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# Integral of Tan 2X
Integral of tan2x along with its formula and proof with examples. Also learn how to calculate integration of tan 2x with step by step examples.
Alan Walker-
Published on 2023-04-21
## Introduction to the integral of tan 2x
In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.
Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function sine. You will also understand how to compute the integral of tan of 2x by using different integration techniques.
## What is the integration of tan 2x?
The integration tan2x is an antiderivative of the tangent function which is equal to (-½ )ln|cos 2x|. It is also known as the reverse derivative of tan 2x function which is a trigonometric identity.
The tangent function is the ratio of sine to the cosine function which is written as:
Tan = sine/cosine
### Integral of tan2x formula
The formula of the integral tan2x contains the integral sign, coefficient of integration, and the function as tan 2x. It is denoted by ∫(tan 2x)dx. In mathematical form, we can write it as:
$∫\tan 2xdx = -½ \ln|\cos 2x| + c$
Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral.
## How to calculate the integral of tan(2x)?
The integration of tan2x is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of the tangent by using:
1. Substitution method
2. Definite integral
## Integral of tan2x by using u substitution
The u-substitution is a method of integration in calculus. It is used to calculate the integral of a function that is complex to be calculated by usual integration. Let’s discuss calculating the integral tan2x by using u-substitution.
### Proof of integral of tan 2x by using u-substitution
Since we know that a function can be replaced by u in the u-substitution calculator. Therefore, we can calculate the integration of tan2x by using this technique.
Assume that,
$I = ∫\tan 2xdx$
By using trigonometric identities, tan 2x can be written as sin 2x/cos 2x. Then,
$I = ∫\frac{\sin 2x}{\cos 2x}dx$
Now, suppose that u = cos 2x and du = -2sin 2x dx
Therefore,
$I = -(½ ) ∫\frac{1}{u}du$
Integrating,
$I = -(½ ) \ln|u|$
Substituting the value of u, we get
$I =-½ \ln|\cos 2x| + c$
Which is the calculation of the integral of tan 2x.
## Integration of tan2x by using definite integral
The definite integral is a type of integration that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as:
$∫^b_a f(x) dx = F(b) – F(a)$
Let’s understand the verification of the integration of tan 2x by using the definite integral.
### Proof of integral of tan2x by using definite integral
To compute the integral tan2x by using a definite integral calculator, we can use the interval from 0 to π. Let’s integrate tan 2x from 0 to π. For this, we can write the integral as:
$∫^π_0 \tan 2x dx = -½ \ln|\cos 2x|^π_0$
Now, substitute the limit in the given function.
$∫^π_0 \tan 2x dx = -½[ \ln \cos 2(π) – \ln \cos 2(0)]$
Since cos 0 is equal to 1 and cos π is equal to -1, therefore,
$∫^π_0 \tan 2x dx = -½ [\ln(1) –\ln(1)]= 0$
Which is the calculation of the definite integral of tan 2x.
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# Definition: Multiplication of Complex Numbers
According to the definition of complex numbers, we can consider the complex numbers $$x,y\in\mathbb C$$ as ordered pairs of some real numbers $$a,b,c,d\in\mathbb R$$, i.e. $$x=(a,b)$$ and $$y=(c,d)$$.
The multiplication of complex numbers "$$\cdot$$" is defined based on the addition, subtraction, and multiplication of real numbers in the corresponding ordered pairs:
$x\cdot y:=(a,b)\cdot (c,d):=(ac-bd,ad+bc),$
Note that this kind of multiplication operation always produces a new ordered pair of real numbers $$(ac-bd,ad+bc)$$, which is a new complex number, called the product of the complex numbers $$x$$ and $$y$$. Thus, the set of complex numbers $$\mathbb C$$ is closed under this kind of multiplication operation.
The multiplication of two complex numbers (points in the complex plane) $$x$$ and $$y$$ can be interpreted/constructed as follows:
1. Connect $$x$$ and $$y$$ with the origin of the complex plane.
2. Measure the angles of the segments with respect to the positive horizontal axis.
3. Measure the lengths of the segments.
4. Add the two angles.
5. Multiply the two lengths.
6. The point $$x\cdot y$$ lies on the segment with the length gained in step 5 with an angle to the positive horizontal axis gained in step 4.
In the following figure, you can drag the complex numbers $$x$$ and $$y$$ and see, how the position of $$x\cdot y$$ changes with the position of the respective factors $$x$$ and $$y$$:
### Fun questions:
• Why is the product of two negative real numbers positive?
• Can you find point $$x$$ in the complex plain, for which $$x=\sqrt{-1}$$? (See also imaginary unit)
Chapters: 1
Parts: 2
Proofs: 3 4 5 6 7 8 9 10 11 12 13
Propositions: 14 15 16 17 18 19 20 21 22
Thank you to the contributors under CC BY-SA 4.0!
Github:
### References
#### Bibliography
1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013
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### Theory:
Fibonacci number
$$1 + 1 = 2$$
$$1 + 2 = 3$$
$$2 + 3 = 5$$
$$5 + 8 = 13$$
$$1$$, $$1$$, $$2$$, $$3$$, $$5$$, $$8$$, $$13$$, $$21$$, $$34$$, …
Each number in a Fibonacci sequence is called a Fibonacci number.
Construction of Fibonacci sequence:
Step 1: The first two numbers are $$1$$.
Step 2: The next numbers are found by adding up the two numbers before it.
Fibonacci spiral:
Fibonacci numbers in nature:
Lucas numbers
Lucas numbers are closely related to Fibonacci numbers. Fibonacci numbers starts with two $$1$$'s, but Lucas numbers start with $$1$$ and $$3$$, and the next numbers are obtained by adding the numbers before it.
Lucas sequence:
$$1 + 3 = 4$$
$$3 + 4 = 7$$
$$4+ 7 = 11$$
$$7 + 11 = 18$$
$$1$$, $$3$$, $$4$$, $$7$$, $$11$$, $$18$$, …
Each number in a Lucas sequence is called a Lucas number.
Golden Ratio
The golden ratio is denoted by the Greek letter $$\phi$$, and the value is $$1.618$$.
$$\phi = 1.681$$
It is surprising that if the sum of the two consecutive numbers in the Fibonacci sequence is divided by the larger number from the two successive number, you will get the number which is nearly equal to the golden ratio.
$\frac{1.2}{2}=1.5,\phantom{\rule{0.147em}{0ex}}\frac{2.3}{3}=1.66,\phantom{\rule{0.147em}{0ex}}\frac{3.5}{5}=1.6,\frac{5.8}{8}=1.625,...$
This spiral stairs follows a Fibonacci spiral pattern. This pattern is one of the reasons behind the elegant look of the stairs.
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# How to Solve a 2x2 Rubik's Cube
112,745
2
14
## Introduction: How to Solve a 2x2 Rubik's Cube
I will show you step by step how to solve a pocket cube. Hope fully it will help. Any questions please ask.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Notation
If you already know how to solve a 3x3 you can skip this part. This is notation when i say R the layer to your right is to be moved once clockwise. L is the left layer. U is the top side. D is the bottom layer.
If there is an (') after a letter you turn the side counterclockwise.
If you still do not explain look it up on google their should be a more thorough definition on Notation.
good 3x3 notation video pratically the same as a 2x2
Below is an example
## Step 2: First Layer
First you must choose a layer. I will choose the Green side.
First find a green piece.
Then you find the piece that belongs beside it in my case I picked the green,white, and red piece. I must find a piece that has both green and red. There will only be one. Put this picture in the lower layer. Below where it actually belongs as in the picture. Then put it on the right side and do this algorithm (notations) R' D' R D
Find the next piece and finish the top layer.
## Step 3: Next Side
Now we will put what we have finished on the bottom. And work on the next side. Take note I say side here not layer. Because we will be just working on the top colors (blue for me) There is four total ways this can be arranged. If you have no blue colors on the top dont worry how you hold it just do this algorithm R U R' U R U2 R'
If you have 1 blue on top hold it like in the picture and then do the algorithm from above.
If you have 2 blue on top hold it like in the picture and then do the algorithm from above. Or the other way for 2.
If you have 4 on top you are lucky you have to do nothing
Below i have all the examples there are.
## Step 4: Finish Cube
Now to finish take a look at the last layer. Find two colors that match up like my black ones here. Put them in the back then do the algorithm R' F R' B2 R F' R' B2 R2
Now the cube is solved. Any questions then message.
## Recommendations
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## 14 Discussions
I'm sorry but I can only ever solve the top layer I get lost after that
what do you mean 'like my black ones here' there are no black ones in the photo
They are on the right side if the white (or yellow i can't tell) side is the up side
I'm sorry I dont understand what you mean by B please clarify
B as in agorithm means the back face if thats what you mean
I must say this step isn't at all clear. let me quite your directions from this step "Now to finish take a look at the last layer. Find two colors that match up like my black ones here. Put them in the back then do the algorithm R' F R' B2 R F' R' B2 R2" "the last layer" not completely clear but I get it. you have solved the opposite side of the cube but each block isn't in the right place. we are trying for that. I'm with you so far. "find two colors that match up like my black ones here" What black ones? you show a single yellow side. and offer no other pictures to go from. but even with that i think I can get some of it. we are looking for two of the same color that are already next to each other, and then we wan't to turn the cube (layer?) so they are n the back. presumably i can then do the algorythm once and it will be solved, (at least that's the way it seems it should work from your description) but instead of solving it it just messes the whole thing back up again. do i need to do it more than once? you don't say. if you could clarify this for me (and for everyone else I'm sure it would be much appreciated
Sorry, I'm lost here? Last step was solving of the two opposite layers, and now you say it is the last layer? Did I miss a step? Also you say we have to find two colors there match up like your black ones here. But the picture show a completed cube? And then we have to put then in the back? How shall the orientate? Because when i do it, I scramble the hole cube again. Sorry if my questions is stupid. Steps 1 to 3 was really helpful, and thats why I wounder what happens here on step 4, where a lot of information is lost :s Thanks in advance...
thanks, i dint have a 2x2 but i have a 4x4 and ive been trying to work out how to do it like a 2x2 ever since i could solve it as a 4x4, once again many thanks
Is a 2x2 rubiks cube harder or easier than a 3x3? because what I'm seeing here it actually looks hard to solve...
TYVM!!!!!!! I learned how to solve it, YAY!
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# 6.4 Relationship between outer product and determinants
The and determinants are closely linked in geometric algebra. They both measure the volume of spaces formed by vectors, with the outer product giving a geometric interpretation and the providing a numerical value. This connection helps us visualize abstract mathematical concepts.
Understanding this relationship is crucial for grasping how vectors interact in higher dimensions. It bridges the gap between algebra and geometry, showing how mathematical operations relate to physical spaces. This knowledge is key for solving problems in physics, engineering, and computer graphics.
## Outer Product and Determinant Connection
### Geometric Interpretation of Outer Product and Determinant
• The outer product of n vectors in an n-dimensional space produces an n-vector representing a signed volume element
• The magnitude of the resulting n-vector equals the absolute value of the determinant of the matrix formed by the vectors
• Magnitude represents the unsigned volume of the parallelotope formed by the vectors
• The sign of the resulting n-vector depends on the orientation of the vectors, corresponding to the sign of the determinant
• Negative sign indicates a left-handed orientation (opposite of standard basis orientation)
### Relationship between Outer Product and Matrix Determinant
• The outer product of the column vectors or row vectors of a square matrix yields a value proportional to the determinant
• The determinant of a matrix can be interpreted as the signed volume of the parallelotope formed by its column or row vectors
• Column vectors represent the edges of the parallelotope emanating from the origin
• Row vectors represent the edges of the parallelotope in a different coordinate system
• The outer product and determinant provide equivalent ways to calculate the signed volume of a parallelotope
## Determinant Calculation with Wedge Product
### Definition and Properties of Wedge Product
• The wedge product (∧) is the antisymmetric part of the outer product
• Antisymmetry means a ∧ b = -b ∧ a for any vectors a and b
• The wedge product is associative: (a ∧ b) ∧ c = a ∧ (b ∧ c) for any vectors a, b, and c
• The wedge product is anticommutative: a ∧ b = -b ∧ a for any vectors a and b
• Swapping the order of vectors in a wedge product changes the sign of the result
### Determinant Calculation using Wedge Product
• For a 2x2 matrix [a b; c d], the determinant equals the wedge product of its column vectors:
• det([a b; c d]) = a ∧ d - b ∧ c
• Geometrically, this represents the signed area of the parallelogram formed by the column vectors
• For a 3x3 matrix [a b c; d e f; g h i], the determinant equals the triple wedge product of its column vectors:
• det([a b c; d e f; g h i]) = a ∧ e ∧ i + b ∧ f ∧ g + c ∧ d ∧ h - a ∧ f ∧ h - b ∧ d ∧ i - c ∧ e ∧ g
• Geometrically, this represents the signed volume of the parallelepiped formed by the column vectors
• The determinant of a matrix can be calculated by taking the wedge product of its column vectors in order or its row vectors in order, with appropriate sign changes based on the permutation of the vectors
• Permuting the vectors in the wedge product changes the sign according to the parity of the permutation (even permutations maintain the sign, odd permutations change the sign)
## Equivalence of Formulations for Volume
### Oriented Volume Calculation using Outer Product
• The oriented volume of a parallelotope formed by n vectors in an n-dimensional space can be calculated using the outer product
• Consider n vectors a₁, a₂, ..., a_n in an n-dimensional space forming a matrix A = [a₁ a₂ ... a_n]
• The outer product of these vectors, a₁ ∧ a₂ ∧ ... ∧ a_n, results in an n-vector representing the oriented volume of the parallelotope
• The magnitude of the n-vector equals the unsigned volume of the parallelotope
• The sign of the n-vector indicates the orientation of the parallelotope (positive for right-handed, negative for left-handed)
### Oriented Volume Calculation using Determinant
• The oriented volume of a parallelotope can also be calculated using the determinant of the matrix formed by its edge vectors
• For the matrix A = [a₁ a₂ ... a_n], the determinant det(A) represents the signed volume of the parallelotope
• The absolute value |det(A)| equals the unsigned volume of the parallelotope
• The sign of det(A) indicates the orientation of the parallelotope (positive for right-handed, negative for left-handed)
### Proof of Equivalence
• To prove the equivalence of the outer product and determinant formulations for oriented volume:
1. Expand the outer product a₁ ∧ a₂ ∧ ... ∧ a_n using the antisymmetry and associativity properties
2. Expand the determinant det(A) using the Laplace expansion or cofactor expansion
3. Show that the expanded outer product and determinant yield the same result, up to a sign depending on the orientation of the vectors
• The equivalence demonstrates that both the outer product and determinant provide consistent measures of the oriented volume of a parallelotope
## Applications of Outer Products and Determinants
### Geometry Applications
• Calculate the signed volume of parallelotopes and simplices in n-dimensional spaces
• Parallelotope: generalization of a parallelogram (2D) or parallelepiped (3D) to higher dimensions
• Simplex: generalization of a triangle (2D) or tetrahedron (3D) to higher dimensions
• Find the area of a parallelogram or the volume of a parallelepiped formed by vectors in 2D or 3D spaces
• : $A = |\vec{a} \wedge \vec{b}|$, where $\vec{a}$ and $\vec{b}$ are the edge vectors
• Volume of parallelepiped: $V = |(\vec{a} \wedge \vec{b}) \cdot \vec{c}|$, where $\vec{a}$, $\vec{b}$, and $\vec{c}$ are the edge vectors
• Calculate the signed area of a triangle or the signed volume of a tetrahedron using the coordinates of their vertices
• Signed area of triangle: $A = \frac{1}{2}|(x_1 \wedge y_2 - x_2 \wedge y_1) + (x_2 \wedge y_3 - x_3 \wedge y_2) + (x_3 \wedge y_1 - x_1 \wedge y_3)|$, where $(x_i, y_i)$ are the vertex coordinates
• Signed volume of tetrahedron: $V = \frac{1}{6}|a \wedge b \wedge c|$, where $a$, $b$, and $c$ are the edge vectors emanating from a common vertex
### Linear Algebra Applications
• Determine the linear independence of vectors using the outer product or determinant
• Linearly dependent vectors will have a zero outer product and a zero determinant
• Linearly independent vectors will have a non-zero outer product and a non-zero determinant
• Use the outer product or determinant to determine the orientation of a set of vectors or a geometric object
• Determine whether a point is on the left or right side of a directed line segment in 2D
• Determine the handedness of a coordinate system in 3D (right-handed or left-handed)
• Apply the relationship between outer products and determinants to solve systems of linear equations
• Cramer's rule: express the solution of a system of linear equations using determinants of the coefficient matrix and augmented matrices
• Gaussian elimination: use the outer product to perform row operations and reduce the system to row echelon form
## Key Terms to Review (12)
Area Interpretation: Area interpretation refers to the geometric understanding of the outer product and how it relates to determining the area of figures formed by vectors in a space. This concept connects the magnitude of the outer product of two vectors to the area of the parallelogram they define, emphasizing the role of orientation and direction. Additionally, area interpretation is crucial when relating the outer product to determinants, as both provide insights into spatial relationships and transformations.
Area of Parallelogram: The area of a parallelogram is a measure of the space contained within its boundaries and is calculated as the product of its base length and height. This relationship highlights how the orientation of the parallelogram does not affect its area, emphasizing the geometric properties that are inherent to these shapes. The area can also be represented using the determinant of a matrix formed by its vertex coordinates, linking it to broader concepts such as linear transformations and geometric interpretations of algebraic operations.
Calculating Areas: Calculating areas refers to the process of determining the size of a two-dimensional surface or shape, typically expressed in square units. This concept is crucial in understanding the relationship between geometric figures and their properties, particularly through the use of outer products and determinants which provide algebraic ways to compute areas based on the coordinates of vertices.
Cross Product Relation: The cross product relation describes a mathematical operation between two vectors in three-dimensional space that results in a third vector perpendicular to both of the original vectors. This operation is closely connected to the outer product and determinants, as it can be interpreted as a specific case of the outer product that captures the area of the parallelogram formed by the two vectors, and its magnitude relates directly to the determinant of a matrix formed by these vectors.
Determinant: A determinant is a scalar value that is derived from a square matrix, which provides important information about the matrix, such as whether it is invertible and its volume scaling factor in linear transformations. Determinants are closely tied to concepts such as area, volume, and the solvability of linear equations, making them essential for understanding matrix behavior and properties.
Finding Volumes: Finding volumes refers to the mathematical process of determining the amount of three-dimensional space occupied by a solid figure. This concept is deeply connected to the geometric principles of area and dimension, where techniques such as the outer product and determinants are utilized to derive the volume of various shapes, especially in higher dimensions.
Geometric Product: The geometric product is a fundamental operation in geometric algebra that combines vectors to produce a multivector, encapsulating both the inner and outer products. This operation not only defines how vectors relate to each other but also provides insights into angles and areas, making it essential for understanding more complex structures like multivectors. It's a powerful tool that serves as the basis for building further algebraic expressions involving vectors and their relationships.
Levi-Civita symbol: The Levi-Civita symbol, often denoted as $\\epsilon_{ijk}$, is a mathematical object used in tensor calculus and differential geometry. It is a completely antisymmetric tensor that plays a crucial role in the computation of determinants and cross products, providing a way to encode orientation and volume in multi-dimensional spaces. This symbol helps relate the outer product of vectors to the determinant of matrices, emphasizing the geometric interpretation of these operations.
Matrix determinant: The matrix determinant is a scalar value that is computed from the elements of a square matrix and encapsulates key properties of the matrix, including its invertibility and volume scaling factor in transformations. It provides crucial information about the linear transformation represented by the matrix, such as whether the transformation is singular or non-singular. Understanding how the determinant relates to other mathematical operations, particularly the outer product, deepens the comprehension of its role in geometry and linear algebra.
Multilinearity: Multilinearity refers to a property of functions that are linear in each of their arguments separately. This means that if you change one argument while keeping the others fixed, the output behaves like a linear function, and similarly for each argument. In various mathematical contexts, such as vector spaces and tensor algebra, multilinearity is crucial for understanding relationships between multiple dimensions and how they interact, particularly when exploring concepts like outer products and determinants.
Outer Product: The outer product is an operation in geometric algebra that takes two vectors and produces a bivector, encapsulating the notion of area and orientation. This operation extends the idea of multiplying vectors, enabling us to capture geometric relationships such as areas and volumes in higher dimensions.
Volume interpretation: Volume interpretation refers to the geometric understanding of the outer product, specifically how it relates to calculating volumes in higher-dimensional spaces. This concept allows us to visualize and comprehend how the outer product of vectors can represent the area, volume, or hyper-volume of geometric shapes formed by those vectors. This interpretation is crucial in understanding the relationship between the outer product and determinants, as both concepts capture how space is occupied and quantified by vectors in various dimensions.
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# Necessary and Sufficient Conditions: If, or Only If?
Sometimes in math, we trip over words, especially when they are used in ways that differ from everyday usage, or when the associated grammar is complicated. This set of three answers from our archive, each of which is referred to by the next one, look at relationships among the ideas of “necessary and sufficient conditions”, and “if and only if”.
## Does “necessity” mean “if” or “only if”?
First, from 1999, we have a question about the words “necessary” and “sufficient” in the statement of a theorem to be proved; such a statement is also called a “biconditional”, as we have conditions in both directions. The geometrical theorem here is simple, probably intended just to demonstrate the form of this sort of theorem. It happens to tie in to our recent discussions of inclusive definitions:
Parts of a Biconditional Statement
I came across the following arguments (in a book) involving the biconditional and the author's proof confused me. The author stated the following theorem:
A quadrilateral is a square if and only if
it is both a rhombus and a rectangle,
and proved the theorem in two steps as follows:
Step 1 (the "if" part): Let Q be a quadrilateral which is both a rhombus and a rectangle. From the definition of a rhombus, all four sides are equal in length. From the definition of a rectangle, all four angles are right. A square is a four-sided figure in which all angles are right and all sides are equal. Therefore, Q is also a square.
Step 2 (the "only if" part): Let Q be a square. By definition, all four sides are equal, so Q is also a rhombus. Again by definition, all angles are right, so Q is a rectangle.
Until here I have followed the author's proof, but further he re-proves the same biconditional using "necessary" and "sufficient"; and there he wrote that:
step 1 (the "necessity" part): the same as "if" part in the last proof.
Step 2 (the "sufficiency" part): same as the "only if" part above.
I believe it's just the other way around: that is, "necessity" corresponds to "only if" and "sufficient" corresponds to "if."
Could anybody be of some help? Am I right? More explanation to make the idea clearer is very welcome.
Unfortunately, Abdellah didn’t quote the “necessary and sufficient” formulation of the theorem; there are two possibilities, and if the book itself didn’t state it explicitly, that may be the source of confusion.
My first thought was that the restatement of the theorem would most naturally be something like this, where the condition is the more complicated statement:
“A necessary and sufficient condition for a quadrilateral to be a square is that it is both a rhombus and a rectangle.”
If so, then Abdellah is arguing that the statement that being both a rhombus and a rectangle is necessary for being a square is equivalent to “A quadrilateral is a square only if it is both a rhombus and a rectangle.” He would be right, though honestly it took me a while to convince myself of this, because the words are so convoluted!
On the other hand, it could be this, swapping the roles of the clauses to put “necessary and sufficient condition” where “if and only if” was:
“For a quadrilateral, being a square isnecessary and sufficient condition for it to be both a rhombus and a rectangle.”
If that is what they meant, the book is right!
Doctor Mike properly assumes that the book is correct (taking the first clause to be the “condition” for the second, as in my second version), and explains the meaning of the words:
Maybe it would help to actually write out the sentences, but let's use "S" to mean Square and "R+R" to mean Rhombus and Rectangle.
The first one is "S is a sufficient condition for R+R." This means that if you are given S, then you have "sufficient" (or "enough") information to prove R+R. That's exactly what was done for the "only if" part.
Next, look at "S is a necessary condition for R+R". This should be the other direction, so let's see why. If you focus on the first part, "S is a necessary condition," then you see we really are talking about S being a necessary and logical consequence of something, namely R+R. Assuming R+R and proving S is what was done above in the "if" step.
