f(x) = 0 x2 + 7x + 12 = 0 (x + 4) (x + 3) = 0 x + 4 = 0 or, x + 3 = 0 x = -4 , -3
### Slide 11:
GRAPHS OF POLYNOMIALS
### Slide 12:
GENERAL SHAPES OF POLYNOMIAL : 1) f(x) = 3 CONSTANT FUNCTION DEGREE = 0 MAX. ZEROES = 0 1 Y -Y -X X 0
### Slide 13:
Cont…. 2) f(x) = x + 2 LINEAR FUNCTION DEGREE = 1 MAX. ZEROES = 1 Y -X X -Y 0
### Slide 14:
Cont…. 3) f(x) = x 2 + 3x + 2 QUADRATIC FUNCTION DEGREE = 2 MAX. ZEROES = 2
### Slide 15:
Cont…. 4) f(x) = x 3 + 4x 2 + 2 CUBIC FUNCTION DEGREE = 3 MAX. ZEROES = 3
### Slide 16:
RELATIONSHIP B/W ZEROES AND COEFFICIENT OF A POLYNOMIAL
### Slide 17:
QUADRATIC : α + β = - Coefficient of x Coefficient of x2 = - b a _______________ _ αβ = Constant term Coefficient of x2 = c a _________________ _
### Slide 18:
RELATIONSHIPS : ON VERYFYING THE RELATIONSHIP BETWEEN THE ZEROES AND COEFFICIENTS. ONFINDING THE VALUES OF EXPRESSIONS INVOLVING ZEROES OF QUADRATIC POLYNOMIAL. ON FINDING AN UNKNOWN WHEN A RELATION BETWEEEN ZEROES AND COEFFICIENTS ARE GIVEN OF ITS A QUADRATIC POLYNOMIAL WHEN THE SUM AND PRODUCT OF ITS ZEROES ARE GIVEN.
### Slide 19:
DIVISION ALGORITHM
### Slide 20:
If f(x) and g(x) are any two polynomials with g(x) ≠ 0,then we can always find polynomials q(x), and r(x) such that: F(x) = q(x) g(x) + r(x) where r(x) = 0 or degree r(x) < degree g(x) ON VERYFYING THE DIVISION ALGORITHM FOR POLYNOMIALS. Cont…. ON FINDING THE QUOTIENT AND REMAINDER USING DIVISION ALGORITHM. ON CHECKING WHETHER A GIVEN POLYNOMIAL IS A FACTOR OF THE OTHER POLYNIMIAL BY APPLYING THEDIVISION ALGORITHM. ON FINDING THE REMAINING ZEROES OF A POLYNOMIAL WHEN SOME OF ITS ZEROES ARE GIVEN.
### Slide 21:
Thanks for being Patient…
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# Multiplying Monomials
There are only three basic steps to be taken when multiplying monomials. However it is assumed that your child is comfortable with the basic structure of monomials, and that they have already mastered the skill of adding and subtracting monomials.
### Get Prepared For Summer.
If you want to learn how to put a stop to the dreaded Summer Math Loss and send your child back to school in September feeling like a math super-hero then watch this free video training.
So before continuing be sure they are familiar with the language that will be used here. Your child must also be both familiar and comfortable with both the commutative and associative properties of arithmetic.
## The Steps involved in Multiplying Monomials
1. Apply your knowledge of commutative and associative properties of arithmetic.
2. Multiply Coefficients
3. Multiply variables by adding their exponents.
## Problems with Exponents?
Do NOT continue unless your child fully understands how to deal with Exponential Numbers! Your kid will get completely lost.
Below I work through some examples of multiplying monomials. Follow along, ensuring you fully understand what happens from one step to the next.
## Worked Examples - Multiplying Monomials
### Example 1
Apply commutative and associative properties. Please note I have changed the color, but used the X symbol to identify the multiplication operation.Multiply CoefficientsMultiply variables by adding Exponents.
### Example 2
Apply commutative and associative properties. Please note I have changed the color, but used the customary '.' symbol to identify the multiplication operation. Multiply CoefficientsMultiply variables by adding Exponents
As you can see, when it comes to multiplying monomials it is simply a case of applying the laws of arithmetic you already know and applying the laws of exponents.