Here is a summary. The following 4 sentences mean the same.
S ----> R+R
S implies R+R
S only if R+R
S is sufficient for R+R
The other direction also has the 4 similar variants.
S <---- R+R
S is implied by R+R
S if R+R
S is necessary for R+R
He left out two forms that can help clarify (or confuse): $$\text{S} \rightarrow \text{R+R}$$ can be read as “if S, then R+R”, Similarly, “S if R+R” means the same thing as “if R+R, then S”. Often, putting the “if” first clarifies the meaning.
As I think about the use of “necessary” and “sufficient” in logic, I realize that these are, in a sense, two different senses of what we think of as a “condition” in everyday life, and part of the confusion may be the ambiguity of that ordinary usage.
When I look up “condition” in a dictionary, the relevant definition is “prerequisite”; in the Merriam-Webster dictionary, an example given is “Available oxygen is an essential condition for animal life”. A condition in this sense is something without which something wouldn’t happen – it will happen only if the condition is satisfied. That’s exactly what a “necessary condition” is. As Doctor Mike said, we can just as well think of this as a logical consequence: If someone is alive, then we know he must have oxygen. But this is not to say that life causes the oxygen! Nor is oxygen a sufficient condition for life; you need other things as well, such as food.
On the other hand, “condition” also means, in grammar, the “if clause” in a conditional sentence like “If A, then B”; in this sense a “condition” is only sufficient, giving a condition under which B will be true, but saying nothing about what happens if A is false. So when we connect the words “condition” and “if”, we tend to think of a sufficient condition (a fact from which we can conclude that something else is true), not a necessary condition (a fact that is required in order for something else to be true).
The same is true in logic: When we talk about a “conditional statement”, we mean $$\text{A} \rightarrow \text{B}$$, or “If A, then B”, where again A is a sufficient, not necessary, condition for B. If A is true, we can be sure that B is true. (In logic, A is called the antecedent and B the consequent – the word “condition” is not used.)
When we use the words “necessary” or “sufficient” with “condition”, we are overriding these uses, and taking a “condition” merely as any statement, which has whichever relation we specify with the other statement.
## Necessary or sufficient
What do we mean by 'necessary condition' and 'sufficient condition' (and sometimes we call a condition both 'necessary and sufficient')? I am very much confused. Help!
I used an example (unlike Abdellah’s question above) in which only one part is true, which makes it a little easier to see the distinctions:
Let's look at the two statements (predicates), "X is a mammal" and "X is a dog". Call the first statement A, and the second B.
Now, A is a _necessary_ condition for B, because A _must_ be true in order for B to be true. B can only be true if A is true; if A is not true, then B can't be true. We can say this in several ways:
A is a necessary condition for B
A <== B (A is implied by B)
B ==> A (B implies A)
A if B (whenever B is true, A will be true)
B only if A (B is true only when A is true)
On the other hand, A is not a _sufficient_ condition for B, which would mean that in order to know that B is true, it is _enough_ to know that A is true. It is not enough to know that X is a mammal, because there are other mammals besides dogs. But if we reverse the two statements, we find that B is a sufficient condition for A: if we know that X is a dog, we know that it is a mammal. So these statements are equivalent:
A is a sufficient condition for B
B <== A (B is implied by A)
A ==> B (A implies B)
B if A (whenever A is true, B will be true)
A only if B (A is true only when B is true)
I needed a different example to illustrate “necessary and sufficient”:
Note that "necessary condition" and "sufficient condition" are opposites; "A is a necessary condition for B" means the same thing as "B is a sufficient condition for A".
Now, if A is a necessary AND sufficient condition for B, then the implication works both ways; it can be expressed as
A <==> B (A is equivalent to B)
A iff B (A if and only if B)
This means that if A is true, B must be true, and if B is true, A must be true. That is not the case in our example statements; but it would be true, for example, if A were "X is less than Y" and B were "Y is greater than X". These two statements mean the same thing; if one is true, then the other is true. So if we want to prove B, it is necessary for A to be true, and it is sufficient to prove that A is true.
## What does “only if” mean?
I think some additional explanation is needed for the meaning of “only if”; but I couldn’t find a good discussion of this in the archive. What follows is an unarchived question from 2011:
The statement "A if and only if B" can be taken apart as "A if B" and "A only if B". Both parts need to be either true or false for the biconditional to be true. Shown with a truth table, easily proven.
Wanting to deeply understand the meaning of the parts, I ran into some struggle. Since I'm not native English speaking, probably I have some problems with the words...but I don't know if that is really the case.
"A only if B"
This is the easy one for me. It says, only if B is true, A is true. So, if B is true and A is true, the formula is true. And if B is true but A is false, the formula is false. And for the rest of possibilities, there is nothing we can say. So this is equivalent to "B --> A".
"A if B"
This is the difficult one for me. Since 'B' comes syntactically after the 'if', this looks for me like "B --> A". But this case is already given above. So how can I rephrase this in a way making sense in English? Or is this just some sloppiness of the definition and I have to get over it? Or did I not understand the real meaning? Can you explain on this? Thank you.
Panny actually got the “if” case right, but wrongly thinks that “only if” means the same thing. That’s easy to do! I replied, first about the language issue:
I think most native English speakers have trouble with this! In fact, they may have more trouble, because they are somewhat familiar with the phrases but have never stopped to think about exactly what they mean. When you think you understand something, you are in greater danger of fooling yourself!
Then, about the “only if” case:
This is actually the harder part; you got it wrong, as most of us do at first. Even I have to think about it a moment to be sure I'm explaining it correctly.
Take an example: "The road is wet only if it rains". What does that statement claim? It says that the only time the road will be wet is if it rains. (This is not really true, of course; snow may have melted on the road.) So if you see that the road is wet, it must be because it rained. Therefore, we can restate this as, "if the road is wet, then it has rained".
Do you see what this means? The statement "A only if B" turns out to mean "if A, then B", or "A -> B". (In my example, A is "the road is wet" and B is "it has rained".)
Again, it is saying that A is true ONLY if B is true, which means that if A is true, you know that B must be true!
Finally, the “if” case:
Here your reasoning is correct; "A if B" means "if B then A", which means "B -> A".
So (a) "A if B" and (b) "A only if B" mean
(a) "B -> A" and (b) "A -> B" respectively.
## Another example
I will quote only parts of one last question and answer on this, because the example is from modular arithmetic, which not everyone will want to get into. Read the whole thing if you do!
Necessary and/or Sufficient Conditions with Modular Math
... identify which of the conditions below are "sufficient", "necessary", "necessary and sufficient" or none of these ...
I don't understand what is meant by sufficient, necessary, and sufficient and necessary.
Here is the generally useful part of the answer, which touches on the “only if” question as well:
A condition A is "necessary" for a result B if B is true ONLY if A is true; that is, A HAS TO be true in order for B to be true. We say that B implies A, or "B only if A".
A condition A is "sufficient" for a result B if B is true WHENEVER A is true; that is, A is ENOUGH to force B to be true. We say that A implies B, or "B if A".
A condition A is "necessary AND sufficient" for a result B if both of the above are valid; we say that A is true IF AND ONLY IF B is true. A and B are equivalent.
As an example, suppose I asked whether "x and y are both even" is a necessary and/or sufficient condition for the product xy to be even. Since IF x and y are both even, THEN xy is even, it is a sufficient condition; knowing they are both even is enough to be certain that the product is even. But xy will also be even if only one of the factors is even; so having both even is NOT necessary.
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# Thread: Parallel and Perpendicular lines
1. ## Parallel and Perpendicular lines
Write an equation in slope-intercept form of the line in parallel to the graph of each equation and passes through the given point.
1. y=4x+5; (2,-3)
2. y=-1/2x-3; (0,0)
3. y=4; (2, 5)
4. 3x+y=3; (3, 5)
_________________________________________________
Write an equation in slope-intercept form of the line that is perpendicular to the graph of each equation and passes through the given point.
1. y=2x+3; (3, -4)
2. y=-4x+5; (1, 1)
3. 2x-5y=3; (-2,7)
4. x=6; (4,2)
_________________________________________________
Determine whether AB or CD are parallel, perpendicular, or neither.
1. A(4, 3), B(5, 2), C(8, 3), D(9,2)
2. A(-2, 3), B(4,0), C(2, 5), D(-1, 11)
_________________________________________________
Wirte an equation in slop intercept form of each line.
• passes through the point at (4, -2) and is parallel to the graph of 5x-2y=6
• perpendicular to the line through points at (1,2) and (8,6) and passes through the point at (0, 5)
2. Consider the slope of each line, for a parallel line they will have the same slope. Parallel line have the same slope because they have to change at the same rate or else they would cross each other eventually right?
Perpendicular lines have slopes that are negative reciprocals of eachother i.e.
the negative reciprocal of $\frac{1}{2} is \frac{-2}{1}$ If we multiplied these together we would get -1 that would be the case with all perpendicular lines, if you multiply the slopes together you should get -1
3. So since we are worried about slopes putting them in the standard slope intercept form would be a good place to start. y=mx+b
4. Originally Posted by Temperamental
Write an equation in slope-intercept form of the line in parallel to the graph of each equation and passes through the given point. Note the parallel part.
1. y=4x+5; (2,-3)
2. y=-1/2x-3; (0,0)
3. y=4; (2, 5)
4. 3x+y=3; (3, 5)
Because it's parallel, the equation of the second line will have the same slope.
1) $y_2 = 4x + c$
Substitute $x = 2 ; y = -3$ (Given point)
$(-3) = (4)(2) + c$
Solve for c.
$c = -11$
Thus: $y_2 = 4x - 11$
=======
Same story. Gradient is the same, but we don't know $c$.
2) $y_2 = - \frac{1}{2} x + c$
Substitute $x = 0 ; y = 0$ (Given point)
$0 = 0 + c$
Solve for c.
$c = 0$
Thus: $y_2 = - \frac{1}{2} x + c$
=======
3) $y=4$
Try to envision this line. It's a horizontal line.
Basically any $y = c$ line would be parallel to this, but they specifically ask for the one which goes through the point where y = 5.
So $y_2 = 5$
=======
Try number 4 for yourself.
_________________________________________________
Originally Posted by Temperamental
Write an equation in slope-intercept form of the line that is perpendicular to the graph of each equation and passes through the given point.
1. y=2x+3; (3, -4)
2. y=-4x+5; (1, 1)
3. 2x-5y=3; (-2,7)
4. x=6; (4,2)
Same as the first ones we did. Except here the gradients are not equal, but their products equal -1. Always. If the product of lines' gradients equal -1 then they are perpendicular.
I'll do number 1.
1.) $y=2x+3$
Now the gradient of $y_2$ should be the negative reciprocal of $y$.
$m_{y_{2}} = - \frac{1}{2}$
$y_2 = - \frac{1}{2} x + c$
You know the drill, substitute $(3, -4)$ and solve for $c$.
_________________________________________________
Originally Posted by Temperamental
Determine whether AB or CD are parallel, perpendicular, or neither.
1. A(4, 3), B(5, 2), C(8, 3), D(9,2)
2. A(-2, 3), B(4,0), C(2, 5), D(-1, 11)
You know now that when gradients are equal they're parallel and when their product is -1 they're perpendicular. They could also be neither.
_________________________________________________
Originally Posted by Temperamental
Wirte an equation in slop intercept form of each line.
• passes through the point at (4, -2) and is parallel to the graph of 5x-2y=6
• perpendicular to the line through points at (1,2) and (8,6) and passes through the point at (0, 5)
Same drill as all the other stuff.
5. Originally Posted by janvdl
Because it's parallel, the equation of the second line will have the same slope.
1) $y_2 = 4x + c$
Substitute $x = 2 ; y = -3$ (Given point)
$(-3) = (4)(2) + c$
Solve for c.
$c = -11$
Thus: $y_2 = 4x - 11$
What happened to the 5? why did you replace it with c?
6. Originally Posted by janvdl
...
2) $y_2 = - \frac{1}{2} x + c$
Substitute $x = 0 ; y = 0$ (Given point)
$0 = 0 + c$
Solve for c.
$c = 0$
Thus: $\boxed{\color{blue}y_2 = - \frac{1}{2} x }$
Originally Posted by Temperamental
What happened to the 5? why did you replace it with c?
The equation of the given line is: $y = 4x+5$
A parallel to this line has the same slope (here m = 4) but a different constant summand. This summand is unknown to us and therefore janvdl labeled it c. And you were asked to calculate the value of c using the fact that the parallel has to pass through the point P(2, -3), that means the coordinates of the point must satisfy the equation of the parallel:
To satisfy an equation means: If you plug in the x- and y-values the equation must be true. And that's the reason why you can use it to calculate the value of c.
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# DETERMINE IF THE LINES ARE PARALLEL PERPENDICULAR OR NEITHER
How to decide whether the lines are parallel, perpendicular or neither ?
Parallel lines :
Parallel lines will have same slope and different y-intercepts.
m1 = m2
m1 and m2 are the slope of those two lines.
Perpendicular lines :
If two lines are perpendicular then product of their slopes will be equal to -1.
m1.m2 = -1
Coincident lines :
Coincident lines will have same slope and same y-intercepts.
The lines which has no relationship in between the slopes and y-intercepts can be considered neither.
Tell whether the lines are parallel, perpendicular, or neither.
Problem 1 :
Line 1 : through (-3, -7) and (1, 9)
Line 2 : through (-1, -4) and (0, -2)
Solution :
Slope of line 1 :x1 = -3, y1 = -7 x2 = 1, y2 = 9m1 = (y2 – y1)/(x2 – x1)= (9 + 7)/(1 + 3)= 16/4= 4 Slope of line 2 :x1 = -1, y1 = -4x2 = 0, y2 = -2m = (y2 – y1)/(x2 – x1)= (-2 + 4)/(0 + 1)= 2/1 = 2
They are not parallel, they are not perpendicular. So, the lines are neither.
Problem 2 :
Line 1 : through (2, 7) and (-1, -2)
Line 2 : through (3, -6) and (-6, -3)
Solution :
Slope of line 1 :x1 = 2, y1 = 7 x2 = -1, y2 = -2m1 = (y2 – y1)/(x2 – x1)= (-2 - 7)/(-1 - 2)= -9/(-3)= 3 Slope of line 2 :x1 = 3, y1 = -6x2 = -6, y2 = -3m2 = (y2 – y1)/(x2 – x1)= (-3 + 6)/(-6 - 3)= 3/-9 = -1/3
m1 x m2 = 3 (-1/3)
m1 x m2 = -1
So, the lines are perpendicular.
Determine whether the graphs of each pair of equations are parallel, perpendicular, or neither.
Problem 3 :
y = 3x + 4
y = 3x + 7
Solution :
y = 3x + 4 ----- (1)
y = 3x + 7 ------(2)
Comparing the above equations with slope intercept form
y = mx + b, we get
m1 = 3, m2 = 3
b1 = 4, b2 = 7
Both lines are having same slope. but different y – intercepts. So, the given lines are parallel.
Problem 4 :
y = -4x + 1
4y = x + 3
Solution :
y = -4x + 1 ------(1)
4y = x + 3
y = x/4 + 3/4 ------(2)
m = -4, m = 1/4
m1 m2 = -4 (1/4)
m1 m2 = -1
Since the product of the slopes is equal to -1. The lines are perpendicular.
Problem 5 :
y = 2x - 5
y = 5x - 5
Solution :
y = 2x – 5 -----(1)
y = 5x - 5 -----(2)
Comparing (1) and (2)
y = mx + b
m = 2, m = 5
b = -5, b = -5
There is no relationship between slopes, so they are neither.
Problem 6 :
y = -1/3x + 2
y = 3x - 5
Solution :
y = -1/3x + 2 -----(1)
y = 3x - 5 -----(2)
y = mx + b
m1 = -1/3, m2 = 3
b = 2, b = -5
m1 = -1/3, m2 = 3
m⋅ m2 = (-1/3) ⋅ 3
Product of the slope is equal to -1. So, the lines are perpendicular.
Problem 7 :
y = 3/5x - 3
5y = 3x - 10
Solution :
y = 3/5x – 3 ------(1)
5y = 3x - 10
y = 3/5x - 10 ------(2)
y = mx + b
m1 = 3/5, m2 = 3/5
b = -3, b = -10
Both lines have the same slope but have different y-intercepts. So, the lines are parallel.
Problem 8 :
y = 4
4y = 6
Solution :
y = 4 ------(1)
4y = 6
y = 6/4
y = 3/2------(2)
Both are horizontal lines and they will not intersect.
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### Document
```Mathematical Induction
Goals
Explain & illustrate construction of proofs of a variety of
theorems using mathematical induction.
Motivation
• Mathematics uses 2 kinds of arguments:
• deductive
• inductive
• Proposition: P( n ): 1 + 2 + … + n = n( n + 1 )/2.
• Observe that P(1), P(2), P(3), & P(4). Conjecture: nN P( n ).
• Mathematical induction is a finite proof pattern for proving
propositions of the form nN P( n ).
The Principle of Mathematical
Induction
•
Let P( n ) be a predicate function: nN P( n ) is a
proposition.
•
To prove nN P( n ), it suffices to prove:
1. P( 1 ) is true.
2. nN ( P( n ) P( n + 1 ) ).
•
This is not magic.
•
It is a recipe for constructing a finite proof for arbitrary nN.
Proving P( 3 )
• Given P( 1 ) n 1 ( P( n ) P( n + 1) ).
• Proof:
1. P( 1 ).
[premise 1]
2. P( 1 ) P( 2 ).
[U.S. of premise 2 for n = 1]
3. P( 2 ).
[step 1, 2, & modus ponens]
4. P( 2 ) P( 3 ).
[U.S. of premise 2 for n = 2]
5. P( 3 ).
[step 2, 3, & modus ponens]
• Construct a finite proof for P( 1,999,765 ).
Mathematical Induction as the
Domino Principle
If
the 1st domino falls over
and
the nth domino falls over implies that the ( n + 1 )st domino
falls over
then
domino n falls over for all n N.
Mathematical Induction as the
Domino Principle
1
2
3
n
n +1
The 3-Step Method
•
The implication in step 2 typically is proved directly.
•
The proof pattern thus has 3 steps:
1.
Prove P( 1 ).
[called the basis]
2.
Assume P( n )
[called the induction hypothesis]
3.
Prove P( n + 1 )
[called the inductive step]
•
The last 2 steps are for arbitrary n N.
•
Using P( n ) to prove P( n + 1 ) implies a recursive formulation of P( n ).
Induction as a Creative Process
• Mathematical induction is similar to, but not identical to,
scientific induction.
• In both cases, a “theory” is created.
• Look at specific cases; perceive a pattern.
• Hypothesizing a pattern, a theory, is a creative process
(only people who are bad at mathematics say otherwise).
• With mathematical induction, a “theory” can be proved.
• Scientific theories cannot be proved.
• They can be disproved.
• A scientific theory can be based on a mathematical model.
• Propositions can be proved within the model.
• Like axioms, the relationship between:
• the mathematical model
• physical reality
cannot be proven correct.
Example
• 1=1
• 3=1+2
• 6=1+2+3
• 10 = 1 + 2 + 3 + 4
• What is a general formula, if any, for 1 + 2 + … + n?
• Let F( n ): 1 + 2 + . . . + n.
• A recursive formulation: F( n ) = F( n - 1 ) + n.
A Geometric Interpretation
• 1:
• 2:
• 3:
• Put these blocks, which represent numbers, together to
form sums:
• 1+2=
• 1+2+3=
n
n
Area is n2/2 + n/2 = n(n + 1)/2
1 + 2 + … + n = n(n + 1)/2
A Mathematical Induction Proof
• F( 1 ) = 1( 1 + 1 )/2 = 1.
• Assume F( n ) = n( n + 1 )/2
• Show F( n + 1 ) = ( n + 1 )( n + 2 )/2.
F( n + 1 ) = 1 + 2 + . . . + n + ( n + 1 ) [Definition]
= F( n ) + n + 1
[Recursive formulation]
= n( n + 1 ) / 2 + n + 1
[Induction hyp.]
= n( n + 1 ) / 2 + ( n + 1 ) (2/2)
= ( n + 1 ) ( n + 2 ) / 2.
• In finding a recursive formulation, we focused on the:
• similarities
• differences
for successive values of n.
• Sometimes, it is useful to:
• Note the difference between F( n ) & F( n – 1 ).
• Find a pattern in this sequence of differences.
Example:
3
1
+
3
2
+...+
• Let F( n ) = 13 + 23 + . . . + n3.
• What is a formula for F(n)?
•
1
= 13
•
9
= 13 + 23
•
36 = 13 + 23 + 33
•
100 = 13 + 23 + 33 + 43
• Do you see a pattern?
3
n
=?
Prove that n F( n ) = [ n( n + 1 )/2 ]2
1. F( 1 ) = 13 = 1 = [ 1( 2 ) / 2 ]2.
2. Assume F( n ) = [ n( n + 1 ) / 2 ]2 .
I.H.
3. Prove F( n + 1 ) = [ ( n + 1 )( n + 2 ) / 2 ]2.
F ( n + 1 ) = 13 + 23 + . . . + n3 + ( n +1 )3
Defn. of F( n + 1 )
= F( n ) + ( n + 1 )3
Recursive formulation
= [ n( n + 1 )/ 2 ]2 + ( n + 1 )3
Use I. H.
= ( n + 1 )2[ ( n / 2 )2 + ( n + 1 ) ]
= ( n + 1 )2[ n2 / 4 + ( 4 / 4 )( n + 1 ) ]
= ( n +1 )2[ ( n2 + 4n + 4 ) / 4 ] = [ ( n + 1) (n + 2) / 2 ]2.
Translating the starting point
•
If we:
• know P( n ) is false for 1 n 9
• think P( n ) is true for n > 9.
•
Then define Q( n ) = P( n + 9 ).
• Use mathematical induction to show that n N Q( n ).
• We thus can start the induction at any natural number.
Example: Stamps
• Suppose the US Post Office prints only 5 & 9 cent stamps.
• Prove n > 34, you can make postage for n cents, using
only 5 & 9 cent stamps.
Let S( n ) denote the statement: You can make postage for
n cents using only 5-cent & 9-cent stamps.
1. Basis: For n = 35: Use 7 5-cent stamps.
2. I.H.: Assume S( n ).
3. Prove S( n + 1 ).
Case: For S( n ), # of 9-cent stamps used = 0:
Only 5-cent stamps are used for S( n ).
# of 5-cent stamps ≥ 7.
Replace 7 5-cent stamps with 4 9-cent stamps.
Case: For S( n ), # of 9-cent stamp used > 0:
Replace 1 9-cent stamp with 2 5-cent stamps.
Generalizing the Basis
• To prove nN P( n ), if suffices to show:
• P( 1 ) P( 2 ).
• nN( [ P( n ) P( n + 1 ) ] P( n + 2 ) )
• If:
• We can push over the first 2 dominos;
• Pushing over any 2 adjacent dominos implies pushing
over the next domino.
then we can push over all the dominos.
The Fibonacci Formula
• Define the nth Fibonacci number, F( n ), as:
• F( 0 ) = 0, F( 1 ) = 1,
• F( n ) = F( n – 1 ) + F( n – 2 ).
• Prove
F( n ) = 5-1/2 ( [ ( 1 + 51/2 ) / 2]n - [ ( 1 - 51/2 ) / 2 ]n ).
Basis: F( 0 ) = 5-1/2 ( [ ( 1 + 51/2 ) /2 ]0 - [ (1 - 51/2) / 2 ]0 )
= 0.
F( 1 ) = 5-1/2 ( [ ( 1 + 51/2 ) / 2 ]1 - [ ( 1 - 51/2 ) / 2 ]1 )
= ( 5-1/2 / 2 ) ( 1 + 51/2 - 1 + 51/2 )
= 1.
The Fibonacci Formula
F( n ) = 5-1/2 ( [ ( 1 + 51/2 ) / 2]n - [ ( 1 - 51/2 ) / 2 ]n )
• Let a = ( 1 + 51/2 ) / 2
b = ( 1 - 51/2 ) / 2.