If there is a weakness in either of these areas, algebra is just going to get more confusing, frustrating and HARD! Make sure the basics are completely understood!
If you need to brush up on any of the prerequisites for this section, the best place to start is at our Basic Algebra section.
Now that you know the steps involved in multiplying monomials, check out how to divide monomials.
Printable Math Worksheets has a lot to offer. Be sure to check it all out.
MATH MOMENTUM ACADEMY If your child is struggling with mathematics and you are a proactive parent, who wants to sort their math problems out, once and for all, then I invite you to explore what Math Momentum Academy can do for you.
## Keep In Touch!
You can send me a quick message, subscribe to K6Math Fun & Update, or join my Facebook Page - K6Math. Choose all the options so you don't miss any of the new material added to this site.
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Greater Number
Mathematical Relational Operator: Greater Than
Last updated date: 19th Mar 2023
Total views: 83.1k
Views today: 0.63k
Namya brought ladies handbags of various sizes starting from size 9 to 21. Suppose bag 1 is of size 12, bag 2 of size 11, bag 3 of size 18, and bag 4 of size 20. Now, to arrange these bags starting from a bigger size to a smaller one, we have the following order:
Size 20 is greater than 18, 18 is greater than 12, 12 is greater than 11. So, here greater than shows the relationship of four numbers with each other, like we did above. We see that greater than is a mathematical operator.
Here, on this page, we will discuss one of the renowned operators “greater than” with certain examples.
Examples of Greater Than: Comparing Numbers
Example 1:
Let us assume that class A has students with the following roll numbers:
• 23
• 12
• 14
• 18
• 25
• 26
Now, let us arrange the numbers by using the concept of ‘greater than’ operator:
Here, the order is:
26 is greater than 25, 25 is greater than 23, 23 is greater than 18, 18 is greater than 14, and 14 is greater than 12.
Now, to minimise our efforts of writing, we replace greater than with ‘>’ sign, which also means the same, so let us write it as:
26 > 25 > 23 > 18 > 14 > 12
Example 2: Niyama bought chocolates at various prices, such as c1 of Rs. 45, c2 of Rs. 34, c3 of Rs. 30, and c4 of Rs. 90.
Now, she arranges these chocolates as the size starting from high price to a lower price, which is:
c4 > c1 > c2 > c3
Here, on changing the order of price, the sign also changes in the following way:
c3 < c2 < c1 < c4
So, this is how we can arrange any item/object in terms of greater than ‘>’ by taking various real-life examples.
FAQs on Greater Number
1. What does greater than or equal mean?
The sign of greater than or equal represents inequality in Maths. For example, Class A has two students with roll numbers 20 and 15. Similarly, Class B has two students of roll number 10 and 15.
Here, on comparing roll numbers of Class A with B, we notice that Class A has 20 greater than both 10 and 15. However, the second roll number 15 > 10 and 15 is equal to 15. This ‘15’ is greater than 10 and equal to 15.
2. How do you use greater than symbol in a sentence?
Let us say that 4 > 2. Here, we know that 4 is actually larger than 2, so we write our symbol with the open mouth of an alligator pointing at the four. So, we read the statement as '4 is greater than 2.