•
Note: a + 1 = a2 & b + 1 = b2.
• Induction hypotheses:
F( n )
= 5-1/2 ( an – bn )
F( n +1 ) = 5-1/2 ( an + 1 – bn + 1 ).
• Induction step: Show F( n + 2 ) = 5-1/2 ( an + 2 - bn + 2 ).
F( n ) = 5-1/2 ( [ ( 1 + 51/2 ) / 2]n - [ ( 1 - 51/2 ) / 2 ]n )
Proof of Induction Step
F( n + 2 ) = F ( n + 1 ) + F ( n )
[Definition]
= 5-1/2 ( an + 1 – bn + 1 ) + 5-1/2 ( an – bn )
= 5-1/2 ( an + 1 + an – bn + 1 – bn )
= 5-1/2 ( an ( a + 1 ) – bn ( b + 1 ) ).
= 5-1/2 ( an + 2 – bn + 2 )
[I.H.]
Generalizing this ...
• If
• P( 1 ) P( 2 ) . . . P( k )
• n { [ P( n + 1 ) P( n + 2 ) . . . P( n + k ) ] P( n + k + 1 ) }
• then n N P( n ).
End
Strong Mathematical Induction
• If
• P( 1 ) P( 2 ) . . . P( k ) and
• for n k,
[ P( 1 ) P( 2 ) . . . P( n ) ] P( n + 1 )
• then, n N P(n).
Example: Fundamental Theorem
of Arithmetic
• Prove that all natural numbers 2 can be
represented as a product of primes.
• Basis: 2: 2 is a prime.
• Assume that 1, 2, . . . , n can be represented
as a product of primes.
• Show that n + 1can be represented as a
product of primes.
• Case n + 1 is a prime: It can be represented as a
product of 1 prime, itself.
• Case n + 1 is composite: n = ab, for some a,b < n.
• Therefore, a = p1p2 . . . pk & b = q1q2 . . . ql, where the
pis & qis are primes.
• Represent n = p1p2 . . . pkq1q2 . . . ql.
|
# INTERPRETING THE UNIT RATE AS SLOPE
A rate is a comparison of two quantities that have different units, such as miles and hours. A unit rate is a rate in which the second quantity in the comparison is one unit.
Example 1 :
A storm is raging on Misty Mountain. The graph shows the constant rate of change of the snow level on the mountain.
A. Find the slope of the graph using the points (1, 2) and (5, 10). Remember that the slope is the constant rate of change.
Change in y-value/Change in x-value :
= (10 - 2)/(5 - 1)
= 8/4
= 2
B. Find the unit rate of snowfall in inches per hour. Explain your method.
2 inches per hour; Sample answer : The point (1, 2) is on the line, and represents 2 inches snowfall in 1 hour.
C. Compare the slope of the graph and the unit rate of change in the snow level. What do you notice ?
They are the same.
D. Which unique point on this graph can represent the slope of the graph and the unit rate of change in the snow level ? Explain how you found the point.
(1, 2) ; Sample answer : the unit rate is the amount of snow in 1 hour. So I found the point with an x-coordinate of 1. That point is (1, 2), which, along with another point on the graph, gives 2 as the slope.
Example 2 :
The equation y = 2.75x represents the rate, in barrels per hour, that oil is pumped from Well A. The graph represents the rate that oil is pumped from Well B. Which well pumped oil at a faster rate ?
Solution :
Step 1 :
Use the equation y = 2.75x to make a table for Well A’s pumping rate, in barrels per hour.
Step 2 :
Use the table to find the slope of the graph of Well A.
Slope = Unit rate
= (5.5 - 2.75)/(2 - 1)
= 2.75/1
= 2.75 barrels/hour.
Step 3 :
Use the graph to find the slope of the graph of Well B.
Slope = Unit rate
= rise/run
= 10/4
= 2.5 barrels/hour.
Step 4 :
Compare the unit rates.
2.75 > 2.5
So Well A’s rate, 2.75 barrels/hour, is faster.
Reflect :
Describe the relationships among the slope of the graph of Well A’s rate, the equation representing Well A’s rate, and the constant of proportionality
The slope and the constant of proportionality equal the value 2.75 in the equation y = 2.75x.
Kindly mail your feedback to [email protected]
We always appreciate your feedback.
## Recent Articles
1. ### Finding Two Numbers with the Given Difference and Product
Sep 29, 23 10:55 PM
Finding Two Numbers with the Given Difference and Product
2. ### Finding Two Numbers with the Given Sum and Product
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# An unbiased die is thrown. What is the probability of getting:(1) an even number and (2) a multiple of 3.
Last updated date: 15th Jul 2024
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Hint: This question is simply based on the definition of probability. Here we will first find the favourable events and then use a formula for finding the probability. In such a type of question where an occurrence of one event is a subset of the other event.
Complete step-by-step solution -
In the question it is given that an unbiased coin is thrown and we have to find the probability of getting an even number in the first part and a multiple of 3 in the second part.
Sample space when a dice is thrown is $\left\{ {1,2,3,4,5,6} \right\}$
$\therefore$ On throwing a dice total possible events =6
(1)
Here the favourable event is the occurrence of an even number.
Sample space of favourable event is $\left\{ {2,4,6} \right\}$
So total number of favourable events = 3.
Now, we know that the formula for finding probability is given as:
Probability = $\dfrac{{{\text{Total number of favourable events}}}}{{{\text{Total number of possible events}}}}$ .
Putting the values in above formula, we get:
Probability = $\dfrac{3}{6} = \dfrac{1}{2}$
(2)
Here the favourable event is the occurrence of a multiple of 3.
Sample space of favourable event is $\left\{ {3,6} \right\}$ .
So total number of favourable events = 2
Now, we know that the formula for finding probability is given as:
Probability = $\dfrac{{{\text{Total number of favourable events}}}}{{{\text{Total number of possible events}}}}$ .
Putting the values in above formula, we get:
Probability = $\dfrac{2}{6} = \dfrac{1}{3}$ .
Note: Before solving this type of question you should remember the formula for finding the probability. Here a dice is thrown. You should know that on throwing a dice the total number of possible events is equal to 6. And depending on what is given is a question you have to calculate the number of favourable events and then use the formula for finding the probability.
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# 5.8 Modeling using variation (Page 5/14)
Page 5 / 14
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies jointly as the square of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and of $\text{\hspace{0.17em}}z\text{\hspace{0.17em}}$ and inversely as the square root of $\text{\hspace{0.17em}}w\text{\hspace{0.17em}}$ and of $\text{\hspace{0.17em}}t\text{\hspace{0.17em}.}$ When $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ $z=3,\text{\hspace{0.17em}}$ $w=16,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}t=3,\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}y=1.\text{\hspace{0.17em}}$ Find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x=3,\text{\hspace{0.17em}}$ $z=2,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}t=5.$
## Technology
For the following exercises, use a calculator to graph the equation implied by the given variation.
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly with the square of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and when
$y=\frac{3}{4}{x}^{2}$
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly as the cube of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and when
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly as the square root of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and when
$y=\frac{1}{3}\sqrt{x}$
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies inversely with $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and when
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies inversely as the square of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and when
$y=\frac{4}{{x}^{2}}$
## Extensions
For the following exercises, use Kepler’s Law, which states that the square of the time, $\text{\hspace{0.17em}}T,\text{\hspace{0.17em}}$ required for a planet to orbit the Sun varies directly with the cube of the mean distance, $\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}$ that the planet is from the Sun.
Using Earth’s time of 1 year and mean distance of 93 million miles, find the equation relating $\text{\hspace{0.17em}}T\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$
Use the result from the previous exercise to determine the time required for Mars to orbit the Sun if its mean distance is 142 million miles.
1.89 years
Using Earth’s distance of 150 million kilometers, find the equation relating $\text{\hspace{0.17em}}T\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}a.$
Use the result from the previous exercise to determine the time required for Venus to orbit the Sun if its mean distance is 108 million kilometers.
0.61 years
Using Earth’s distance of 1 astronomical unit (A.U.), determine the time for Saturn to orbit the Sun if its mean distance is 9.54 A.U.
## Real-world applications
For the following exercises, use the given information to answer the questions.
The distance $\text{\hspace{0.17em}}s\text{\hspace{0.17em}}$ that an object falls varies directly with the square of the time, $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ of the fall. If an object falls 16 feet in one se c ond, how long for it to fall 144 feet?
3 seconds
The velocity $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ of a falling object varies directly to the time, $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ of the fall. If after 2 seconds, the velocity of the object is 64 feet per second, what is the velocity after 5 seconds?
The rate of vibration of a string under constant tension varies inversely with the length of the string. If a string is 24 inches long and vibrates 128 times per second, what is the length of a string that vibrates 64 times per second?
48 inches
The volume of a gas held at constant temperature varies indirectly as the pressure of the gas. If the volume of a gas is 1200 cubic centimeters when the pressure is 200 millimeters of mercury, what is the volume when the pressure is 300 millimeters of mercury?
The weight of an object above the surface of Earth varies inversely with the square of the distance from the center of Earth. If a body weighs 50 pounds when it is 3960 miles from Earth’s center, what would it weigh it were 3970 miles from Earth’s center?
49.75 pounds
The intensity of light measured in foot-candles varies inversely with the square of the distance from the light source. Suppose the intensity of a light bulb is 0.08 foot-candles at a distance of 3 meters. Find the intensity level at 8 meters.
The current in a circuit varies inversely with its resistance measured in ohms. When the current in a circuit is 40 amperes, the resistance is 10 ohms. Find the current if the resistance is 12 ohms.
33.33 amperes
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
|
# Subtraction With Regrouping Worksheets
Once students master simple subtraction, they quickly move to 2-digit subtraction, which often requires students to apply the concept of “borrow one” to subtract correctly without producing a negative number. The best way to demonstrate this concept to young mathematicians is to illustrate the process of subtracting each number from a 2-digit number in the equation by splitting it into individual columns where the first number of the number to be subtracted lines up with the first number of its subtraction number.
So-called manipulative tools such as number lines or counters can also help students understand the concept of regrouping, which is a technical term for “borrowing one”, in which they can use one to avoid negative numbers in the process of subtracting 2 digit numbers from one another.
## Describe Linear Subtraction of 2-Digit Numbers
These simple subtraction worksheets (# 1, # 2, # 3, # 4, and # 5) help guide students through the process of subtracting 2-digit numbers from each other, which often requires regrouping if the reduced numbers require students to ” borrow one “from the greater decimal point.
The concept of borrowing one in simple subtraction comes from the process of subtracting each number in a 2-digit number from the one directly above when structured like Question 13 on worksheet # 1:
24-16
In this case, 6 cannot be subtracted from 4, so students have to “borrow one” from 2 in 24 to subtract 6 from 14, making the answer to this problem 8. Neither of the problems on this worksheet resulted in a negative number, which must be handled after students understand the core concept of subtracting positive numbers from each other, often first illustrated by presenting the sum of an item such as an apple and asking what happens when x the sum of them is taken away.
## Samples
3 Digit Subtraction With Regrouping Worksheets
Darlene Young
February 18, 2021
Darlene Young
February 18, 2021
Number Subtraction With Regrouping Worksheets
Darlene Young
February 18, 2021
Remember when you challenge your students with worksheets # 6, # 7, # 8, # 9, and # 10 that some children will need manipulatives like number lines or counters.
This visual tool helps explain a regrouping process where they can use a number line to keep track of the number that is subtracted as it “gets one” and jumps up by 10 then the natural number under is subtracted.
As with 78 – 49, a student would use a number line to check 9 out of 49 individually which was subtracted from 8 in 78, regrouped to 18 – 9, then subtracted the 4 from the remaining 6 after regrouping 78 to 60 + (18 – 9) – 4.
Again, it is easier to explain to students when you allow them to cross numbers and practice on questions like the one on the worksheet above. By presenting the equation linearly with the decimal places of each 2 digit number which is aligned with the number below, students will be better able to understand the concept of regrouping.
To solve problems faced by students, teacher solutions, so that students grow enthusiastic about learning mathematics, in this case the teacher’s division material applies the division method with repeated subtraction until the result is zero.
First of all the teacher gave the method to solve questions 35: 7 ……. ? the child observes the numbers written by the teacher. After observing the leading number which is 35, the division symbol then the number 7. The number in front, the number that is subtracted. The number that follows the number used to subtract and is repeated until the result is zero.
After students understand the method conveyed by the teacher, students begin to practice and try to operate the calculation of division with repeated subtraction 35: 7 (35-7-7-7-7-7) = 0. Number 7 how many there are? Of course, the child will answer there 5 times so that it can be reversed 35: 5 (35-5-5-5-5-5-5) = 0 number 5 totaling 7. In addition, the teacher proves by bringing a can of 100 candy seeds, then divided into one class containing 25 students each student takes one by one until it runs out. Each child gets 4 candies and sees that the can is empty so that 100: 25 = 4 and 100: 25 (100-25-25-25-25) = 0.
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# 数学代写|线性代数代写linear algebra代考|МАTH1014
## 数学代写|线性代数代写linear algebra代考|THE COMPOSITION OF LINEAR TRANSFORMATIONS
In this section, we deal with the composition of linear transformations. If $F$ and $G$ are linear transformations respectively associated with two matrices $A$ and $B$ (with respect to the canonical bases), we have that the composition $F \circ G$ is a linear transformation and we want to know which matrix is associated with it.
We first recall the definition of composition of functions. If $g: A \rightarrow B$ and $f: B \rightarrow C$ are functions, and the domain of $f$ coincides with the codomain of $g$, we define the composition as $f \circ g: A \rightarrow C$, which corresponds to applying first $g$, then f. Formally:
$$\begin{array}{rlcc} f \circ g: A & \longrightarrow & C \ a & \mapsto & f(g(a)) . \end{array}$$
We observe that the composition is associative, that is, if we have three functions $h: A \rightarrow B, g: B \rightarrow C, f: C \rightarrow D$, we have that $f \circ(g \circ h)=(f \circ g) \circ h$. In fact, for every $a \in A$ :
$$(f \circ(g \circ h))(a)=f((g \circ h)(a))=f(g(h(a)))=(f \circ g)(h(a))=((f \circ g) \circ h)(a) .$$
The composition operation is not commutative, i.e. general $f \circ g \neq g \circ f$ and it might even happen that $g \circ f$ is not defined.
Example 5.3.1 Let
We compute $f \circ g: \mathbb{R} \rightarrow \mathbb{R}$ and $g \circ f: \mathbb{R} \rightarrow \mathbb{R}$
$$f \circ g: x \mapsto g(x)=x+2 \mapsto f(x+2)=(x+2)^2-1=x^2+2 x+3,$$
$$g \circ f: y \mapsto f(y)=y^2-1 \mapsto g\left(y^2-1\right)=\left(y^2-1\right)+2=y^2+1 \text {. }$$
In this case, $f \circ g \neq g \circ f$.
Let us now see an important example in linear algebra.
Example 5.3.2 Consider the two linear transformations $L_A: \mathbb{R}^3 \longrightarrow \mathbb{R}^2, L_B$ : $\mathbb{R}^2 \longrightarrow \mathbb{R}^2$ associated with the matrices:
$$A=\left(\begin{array}{ccc} -1 & 1 & 2 \ 3 & 1 & 0 \end{array}\right), \quad B=\left(\begin{array}{ll} 2 & 1 \ 1 & 3 \end{array}\right) \text {, }$$
with respect to the canonical bases in $\mathbb{R}^2$ and $\mathbb{R}^3$. We see immediately that $L_B \circ L_A$ is defined, while $L_A \circ L_B$ it is not defined. This is because $L_A$ must have as argument a vector in $\mathbb{R}^3$, while for every $\mathbf{v} \in \mathbb{R}^2$ we have that $L_B(\mathbf{v}) \in \mathbb{R}^2$, then $L_A\left(L_B(\mathbf{v})\right)$ does not make sense.
Let us now see that, whenever we can take the composition of two linear transformations, we still obtain a linear transformation.
## 数学代写|线性代数代写linear algebra代考|KERNEL AND IMAGE
We now want to get into the theory of linear transformation and introduce the concepts of kernel and image, which are respectively subspaces of the domain and codomain of a linear transformation.
Definition 5.4.1 Let $V$ and $W$ be two vector spaces and $L: V \rightarrow W$ be a linear transformation. We call kernel of $L$, the set of vectors in $V$ whose image is the zero vector of $W$. This set is denoted by $\operatorname{Ker} L$.
We call image of $L$, the set of vectors in $W$ which are images of some vectors of $V$, that is,
$$\operatorname{Im}(L)={\mathbf{w} \in W \quad \mid \mathbf{w}=L(\mathbf{v}) \text { for some } \mathbf{v} \in V} .$$
Let us see some examples.
Example 5.4.2 1. Consider the derivation $D: \mathbb{R}[x] \longrightarrow \mathbb{R}[x]$, defined by $D(p(x))=p^{\prime}(x)$. As we have seen $D$ is a linear transformation. We want to know which are the polynomial $p(x)$ whose image is zero, i.e. such that $D(p(x))=0$. From calculus, we know they are all the constant polynomials. So $\operatorname{Ker}(D)={c \mid c \in \mathbb{R}}$. Let us look at the image of $D$. We ask which polynomials are derivatives of other polynomials. From calculus we know they are all the polynomials (we are in fact integrating), so $\operatorname{Im}(D)=\mathbb{R}[x]$.
1. Consider now the linear transformation $L: \mathbb{R}^3 \longrightarrow \mathbb{R}^2$ defined by: $L\left(\mathbf{e}_1\right)=2 \mathbf{e}_1-\mathbf{e}_2$, $L\left(\mathbf{e}_2\right)=\mathbf{e}_1, L\left(\mathbf{e}_3\right)=\mathbf{e}_1+2 \mathbf{e}_2$.
As seen in Theorem $5.2 .2$, we know that $L(x, y, z)=(2 x+y+z,-x+2 z)$. We want to determine kernel and image of $L$. We have that $\operatorname{Ker}(L)$ is the set of vectors whose image is the zero vector, that is,
\begin{aligned} \operatorname{Ker}(L) &={(x, y, z) \mid(2 x+y+z,-x+2 z)=(0,0)} \ &={(x, y, z) \mid 2 x+y+z=0,-x+2 z=0} \ &=\left{(x, y, z) \mid z=\frac{1}{2} x, y=-2 x-z=-\frac{5}{2} x\right}=\left{\left(x,-\frac{5}{2} x, x\right) \mid x \in \mathbb{R}\right} \ &=\langle(2,-5,1)\rangle \end{aligned}
# 抽象代数代考
## 数学代写|线性代数代写linear algebra代考|THE COMPOSITION OF LINEAR TRANSFORMATIONS
$$f \circ g: A \longrightarrow C a \mapsto f(g(a)) \text {. }$$
$$(f \circ(g \circ h))(a)=f((g \circ h)(a))=f(g(h(a)))=(f \circ g)(h(a))=((f \circ g) \circ h)(a) .$$
$$f \circ g: x \mapsto g(x)=x+2 \mapsto f(x+2)=(x+2)^2-1=x^2+2 x+3,$$
$$g \circ f: y \mapsto f(y)=y^2-1 \mapsto g\left(y^2-1\right)=\left(y^2-1\right)+2=y^2+1 .$$
$$A=\left(\begin{array}{llllll} -1 & 1 & 2 & 3 & 1 & 0 \end{array}\right), \quad B=\left(\begin{array}{llll} 2 & 1 & 1 & 3 \end{array}\right),$$
## 数学代写|线性代数代写linear algebra代考|KERNEL AND IMAGE
$$\operatorname{Im}(L)=\mathbf{w} \in W \quad \mid \mathbf{w}=L(\mathbf{v}) \text { for some } \mathbf{v} \in V .$$
1. 现在考虑线性变换 $L: \mathbb{R}^3 \longrightarrow \mathbb{R}^2$ 被定义为: $L\left(\mathbf{e}_1\right)=2 \mathbf{e}_1-\mathbf{e}_2, L\left(\mathbf{e}_2\right)=\mathbf{e}_1, L\left(\mathbf{e}_3\right)=\mathbf{e}_1+2 \mathbf{e}_2$.
如定理所示 $5.2 .2$ ， 我们知道 $L(x, y, z)=(2 x+y+z,-x+2 z)$. 我们要确定内核和图像 $L$. 我们有那个 $\operatorname{Ker}(L)$ 是其图像为雴向量的向量集合，即
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# Definition of Continuity of a Function
This tutorial is for you if you are searching for – “Definition of Continuity of a function, Discontinuity and Missing Point Discontinuity.”
Let’s begin –
## Continuous Functions
A Function for which a small change in the independent variable causes only a small change and not a sudden jump in the dependent variable are called continuous functions. Naively, we may say that a function is continuous at a fixed point if we can draw the graph of the function around that point without lifting the pen from the plane of the paper.
## Continuity of a function at a point
A function f(x) is said to be continuous at x = a, if
$$\displaystyle{\lim_{x \to a}}$$ f(x) = f(a).
Symbolically f is continuous at x = a if
$$\displaystyle{\lim_{h \to 0}}$$ f(a – h) = $$\displaystyle{\lim_{h \to 0}}$$ f(a + h) = f(a), h > 0
i.e. $$LHL_{x = a}$$ = $$RHL_{x = a}$$ equals value of ‘f’ at x = a.
Example : If f(x) = {$$sin{\pi x\over 2}$$, x < 1 and [x], x $$\geq$$ 1} then find whether f(x) is continuous or not at x = 1, where [ ] denotes greatest integer function.
Solution : For continuity at x = 1, we determine, f(1), $$\displaystyle{\lim_{x \to {1^-}}}$$ f(x) and $$\displaystyle{\lim_{x \to {1^+}}}$$ f(x)
Now, f(1) = [1] = 1
$$\displaystyle{\lim_{x \to {1^-}}}$$ f(x) = $$\displaystyle{\lim_{x \to {1^-}}}$$ $$sin{\pi x\over 2}$$ = $$sin{\pi\over 2}$$ = 1 and $$\displaystyle{\lim_{x \to {1^+}}}$$ f(x) = $$\displaystyle{\lim_{x \to {1^+}}}$$ [x] = 1
so f(1) = $$\displaystyle{\lim_{x \to {1^-}}}$$ f(x) = $$\displaystyle{\lim_{x \to {1^+}}}$$ f(x)
$$\therefore$$ f(x) is continuous at x = 1.
## Continuity of a function in an interval
(a) A function is said to be continuous in (a,b) if f is continuous at each & every point belonging to (a, b).
(b) A function is said to be continuous in a closed interval [a,b] if :
(i) f is continuous in the open interval (a,b)
(ii) f is right continuous at ‘a’ i.e. $$\displaystyle{\lim_{x \to {a^+}}}$$ f(x) = f(a) = a finite quantity.
(iii) f is left continuous at ‘b’ i.e. $$\displaystyle{\lim_{x \to {b^-}}}$$ f(x) = f(b) = a finite quantity.
Note :
(i) All polynomials, trigonometrical functions, exponential & logarithmic functions are continuous in their domains.
(ii) If f(x) & g(x) are two functions that are continuous at x = c then the function defined by:
$$F_1(x)$$ = f(x) + g(x); $$F_2(x)$$ = Kf(x), where K is any real number; $$F_3(x)$$ = f(x).g(x) are also continuous at x = c.
Further, if g(c) is not zero, then $$F_4(x)$$ = $$f(x)\over g(x)$$ is also continuous at x = c.
Hope, you learnt definition of continuity of a function and continuity of a function at a point and over an interval. Practice more question on continuity of a function to learn more and get ahead in competition. Good Luck!
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# Know Addition and Subtraction Facts to 100
In this worksheet, students will practise their addition and subtraction number bonds to 20 and use them to derive related facts to 100.
Key stage: KS 1
Curriculum topic: Number: Addition and Subtraction
Curriculum subtopic: Use Addition/Subtraction Facts to 100
Popular topics: Subtraction worksheets, Addition worksheets
Difficulty level:
#### Worksheet Overview
Look at this number square, which shows all the bonds you should know by heart up to 20.
You can use this table to help to work out additions and subtractions right up to 100.
Example 1
Method 1:
Split 36 into 3 tens and 6 ones.
Add the ones together: 6 + 9 = 15 This is where the number square above would help you!
Now add 30 + 15 together. You could split 15 into 1 ten and 5 ones: 30 + 10 = 40 and then 40 + 5 = 45
Method 2:
Can you make the calculation easier by slightly changing it?
In this example, you are adding 9. How much easier would it be if you added 10?
Let's try: 36 + 10 = 46. Easy right?
But, we were supposed to add only 9! So, simply subtract 1 and you've got your answer: 46 - 1 = 45
Example 2
Do this subtraction: 15 - 9
Method 1:
Look at the number square. When you subtract, the starting number is inside the square.
Find 15 in the square.
Follow the lines and columns to the numbers that add together to make 15, and you get 6 and 9.
So, 15 - 9 = 6
Method 2:
Can you make the calculation easier by slightly changing it?
In this example, you are subtracting 9. How much easier would it be if you subtracted 10?
Let's try: 15 - 10 = 5
But, we were supposed to subtract 9! So, simply add 1 and you've got your answer: 5 + 1 = 6
So, 15 - 9 = 6
Can you see that there is often more than one way of working out the answer to addition and subtraction problems? Just find the way that works best for each question.
Let's have a go at some questions now.
If you want to look at the 100 square while you're doing the questions, just click on the red help button on the screen.
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# How to Find the Surface Area of a Pyramid
The surface area of any pyramid can be found by adding the surface area of the base to the surface area of the lateral faces. When working with regular pyramids, you can find the surface area using a formula, as long as you know how to find the area of the base of the pyramid. Since the base can be any polygon, it is helpful to know how to find the area of shapes such as pentagons and hexagons. When working with the common, regular square pyramid, however, calculating the total surface area is a simple calculation, provided you know the slant height of the pyramid and the side length of the square base.
Method 1
Method 1 of 2:
### Finding the Surface Area of Any Regular Pyramid
1. 1
Set up the formula for the surface area of a regular pyramid. The formula is ${\displaystyle SA={\frac {p\times h}{2}}+B}$, where ${\displaystyle SA}$ equals the total surface area of the pyramid, ${\displaystyle p}$ equals the perimeter of the base, ${\displaystyle h}$ equals the slant height of the pyramid, and ${\displaystyle B}$ equals the area of the base.[1]
• The basic formula for the surface area of any pyramid, regular or irregular, is Total Surface Area = Base Area + Lateral Area.[2]
• Don’t confuse “slant height” with “height.” The “slant height” is the diagonal distance from the apex of the pyramid to the edge of the base.[3] The “height” is the perpendicular distance from the vertex to the base.
2. 2
Plug the perimeter of the base into the formula. If you aren’t given the perimeter but know the length of one edge of the base, you can calculate the perimeter by multiplying the length of one edge by the number of edges.
• For example, If you are finding the surface area of a hexagonal pyramid, and you know that the length of one edge of the base is 4 cm, you would calculate ${\displaystyle 4\times 6=24}$ to find the perimeter of the base, since a hexagon has six edges, or sides. Thus, the perimeter of the base is 24 cm, so your surface area formula will look like this: ${\displaystyle SA={\frac {24\times h}{2}}+B}$.
3. 3
Plug the value of the slant height into the formula. Make sure you are using the slant height, not the perpendicular height. The problem should provide the slant height. If you don’t know the slant height, you cannot use this method.
• For example, if the slant height of a hexagonal pyramid is 12 cm, your formula will look like this: ${\displaystyle SA={\frac {24\times 12}{2}}+B}$.
4. 4
Calculate the area of the base. How you do this will depend on the shape of the base. To learn more about finding the area of a polygon, read Find the Area of Regular Polygons.
• For example, if you are working with a hexagonal pyramid, the base is a hexagon. To find out how to calculate the area of the base, you can read Calculate the Area of a Hexagon. The formula is ${\displaystyle A={\frac {3{\sqrt {3}}\times s^{2}}{2}}}$, where ${\displaystyle s}$ is the length of one side of the hexagon. Since the length of one side of the hexagon is 4 cm, you would calculate:
${\displaystyle A={\frac {3{\sqrt {3}}\times 4^{2}}{2}}}$
${\displaystyle A={\frac {3{\sqrt {3}}\times 16}{2}}}$
${\displaystyle A={\frac {48{\sqrt {3}}}{2}}}$
${\displaystyle A={\frac {83.14}{2}}}$
${\displaystyle A=41.57}$.
So the area of the base is 41.57 square centimeters.
EXPERT TIP
Math Instructor, City College of San Francisco
Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University.
Grace Imson, MA
Math Instructor, City College of San Francisco
Our Expert Agrees: The surface area of a pyramid is equal to the sum of the areas of all of the faces. First, you have to get the area of the base, then add the area of the lateral sides, which is one face times the number of sides.
5. 5
Plug the area of the base into the formula. Make sure you substitute for the variable ${\displaystyle B}$.
• For example, if the area of the hexagonal base is 41.57 sq. cm., your formula for surface area will now look like this: ${\displaystyle SA={\frac {24\times 12}{2}}+41.57}$.
6. 6
Multiply the perimeter of the base and the slant height of the pyramid. Then, divide by two. This will give you the lateral surface area of the pyramid.
• For example:
${\displaystyle SA={\frac {24\times 12}{2}}+41.57}$
${\displaystyle SA={\frac {288}{2}}+41.57}$
${\displaystyle SA=144+41.57}$
7. 7
Add the two values together. The sum will be the lateral surface area, plus the base surface area, providing you with the total surface area for the pyramid, in square units.
• For example:
${\displaystyle SA=144+41.57}$
${\displaystyle SA=185.57}$
So, the total surface area of a hexagonal pyramid, given a base edge length of 4 cm and a slant height of 12 cm, is 185.57 square centimeters.
Method 2
Method 2 of 2:
### Finding the Surface Area of a Square Pyramid
1. 1
Set up the formula for surface area of a square pyramid. The formula is ${\displaystyle SA=b^{2}+4({\frac {bh}{2}})}$, where ${\displaystyle b}$ is equal to the length of one side of the base, and ${\displaystyle h}$ is equal to the slant height of the pyramid.
• Don’t confuse “slant height” with “height.” The “slant height” is the diagonal distance from the apex of the pyramid to the edge of the base.[4] The “height” is the perpendicular distance from the vertex to the base.
• Note that this formula is just another way of writing Total Surface Area = Base Area (${\displaystyle b^{2}}$) + Lateral Area (${\displaystyle 4({\frac {bh}{2}})}$). This formula only works for regular square pyramids.
2. 2
Plug in the values for the side length and slant height into the formula. Make sure you substitute the side length of the base for ${\displaystyle b}$ and the slant height for ${\displaystyle h}$.
• For example, if the length of one side of the base of a square pyramid is 4 cm, and the slant height is 12 cm, the formula will look like this: ${\displaystyle SA=4^{2}+4({\frac {(4)(12)}{2}})}$.
3. 3
Square the side length of the base. This will give you the surface area of the base.
• For example:
${\displaystyle SA=4^{2}+4({\frac {(4)(12)}{2}})}$
${\displaystyle SA=16+4({\frac {(4)(12)}{2}})}$
4. 4
Multiply the side length of the base by the slant height and divide by two. Then, multiply by 4. This will give you the lateral surface area of the pyramid.
• For example:
${\displaystyle SA=16+4({\frac {(4)(12)}{2}})}$
${\displaystyle SA=16+4({\frac {48}{2}})}$
${\displaystyle SA=16+4(24)}$
${\displaystyle SA=16+96}$
5. 5
Add the base surface area and the lateral surface area. This will give you the total surface area of the pyramid, in square units.
• For example:
${\displaystyle SA=16+96}$
${\displaystyle SA=112}$
So, the total surface area of a square pyramid, with a base side length of 4 cm and a slant height of 12 cm, is 112 square centimeters.
## Community Q&A
Search
• Question
How do you find the lateral area of hexagonal pyramid, given the height and length of each side?
Use the base times height for the rectangles, and the altitude times base of the hexagonal face times three.
• Question
How would you calculate the surface area of pyramid that does not have a square base?
Use the formula (p x h/2) + (B), where p is the perimeter of the base, h is the slant height of the pyramid, and B is the area of the base. Below are some articles on finding the area of a pentagon and hexagon, two common pyramid bases: http://www.wikihow.com/Find-the-Area-of-a-Pentagon http://www.wikihow.com/Calculate-the-Area-of-a-Hexagon
• Question
How do I double the lateral surface area of a square pyramid?
Donagan
One way would be to double either the length of the sides of the base or the slant height (but not both).
200 characters left
## Things You'll Need
• Pencil
• Paper
• Calculator (optional)
• Ruler (optional)
Co-authored by:
Math Instructor, City College of San Francisco
This article was co-authored by Grace Imson, MA. Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been viewed 382,817 times.
Co-authors: 22
Updated: September 16, 2021
Views: 382,817
Article SummaryX
To find the surface area of a pyramid, start by multiplying the perimeter of the pyramid by its slant height. Then, divide that number by 2. Finally, add the number you get to the area of the pyramid's base to find the surface area. To learn how to find the surface area of a square pyramid, scroll down!
Thanks to all authors for creating a page that has been read 382,817 times.
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# Sin Theta Calculator
Created by Luciano Miño
Reviewed by Luis Hoyos
Last updated: Feb 02, 2023
The sin theta calculator can aid you in your math problems and obtain the sine value of any angle.
We've paired this calculator with a few paragraphs covering:
• What is the definition of sin (sine definition);
• What sin of 0 is;
• Properties of sine; and
## What is the definition of sin/sine?
Before we start with the sine function definition, we need to introduce the unit circle. This circle is centered at the origin, and its radius equals one.
If we draw a line from the origin to any point on this unit circle, an angle theta $\theta$ will be formed between this radius and the horizontal axis.
The sine, or sin, is the y-axis coordinate of this radius as the angle changes describing any point on the unit circle.
Like cosine, sine is a periodic function with a period of . This means that for any argument $\theta$: $\sin(\theta + 2k\pi) = \sin(\theta)$ where $k$ is any integer.
💡 Test it out! Input any angle in our sin theta calculator and write down the sine result. Now try again with the same angle, but add 2*π (or 360°, if you're using degrees) to it and see if the results match.
## Properties of sine
Here's a list of some of the sine properties and trigonometric identities involving the sine function:
$\sin(-x)$ $= -\sin(x)$ $\sin^2(x)$ $= 1 - \cos^2(x)$ $\sin(2x)$ $= 2\cdot \sin(x) \cdot \cos(x)$ $\sin(\frac{x}{2})$ $= \pm \sqrt{\frac{1 -\cos(x)}{2}}$ $\sin(x+y)$ $= \sin(x)\cos(y) + \cos(x)\sin(y)$ $\sin(x-y)$ $= \sin(x)\cos(y) - \cos(x)\sin(y)$ $\frac{d}{dx}\sin(x)$ $= \cos(x)$ (*)
🙋 You can input the sine argument in either degrees, radians, or π radians with our sin theta calculator!
## Other sine calculators
Feel free to check our other tools related to the sin theta calculator:
## FAQ
### What is the sin of 2 theta?
The sine of 2 theta (2θ) equals 2 sin(θ)cos(θ). According to one of the sine properties: sin(2x) = 2sin(x)cos(x). So, if you know the values for the sine and cosine of theta, you can easily find the sine of 2 theta.
### How do I find the sin of theta/2?
To find the sin of theta/2:
1. Write down the sine half-angle equation: sin(θ/2) = ±√[(1-cos(θ))/2].
2. Replace theta θ within the equation and solve the square root.
3. To choose the sign, follow this rule:
• The result is positive (+) if the half angle lies in the I or the II quadrant; or
• Negative (-) if it lies on the III or IV quadrant.
### What is the sin of 0?
0. The value for the sine function at 0 is 0. The y-axis component of the radius in the unit circle is 0 when the radius lies on the horizontal axis (which corresponds to angle zero).
Luciano Miño
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# Thales Ship At Sea Activity Purpose: The Purpose Of The Activity Was T Essay
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Thales “Ship at Sea” Activity Purpose: The purpose of the activity was to learn that the Corresponding Parts of Congruent Triangles are Congruent (CPCTC), and how you can use it in different situations. We familiarized ourselves with the corresponding parts of congruent triangles. We also were supposed to find the distance to an object without actually measuring the distance to that object directly. Step one: Suzie Pipperno and I had to pick a concrete block about forty feet away from the sidewalk in back of the school. Step two: We then tried to align a cone with the cement block without getting close to it. Step three: We had to pace out a certain distance, 10 steps, from the cone, place a flag, the pace the same distance again, in a continuous segment, and place another cone. Step four: We walked at a right angle to the second cone until we had the cement block and the flag perfectly in line. Step five: We took a string and stretched it the distance from the second cone to the place we stopped walking. Step six: We placed the string against a tape measure and found that the approximate distance from the cement block to the first cone was thirty eight feet-two inches. Step seven: We used the string to measure the exact distance from the cement block the first cone using the tape measure to measure the string, which was forty two feet-one inch. Step Eight: We used the string to get an exact measurement from the first cone to the flag. Then used the string to correct the distance of the second cone from the flag. Step nine: We walked at a right angle from the second cone until the flag and the cement block are lined up again. Step ten: We used the string and tape measure to measure the distance of the path we walked and came up with forty one feet-two inches. Conclusion: We were able to conclude, without directly measuring the distance to the cement block, that the distance to the block was approximately forty one feet-two inches. Relation: The way this activity relates to our mathematical studies is that it familiarizes us with the congruent parts of congruent triangles, and teaches us that you can use the congruence of triangles in real life. How we proved the triangles congruent: If you look at the attached diagram you will see that there are 2 sides with a | through them. That means that those sides, or line segments, are congruent. You will also notice two angles with?s spanning their angle measure. That means that that those two angles are congruent. Also you will see two sides with a || through them. That means the same thing as the first pair of segments with the | through them, but it signifies that those two line segments are congruent with each other and not the other two. These triangles are congruent by a postulate SAS (Side-Angle-Side). Which states that if two triangles have a Side an Angle and a Side Congruent then both of the triangles are totally congruent. Comments on Activity: I think that the activity was worthwhile, because I learned how errors in measurement and sighting can cause inaccuracies in measured distences, and the larger the distances you are working with, the larger the errors. Idea to Improve or Extend: My idea is to do the activity three times, and in each have the block at a different distance. This would enable you to see how distance effects accuracy. Glossary Angle- an angle consists of two different rays that have the same initial point, the vertex.Congruent angles- two angles that share the same measure Congruent segments- two segments that share the same measure CPCTC- abbreviation for corresponding parts of congruent triangles are congruent Postulate-A statement accepted without proof as true SAS Postulate-If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent Triangle- A polygon with three sides
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KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4
Question 1.
Given a cylindrical tank, in which situation will you find the surface area and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.
Solution:
(a) In this situation volume is to find out.
(b) Surface area.
(c) Volume.
Question 2.
The diameter of cylinder A is 7 cm, and the height is 14 cm. The diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both cylinders. Check whether the cylinder with greater volume also has greater surface area?
Solution:
Cylinder B has a greater volume
Radius of cylinder (A) = $$\frac{7}{2}$$ cm
Height of cylinder = 14 cm
Volume of cylinder (A) = πr2h
= $$\frac{22}{7} \times\left(\frac{7}{2}\right)^{2} \times 14$$
= $$\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 14$$
= 11 × 49
= 539 cm3
Radius of cylinder B = 7 cm
Height of cylinder = 7 cm
Volume of cylinder B = πr2h
= $$\frac{22}{7}$$ × 7 × 7 × 7
= 22 × 49
= 1078 cm3
Surface area of cylinder (A) = 2πrh + 2πr2
= 2πr(r + h)
= $$2 \times \frac{22}{7} \times \frac{7}{2}\left[\frac{7}{2}+14\right]$$
= $$22\left[\frac{7+28}{2}\right]$$
= 11 × 35
= 385 cm2
Surface area of cylinder B = 2πrh + 2πr2
= 2πr(r + h)
= 2 × $$\frac{22}{7}$$ × 7[7 + 7]
= 44 × 14
= 616 cm2
a
Question 3.
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?
Solution:
Volume of cuboid = 900 cm3
⇒ (Area of base) × height = 900
⇒ 180 × height = 900
⇒ height = $$\frac{900}{180}$$ = 5 cm
Hence, height of cuboid = 5 cm
Question 4.
A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with a side of 6 cm can be placed in the given cuboid?
Solution:
Volume of cuboid = 60 × 54 × 30
= 1800 × 54
= 97200 cm3
Edge of cube = 6 cm
Volume of cube = (Side)3
= (6)3
= 6 × 6 × 6
= 216 cm3
Number of cubes = $$\frac{\text { Volume of cuboid }}{\text { Volume of cube }}$$
= $$\frac{97200}{216}$$
= 450
Question 5.
Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm?
Solution:
Volume of cylinder = 1.54 m3
Diameter of base = 140 cm
radius of base = $$\frac{140}{2}$$
= 70 cm
= $$\frac{70}{100}$$ m
= 0.70 m
πr2h = 1.54
$$\frac{22}{7}$$ × 0.7 × 0.7 × h = 1.54
h = $$\frac{1.54 \times 7}{22 \times 0.7 \times 0.7}$$
= $$\frac{154 \times 7}{22 \times 7 \times 7}$$
= 1
∴ Height of cylinder = 1 mts.
Question 6.
A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?
Solution:
Radius of cylinder = 1.5 m
Height of cylinder = 7 m
Volume of tank = πr2h
= $$\frac{22}{7}$$ × 1.5 × 1.5 × 7
= 22 × 2.25
= 49.50 m3
1 m3 = 1000 lts
49.5 m3 = 49.5 × 1000 = 49500 lts
Question 7.
If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
Solution:
Let edge of cube = x unit
(i) Surface area of cube = 6(Side)2 = 6x2
Volume of cube = (Side)3 = x3
If edge of cube is doubled = 2x unit
(ii) Surface area of cube = 6(Side)2
= 6(2x)2
= 6 × 4x2
= 24x2
Volume of cube = (Side)3
= (2x)3
= 8x3
Original surface area = 6x2
Increased surface area = 24x2
Hence, Surface area increased = $$\frac{24 x^{2}}{6 x^{2}}$$ = 4 times increased
Volume increased = $$\frac{8 x^{3}}{x^{3}}$$ = 8 times increased
Question 8.
Water is pouring into a cubiodal reservoir at the rate of 60 liters per minute. If the volume of reservoir is 108 m3, find the number of hours it will take to fill the reservoir.
Solution:
Volume of reservoir = 108 m3
= 108 × 1000
= 108000 lts
Since, water is pouring into reservoir at the rate of 60 lts per min.
Time taken to fill the reservoir = $$\frac{108000}{60}$$
= 1800 mins
= $$\frac{1800}{60}$$ hrs
= 30 hrs
a
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# Algebraic Operations on Complex Numbers
## Algebraic Of Complex Numbers
Have you ever heard of complex numbers? Do you know what an iota is? What kind of number is $\sqrt{-2}$? Does the number even exist? To get all your answers, let’s first understand the entire number system.
### Number System
The number system is broadly divided into two parts: Real Numbers and Complex Numbers.
1. ### Real numbers
Real numbers are those which can be shown on a number line. On the other hand, complex numbers are those which can not be expressed on a number line or be experienced in real life. Real Numbers are further divided into two categories called rational and irrational numbers.
Rational numbers are numbers which can be expressed as fractions and their denominators are not equal to 0. All the real numbers which are not rational are Irrational numbers. Rational numbers are made by dividing two integers. Integers include all negative and positive natural numbers along with zero. Integers are further divided into two sub-categories: whole numbers and natural numbers. Whole numbers are positive counting numbers along with 0. When you remove 0 from whole numbers, we obtain positive counting numbers which are known as natural numbers.
1. ### Complex Numbers
Complex numbers are also known as Imaginary numbers. Now that we know the definition of complex numbers and that complex numbers are the part of the Number System, let’s see some examples.
All the negative numbers under root are imaginary numbers.
$\sqrt{-2}$ and $\sqrt{-2}$ are two very different things. The first one is a real number. Since it’s a negative number under root the second one is a complex number. A complex number is represented in the following way: a+bi, where a is the real part and b is the imaginary part.
You can write the complex number $\sqrt{-2}$ in a+bi form. 0+2i is equal to $\sqrt{-2}$. You must be wondering why are we using the symbol ’i’? What does it mean? Well, it is iota. Have you ever heard of iota? If not, then is all that you need to know about iota.
### IOTA
Iota is a greek letter which is used to represent the imaginary part of a complex number. Iota(i) is considered to be the square root of -1. It may also be defined as a number whose square is -1.
i=$\sqrt{-1}$
i²=-1
i³=i
i⁴=1
### Algebraic Operations On Complex Numbers:
Four types of algebraic operations can be done on complex numbers. These four algebra of complex numbers are:
• Subtraction
• Multiplication
• Division
There are several properties that algebra on imaginary numbers follow:
### Closure law
The sum or product of two imaginary numbers will always get you an imaginary number.
### Commutative Law
If you change the order of imaginary number while adding or multiplying the result will not change that is the answer you get will always be the same.
### Associative Law
If you add or multiply any three complex numbers in any order the result will always remain the same.
This property tells us that if we add zero to any complex we will get the same complex number. This shows that there’s a number that can be added to get the same number back. It is also known as zero complex number and is denoted as 0 (or 0 + i0).
A complex number has the opposite sign for its both real and imaginary parts. This is known as the Existence of Additive inverse.
### Multiplicative Identity
Multiplicative Identity is a property which talks about the existence of a complex number that when multiplied to another will get the same result. it is denoted as 1 (or 1 + i0)
### Multiplicative Inverse
It is a property of any non- zero complex number to have a reciprocal. This is known as the multiplicative inverse.
### Distributive Property
When you split the multiplication of a complex number by another term this property is known as the distributive property.
Note: Subtraction follows all the properties followed by addition.
### Fun Facts:
1. Both real and imaginary parts are present in the square root of i.
2. The N-th root can have N number of unique solutions and any root of i has multiple unique solutions
3. The result may vary depending on whether i is present in the numerator or denominator in an imaginary fraction.
4. When you raise i to the i power, the number you get is a real number.
5. Numbers like $\pi$, i and e are all related to one another.
1. What Is A Complex Number? Discuss About The Powers Of IOTA.
Complex numbers are those which can not be expressed on a number line or be experienced in real life. Complex numbers are also known as Imaginary numbers. All the negative numbers under root are imaginary numbers. A complex number is represented in the following way: a+bi, where a is the real part and b is the imaginary part.
For Example :
1. 4+5i
2. 10-9i
3. √-2
Iota(i) is considered to be the square root of -1. It may also be defined as a number whose square is -1.
i=√-1
i²=-1
i³=i
i⁴=1
2. What All Algebraic Operations Can Be Done On Complex Numbers? What Is Argand Plane?
Four types of algebraic operations can be done on complex numbers. These four operations on the complex numbers are:
• Subtraction
• Multiplication
• Division
(w + ix) + (y + iz) = (w + y) + i(x + z)
Subtraction
(w + ix) – (y + iz) = (w – y) + i(x – z)
Multiplication
(w + ix). (y + iz) = (wy – xz) + i(wz + xy)
Division
(w + ix) / (y + iz) = (wy+xz)/ (y2 + z2) + i(xy – wz) / (y2 + z2)
The way we have the cartesian plane and the XY plane in it to represent real numbers. Similarly, we have the Argand plane which has a system of rectangular coordinates to represent the complex numbers which are written in the form a+ib. Here a and b are the coordinates.
|
# How do you implicitly differentiate 3y + y^4/x^2 = 2?
Aug 5, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{4}}{3 {x}^{4} + 4 {y}^{3}}$
#### Explanation:
When we implicitly differentiate a function $f \left(x , y\right) = 0$, whenever we differentiate w.r.t. we use chain rule and when we differentiate w.r.t. $y$ and then multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.