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Miscellaneous
Chapter 9 Class 11 Straight Lines
Serial order wise
### Transcript
Misc 18 If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m. Let line AB be y = 3x + 1 ,line CD be 2y = x + 3 & line PQ be y = mx + 4 Lines AB & CD are equally inclined to the line PQ First we find slopes of lines Slope of line AB y = 3x + 1 The above equation is of the form y = m1x + c1 where slope of line AB = m1 ∴ Slope of line AB = m1 = 3 Slope of line CD 2y = x + 3 y = (𝑥 + 3)/2 y = (1/2)x + 3/2 The above equation of the form y = m2x + c2 where slope of line CD = m2 ∴ Slope of CD = m2 = 1/2 Slope of line PQ y = mx + 4 The above equation of the form y = mx + c where m is slope Thus, Slope of line PQ = m Now, given that lines AB & CD are equally inclined to the line PQ i.e. line AB & line CD make equal angle with line PQ Angle between AB & PQ = angle between CD & PQ We know that angle between two lines whose slope are m1 & m2 is tan θ = |(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Finding angle between AB & PQ Slope of AB = m1 = 3 & slope of PQ = m tan θ = |(𝑚 − 3)/(1 + 3𝑚)| |(𝑚 − 3)/(1 + 3𝑚)| = tan θ Finding angle between CD & PQ Slope of PR = 1/2 & slope of PQ = m tan θ = |(𝑚 − 1/2)/(1 + 1/2 𝑚)| tan θ = |((2𝑚 − 1)/2)/((2 + 𝑚)/2)| = |(2𝑚 − 1)/(2 + 𝑚)| Since angles (θ) are equal So, tan θ must be equal From (A) & (B) |(𝑚 − 3)/(1 + 3𝑚)| = |(2𝑚 − 1)/(2 + 𝑚)| So, ((𝑚 − 3)/(1 + 3𝑚)) = ((2𝑚 − 1)/(2 + 𝑚)) or ((𝑚 − 3)/(1 + 3𝑚)) = − ((2𝑚 − 1)/(2 + 𝑚)) Solving ((𝒎 − 𝟑)/(𝟏 + 𝟑𝒎))= ((𝟐𝒎 − 𝟏)/(𝟐 + 𝒎)) (𝑚 − 3)/(1 + 3𝑚) = (2𝑚 − 1)/(2 + 𝑚) (m − 3)(2 + m) = (2m − 1) (1 + 3m) m (2 + m) − 3(2 + m) = 2m + 6m2 − 1 + 3m 5m + m2 − 6 = 5m + 6m2 − 1 5m − 5m + m2 − 6m2 − 6 + 1 = 0 0 − 5m2 − 5 = 0 − 5m2 = 5 m2 = 5/( − 5) m2 = − 1 This is not possible as square of a number cannot be negative Solving (𝒎 − 𝟑)/(𝟏 + 𝟑𝒎) = − ((𝟐𝒎 − 𝟏)/(𝟐 + 𝒎)) (𝑚 − 3)/(1 + 3𝑚) = (1 − 2𝑚)/(2 + 𝑚) (m − 3) (2 + m) = (1 − 2m)(1 + 3m) m(2 + m) − 3(2 + m) = 1(1 + 3m) − 2m(1 + 3m) 2m + m2 − 6 − 3m = 1 + 3m − 2m − 6m2 m2 − 6 − m = − 6m2 + m + 1 m2 + 6m2 − m − m − 6 − 1 = 0 7m2 − 2m − 7 = 0 The above equation is of the form ax2 + 3x + c = 0 Where solution is x = ( − 𝑏 ± √(𝑏^2 − 4𝑎𝑐))/2𝑎 Here a = 7, b = − 2,c = –7 & x = m So, m = ( −(−2) ± √(〖(−2)〗^2 − 4(7)(−7)))/(2(7)) m = (2 ± √(4 (1 + (7) × (−7)))/14 m = (2 ± 2 √((1 + 49)))/14 m = (2(1 ± √50 ))/14 m = (2(1± √(5 × 5 × 2)))/14 m = 1/7(1 ± 5√2) m = (1 ± 5√2)/7 Thus, the required value of m is (𝟏 + 𝟓√𝟐)/𝟕 & (𝟏 − 𝟓√𝟐)/𝟕
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10. A Methodical Approach to Problem Solving
Student: Well, what about word problems? I can work it out once I know what to do, but how can I figure out the equation?
Dr. O: You need to know and use Polya’s Method5 for problem solving.
Step 1: Understand the Problem
Step 2: Devise a Plan
Step 3: Carry Out the Plan
Step 4: Check
Step 1: To understand what you are required to do in the problem, ask yourself some questions:
• What is my goal (the main questions being asked)?
• What information am I told in the item?
• What constraints or conditions are imposed by the problem?
• What will the results look like?
• How can I state the goal in mathematical terms and symbols?
• What additional information do I need, such as a formula, a definition, or a concept?
• Do I need to make a sketch?
• If I don’t know the additional information, where do I find it?
• What do I need to determine first to answer the question that is asked?
• How can I organize the information to reach my goal, that is, to answer the question?