As such as we have $3 y + {y}^{4} / {x}^{2} = 2$
$3 \times 1 \times \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{4 {y}^{3} \times \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{4} \times 2 x}{x} ^ 4 = 0$ or
$3 \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x {y}^{4}}{x} ^ 4 = 0$ or
$3 \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {y}^{3} / {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {y}^{4} / {x}^{3} = 0$ or
$\frac{\mathrm{dy}}{\mathrm{dx}} \left[3 + \frac{4 {y}^{3}}{x} ^ 4\right] = \frac{2 {y}^{4}}{x} ^ 3$ or
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{2 {y}^{4}}{x} ^ 3}{3 + \frac{4 {y}^{3}}{x} ^ 4}$ or
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{4}}{3 {x}^{4} + 4 {y}^{3}}$
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# Introducing Fractions:
## Teacher (person planning lesson): Clarisen Lacuata & Bobbie Bautista
Subject: Introduction to Fractions
Purpose Statement: The purpose of this lesson is to introduce fractions
Instructional Context:
__X__Beginning Lesson (introduce concept)
_____Developmental Lesson
_____Culminating Lesson
_____Other: __________________________________________________________
Teacher Content Knowledge/ Big Ideas and Essential Questions:
Include the Big Ideas and core concepts of the discipline the teacher needs
in order to teach the lesson
"A fraction tells us only about the relationship between the part and the
whole" (page 295)
Part- Whole is an effective way to teach children the meaning of fractions.
Allowing students to look at groups of people such as 3/5 of a class went on a
field trip or through part of a length such as 2 1/2 miles.
"A fraction tells us only about the relationship between the part and the
whole" (page 295)
Understanding equivalent fractions is critical. Two equivalent fractions are
two ways of describing the same amount by using different- sized fractional
parts. (Van De Walle, 2013, p. 290)
Fraction can represent parts of regions, parts of sets, parts of measures,
division or ratio.
A fraction is not meaningful without knowing what the whole is.
Taken from Van De Wall, John. Elementary and Middle School Mathematics.
8th Pearson Education Inc, 290-312. Print.
Common Core State Standards:
CC-MA-2010.3.NF
Number and Operations: Fractions
CC-MA-2010.3.NF.1
Understand a fraction 1/b as the quantity formed by 1 part when a whole is
partitioned into b equal parts; understand a fraction a/b as the quantity
formed by a parts of size 1/b.
## Student Learning Outcomes:
Defines what the student should know and be able to do upon completion of
the lesson; usually more specific than the benchmarks.
Knowledge:
What a numerator (part) and a denominator (whole) is
Skill:
## Recognizing and naming fractions
Assessments:
Students must contribute to the discussion as a group and in whole class
discussions.
Answering questions correctly for the candy bar worksheet will prove that
students know and understand what a numerator and a denominator is.
Materials:
Area models:
Manipulatives
Candy bar
Candy bar worksheet
Set-Up:
Students will be in groups for the manipulatives and candy bar for discussion
and reasoning.
They will work on their worksheets individually.
Procedures:
a: Introduction (expected time, e.g., 7 minutes)
Every student will begin with a paper candy bar that has 12 parts. The
teacher will have an actual candy bar and follow what the children are doing.
Discussion:
The candy bar is a whole.
If you cut the candy bar in half what do you notice?
1 There are 6 pieces in 1 half of the candy bar. 6
2 6 pieces is 1 half of 12 = 6/12
3 We will talk about how if you split the candy bar in half with a
friend, you both would have EQUAL PARTS.
Then they will have to divide these parts in half again (two other friends
come over, so they have to split each half into equal parts again.)
1 We will label the parts with the fraction name as they are being
split.
We will keep spitting the parts until there are 12 1/12 pieces.
We will go over again how we got from 1 whole to 1/12.
Using magnetic fraction bars, we will compare the sizes of the parts.
The students will enjoy the candy bar before the next activity
The students will be paired up and be given a small bag of M&Ms (prebagged on purpose by the teacher).
They will divide their M&M into groups according to color.
Students will figure out what fraction of the whole bag that color is
For example: There are 24 M&Ms in all. 8 are red, so red is 8/24 of the whole
bag.
Students are most likely to tell us that red is 1/3 of the bag.
After figuring out the fractions of each color, they will draw them out with
## crayons, label them, and eat their fractions.
b. Developing
Expected time: 30 minutes
After eating their fractions, the students will be put into partners. Each pair
will recieve pattern blocks. They will be instructed to show what the teacher
How many green triangles are in a blue rhombus?
How many green triangles are in a yellow hexagon?
How many blue rhombuses are in a yellow hexagon?
How many red trapezoids are in a yellow hexagon?
How many green triangles are in a red trapezoid?
Between these questions we will ask thought provoking questions: "What
does a green triangle represent in a blue rhombus?" The answer we are
looking for is "One green triangle represents one out of two (1/2)" etc.
After doing this partner/whole class discussion. The students will have 15
minutes to work individually on the Candy Bar Worksheet (attached) and turn
it in to the teacher.
c. Concluding (expected time, e.g., 8 minutes)
The students will write a short journal entry about what they've learned
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# Difference between revisions of "2011 AMC 10A Problems/Problem 21"
## Problem 21
Two counterfeit coins of equal weight are mixed with $8$ identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the $10$ coins. A second pair is selected at random without replacement from the remaining $8$ coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all $4$ selected coins are genuine?
$\textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16}$
## Solution 1
Note that we are trying to find the conditional probability $P(A \vert B) = \frac{P(A \cap B)}{P(B)}$ where $A$ is the $4$ coins being genuine and $B$ is the sum of the weight of the coins being equal. The only possibilities for $B$ are (g is abbreviated for genuine and c is abbreviated for counterfeit, and the pairs are the first and last two in the quadruple) $(g,g,g,g),(g,c,g,c),(g,c,c,g),(c,g,g,c),(c,g,c,g)$. We see that $A \cap B$ happens with probability $\frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} \times \frac{5}{7} = \frac{1}{3}$, and $B$ happens with probability $\frac{1}{3}+4 \times \left( \frac{8}{10} \times \frac{2}{9} \times \frac{7}{8}\times \frac{1}{7}\right) = \frac{19}{45}$, hence $P(A \vert B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{19}{45}}=\boxed{\frac{15}{19} \ \mathbf{(D)}}$.
## Solution 2
If we pick $4$ distinguishable real coins from the set of $8$ real coins, there are $\binom{8}{4}$ ways to pick the coins. If we then place the coins in four distinguishable slots on the scale, there are $4!$ ways to arrange them, giving $4!\cdot \binom{8}{4}$ ways to choose and place $8$ real coins. This gives $1680$ desirable combinations.
If we pick $2$ real coins and $2$ fake coins, there are $\binom{8}{2}\binom{2}{2}$ ways to choose the coins. There are $4$ choices for the first slot on the left side of the scale. Whichever type of coin is placed in that first slot, there are $2$ choices for the second slot on the left side of the scale, since it must be of the opposite type of coin. There are $2$ choices for the first slot on the right side of the scale, and only $1$ choice for the last slot on the right side.
Thus, there are $4\cdot 2\cdot 2\cdot 1 = 16$ ways to arrange the coins, and $\binom{8}{2}\binom{2}{2} = 28$ sets of possible coins, for a total of $16\cdot 28 = 448$ combinations that are legal, yet undesirable.
The overall probability is thus $\frac{1680}{1680 + 440} = \boxed{\frac{15}{19} \ \mathbf{(D)}}$.
Note that in this solution, all four slots on the scale are deemed to be distinguishable. In essense, this "overcounts" all numbers by a factor of $8$, since you can switch the coins on the left side, you can switch the coins on the right side, or you can switch sides of the scale. But since all numbers are increased 8-fold, it will cancel out when calculating the probability.
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# 3.2: The Derivative as a Function
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##### Learning Objectives
• Define the derivative function of a given function.
• Graph a derivative function from the graph of a given function.
• State the connection between derivatives and continuity.
• Describe three conditions for when a function does not have a derivative.
• Explain the meaning of a higher-order derivative.
As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious. In this section we define the derivative function and learn a process for finding it.
## Derivative Functions
The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows.
##### Definition: Derivative Function
Let $$f$$ be a function. The derivative function, denoted by $$f'$$, is the function whose domain consists of those values of $$x$$ such that the following limit exists:
$f'(x)=\lim_{h→0}\frac{f(x+h)−f(x)}{h}. \label{derdef}$
A function $$f(x)$$ is said to be differentiable at $$a$$ if $$f'(a)$$ exists. More generally, a function is said to be differentiable on $$S$$ if it is differentiable at every point in an open set $$S$$, and a differentiable function is one in which $$f'(x)$$ exists on its domain.
In the next few examples we use Equation \ref{derdef} to find the derivative of a function.
##### Example $$\PageIndex{1}$$: Finding the Derivative of a Square-Root Function
Find the derivative of $$f(x)=\sqrt{x}$$.
Solution
Start directly with the definition of the derivative function.
Substitute $$f(x+h)=\sqrt{x+h}$$ and $$f(x)=\sqrt{x}$$ into $$f'(x)= \displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}$$.
$$f'(x)=\displaystyle \lim_{h→0}\frac{\sqrt{x+h}−\sqrt{x}}{h}$$ $$=\displaystyle\lim_{h→0}\frac{\sqrt{x+h}−\sqrt{x}}{h}⋅\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$$ Multiply numerator and denominator by $$\sqrt{x+h}+\sqrt{x}$$ without distributing in the denominator. $$=\displaystyle\lim_{h→0}\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}$$ Multiply the numerators and simplify. $$=\displaystyle\lim_{h→0}\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)}$$ Cancel the $$h$$. $$=\dfrac{1}{2\sqrt{x}}$$ Evaluate the limit
##### Example $$\PageIndex{2}$$: Finding the Derivative of a Quadratic Function
Find the derivative of the function $$f(x)=x^2−2x$$.
Solution
Follow the same procedure here, but without having to multiply by the conjugate.
Substitute $$f(x+h)=(x+h)^2−2(x+h)$$ and $$f(x)=x^2−2x$$ into $$f'(x)= \displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}.$$
$$f'(x)=\displaystyle\lim_{h→0}\frac{((x+h)^2−2(x+h))−(x^2−2x)}{h}$$ $$=\displaystyle\lim_{h→0}\frac{x^2+2xh+h^2−2x−2h−x^2+2x}{h}$$ Expand $$(x+h)^2−2(x+h)$$. $$=\displaystyle\lim_{h→0}\frac{2xh−2h+h^2}{h}$$ Simplify $$=\displaystyle\lim_{h→0}\frac{h(2x−2+h)}{h}$$ Factor out $$h$$ from the numerator $$=\displaystyle\lim_{h→0}(2x−2+h)$$ Cancel the common factor of $$h$$ $$=2x−2$$ Evaluate the limit
##### Exercise $$\PageIndex{1}$$
Find the derivative of $$f(x)=x^2$$.
Hint
Use Equation \ref{derdef} and follow the example.
$$f'(x)=2x$$
We use a variety of different notations to express the derivative of a function. In Example $$\PageIndex{2}$$ we showed that if $$f(x)=x^2−2x$$, then $$f'(x)=2x−2$$. If we had expressed this function in the form $$y=x^2−2x$$, we could have expressed the derivative as $$y′=2x−2$$ or $$\dfrac{dy}{dx}=2x−2$$. We could have conveyed the same information by writing $$\dfrac{d}{dx}\left(x^2−2x\right)=2x−2$$. Thus, for the function $$y=f(x)$$, each of the following notations represents the derivative of $$f(x)$$:
$$f'(x), \quad \dfrac{dy}{dx}, \quad y′,\quad \dfrac{d}{dx}\big(f(x)\big)$$.
In place of $$f'(a)$$ we may also use $$\dfrac{dy}{dx}\Big|_{x=a}$$. Use of the $$\dfrac{dy}{dx}$$ notation (called Leibniz notation) is quite common in engineering and physics. To understand this notation better, recall that the derivative of a function at a point is the limit of the slopes of secant lines as the secant lines approach the tangent line. The slopes of these secant lines are often expressed in the form $$\dfrac{Δy}{Δx}$$ where $$Δy$$ is the difference in the $$y$$ values corresponding to the difference in the $$x$$ values, which are expressed as $$Δx$$ (Figure $$\PageIndex{1}$$). Thus the derivative, which can be thought of as the instantaneous rate of change of $$y$$ with respect to $$x$$, is expressed as
$$\displaystyle \frac{dy}{dx}= \lim_{Δx→0}\frac{Δy}{Δx}$$.
## Graphing a Derivative
We have already discussed how to graph a function, so given the equation of a function or the equation of a derivative function, we could graph it. Given both, we would expect to see a correspondence between the graphs of these two functions, since $$f'(x)$$ gives the rate of change of a function $$f(x)$$ (or slope of the tangent line to $$f(x)$$).
In Example $$\PageIndex{1}$$, we found that for $$f(x)=\sqrt{x}$$, $$f'(x)=\frac{1}{2\sqrt{x}}$$. If we graph these functions on the same axes, as in Figure $$\PageIndex{2}$$, we can use the graphs to understand the relationship between these two functions. First, we notice that $$f(x)$$ is increasing over its entire domain, which means that the slopes of its tangent lines at all points are positive. Consequently, we expect $$f'(x)>0$$ for all values of x in its domain. Furthermore, as $$x$$ increases, the slopes of the tangent lines to $$f(x)$$ are decreasing and we expect to see a corresponding decrease in $$f'(x)$$. We also observe that $$f(0)$$ is undefined and that $$\displaystyle \lim_{x→0^+}f'(x)=+∞$$, corresponding to a vertical tangent to $$f(x)$$ at $$0$$.
In Example $$\PageIndex{2}$$, we found that for $$f(x)=x^2−2x,\; f'(x)=2x−2$$. The graphs of these functions are shown in Figure $$\PageIndex{3}$$. Observe that $$f(x)$$ is decreasing for $$x<1$$. For these same values of $$x$$, $$f'(x)<0$$. For values of $$x>1$$, $$f(x)$$ is increasing and $$f'(x)>0$$. Also, $$f(x)$$ has a horizontal tangent at $$x=1$$ and $$f'(1)=0$$.
##### Example $$\PageIndex{3}$$: Sketching a Derivative Using a Function
Use the following graph of $$f(x)$$ to sketch a graph of $$f'(x)$$.
Solution
The solution is shown in the following graph. Observe that $$f(x)$$ is increasing and $$f'(x)>0$$ on $$(–2,3)$$. Also, $$f(x)$$ is decreasing and $$f'(x)<0$$ on $$(−∞,−2)$$ and on $$(3,+∞)$$. Also note that $$f(x)$$ has horizontal tangents at $$–2$$ and $$3$$, and $$f'(−2)=0$$ and $$f'(3)=0$$.
##### Exercise $$\PageIndex{2}$$
Sketch the graph of $$f(x)=x^2−4$$. On what interval is the graph of $$f'(x)$$ above the $$x$$-axis?
Hint
The graph of $$f'(x)$$ is positive where $$f(x)$$ is increasing.
$$(0,+∞)$$
## Derivatives and Continuity
Now that we can graph a derivative, let’s examine the behavior of the graphs. First, we consider the relationship between differentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there; however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuous at a point and fail to be differentiable at the point for one of several reasons.
##### Differentiability Implies Continuity
Let $$f(x)$$ be a function and $$a$$ be in its domain. If $$f(x)$$ is differentiable at $$a$$, then $$f$$ is continuous at $$a$$.
##### Proof
If $$f(x)$$ is differentiable at $$a$$, then $$f'(a)$$ exists and, if we let $$h = x - a$$, we have $$x = a + h$$, and as $$h=x-a\to 0$$, we can see that $$x\to a$$.
Then
$f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\nonumber$
can be rewritten as
$$f'(a)=\displaystyle \lim_{x→a}\frac{f(x)−f(a)}{x−a}$$.
We want to show that $$f(x)$$ is continuous at $$a$$ by showing that $$\displaystyle \lim_{x→a}f(x)=f(a).$$ Thus,
\begin{align*} \displaystyle \lim_{x→a}f(x) &=\lim_{x→a}\;\big(f(x)−f(a)+f(a)\big)\\[4pt] &=\lim_{x→a}\left(\frac{f(x)−f(a)}{x−a}⋅(x−a)+f(a)\right) & & \text{Multiply and divide }(f(x)−f(a))\text{ by }x−a.\\[4pt] &=\left(\lim_{x→a}\frac{f(x)−f(a)}{x−a}\right)⋅\left( \lim_{x→a}\;(x−a)\right)+\lim_{x→a}f(a)\\[4pt] &=f'(a)⋅0+f(a)\\[4pt] &=f(a). \end{align*}
Therefore, since $$f(a)$$ is defined and $$\displaystyle \lim_{x→a}f(x)=f(a)$$, we conclude that $$f$$ is continuous at $$a$$.
We have just proven that differentiability implies continuity, but now we consider whether continuity implies differentiability. To determine an answer to this question, we examine the function $$f(x)=|x|$$. This function is continuous everywhere; however, $$f'(0)$$ is undefined. This observation leads us to believe that continuity does not imply differentiability. Let’s explore further. For $$f(x)=|x|$$,
$$f'(0)=\displaystyle \lim_{x→0}\frac{f(x)−f(0)}{x−0}= \lim_{x→0}\frac{|x|−|0|}{x−0}= \lim_{x→0}\frac{|x|}{x}$$.
This limit does not exist because
$$\displaystyle \lim_{x→0^−}\frac{|x|}{x}=−1$$ and $$\displaystyle \lim_{x→0^+}\frac{|x|}{x}=1$$.
See Figure $$\PageIndex{4}$$.
Let’s consider some additional situations in which a continuous function fails to be differentiable. Consider the function $$f(x)=\sqrt[3]{x}$$:
$$f'(0)=\displaystyle \lim_{x→0}\frac{\sqrt[3]{x}−0}{x−0}=\displaystyle \lim_{x→0}\frac{1}{\sqrt[3]{x^2}}=+∞$$.
Thus $$f'(0)$$ does not exist. A quick look at the graph of $$f(x)=\sqrt[3]{x}$$ clarifies the situation. The function has a vertical tangent line at $$0$$ (Figure $$\PageIndex{5}$$).
The function $$f(x)=\begin{cases} x\sin\left(\frac{1}{x}\right), & & \text{ if } x≠0\\0, & & \text{ if } x=0\end{cases}$$ also has a derivative that exhibits interesting behavior at $$0$$.
We see that
$$f'(0)=\displaystyle \lim_{x→0}\frac{x\sin\left(1/x\right)−0}{x−0}= \lim_{x→0}\sin\left(\frac{1}{x}\right)$$.
This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure $$\PageIndex{6}$$).
In summary:
1. We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a function is continuous, it may still fail to be differentiable.
2. We saw that $$f(x)=|x|$$ failed to be differentiable at $$0$$ because the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at $$0.$$ From this we conclude that in order to be differentiable at a point, a function must be “smooth” at that point.
3. As we saw in the example of $$f(x)=\sqrt[3]{x}$$, a function fails to be differentiable at a point where there is a vertical tangent line.
4. As we saw with $$f(x)=\begin{cases}x\sin\left(\frac{1}{x}\right), & & \text{ if } x≠0\\0, & &\text{ if } x=0\end{cases}$$ a function may fail to be differentiable at a point in more complicated ways as well.
##### Example $$\PageIndex{4}$$: A Piecewise Function that is Continuous and Differentiable
A toy company wants to design a track for a toy car that starts out along a parabolic curve and then converts to a straight line (Figure $$\PageIndex{7}$$). The function that describes the track is to have the form $$f(x)=\begin{cases}\frac{1}{10}x^2+bx+c, & & \text{ if }x<−10\\−\frac{1}{4}x+\frac{5}{2}, & & \text{ if } x≥−10\end{cases}$$ where $$x$$ and $$f(x)$$ are in inches. For the car to move smoothly along the track, the function $$f(x)$$ must be both continuous and differentiable at $$−10$$. Find values of $$b$$ and $$c$$ that make $$f(x)$$ both continuous and differentiable.
Solution
For the function to be continuous at $$x=−10$$, $$\displaystyle \lim_{x→10^−}f(x)=f(−10)$$. Thus, since
$$\displaystyle \lim_{x→−10^−}f(x)=\frac{1}{10}(−10)^2−10b+c=10−10b+c$$
and $$f(−10)=5$$, we must have $$10−10b+c=5$$. Equivalently, we have $$c=10b−5$$.
For the function to be differentiable at $$−10$$,
$$f'(10)=\displaystyle \lim_{x→−10}\frac{f(x)−f(−10)}{x+10}$$
must exist. Since $$f(x)$$ is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other:
\displaystyle \begin{align*} \lim_{x→−10^−}\frac{f(x)−f(−10)}{x+10} &= \lim_{x→−10^−}\frac{\frac{1}{10}x^2+bx+c−5}{x+10}\\[4pt] &= \lim_{x→−10^−}\frac{\frac{1}{10}x^2+bx+(10b−5)−5}{x+10} & & \text{Substitute }c=10b−5.\\[4pt] &= \lim_{x→−10^−}\frac{x^2−100+10bx+100b}{10(x+10)}\\[4pt] &= \lim_{x→−10^−}\frac{(x+10)(x−10+10b)}{10(x+10)} & & \text{Factor by grouping}\\[4pt] &=b−2 \end{align*}.
We also have
\displaystyle \begin{align*} \lim_{x→−10^+}\frac{f(x)−f(−10)}{x+10} &= \lim_{x→−10^+}\frac{−\frac{1}{4}x+\frac{5}{2}−5}{x+10}\\[4pt] &= \lim_{x→−10^+}\frac{−(x+10)}{4(x+10)}\\[4pt] &=−\frac{1}{4} \end{align*}.
This gives us $$b−2=−\frac{1}{4}$$. Thus $$b=\frac{7}{4}$$ and $$c=10(\frac{7}{4})−5=\frac{25}{2}$$.
##### Exercise $$\PageIndex{3}$$
Find values of a and b that make $$f(x)=\begin{cases}ax+b, & & \text{ if } x<3\\x^2, & & \text{ if } x≥3\end{cases}$$ both continuous and differentiable at $$3$$.
Hint
Use Example $$\PageIndex{4}$$ as a guide.
$$a=6$$ and $$b=−9$$
## Higher-Order Derivatives
The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as higher-order derivatives. The notation for the higher-order derivatives of $$y=f(x)$$ can be expressed in any of the following forms:
$$f''(x),\; f'''(x),\; f^{(4)}(x),\; …\; ,\; f^{(n)}(x)$$
$$y''(x),\; y'''(x),\; y^{(4)}(x),\; …\; ,\; y^{(n)}(x)$$
$$\dfrac{d^2y}{dx^2},\;\dfrac{d^3y}{dy^3},\;\dfrac{d^4y}{dy^4},\;…\;,\;\dfrac{d^ny}{dy^n}.$$
It is interesting to note that the notation for $$\dfrac{d^2y}{dx^2}$$ may be viewed as an attempt to express $$\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$$ more compactly.
Analogously, $$\dfrac{d}{dx}\left(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\right)=\dfrac{d}{dx}\left(\dfrac{d^2y}{dx^2}\right)=\dfrac{d^3y}{dx^3}$$.
##### Example $$\PageIndex{5}$$: Finding a Second Derivative
For $$f(x)=2x^2−3x+1$$, find $$f''(x)$$.
Solution
First find $$f'(x)$$.
Substitute $$f(x)=2x^2−3x+1$$ and $$f(x+h)=2(x+h)^2−3(x+h)+1$$ into $$f'(x)=\displaystyle \lim_{h→0}\dfrac{f(x+h)−f(x)}{h}.$$
$$f'(x)=\displaystyle \lim_{h→0}\frac{(2(x+h)^2−3(x+h)+1)−(2x^2−3x+1)}{h}$$ $$=\displaystyle \lim_{h→0}\frac{4xh+2h^2−3h}{h}$$ Simplify the numerator. $$=\displaystyle \lim_{h→0}(4x+2h−3)$$ Factor out the $$h$$ in the numerator and cancel with the $$h$$ in the denominator. $$=4x−3$$ Take the limit.
Next, find $$f''(x)$$ by taking the derivative of $$f'(x)=4x−3.$$
$$f''(x)=\displaystyle \lim_{h→0}\frac{f'(x+h)−f'(x)}{h}$$ Use $$f'(x)=\displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}$$ with $$f ′(x)$$ in place of $$f(x).$$ $$=\displaystyle \lim_{h→0}\frac{(4(x+h)−3)−(4x−3)}{h}$$ Substitute $$f'(x+h)=4(x+h)−3$$ and $$f'(x)=4x−3.$$ $$=\displaystyle \lim_{h→0}4$$ Simplify. $$=4$$ Take the limit.
##### Exercise $$\PageIndex{4}$$
Find $$f''(x)$$ for $$f(x)=x^2$$.
Hint
We found $$f'(x)=2x$$ in a previous checkpoint. Use Equation \ref{derdef} to find the derivative of $$f'(x)$$
$$f''(x)=2$$
##### Example $$\PageIndex{6}$$: Finding Acceleration
The position of a particle along a coordinate axis at time $$t$$ (in seconds) is given by $$s(t)=3t^2−4t+1$$ (in meters). Find the function that describes its acceleration at time $$t$$.
Solution
Since $$v(t)=s′(t)$$ and $$a(t)=v′(t)=s''(t)$$, we begin by finding the derivative of $$s(t)$$:
\displaystyle \begin{align*} s′(t) &= \lim_{h→0}\frac{s(t+h)−s(t)}{h}\\[4pt] &=\lim_{h→0}\frac{3(t+h)^2−4(t+h)+1−(3t^2−4t+1)}{h}\\[4pt] &=6t−4. \end{align*}
Next,
\displaystyle \begin{align*} s''(t)&= \lim_{h→0}\frac{s′(t+h)−s′(t)}{h}\\[4pt] &=\lim_{h→0}\frac{6(t+h)−4−(6t−4)}{h}\\[4pt] &=6. \end{align*}
Thus, $$a=6 \;\text{m/s}^2$$.
##### Exercise $$\PageIndex{5}$$
For $$s(t)=t^3$$, find $$a(t).$$
Hint
Use Example $$\PageIndex{6}$$ as a guide.
$$a(t)=6t$$
## Key Concepts
• The derivative of a function $$f(x)$$ is the function whose value at $$x$$ is $$f'(x)$$.
• The graph of a derivative of a function $$f(x)$$ is related to the graph of $$f(x)$$. Where $$f(x)$$ has a tangent line with positive slope, $$f'(x)>0$$. Where $$f(x)$$ has a tangent line with negative slope, $$f'(x)<0$$. Where $$f(x)$$ has a horizontal tangent line, $$f'(x)=0.$$
• If a function is differentiable at a point, then it is continuous at that point. A function is not differentiable at a point if it is not continuous at the point, if it has a vertical tangent line at the point, or if the graph has a sharp corner or cusp.
• Higher-order derivatives are derivatives of derivatives, from the second derivative to the $$n^{\text{th}}$$ derivative.
## Key Equations
• The derivative function
$$f'(x)=\displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}$$
## Glossary
derivative function
gives the derivative of a function at each point in the domain of the original function for which the derivative is defined
differentiable at $$a$$
a function for which $$f'(a)$$ exists is differentiable at $$a$$
differentiable on $$S$$
a function for which $$f'(x)$$ exists for each $$x$$ in the open set $$S$$ is differentiable on $$S$$
differentiable function
a function for which $$f'(x)$$ exists is a differentiable function
higher-order derivative
a derivative of a derivative, from the second derivative to the $$n^{\text{th}}$$ derivative, is called a higher-order derivative
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Paul's Online Notes
Home / Calculus I / Applications of Derivatives / The Shape of a Graph, Part II
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### Section 4-6 : The Shape of a Graph, Part II
15. Determine the minimum degree of a polynomial that has exactly one inflection point.
Show All Steps Hide All Steps
Hint : What is the simplest possible form of the 2nd derivative that we can have that will guarantee that we have a single inflection point?
Start Solution
First, let’s suppose that the single inflection point occurs at $$x = a$$ for some number $$a$$. The value of $$a$$ is not important, this only allows us to discuss the problem.
Now, if we start with a polynomial, call it $$p\left( x \right)$$, then the 2nd derivative must also be a polynomial and we have to have $$p''\left( a \right) = 0$$. In addition, we know that the 2nd derivative must change signs at $$x = a$$.
The simplest polynomial that we can have that will do this is,
$\underline {p''\left( x \right) = x - a}$
This clearly has $$p''\left( a \right) = 0$$ and it will change sign at $$x = a$$. Note as well that we don’t really care which side is concave up and which side is concave down. We only care that the 2nd derivative changes sign at $$x = a$$ as it does here.
Hint : We saw how to “undo” differentiation in the practice problems in the previous section. Here we simply need to do that twice and note that we don’t actually have to undo the derivatives here, just think about what they would have to look like.
Show Step 2
Okay, saw how to “undo” differentiation in the practice problems of the previous section. We don’t actually need to do that here, but we do need to think about what undoing differentiation will give here.
The 2nd derivative is a 1st degree polynomial and that means the 1st derivative had to be a 2nd degree polynomial. This should make sense to you if you understand how differentiation works.
We know that we have to differentiate the 1st derivative to get the 2nd derivative. Therefore, because the highest power of $$x$$ in the 2nd derivative is 1 and we know that differentiation lowers the power by 1 the highest power of $$x$$ in the 1st derivative must have been 2.
Okay, we’ve figured out that the 1st derivative must have been a 2nd degree polynomial. This in turn means that the original function must have been a 3rd degree polynomial. Again, differentiation lowers the power of $$x$$ by 1 and if the highest power of $$x$$ in the 1st derivative is 2 then the highest power of x in the original function must have been 3.
So, the minimum degree of a polynomial that has exactly one inflection point must be three (i.e. a cubic polynomial).
Note that we can have higher degree polynomials with exactly one inflection point. This is simply the minimal degree that will give exactly one inflection point.
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# we have to take the measure of inner scale or outer scale?????????
### Measuring Angles
A protractor is used to measure angles. In this section, we will consider the use of a protractor that has the shape of a semi-circle and two scales marked from 0º to 180º.
The two scales make it easy for us to measure angles facing different ways.
To measure the size of angle ABC, place the protractor over the angle so that the centre of the protractor is directly over the angle's vertex, B; and the base line of the protractor is along the arm, BA, of the angle.
We use the inner scale to measure the angle ABC, as the arm AB passes through the zero of the inner scale. Following the inner scale around the protractor, we find that the other arm, BC, passes through the inner scale at 60º. So, the size of angle ABC is 60 degrees. We write this as follows:
To measure the size of angle PQR, place the protractor over the angle so that the centre of the protractor is directly over the angle's vertex, Q; and the base line of the protractor is along the arm, PQ, of the angle.
We use the outer scale to measure the angle PQR, as the arm PQ passes through the zero of the outer scale. Following the outer scale around the protractor, we find that the other arm, QR, passes through the outer scale at 120º. So, the size of angle PQR is 120 degrees. We write this as follows:
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
if the angle faces the left side then it is the outer scale.
if the angles faces the right side then it is the inner scale.
• 0
What are you looking for?
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# 2018 AMC 10A Problems/Problem 25
## Problem
For a positive integer $n$ and nonzero digits $a$, $b$, and $c$, let $A_n$ be the $n$-digit integer each of whose digits is equal to $a$; let $B_n$ be the $n$-digit integer each of whose digits is equal to $b$, and let $C_n$ be the $2n$-digit (not $n$-digit) integer each of whose digits is equal to $c$. What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$?
$\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}$
## Solution 1
Observe $A_n = a(1 + 10 + \dots + 10^{n - 1}) = a \cdot \tfrac{10^n - 1}{9}$; similarly $B_n = b \cdot \tfrac{10^n - 1}{9}$ and $C_n = c \cdot \tfrac{10^{2n} - 1}{9}$. The relation $C_n - B_n = A_n^2$ rewrites as $$c \cdot \frac{10^{2n} - 1}{9} - b \cdot \frac{10^n - 1}{9} = a^2 \cdot \left(\frac{10^n - 1}{9}\right)^2.$$Since $n > 0$, $10^n > 1$ and we may cancel out a factor of $\tfrac{10^n - 1}{9}$ to obtain $$c \cdot (10^n + 1) - b = a^2 \cdot \frac{10^n - 1}{9}.$$This is a linear equation in $10^n$. Thus, if two distinct values of $n$ satisfy it, then all values of $n$ will. Now we plug in $n=0$ and $n=1$ (or some other number), we get $2c - b = 0$ and $11c - b= a^2$ . Solving the equations for $c$ and $b$, we get $$c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.$$To maximize $a + b + c = a + \tfrac{a^2}{3}$, we need to maximize $a$. Since $b$ and $c$ must be integers, $a$ must be a multiple of $3$. If $a = 9$ then $b$ exceeds $9$. However, if $a = 6$ then $b = 8$ and $c = 4$ for an answer of $\boxed{\textbf{(D)} \text{ 18}}$.
## Solution 2
Immediately start trying $n = 1$ and $n = 2$. These give the system of equations $11c - b = a^2$ and $1111c - 11b = (11a)^2$ (which simplifies to $101c - b = 11a^2$). These imply that $a^2 = 9c$, so the possible $(a, c)$ pairs are $(9, 9)$, $(6, 4)$, and $(3, 1)$. The first puts $b$ out of range but the second makes $b = 8$. We now know the answer is at least $6 + 8 + 4 = 18$.
We now only need to know whether $a + b + c = 20$ might work for any larger $n$. We will always get equations like $100001c - b = 11111a^2$ where the $c$ coefficient is very close to being nine times the $a$ coefficient. Since the $b$ term will be quite insignificant, we know that once again $a^2$ must equal $9c$, and thus $a = 9, c = 9$ is our only hope to reach $20$. Substituting and dividing through by $9$, we will have something like $100001 - \frac{b}{9} = 99999$. No matter what $n$ really was, $b$ is out of range (and certainly isn't $2$ as we would have needed).
The answer then is $\boxed{\textbf{(D)} \text{ 18}}$.
## Solution 3 (Cheating)
Notice that $(0.\overline{3})^2 = 0.\overline{1}$ and $(0.\overline{6})^2 = 0.\overline{4}$. Setting $a = 3$ and $c = 1$, we see $b = 2$ works for all possible values of $n$. Similarly, if $a = 6$ and $c = 4$, then $b = 8$ works for all possible values of $n$. The second solution yields a greater sum of $\boxed{\textbf{(D)} \text{ 18}}$.
~ dolphin7
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# RS Aggarwal Solutions Chapter 16 Coordinate Geometry Exercise 16A Class 10 Maths
Chapter Name RS Aggarwal Chapter 16 Coordinate Geometry Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 16BExercise 16CExercise 16DMCQ Related Study NCERT Solutions for Class 10 Maths
### Exercise 16A Solutions
1. Find the distance between the points
(i) A(9, 3) and B(15, 11)
(ii) A (7,- 4) and B(-5, 1)
(iii) A(-6, - 4) and B(9, -12)
(iv) A(1, -3) and B (4, -6)
(v) P(a + b, a – b) and Q(a – b, a + b)
(vi) P(a sin α, α cos α) and Q(α cos α, - α sin α)
Solution
(i) A(9, 3) and B(15, 11)
The given points are A (9, 3) and B (15, 11)
Then (x1 = 9, y1 = 3) and (x2 = 15, y2 = 11)
(ii) A(7, -4) and B(-5, 1)
The given points are A(7, -4) and B(-5, 1)
Then, (x1 = 7, y1 = - 4) and (x2 = -5, y2 = 1)
(iii) A(-6, -4) and B(9, -12)
The given points are A(-6, -4) and B(9, -12)
Then (x1 = - 6, y1 = -4) and (x2 = 9, y2 = - 12)
(iv) A(1, -3) and B(4, -6)
The given point are A(1, -3) and B(4, -6)
Then (x1 = 1, y1 = -3) and (x2 = 4, y2 = - 6)
(v) P(a + b, a – b) and Q(a – b, a + b)
The given points are P(a + b, a – b) and Q(a – b, a + b)
Then (x1 = a + b, y1 = a – b) and (x2 = a – b, y2 = a + b)
(vi) P(a sin α, a cos α) and Q(a cos α, -a sin α)
The given points are P(a sin α, a cos α) and Q(a cos α, - a sin α)
Then (x1 = a sin α, y1 = a cos α) and (x2 = a cos α, y2 = -a sin α)
2. Find the distance of each of the following points from the origin:
(i) A(5, -12)
(ii) B(-5, 5)
(iii) C(-4, -6)
Solution
(i) A(5, -12)
Let O(0, 0) be the origin
(ii) B(-5, 5)
Let O(0, 0) be the origin.
(iii) C(-4, -6)
Let O(0, 0) be the origin
3. Find all possible values of x for which the distance between the points A(x, -1) and B(5, 3) is 5 units.
Solution
Given AB = 5 units
Therefore, (AB)2 = 25 units
⇒ (5 – a)2 + {3 – (-1)}2 = 25
⇒ (5 – a)2 + (3 + 1)2 = 25
⇒ (5 – a)2 + (4)2 = 25
⇒ (5 – a)2 + 16 = 25
⇒ (5 – a)2 = 25 – 16
⇒ (5 – a)2 = 9
⇒ (5 – a) = ± √9
⇒ 5 – a = ±3
⇒ 5 – a = 3 or 5 – a = - 3
⇒ a = 2 or 8
Therefore, a = 2 or 8.
4. Find all possible values of y for which distance between the points A(2, -3) and B(10, y) is 10 units
Solution
The given points are A(2, -3) and B(10, y)
⇒ 73 + y2 + 6y = 100 [Squaring both sides)
⇒ y2 + 6y – 27 = 0
⇒ y2 + 9y – 3y – 27 = 0
⇒ y(y + 9) – 3(y + 9) = 0
⇒ (y + 9)(y – 3) = 0
⇒ y + 9 = 0 or y – 3 = 0
⇒ y = -9 or y = 3
Hence, the possible values of y are -9 and 3.
5. Find value of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.
Solution
The given points are P(x, 4) and Q(9, 10).
⇒ x2 – 18x + 117 = 100 (Squaring both sides)
⇒ x2 – 18x + 17 = 0
⇒ x2 – 17x – x + 17 = 0
⇒ x(x – 17) – 1(x – 17) = 0
⇒ (x – 17)(x – 1) = 0
⇒ x – 17 = 0 or x – 1 = 0
⇒ x = 17 or x = 1
Hence, the values of x are 1 and 17.
6. If the point A(x, 2) is equidistant from the points B(8, -2) and C(2, -2) , find the value of x. Also, find the value of x. Also, find the length of AB,
Solution
As per the question
AB = AC
7. If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5) find the value of p. Also, find the length of AB.
Solution
As per the question
AB = AC
8. Find the point on the a-axis which is equidistant from the points (2, -5) and (-2, 9).
Solution
Let (x, 0) be the points on the x-axis. Then as per the question, we have
⇒ (x – 2)2 + (5)2 = (x + 2)2 + (-9)2 (Squaring both sides)
⇒ x2 – 4x + 4 + 25 = x2 + 4x + 4 + 81
⇒ 8x = 25 – 81
⇒ x = -(56/8) = - 7
Hence, the point on the x-axis is (-7, 0).
9. Find the points on the a-axis, each of which is at a distance of 10 units from the point A(11, -8).
Solution
Let P(x, 0) be the point on the x-axis. Then as per the question we have
AP = 10
⇒ (x – 11)2 + 82 = 100 (Squaring both sides)
⇒ (x – 11)2 = 100 – 64 = 36
⇒ x – 11 = ± 6
⇒ x = 11 ± 6
⇒ x = 11 – 6, 11 + 6
⇒ x = 5, 17
Here the points on the x-axis are (5, 0) and (17, 0).
10. Find the points on the y-axis which is equidistant from the points A(6, 5) and B(-4, 3)
Solution
Let P (0, y) be a point on the y-axis. Then as per the question, we have
AP = BP
⇒ (6)2 + (y – 5)2 = (4)2 + (y – 3)2 (Squaring both sides)
⇒ 36 + y2 – 10y + 25 = 16 + y2 – 6y + 9
⇒ 4y = 36
⇒ y = 9
Hence, the point on the y-axis is (0, 9).
11. If the points P(x, y) is a point equidistant from the points A(5, 1) and B(-1, 5), Prove that 3x = 2y.
Solution
As per the question, we have
AP = BP
⇒ (x – 5)2 + (y – 1)2 = (x + 1)2 + (y – 5)2 (Squaring both sides)
⇒ x2 – 10x + 25 + y2 – 2y + 1 = x2 + 2x + 1 + y2 – 10y + 25
⇒ - 10x – 2y = 2x – 10y
⇒ 8y = 12x
⇒ 3x = 2y
Hence, 3x = 2y
12. If p(x, y) is point equidistant from the points A(6, -1) and B(2, 3), show that x – y = 3
Solution
The given points are A(6, -1) and B(2, 3). The point P(X, y) equidistant from the points A and B So, PA = PB
Also, (PA)2 = (PB)2
⇒ (6 – x)2 + (-1 – y)2 = (2 – x)2 + (3 – y)2
⇒ x2 – 12x + 36 + y2 + 2y + 1 = x2 – 4x + 4 + y2 – 6y + 9
⇒ x2 + y2 – 12x + 2y + 37 = x2 + y2 – 4x – 6y + 13
⇒ x2 + y2 – 12x + 2y – x2 –y2 + 4x + 6y = 13 – 37
⇒ - 8x + 8y = - 24
⇒ -8(x – y) = - 24
⇒ x – y = -24/-8
⇒ x – y = 3
Hence proved.
13. Find the co-ordinates of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5)
Solution
Let the required point be P(x, y). Then AP = BP = CP
That is, (AP)2 = (BP)2 = (CP)2
This means (AP)2 = (BP)2
⇒ (x – 5)2 + (y – 3)2 = (x – 5)2 + (y + 5)2
⇒ x2 – 10x + 25 + y2 – 6y + 9 = x2 – 10x + 25 + y2 + 10y + 25
⇒ x2 – 10x + y2 – 6y + 34 = x2 – 10x + y2 + 10y + 50
⇒ x2 – 10x + y2 – 6y – x2 + 10x – y2 – 10y = 50 – 34
⇒ -16y = 16
⇒ y = -16/16 = - 1
And (BP)2 = (CP)2
⇒ (x – 5)2 + (y + 5)2 = (x – 1)2 + (y + 5)2
⇒ x2 – 10x + 25 + y2 + 10y + 25 = x2 – 2x + 1 + y2 + 10y + 25
⇒ x2 – 10x + y2 + 10y + 50 = x2 – 2x + y2 + 10y + 26
⇒ x2 – 10x + y2 + 10y – x2 + 2x – y2 – 10y = 26 – 50
⇒ -8x = - 24
⇒ x = -24/-8 = 3
Hence, the required point is (3, -1).
14. If the points A(4, 3) and B(x, 5) lies on a circle with the centre O(2, 3). Find the value of x.
Solution
Given, the points A(4, 3) and B(x, 5) lie on the a circle with center O(2, 3).
Then OA = OB
Also (OA)2 = (OB)2
⇒ (4 - 2)2 + (3 – 3)2 = (x – 2)2 + (5 – 3)2
⇒ (2)2 + (0)2 = (x – 2)2 + (2)2
⇒ 4 = (x – 2)2 + 4
⇒ (x – 2)2 = 0
⇒ x = 2
Therefore, x = 2
15. If the point C(-2, 3) is equidistant from the points A(3, -1) and B(x, 8), find the value of x. Also, find the distance between BC.
Solution
As per the question, we have
AC = BC
16. If the point P(2, 2) is equidistant from the points A(-2, k) and B(-2k, -3), find k. Also, find the length of AP.
Solution
As per the question, we have
AP = BP
17. If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.
Solution
As per the question, we have
⇒ (x – a – b)2 + (y – b + a)2 = (x – a + b)2 + (y – a – b)2 (Squaring both sides)
⇒ x2 + (a + b)2 – 2x(a + b) + y2 + (a – b)2 – 2y(a – b) = x2 + (a – b)2 – 2x(a – b) + y2 + (a + b)2 – 2y(a – b)
⇒ -x(a + b) – y(a – b) = -x(a – b) – y(a + b)
⇒ -xa – xb – ay + by = - xa + bx – ya – by
⇒ by = bx
Hence, bx = ay.
18. Using the distance formula, show that the given points are collinear.
(i) (1, -1), (5, 2) and (9, 5)
(ii) (6, 9), (0, 1) and (-6, -7)
(iii) (-1, -1), (2, -3) and (8, 11)
(iv) (-2, 5), (0, 1) and (2, -3)
Solution
(i) Let A(1, -1), B(5, 2) and C(9, 5) be the give points. Then
∴ AB + BC = (5 + 5) units = 10 units = AC
Hence, the given points are collinear.
(ii) Let A(6, 9), B(0, 1) and C(-6, -7) be the give points. Then
∴ AB + BC = (10 + 10) units = 20 units = AC
Hence, the given points are collinear
(iii) Let A(-1, -1), B(2, 3) and C(8, 11) be the give points. Then
∴ AB + BC = (5 + 10) units = AC
Hence, the given points are collinear.
(iv) Let A(-2, 5), B(0, 1) and C(2, -3) be the given points. Then
∴ AB + BC = (2√5 + 2√5) units = 4√5 units = AC
Hence, the given points are collinear.
19. Show that the points A(7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.
Solution
The given points are A(7, 10), B(-2, 5) and (3, -4)
Thus, (AB)2 + (BC)2 = (AC)2
This show that △ABC is right-angles at B.
Therefore, the points A(7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right-angled triangle.
20. Show that the points A(3, 0) and B(6, 4) and C(-1, 3) are the vertices of an isosceles right triangle.
Solution
The given points are A(3, 0), B(6, 4) and C(-1, 3). Now,
∵ AB = AC and AB2 + AC2 = BC2
Therefore, A(3, 0), B(6, 4) and C(-1, 3) are die vertices of an isosceles right triangle.
21. If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right triangle with B = 90°, then find the value of t.
Solution
∵ ∠B = 90°
∴ AC2 = AB2 + BC2
⇒ (5 + 2)2 + (2 – t)2 = (5 – 2)2 + (2 + 2)2 + (2 + 2)2 + (2 – t)2
⇒ (7)2 + (t – 2)2 = (3)2 + (4)2 + (4)2 + (t + 2)2
⇒ 49 + t2 – 4t + 4 = 9 + 16 + 16 + t2 + 4t + 4
⇒ 8 – 4t = 4t
⇒ 8t = 8
⇒ t = 1
Hence, t = 1.
22. Prove that the points A(2, 4), B(2, 6) and C(2 + √3, 5) are the vertices of an equilateral triangle.
Solution
The given points are A(2, 4), B(2, 6) and C(2 + √3, 5). Now
Hence, the points A(2, 4), B(2, 6) and C(2 + √3, 5) are the vertices of an equilateral triangle.
23. Show that the points (-3, -3), (3, 3) and C(-3√3, 3√3) are the vertices of an equilateral triangle.
Solution
Let the given points be A(-3, -3), B(3, 3) and C(-3√3, 3√3). Now
Hence, the given points are the vertices of an equilateral triangle.
24. Show that the points A(-5, 6), B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area.
Solution
Let the given points be A(-5, 6) , B(3, 0) and C(9, 8).