Step 2: To devise a plan, you start by writing the English version of the answers to some of the questions from step 1, then translating them into mathematical expressions and equations. Once you have an equation that will satisfy the goal, you are ready for the next step.
Step 3: To carry out your plan, solve the equation or equations you devised to give you your final result. Make sure that you follow the order of operations and the correct processes needed.
Step 4: To check your result, make sure that you answer all the questions asked in the problem. Your result needs to satisfy the constraints established by the problem and be reasonable.
As with any practical method, practice makes perfect when you are applying it correctly. Let me demonstrate.
Suppose you were assigned the following application (a.k.a.: word problem):
The measure of the second angle of a triangle is three times the measure of the first and the third is nine degrees less than five times the first. Find the measures of the angles.
##### Step 1: Understand the Problem
Goal: Find the measures of the angles.
Information given: a triangle, which means three angles. I’m told about the measures of angles 2 and 3. What about the measure of angle 1? It is the basis for finding the measures of both angles 2 and 3.
What do I need to know about the measures of the angles of a triangle? The sum of the measures of the three angles is 180°.
Organizing: Because I need to know the measure of angle 1 to find the measures of angles 2 & 3, I represent the measure of angle 1 by the variable a.
Let the measure of angle 1 = a.
Now I must meet the constraints imposed by the problem to write the representations of the measures of angles 2 & 3.
The constraint for angle 2: The measure of the second angle of a triangle is three times the measure of the first, so I must state the measure of angle 2 as 3 times a.
Let the measure of angle 2 = 3a.
The constraint for angle 3: and the third is nine degrees less than five times the first. The key words require me to write the measure of angle 3 as 5a - 9.
Let the measure of angle 3 = 5a - 9.
Sketching and labeling a drawing of a triangle may help me see the situation more clearly.
Now I’m ready to devise a plan.
##### Step 2: Devise a Plan
I need to write an equation that will let me answer the question of the problem. Since the sum of the measures of the 3 angles in a triangle =180°, I can write:
(a) + (3a) + (5a - 9) = 180°
##### Step 3: Carry Out the Plan
Begin by combining like terms in the left side expression (to the left of the equals sign).
9a – 9 = 180 To get the term 9a by itself, I need to add 9 to both sides.
9a – 9 + 9 = 180 + 9 Simplifying produces
9a = 189 My goal is to isolate a, so I’ll reduce the factor of 9 to a 1 by dividing both sides of the equation by 9.
Simplifying both sides gives
a = 21° The measure of angle one is 21°. Have I reached the goal? No, I still need to find the measures of angles 2 & 3.
Angle 2 = 3a Substitute the value of angle one for a and evaluate.
Angle 2 = 3(21) = 63° Now find angle three.
Angle 3 = 5a- 9 Substitute the value of angle one for a and evaluate.
Angle 3 = 5(21) - 9 = 105- 9 = 96°
Angle 1 = 21°, Angle 2 = 63°, and Angle 3 = 96°
Step 4: Check
My angles need to meet the constraints of the problem and have a sum =180°.
First, review the problem and note the constraints.
The measure of the second angle of a triangle is three times the measure of the first and the third is nine degrees less than five times the first. Find the measures of the angles.
Triangle means three angles. No explicit constraint is placed on the first angle except that it is the basis for the other two. Let the measure of angle 1 = a. Since the measure of the second angle of a triangle is three times the measure of the first, I let the measure of angle 2 = 3a. The measure of the third is nine degrees less than five times the first, so I let angle 3 = 5a- 9°.
I need to answer these three equations correctly and have the sum of the measures equal 180° to solve the problem. I calculated the measure of angle 1 as 20°.
Now I write my equations and check to see if my solutions meet the constraints.
Angle 1 = a = 21°
Angle 2 = 3a = 3(21) = 63°
Angle 3 = 5a - 9° = 5(21) - 9 = 105- 9 = 96°
The values I computed for measures of the three angles meet the constraints of the problem. To finish my check, I add the three measures together. If the sum is 180°, then I have the correct measures.
21° + 63° + 96° = 180°, so I have my solutions.
The textbook also introduces a five-step approach that is a variation of Polya’s in Chapter 2.