Therefore, AB = BC = 10 units
Also, (AB)2 + (BC)2 = (10)2 + (10)2 = 200
and (AC)2 = (10√2)2 = 200
Thus, (AB)2 + (BC)2 = (AC)2
This show that △ ABC is right angled at B.
Therefore, the points A(-5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Also, area of a triangle = 1/2 × base × height
If AB is the height and BC is the base,
Area = 1/2 × 10 × 10
= 50 square units
25. Show that the points O(0, 0), A(3, √3) and B(3, -3√3) are the vertices of an equilateral triangle. Find the area of this triangle.
Solution
The given points are O(0, 0), A(3, √3) and B(3, -√3)
Therefore, OA = AB = OB = 2√3 units.
Thus, the points O(0, 0) A(3, √3) and B(3, -√3) are the vertices of an equilateral triangle
Also, the area of the triangle OAB = √3/4 × (side)2
= √3/4 × (2√3)2
= √3/4 × 12
= 3√3 square units.
26. Show that the following points are the vertices of a square:
(i) A(3, 2), B(0, 5), C(-3, 2) and D(0, -1)
(ii) A(6, -2), B(2, 1), C(1, 5) and D(5, 6)
(iii) A(0, -2), B(3, 1), C(0, 4) and D|(-3, 1)
Solution
(i) The given points are A(3, 2), B(0, 5), C(-3, 2) and D(0, -1)
Thus, diagonal AC = diagonal BD
Therefore, the given points from a square.
(ii) The given points are A(6, 2), B(2, 1), C(1, 5) and D(5, 6)
Thus, diagonal AC = diagonal BD
Therefore, the given points from a square.
(iii) The given points are P(0, -2), Q(3, 1), R(0, 4) and S(-3, 1)
Thus, diagonal PR = diagonal QS
Therefore, the given points from a square.
27. Show that the points A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus.
Solution
The given points are A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4)
Thus, diagonal AC is not equal to diagonal BD.
Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals
Hence, ABCD a rhombus.
Area of a rhombus = 1/2 × (product of diagonals)
= 1/2 × (52) × (9√2)
= 45(2)/2
= 45 square units.
28. Show that the points A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) are the vertices of a rhombus. Find its area.
Solution
The given points are A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1)
Therefore, the given points are the vertices of a rhombus.
Area (ABCD) = 1/2 × AC × BD
= 1/2 × 4√2 × 6√2
= 24 sq. units
Hence, the area of the rhombus is 24 sq. units.
29. Show that the points A(6, 1), B(8, 2) and C(9, 4) and D(7, 3) are the vertices of a rhombus. Find its area.
Solution
The given points are A(6, 1), B(8, 2), C(9, 4) and D(7, 3)
∵ AB = BC = CD = AD = √5 and AC ≠ BD
Therefore, the given points are vertices of a rhombus. Now,
Area (ABCD) = 1/2 × AC × BD
= 1/2 × 3√2 × √2 = 3 sq units.
Hence, the area of the rhombus is 3 sq. units.
30. Show that the points A(2, 1), B(5, 2) , C(6, 4) and D(3, 3) are the angular points of a parallelogram. Is this figure a rectangle ?
Solution
The given points are A(2, 1), B(5, 2), C(6, 4) and D(3, 3)
But diagonal AC is not equal to diagonal BD.
Hence, the given points do not form a rectangle.
31. Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not rectangle.
Solution
The given points are A(1, 2), B(4, 3),C(6, 6) and D(3, 5).
Thus, the diagonal AC and BD are not equal and hence ABCD is not a rectangle.
32. Show that the following points are the vertices of a rectangle.
(i) A(-4, -1),B(-2, -4), C(4, 0) and D(2, 3)
(ii) A(2, -2), B(14, 10), C(11, 13) and D(-1, 1)
(iii) A(0, -4), B(6, 2), C(3, 5) and D(-3, -1)
Solution
(i) The given points are A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3)
Also, diagonal AC = diagonal BD
Hence, the given points from a rectangle.
(ii) The given points are A(2, -2), B(14, 10) C(11, 13) and D(-1, 1)
Also, diagonal AC = diagonal BD
Hence, the given points from a rectangle.
(iii) The given points are A(0, -4), B(6, 2), C(3, 5) and D(-3, 1)
Also, diagonal AC = diagonal BD
Hence, the given points from a rectangle.
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Convert pebi to zebi
Learn how to convert 1 pebi to zebi step by step.
Calculation Breakdown
Set up the equation
$$1.0\left(pebi\right)={\color{rgb(20,165,174)} x}\left(zebi\right)$$
Define the prefix value(s)
$$The \text{ } value \text{ } of \text{ } pebi \text{ } is \text{ } 2.0^{50}$$
$$The \text{ } value \text{ } of \text{ } zebi \text{ } is \text{ } 2.0^{70}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(pebi\right)={\color{rgb(20,165,174)} x}\left(zebi\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} 2.0^{50}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 2.0^{70}}}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} 2.0^{50}} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 2.0^{70}}$$
$$\text{Conversion Equation}$$
$$2.0^{50} = {\color{rgb(20,165,174)} x} \times 2.0^{70}$$
Cancel factors on both sides
$$\text{Cancel factors}$$
$${\color{rgb(255,204,153)} \cancel{2.0^{50}}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancelto{2^{20}}{2.0^{70}}}$$
$$\text{Simplify}$$
$$1.0 = {\color{rgb(20,165,174)} x} \times 2^{20}$$
Switch sides
$${\color{rgb(20,165,174)} x} \times 2^{20} = 1.0$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{1.0}{2^{20}}\right)$$
$${\color{rgb(20,165,174)} x} \times 2^{20} \times \dfrac{1.0}{2^{20}} = \times \dfrac{1.0}{2^{20}}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{2^{20}}} \times \dfrac{1.0}{{\color{rgb(255,204,153)} \cancel{2^{20}}}} = \dfrac{1.0}{2^{20}}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{1.0}{2^{20}}$$
Rewrite equation
$$\dfrac{1.0}{2^{20}}\text{ can be rewritten to }2^{-20}$$
$$\text{Rewrite}$$
$${\color{rgb(20,165,174)} x} = 2^{-20}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx0.0000009537\approx9.5367 \times 10^{-7}$$
$$\text{Conversion Equation}$$
$$1.0\left(pebi\right)\approx{\color{rgb(20,165,174)} 9.5367 \times 10^{-7}}\left(zebi\right)$$
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## Complex Multiplication and division
In this page we are going to see complex multiplication and division.
Multiplication:
We are going to use FOIL method to multiply two numbers.
FOIL:
F - Firsts
O - Outers
I - Inners
- Lasts
### Complex multiplication and division - Examples
Examples:
1. Simplify: (2+3i)(3-4i)
Solution: = [2x3]+[(3i)(-4i)]+[2x(-4i)]+[(3i)x3]
= 6 + (-12i.i) -8i +9i
= 6 - 12i2-8i+9i(we know i2=-1)
= 6 - 12(-1) +i
= 6+12+i
= 18 +i
2. Simplify: (2-3i)2
Solution: = (2-3i)(2-3i)
= [2x2]+[(-3i)(-3i)]+[2(-3i)]+[2(-3i)]
= 4+9i2-6i-6i
As i2 = -1
= 4+9(-1)-12i
= -5-12i
= -(5+12i)
There is another easiest way to do the multiplication;
Let us do the the second example using the above rule;
(2-3i)(2-3i) = [(2x2)-(-3)x(-3)]+i[2x(-3)+2x(-3)]
= [4-9]+i[-6-6]
= -5 -12i
We got the same answer. So it is easy to use the above rule to multiply two complex numbers.
Conjugate:
Conjugate of a complex is a number having the same real part but having the negative imaginary part.
Example:
Conjugate of 3+2i = 3-2i
Conjugate of 4-7i = 4+7i
Conjugate of -3+9i = -3-9i
Division:
To divide two numbers we have to use conjugate of the denominator.
Example:
Divide 3+4i/2-3i
Solution:
To do the division first we have to multiply the numerator and denominator by the conjugate of the denominator.
Now let us do one more problem for division.
Example:
Divide: 3-2i/5-2i
Solution:
To do the division we have to multiply the numerator and denominator by the conjugate of the denominator.
Using the above methods we can do complex multiplication and division. Practice the above problems well to master in multiplication and division and try to do the problems given below on your own. Solutions for the practice problems are also given. If you have any doubts please contact us, we will help you to clear the doubts.
Problems for practice:
Simplify:
1. (2-5i)(7+4i) jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} 34-27i.
2. (5+2i)(2-3i) jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} 16-11i.
3. (5+2i)/(2-5i) jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} 1i.
4. (7+i)/(1+i) jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} 4-3i
Mandelbrot set
We will see some interesting fact about complex numbers. The beautiful Mandelbrot set is based on complex numbers. In this set, the sequence does not approach infinity. It is related to Julia sets.
The following image of Mandelbrot set is created by sampling complex numbers. This is named after the great mathematician Benoit Mandelbrot.
Uses of complex numbers:
• Complex numbers are used both in pure math and real world applications.
• Some functions which generate fractals includes complex numbers.
• Complex numbers have practical applications in technical fields.
• Eigen values and Eigen vectors involves complex numbers.
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# Find the probability distribution of number of heads in four tosses of a coin.
This is a multi-part question in the textbook answered separately on Clay6.com
Toolbox:
• Once we write down the sample space for the problem we can find the number of heads or tails for the favorable outcomes in each case and find their probabilities and arrange as a probability distribution.
Given: 4 tosses of a coin, the sample space can be represented as follows: S = $\begin{Bmatrix} HHHH & HHHT & HHTH &HHTT \\ HTHH &HTHT &HTTH &HTTT \\ THHH & THHT &THTH &THTT \\ TTHH &TTHT &TTTH & TTTT \end{Bmatrix}$
All these events are equally likely and have the same probability = $\large\frac{1}{16}$
Let X be the number of heads in those tosses. We can see that X can take the values of 0,1,2,3 or 4. It follows that:
P (X = 0) = P (TTTT) = $\large\frac{1}{16}$
P (X = 1) = P (HTTT, THTT, TTHT, TTTH) = $\large\frac{4}{16} = \frac{1}{4}$
P (X = 2) = P (HHTT, HTHT, HTTH, TTHH, THHT, THTH) = $\large\frac{6}{16} = \frac{3}{8}$
P (X = 3) = P (HHHT, HHTH, HTHH, THHH) = $\large\frac{4}{16} = \frac{1}{4}$
P (X = 4) = P (HHHH) = $\large\frac{1}{16}$
Therefore we can write the probability distribution as follows:
$\begin{matrix} \textbf{X} & 0 &1 &2&3&4 \\ \textbf{P(X)} &\large \frac{1}{16} &\large \frac{1}{4} &\large \frac{3}{8} &\large\frac{1}{4}&\large\frac{1}{16}\end{matrix}$
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# Difference between revisions of "2012 AMC 8 Problems/Problem 20"
## Problem
What is the correct ordering of the three numbers $\frac{5}{19}$, $\frac{7}{21}$, and $\frac{9}{23}$, in increasing order?
$\textbf{(A)}\hspace{.05in}\frac{9}{23}<\frac{7}{21}<\frac{5}{19}\quad\textbf{(B)}\hspace{.05in}\frac{5}{19}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(C)}\hspace{.05in}\frac{9}{23}<\frac{5}{19}<\frac{7}{21}$
$\textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23}$
## Solution 1
The value of $\frac{7}{21}$ is $\frac{1}{3}$. Now we give all the fractions a common denominator.
$\frac{5}{19} \implies \frac{345}{1311}$
$\frac{1}{3} \implies \frac{437}{1311}$
$\frac{9}{23} \implies \frac{513}{1311}$
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 2
Instead of finding the LCD, we can subtract each fraction from $1$ to get a common numerator. Thus,
$1-\dfrac{5}{19}=\dfrac{14}{19}$
$1-\dfrac{7}{21}=\dfrac{14}{21}$
$1-\dfrac{9}{23}=\dfrac{14}{23}$
All three fractions have common numerator $14$. Now it is obvious the order of the fractions. $\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}\implies\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}$. Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 3
Change $7/21$ into $1/3$; $$\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}$$ $$\frac{5}{15}>\frac{5}{19}$$ $$\frac{7}{21}>\frac{5}{19}$$ And $$\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}$$ $$\frac{9}{27}<\frac{9}{23}$$ $$\frac{7}{21}<\frac{9}{23}$$ Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 4
When $\frac{x}{y}<1$ and $z>0$, $\frac{x+z}{y+z}>\frac{x}{y}$. Hence, the answer is ${\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$. ~ ryjs
This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.
## Solution 5
By dividing, we see that 5/19 ≈ 0.26, 7/21 ≈ 0.33, and 9/23 ≈ 0.39. When we put this in order, $0.26$ < $0.33$ < $0.39$. So our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$ ~ math_genius_11
## Solution 6
$\frac{5}{19}$ is very close to $\frac{1}{4}$, so you can round it to that. Similarly, $\frac{7}{21} = \frac{1}{3}$ and $\frac{9}{23}$ can be rounded to $\frac{1}{2}$, so our ordering is 1/4, 1/3, and 1/2, or $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 7
The numbers are in form $\frac{x}{x+14}$. Using quotient rule on $\frac{d}{dx}(\frac{x}{x+14})$ gives $\frac{14}{(x+14)^2}$ and this is positive. Because the derivative is always positive and the values of $x$ given by this question $(5, 7, 9)$ can be put on an interval that does not contain the critical point $x=-14$, a greater $x$ implies a greater $\frac{x}{x+14}$, thus giving us the answer of $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$. ~lopkiloinm
## See Also
2012 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Calculating the Area of a Right Triangle: Step-by-Step Guide
Discover the fundamental principles of calculating the area of a right triangle. In this article, we will explore the step-by-step process and formulas to determine the area of a right triangle, providing a clear understanding for educators and learners alike.
## Understanding the Formula for Finding the Area of a Right Triangle
The formula to find the area of a right triangle is A = 0.5 * base * height. This formula is crucial for students to understand as it forms the basis for calculating the area of a right triangle. Explaining how this formula is derived and its significance in geometry can help students grasp the concept more effectively.
## Practical Applications of Right Triangle Area Calculation
Exploring real-world examples where the calculation of the area of a right triangle is used can help students see the relevance of this mathematical concept. Discussing applications in fields such as architecture, engineering, and design can make the learning experience more engaging and meaningful.
## Visualizing the Area Calculation Process
Utilizing diagrams and visual aids to illustrate the process of calculating the area of a right triangle can enhance students' understanding. By visually representing the base, height, and the resulting area, learners can develop a clearer mental image of the mathematical concept.
## Exploring Alternative Methods for Area Calculation
Introducing alternative approaches to finding the area of a right triangle, such as using trigonometric functions or the Pythagorean theorem, can broaden students' problem-solving skills. Understanding different methods can also deepen their comprehension of the underlying geometric principles.
## Encouraging Critical Thinking and Problem Solving
Engaging students in challenging exercises that require them to apply the area calculation formula to solve complex problems can foster critical thinking and analytical skills. Encouraging them to think creatively and approach problems from different angles can strengthen their mathematical abilities.
## Preguntas Frecuentes
### What is the formula for finding the area of a right triangle?
The formula for finding the area of a right triangle is 1/2 * base * height.
### Are there different methods for calculating the area of a right triangle?
Yes, there are different methods for calculating the area of a right triangle. The most common method is using the formula A = 1/2 * base * height, but you can also use trigonometric functions such as sine and cosine to find the area.
### How can the concept of area be introduced to students when teaching about right triangles?
The concept of area can be introduced to students when teaching about right triangles by showing them how to calculate the area using the formula A = 1/2 * base * height and demonstrating how this formula applies specifically to right triangles.
### What real-life applications can be used to demonstrate the importance of understanding the area of a right triangle in mathematics education?
Real-life applications such as construction, architecture, and engineering can demonstrate the importance of understanding the area of a right triangle in mathematics education.
In conclusion, understanding how to calculate the area of a right triangle is a fundamental skill for students in their Mathematics education. By mastering this concept, students can develop a solid foundation for more advanced geometric and algebraic concepts. It is important for educators to provide clear explanations and engaging activities to help students grasp this essential mathematical skill. With practice and guidance, students can confidently apply the formula to find the area of any right triangle, laying the groundwork for their continued success in mathematics.
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# 1961 AHSME Problems/Problem 20
## Problem
The set of points satisfying the pair of inequalities $y>2x$ and $y>4-x$ is contained entirely in quadrants:
$\textbf{(A)}\ \text{I and II}\qquad \textbf{(B)}\ \text{II and III}\qquad \textbf{(C)}\ \text{I and III}\qquad \textbf{(D)}\ \text{III and IV}\qquad \textbf{(E)}\ \text{I and IV}$
## Solution
$[asy] fill((-1,5)--(5,5)--(5,-1)--cycle,red); fill((2.5,5)--(-5,5)--(-5,-5)--(-2.5,-5)--cycle,cyan); fill((-1,5)--(4/3,8/3)--(2.5,5)--cycle,magenta); draw((-1,5)--(5,-1),dotted,Arrows); draw((-2.5,-5)--(2.5,5),dotted,Arrows); import graph; size(8.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=5.2,ymin=-5.2,ymax=5.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ Graph both equations. The solution to the system of inequalities is the magenta area because it is in both of the solutions of each individual inequality. From the graph, the magenta area is in quadrants $I$ and $II$, so the answer is $\boxed{\textbf{(A)}}$.
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Home > Lessons > Using the Quadratic Formula Updated November 25th, 2016
Introduction
This lesson will teach you how to solve quadratic equations by using the quadratic formula. Here are the sections within this lesson:
The quadratic equations that will be solved within this lesson will be second degree equations of a single variable. The equation must be of the form...
In other words, the equation must be equal to zero. Only then may we be able to identify the "a," the "b," and the "c" within the equation. The "a" is the coefficient of the squared term. The "b" is the coefficient of the linear term and the "c" is the constant term. When quadratics are in this form, they can be solved using this formula.
See our instructional videos section to see how the quadratic formula was derived.
Here is our first example of a quadratic equation.
To get the equation equal to zero, subtract 6 from both sides. Doing so, we get...
For this example, a = 1, b = -1, and c = -6. We will substitute these values into the formula.
We can clean up the formula by taking the opposite of negative one and by cleaning up the expression underneath the radical. We can also simplify the denominator by multiplying two and one.
Simplifying further, we get...
The square root of 25 is 5.
Once we get down to this step, we have to bifurcate (split into two). The "+" means we have two equations -- one equation with a plus and the other with a minus sign.
Now we will solve each separately.
Therefore, the two solutions are 3 and -2.
However, not all equations are this simple to solve. Let us examine another example.
Again, we have to get the equation equal to zero before we use the quadratic equation. We must add 5x to both sides and subtract 4 from both sides to get zero on the right side.
Now, a = 6, b = 5, and c = -4. These values must be substituted into the formula as follows.
We can start cleaning up the formula.
The value under the radical is 25 + 96, which is 121.
The square root of 121 is 11.
Now, we bifurcate and simplify.
This makes the two solutions 1/2 and -4/3.
Instructional Videos Watch our instructional videos: Quizmasters After reading the lessons, try our quizmasters. MATHguide has developed numerous testing and checking programs to solidify these skills: Related Lessons esson: Completing the Square esson: Operations on Polynomials esson: Rational Expressions esson: Factoring esson: Polynomials, Factors, Zeros esson: Conic Sections
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# Ann and Pierre purchased \$37 worth of French fries.
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Ann and Pierre purchased \$37 worth of French fries. [#permalink]
### Show Tags
15 Jul 2014, 23:45
3
00:00
Difficulty:
55% (hard)
Question Stats:
65% (01:48) correct 35% (01:58) wrong based on 180 sessions
### HideShow timer Statistics
Ann and Pierre purchased \$37 worth of French fries. Ann spent \$3 on each order of French fries she bought, and Pierre spent \$5 on each order of French fries he bought. How many orders of French fries did Ann purchase?
(1) Ann and Pierre would have spent a total of \$111 if Ann had spent \$9 on each order of French fries she purchased, and Pierre had spent \$15 on each order of French fries he purchased.
(2) Ann and Pierre purchased a total of 9 orders of French fries.
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Joined: 23 May 2014
Posts: 8
Re: Ann and Pierre purchased \$37 worth of French fries. [#permalink]
### Show Tags
16 Jul 2014, 00:22
1
1
If we let "a" be the number of orders by Anna and "p" be the number of order by Pierre, the information supplied:
* 37 = 3a + 5p
* a and p are integers
(1) tells us that 111 = 9a + 15p
This simplifies to 37 = 3a + 5p
No new information given.
Not enough information to solve the equation (could be either a = 9 and p = 2; or a = 4 and p = 5
(2) tells us that a + p = 9
substituting a = 9 - p into 37 = 3a + 5p yields a = 4.
Sufficient.
Therefore, B.
I would definitely not classify this question as 600-700... more like sub-600...
Math Expert
Joined: 02 Sep 2009
Posts: 58453
Re: Ann and Pierre purchased \$37 worth of French fries. [#permalink]
### Show Tags
16 Jul 2014, 01:56
dbaremberg wrote:
Ann and Pierre purchased \$37 worth of French fries. Ann spent \$3 on each order of French fries she bought, and Pierre spent \$5 on each order of French fries he bought. How many orders of French fries did Ann purchase?
(1) Ann and Pierre would have spent a total of \$111 if Ann had spent \$9 on each order of French fries she purchased, and Pierre had spent \$15 on each order of French fries he purchased.
(2) Ann and Pierre purchased a total of 9 orders of French fries.
If we let "a" be the number of orders by Anna and "p" be the number of order by Pierre, the information supplied:
* 37 = 3a + 5p
* a and p are integers
(1) tells us that 111 = 9a + 15p
This simplifies to 37 = 3a + 5p
No new information given.
Not enough information to solve the equation (could be either a = 9 and p = 2; or a = 4 and p = 5
(2) tells us that a + p = 9
substituting a = 9 - p into 37 = 3a + 5p yields a = 4.
Sufficient.
Therefore, B.
I would definitely not classify this question as 600-700... more like sub-600...
Agree, the question is as simple as it gets.
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Re: Ann and Pierre purchased \$37 worth of French fries. [#permalink]
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05 Aug 2014, 01:18
Let orders of Ann be x and orders of Pierre be y, then
3x + 5y = 37
We are looking for x.
(1) 9x + 15 y = 111. It's a trap! If you combine the equations you will get nothing, since 3 x first equation = second equation. IS.
(2) x + y = 9 --> y = 9-x --> set in equation from question, SUFF.
B.
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Re: Ann and Pierre purchased \$37 worth of French fries. [#permalink]
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06 Aug 2014, 10:55
Gnpth wrote:
Ann and Pierre purchased \$37 worth of French fries. Ann spent \$3 on each order of French fries she bought, and Pierre spent \$5 on each order of French fries he bought. How many orders of French fries did Ann purchase?
(1) Ann and Pierre would have spent a total of \$111 if Ann had spent \$9 on each order of French fries she purchased, and Pierre had spent \$15 on each order of French fries he purchased.
(2) Ann and Pierre purchased a total of 9 orders of French fries.
B
For the question that involves 2 equations and 2 variables, always check to see if you solve it from the equations.
3A + 5P = 37, if you multiply this by 3, you get (1) 9A + 15P = 111
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Re: Ann and Pierre purchased \$37 worth of French fries. [#permalink]
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30 Aug 2018, 10:38
Let a be the number of orders that Ann purchased and let p be the number of orders that Pierre purchased. From the question stem, we can set up the equation 37 = 3a + 5p, which expresses the total cost of the French fries purchased by Ann and Pierre. In order to solve for the number of orders of French fries that Ann bought, we need to be able to set up another distinct equation that contains the variables a and p.
Statement (1): insufficient. From this information, we can write the equation 111 = 9a + 15b. We have another equation relating a and p, but is it distinct? No. Upon closer inspection, this is simply the result of multiplying both sides of the original equation 37 = 3a + 5p by 3. We can eliminate choices (A) and (D).