.
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ICSE Class 6 Maths Important Questions
The Indian Certificate of Secondary Education is an examination conducted by the Council for the Indian School Certificate Examination. The ICSE is responsible for framing the curriculum and designating the topic for Class 6 students. Class 6 Maths important questions have been created with the whole goal of helping students quickly learn mathematical concepts and develop higher logical reasoning and problem-solving skills. ICSE Class 6 Maths Important Questions are prepared by subject experts following the ICSE Class 6 Maths Syllabus prescribed by the Council.
Going through these important questions of Maths, along with other reference study materials, will help students to prepare themselves effectively. Once they complete the entire syllabus, they should practice these important questions to make their concepts more clear. Students of ICSE Class 6 can go through the important questions of maths provided below and should practice them on a regular basis.
Students of Class 6 can practise the ICSE Class 6 Important Questions of Mathematics provided below to revise the entire syllabus in a quick span of time.
1. Find the sum of the following: -146, -78, 124, 69
2. Use the divisibility tests to determine whether the number 378 is divisible by 2, 3, 4, 5, 6, 9 and 11.
3. The monthly income of Arun and Varun are Rs. 18000 and Rs. 27000, respectively. What is the ratio of the income of Varun to that of Arun in its simplest form?
4. Ananya, along with her 2 friends, ordered one sandwich each at their favourite restaurant. They left a tip of 7 rupees for the waiter. If they spent a total of a hundred rupees, find the cost of each sandwich. Frame an equation for the given situation and then solve the same.
5. Let A = {x: x is a letter in the word CHANDIGARH} and B = {x: x is a letter in the word RAJASTHAN}
1. Find A ∩ B and A ∪ B
2. Find n(A), n(B), n(A ∩ B) and n(A ∪ B)
3. Verify: n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
6. If the area of a rectangular plot is 144 sq. m. and its length is 16 m, find the breadth of the plot. Also, find the cost of painting a blue border around the plot if the cost is Rs. 3 per meter.
7. The table represents the number of books sold in a store during four months of the year. Make a bar graph to represent the data.
Months January February March April No. of books 140 120 110 130
8. In the given figure, all the adjacent sides are at right angles. Find:
1. The perimeter of the figure
2. Area of the figure.
9. A painter can paint a wall of area 20 in 10 hours. If he works at a constant speed, how much time will he take to paint a wall of area 30?
10. If the H.C.F. of two numbers is 24 and their product is 5760, find their L.C.M.
11. Write the following numbers in the descending order of their values and represent them on a number line.
1. 8, -6, 2, -12, 0, 3, 15 and -1
2. Integers greater than -6 and less than 2.
12. Express each of the following as an algebraic expression.
1. Sum of y and 7.
2. Number m divided by 23 and added to 5
13. Name the following:
1. Name the two parts of the circle in blue.
1. What is AB, and how does it divide the circle.
14. A vegetable trader buys some tomatoes and onions for Rs. 420 such that the ratio of the total weight of tomatoes to the total weight of onions is 2:3. The total weight of the tomatoes and onions is 60 kg. If the ratio of the total price of tomatoes to the total price of onions is 8:27, then what is the cost of 5 kg of tomatoes and 5 kg of onions?
15. Find the value of: 34+5, where
16. Name the types of the following sets:
1. Set of even numbers which are not divisible by 2.
2. {Number of people in India}
3. Set of odd numbers between 7 and 19.
17. Write each of the following sets in roster form as well as in set builder form:
1. Set of factors of 48
2. Set of integers between -3 and 8
18. What is the central angle of the sector corresponding to the expenditure incurred on Royalty?
19. State whether true or false:
1. {3, 5, 7…} is a finite set.
2. A line has an infinite number of points on it.
3. 0.45 = 45%
20. The sides of a triangle are in the ratio 3:2:4. If the perimeter of the triangle is 27cm, find the length of each side.
Advantages of solving ICSE Class 6 Maths Important Questions
• It helps them to revise the important numerical problems from each chapter.
• The important questions of Maths cover the entire Class 6 Maths syllabus.
• Practising the ICSE important questions will help students in boosting their confidence levels.
• It will help them evaluate their performance level and work on their weak points.
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