Statement (2): sufficient. From this statement, we can set up the equation 9 = a + b. Is this equation distinct from the original equation? Yes. This is therefore sufficient to find the value of a, or the number of orders of French fries that Ann purchased. Choice (B) is correct.
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Re: Ann and Pierre purchased \$37 worth of French fries. [#permalink] 30 Aug 2018, 10:38
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# 10.5 Polar form of complex numbers
Page 1 / 8
In this section, you will:
• Plot complex numbers in the complex plane.
• Find the absolute value of a complex number.
• Write complex numbers in polar form.
• Convert a complex number from polar to rectangular form.
• Find products of complex numbers in polar form.
• Find quotients of complex numbers in polar form.
• Find powers of complex numbers in polar form.
• Find roots of complex numbers in polar form.
“God made the integers; all else is the work of man.” This rather famous quote by nineteenth-century German mathematician Leopold Kronecker sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras , Descartes , De Moivre, Euler , Gauss , and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science.
We first encountered complex numbers in Complex Numbers . In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre’s Theorem.
## Plotting complex numbers in the complex plane
Plotting a complex number $\text{\hspace{0.17em}}a+bi\text{\hspace{0.17em}}$ is similar to plotting a real number, except that the horizontal axis represents the real part of the number, $\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}$ and the vertical axis represents the imaginary part of the number, $\text{\hspace{0.17em}}bi.$
Given a complex number $\text{\hspace{0.17em}}a+bi,\text{\hspace{0.17em}}$ plot it in the complex plane.
1. Label the horizontal axis as the real axis and the vertical axis as the imaginary axis.
2. Plot the point in the complex plane by moving $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units in the horizontal direction and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ units in the vertical direction.
## Plotting a complex number in the complex plane
Plot the complex number $\text{\hspace{0.17em}}2-3i\text{\hspace{0.17em}}$ in the complex plane .
From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. See [link] .
Plot the point $\text{\hspace{0.17em}}1+5i\text{\hspace{0.17em}}$ in the complex plane.
## Finding the absolute value of a complex number
The first step toward working with a complex number in polar form is to find the absolute value. The absolute value of a complex number is the same as its magnitude , or $\text{\hspace{0.17em}}|z|.\text{\hspace{0.17em}}$ It measures the distance from the origin to a point in the plane. For example, the graph of $\text{\hspace{0.17em}}z=2+4i,\text{\hspace{0.17em}}$ in [link] , shows $\text{\hspace{0.17em}}|z|.$
## Absolute value of a complex number
Given $\text{\hspace{0.17em}}z=x+yi,\text{\hspace{0.17em}}$ a complex number, the absolute value of $\text{\hspace{0.17em}}z\text{\hspace{0.17em}}$ is defined as
$|z|=\sqrt{{x}^{2}+{y}^{2}}$
It is the distance from the origin to the point $\text{\hspace{0.17em}}\left(x,y\right).$
Notice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin,
## Finding the absolute value of a complex number with a radical
Find the absolute value of $\text{\hspace{0.17em}}z=\sqrt{5}-i.$
Using the formula, we have
$\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(\sqrt{5}\right)}^{2}+{\left(-1\right)}^{2}}\hfill \\ |z|=\sqrt{5+1}\hfill \\ |z|=\sqrt{6}\hfill \end{array}$
Find the absolute value of the complex number $\text{\hspace{0.17em}}z=12-5i.$
13
## Finding the absolute value of a complex number
Given $\text{\hspace{0.17em}}z=3-4i,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}|z|.$
Using the formula, we have
$\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(3\right)}^{2}+{\left(-4\right)}^{2}}\hfill \\ |z|=\sqrt{9+16}\hfill \\ \begin{array}{l}|z|=\sqrt{25}\\ |z|=5\end{array}\hfill \end{array}$
The absolute value $\text{\hspace{0.17em}}z\text{\hspace{0.17em}}$ is 5. See [link] .
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5
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GreeneMath.com - Introduction to Roots Test #2
In this Section:
In this section, we learn about roots. We begin our process by learning about the most basic of all roots, the square root. The square root of a number such as 9, is a number that when multiplied by itself gives us back the number 9. Two such numbers would be 3, and (-3). If we multiply 3 • 3 we get 9, also if we multiply (-3) • (-3) we get 9. Therefore the square root of 9 is 3 or (-3). We then move into some basic notation issues that occur when dealing with roots in general. We introduce something known as an irrational number, and discuss the difference between a rational and irrational number. We find that any number whose square root is a rational number is called a perfect square. We wrap up the section by showing what to do when we have higher level roots.
Sections:
In this Section:
In this section, we learn about roots. We begin our process by learning about the most basic of all roots, the square root. The square root of a number such as 9, is a number that when multiplied by itself gives us back the number 9. Two such numbers would be 3, and (-3). If we multiply 3 • 3 we get 9, also if we multiply (-3) • (-3) we get 9. Therefore the square root of 9 is 3 or (-3). We then move into some basic notation issues that occur when dealing with roots in general. We introduce something known as an irrational number, and discuss the difference between a rational and irrational number. We find that any number whose square root is a rational number is called a perfect square. We wrap up the section by showing what to do when we have higher level roots.
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# Find angle based on bisector, altitude, median info
In △ABC, CM is the median, CD is the angle bisector of ∠ACB, CH is the altitude. CM, CD, and CH divide ∠ACB by four equal angles. Find angles of △ABC.
I started with finding angle relationships but was not able to use them to proceed. Maybe this is not the right approach. Can you advise? Thanks.
Assume that $$AC. Let $$O$$ be the centre of the circumcircle of $$ABC$$. Prove that (in any triangle) angles $$\sphericalangle{BCO}=\sphericalangle{ACH}$$. Since the angle bisector already divides the angle in halves, we have to have that angle $$\sphericalangle{ACH}$$ is a quater of angle $$C$$ as well as $$\sphericalangle{BCM}$$. But since $$\sphericalangle{BCO}=\sphericalangle{ACH}$$ we must have $$\sphericalangle{BCO}=\sphericalangle{BCM}$$ i.e. points $$C$$, $$M$$, $$O$$ are collinear. Prove that it implies that $$\sphericalangle{C}=90^{\circ}$$ and usethe fact that then $$\sphericalangle{ACH}=\frac{90^{\circ}}{4}$$ to compute the rest.
Let $$\measuredangle ACH=\alpha$$.
Thus, $$\measuredangle A=90^{\circ}-\alpha$$ and $$\measuredangle B=90^{\circ}-3\alpha.$$
Thus, by the law of sines we obtain: $$\frac{\sin3\alpha}{\cos\alpha}=\frac{AM}{MC}=\frac{BM}{MC}=\frac{\sin\alpha}{\cos3\alpha},$$ which gives $$\sin6\alpha=\sin2\alpha$$ or $$6\alpha+2\alpha=180^{\circ}.$$ Can you end it now?
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In a car race, car $\text{A}$ beats car $\text{B}$ by $45 \; \text{km},$ car $\text{B}$ beats car $\text{C}$ by $50 \; \text{km},$ and car $\text{A}$ beats car $\text{C}$ by $90\;\text{km}.$ The distance $\text{(in km)}$ over which the race has been conducted is
1. $500$
2. $475$
3. $550$
4. $450$
Given that, in a car race, car$\text{A}$ beats car$\text{B}$ by $45 \; \text{km},$ and car$\text{B}$ beats car$\text{C}$ by $50 \; \text{km},$ and car$\text{A}$ beats car$\text{C}$ by $90 \; \text{km}.$
Let the distance over which the race has been conducted be $’x’ \; \text{km}.$
Case $1:$
DIAGRAM
Case $2:$
DIAGRAM
Case $3:$
DIAGRAM
We know that, $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$
$\Rightarrow \boxed {\text{Speed} \propto \text{Distance}} \; (\text{Time = constant})$
Now, we can calculate the ratio of speeds:
• In Case $1: \; \frac{S_{A}}{S_{B}} = \frac{x}{x – 45} \; \longrightarrow (1)$
• In Case $2: \; \frac{S_{B}}{S_{C}} = \frac{x}{x – 50} \; \longrightarrow (2)$
• In Case $3: \; \frac{S_{A}}{S_{C}} = \frac{x}{x – 90} \; \longrightarrow (3)$
Now, multiply equation $(1) \& (2),$ we get
$\left( \frac{S_{A}}{S_{B}} \right) \times \left( \frac{S_{B}}{S_{C}} \right) = \left( \frac{x}{x-45} \right) \left( \frac{x}{x-50} \right)$
$\Rightarrow \frac{S_{A}}{S_{C}} = \frac{x^{2}}{(x-45)(x-50)}$
$\Rightarrow \frac{x}{x-90} = \frac{x^{2}}{(x-45)(x-50)}$
$\Rightarrow (x-45)(x-50) = x(x-90)$
$\Rightarrow x^{2} – 50x – 45x + 2250 = x^{2} – 90x$
$\Rightarrow – 95x + 90x = – 2250$
$\Rightarrow -5x = -2250$
$\Rightarrow x = \frac{2250}{5}$
$\Rightarrow \boxed{x = 450 \; \text{km}}$
$\therefore$ The distance (in km) over which the race has been conducted is $450.$
Correct Answer$: \text{D}$
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Step 1: Solve all Centers Home | Pre-Solution Stuff | Step 1 | Step 2 | Step 3 Solution Moves Lists
Part 1: The first step in solving your cube is you have to solve the centers so you can have a point of reference for solving the rest of the cube. The trick to this is that you have to get all the centers in the right spot (if the centers don't correctly line up with each other then the edges and corners won't line up either and it makes a real big mess). The first thing you need to do is solve two opposite centers. If you know of two colors that are opposite on your cube it makes things a little easier. The colors that are opposite on my cube are this, in case yours are the same, blue-white, green-yellow, red-orange. If you're not sure of two opposite colors on your cube then here is a way to double check to make sure you get two opposite colors. Take two corners, each with say red and yellow in their color schemes. You can pick any two corners but make sure each corner cublet has two colors that are the same. You don't have to move them just look at them. Now that you have these two cublets if you were to solve them then they would be on the same edge, only opposite sides of that edge (i.e. one corner would touch the top face and the other would touch the bottom face). Now just look at the other two colors on each of those pieces and you have two opposite colors. Here is an example to clear this up. Let's say I pick two corner pieces, red-yellow-blue, and red-yellow-white. Now if I were to solve these two corners then the red-yellow part would be on the same edge of the cube. This would put them on opposite sides of that edge so that each touched a different face (either bottom or top). Now I know that blue and white are opposites on that particular cube because they are on opposite sides of the cube when the corners are solved.
Top View Bottom View
Part 2: Now that you know two opposite colors you have to solve those two centers. Lets stick with the two opposite colors blue and white for this example (these colors may not be opposite on your cube so if they're aren't just use two colors that are). The first step is to make two rows of each color. Get two pieces into this position first,
In this position do the move l to get the blue piece from the top lined up to the one on the front face.
Now to remeber where blue needs to go do the move F ² to get it into the u slice. When you have a solid row in the u slice think of that as being set as the center color for that face. This helps to remeber where to postion the centers in relation to each other.
Now you have to solve a white row. There are three things that can happen from here. Either you already have a row of white centers solved, in that case get it onto the opposite face as blue and set it (put it in the u slice) If your cube already had a white row solved then after you set it click here, or you will have 2 white centers not on the back or front faces, or you will have 3 whites on the back face, and one white not on the back or front face.
1. If you have two white centers not in the back or front face get your cube to look like this,
From here do the move f ' to solve the white row and then d ' B ² to put it into the back face and set it.
Now you should have 2 white centers in what I call the middle section. The middle section is all the faces except the ones where you're solving the centers, in this case the up face, down face, left face, and right face. If any centers are in the front or the back face then turn the d slice to get them into the middle section. Be careful when doing this to not turn the other white center back into the front or back face at the same time. If it looks like that's going to happen then move one of the whites to a face that hasn't been set with another color, the left or right face, and do the move N ² (N being whatever slice the piece is on) Then turn the other white into the middle section. Now that those two centers are in the "middle section" between the front and back faces get them into this position,
Front View Back View
From here do the moves f ' d '. This will completely solve the white center. Go on to #3.
2. Now if you only had one white center in the middle section and 3 white centers on the back face here is what you do. First of all do the move F. This gets the blue row out of the way for a turn on the l slice. Now get the one white piece in the middle section and get it into the position in the diagrams. Now turn the back face to get the 3 whites into the position on the diagram. Now do the move l. This solves the white row. Now do the moves U ' l ' This will completely solve the white center.
Front View Back View
3. Now your cube should have the white center solved and a blue row on the opposite face. If it doesn't then start over. From here you need to solve the last blue row. There are three things that can happen from here. You will either have the last blue row solved but in the middle section, or you will have 2 blue center pieces in the middle section, or you will have one blue center piece in the middle section and one in the front face.
I. If your cube has a blue row already solved but it is in the middle section then you have to do one of two moves.
If your cube looks like this then do this move L ² d ' L ² d. This move will put the blue row in the u face then move a white row to the opposite face of the cube, restore the blue row to the d face then solve the centers.
Front View Back View
If your cube looks like this then do this move R ² d R ² d '. This move will put the blue row in the u face then move a white row to the opposite face of the cube, restore the blue row to the d face then solve the centers.
Front View Back View
II. If your cube has the last two blue center pieces in the middle section then get your cube to look like this,
Front View Back View
From here do the move f ' L ² d ' L ² d. This will solve the blue row then move it to the u face, move the white row to the opposite side of the cube, move the blue row back onto the d slice, then move them back in place of their centers.
III. If your cube has one blue center piece in the front face and one in the middle section then get your cube to look like this,
Front View Back View
From here do the move F d F ' d '. This move will solve the last blue row and then the last move restores the two centers.
Your cube should now look like this,
Front View Back View
4. Now that you have the first two centers solved you have to do the other four. For the next step blue and white become the top and bottom faces. For these examples I'll use white as the top face and blue as the bottom face. Now you have to solve the front and back centers. Do the same thing you did with two corners that you did in the beginning to find two more opposite centers on your cube. For this step you don't need to worry about using the corners to position the centers right it will still work out. Now that you have two more opposite colors on your cube they become the front and back centers. For this example I'll use green and yellow as the next opposite centers to solve.
There are 4 things that can happen at this point. Either both of these centers will be solved and in the right spot (go to #5), they will be solved and in the wrong spots, you will have solid rows of each color but the centers not solved, the pieces for the centers will be scrambled everywhere with few or no solid rows.
I. If your cube has the centers solved but not in the right spots (not on opposite sides of the cube) then hold your cube as in the diagram below and do the move d ' B ² d ² L ² d '. The colors on your cube can be switched from this diagram. Instead of having green in the front face and yellow in the left face you can have yellow in the front face and green in the left face. In that case do the same move it will still work.
II. If your cube has 2 rows of each color solved then get your cube to look like this
Front View Back View
The black dot on the U face in both diagrams is in the same spot on both diagrams it is just used to help show how the cube is oriented
From here do the move d to solve the green and yellow centers.
III. If your cube has few or no solid rows of green or yellow then here is what you do. First you have to set any rows that you do have. If you have one green row and one yellow row make sure to set them on opposite sides of the cube. Now you have to solve the remaining pieces. You do this basically the same way you solved the blue and white rows in the beginning, only now you do it without using the top or bottom faces. Here is an example,
In this example you would do the move d F ' d '. This will solve the yellow row then set it in the front face. Once you have 4 rows do the same thing as in II. to solve the centers. Here is an example for if you already have a row set
Here you would do the move d ' L d. This will solve the green row on the L face then move it into the front face to solve the center. Just mess around with these moves to solve the centers. Maybe you'll even come up with a few new moves. A hint on if you have one center solved and only one row or no rows solved of the other color. Hold that center so that it is on the front face and blue and white are on the bottom and top faces then do the move d ' L ² This will set one of the rows in the L face. This helps you to use the d face to help you solve the other center. Once you have two rows of each color then refer to II. to finish the two centers. Basically you have to think in rows. Once you do have a row completed remember to put it in the u face. That way you have room to use the d face to do all your moves. I hardly every use the u face to do the work in any of these steps. You can adjust this to your liking but I prefer to do everything on the lower half of the cube.
5. Now you should have 4 centers solved. Using the corners double check to make sure both pairs of solved centers are opposite colors. Now you have two centers left to solve and they should be on opposite sides of the cube. If you do not have this on your cube then go back to whichever step best resembles your cube and try again. Now the two unsolved faces become the front and back faces. For these examples I'm going to use orange as the front face and red as the back face. Before you start solving these centers you have to make sure you're solving them into the correct face. To do this you have to solve one of the corners that touches two of the solved centers. From here you know the color of the face with the unsolved center. Here is an example,
Now you would know that the dark gray area is orange and the opposite face (the back face) is red. There are 6 positions your cube can be in now. Either all the centers are now solved and in the right places (Go on to step 2), all the centers are solved but the red and orange ones are switched, you have two solid rows of each color but they are not all on the right face, you have a checkered pattern on both faces, you have 3 of each color on one face and one of the other color on the same face, or you have a checkered pattern on one face and two rows on the other.
I. If all of your centers are solved but the red and orange ones are switched then hold your cube so that the centers that need to be switched are on the front and back faces like in the diagram
Front View Back View
From here do the move d ² F ² B ² d ². This will switch the centers by rows.
II. If your cube has a solid row of each color on both the front and back faces then get it to look like this
Front View Back View
From here do the move d ² B ² d ² This will solve the red and orange centers.
III. You may also have a checkered pattern of both faces like this, *NOTE* Before doing this move check to make sure that your cube looks like these diagrams. If it does not then turn only the back and/or front faces to get it into this position.
Front View Back View
Now do the move d ² F B d ². This will solve both the orange and red centers and get them in the right place. Once again note that this move only works when your cube is in the position of the diagrams. If the move did not work then make sure that the centers look exactly like they do in the diagram.
IV. If your cube has three of one color and one of the opposite color on both the front and the back faces then there are two things that can happen from here. You will either have three orange and one red on the front face or three red and one orange on the front face. Basically in this step you either have most of the front face colors on the front face or most of the back face colors on the front face. Solve a corner that connects the front face, left face, and right face in relation to how you're holding the cube now. Now you know the color of the front face.
If the three center pieces on the front are supposed to be on the front (orange in my example) then get your cube to look like this
Front View Back View
From here do the move d ² F ' d ². This will make the last two rows for the centers then solve them. This will solve the red and orange centers.
If you're cube has three center pieces from the back face on the front (red in my example) then get your cube to look like this
Front View Back View
From here do the move d ² F ' B ² d ². This will get all the pieces into solid rows of their color then solve the centers.
V. Your cube may have a checkered pattern on one face and solid rows on the other face. If it does turn the F and/or B faces to get your cube like the diagrams, (*NOTE* Before doing this move check to make sure that your cube looks exactly like these diagrams. If it does not then turn only the back and/or front faces to get it into this position).
Front View Back View
Now do the move d ² B' d ² B d ² B ² d ². This will get the pieces into the position from number IV and then solve them the same way. If the move did not work then you might not have had the pieces in the exact same spot as the diagram. Try to find the step that most closely resembles your cube and try again.
Overview
Now your cube should look like this,
Front-Top View Back-Bottom View
Now you have solved your cube 1/3 of the way! Only 2 more steps to go! Before going on to the next step solve one corner with it's three centers and double check to make sure those three are on the right spot. Then, if you know your opposite colors, you can look at the opposite face of each of those front faces to make sure the centers are solved. If your centers are not lined up then you're going to have a very hard time in step 3.
Don't forget to double check!
Home | Pre-Solution Stuff | Step 1 | Step 2 | Step 3
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# 4th Class Mathematics Geometrical Figures Quadrilateral
Quadrilateral
Category : 4th Class
### Quadrilateral
The geometrical figure having four sides are called quadrilateral. Look at the following figures:
Properties of Quadrilaterals
• A quadrilateral has four sides.
• A quadrilateral has four vertices.
• A quadrilateral has four angles.
• Sum of all four angles of a quadrilateral is $\text{36}0{}^\circ$
• Sides of the quadrilateral ABCD are AB, BC, CD, and AD
• Vertexes of the quadrilateral ABCD are point A, point B, point C and point D
• Angles of the quadrilateral are$\angle \text{ABC}$, $\angle \text{BCD},\text{ }\angle \text{CDA},$and$\angle \text{DAB}$
• Sum of all the four angles of the quadrilateral ABCD , $\angle \text{A}+\angle \text{B}+\angle \text{C}+\angle \text{D}=$$\text{6}0{}^\circ +\text{12}0{}^\circ +\text{8}0{}^\circ +\text{1}00{}^\circ =\text{36}0{}^\circ$
Types of Quadrilateral
In this chapter we will study about two types of quadrilateral:
• Rectangle
• Square
Rectangle
Rectangle is a quadrilateral in which (i) all angles are of$\text{9}0{}^\circ$and (ii) opposite sides are equal in length.
ABCD is a rectangle in which
(i) $\angle \text{A}=\angle \text{B}=\angle \text{C}=\angle \text{D}=\text{9}0{}^\circ$
(ii) AB = CD = 6 cm, and BC = AD = 4 cm.
Square
Square is a quadrilateral whose (i) All angles are of$\text{9}0{}^\circ$and (ii) All sides are equal
ABCD is a rectangle in which (i) $\angle \text{A}=\angle \text{B}=\angle \text{C}=$$\angle \text{D}=\text{9}0{}^\circ$(ii)$\text{AB}=\text{BC}=\text{CD}$$=\text{DA}=\text{5 cm}$
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# 5th grade word problems and fractions pd
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• Activity 15.8Slicing SquaresGive students a worksheet with four squares in a row, each approximately 3 cm on a side. Have them shade in the same fraction in each square using vertical dividing line. You can use the context of a garden or farm. For example, slice each square in fourths and shade three-fourths as in Figure 15.20. Next, tell students to slice each square into equal-sized horizontal slices. Each square must be partitioned differently, using from one to eight slices. For each sliced square, they record an equations showing the equivalent fractions. Have them examine their equations and drawings to look for any patterns. You can repeat this with four more squares and different fractions.What product tells how many parts are shaded?What product tells how many parts in the whole?Notice that the same factor is used for both part and whole
• Give students an equation expressing an equivalence between two fraction but with one of the numbers missing and ask them to draw a picture to solve. Here are four different examples:5/3 = _/62/3 = 6/_8/12 = _/39/12 = 3/_The missing number can be either a numerator or a denominator. Furthermore, the missing number can either be larger or smaller that the corresponding part of the equivalent fraction. (All four possibilities are represented in the examples.) The examples shown involve simple whole-number multiples between equivalent fractions. Next, consider pairs such as 6/8 = _/12 or 9/12 = 6/_. In these equivalences, one denominator or numerator is not a whole number multiple of the other.
• ### 5th grade word problems and fractions pd
1. 1. 5th Grade Fractions &Word Problems Laura Chambless RESA Consultantwww.protopage.com/lchambless
2. 2. CCSS and Gaps What are your gaps in curriculum?1. Review CCSS for Fractions2. Think about your resources3. Think about your teaching – Highlight anything your resources covers well in YELLOW. – Highlight any part of the standard you would like more clarification on in BLUE.
3. 3. Learning TargetUse equivalent fractions as a strategy to add and subtract fractions. 5.NF.1, 5.NF.2Apply and extend previous understandings of multiplication and division to multiply and divide fractions. 5.NF.3, 5.NF.4, 5.NF.5, 5.NF.6, 5.NF.7
4. 4. Fraction Word Problem40 students joined the soccer club. 5/8 of the students were boys. How many girls joined the soccer club? Draw a picture and solve it.1. 2 min. working problem on own2. 5 min. sharing with group3. Class discussionFound at: http://www.mathplayground.com/wpdatabase/Fractions1_3.htm
5. 5. Problem Solving with Bar Diagrams1. Understand: Identify what is known and what is unknown. Draw the bar diagram to promote comprehension and demonstrates understanding. (Situation vs. Solution Equation)2. Plan: Decide how you will solve the problem (find the unknown). Analyze the bar diagram to find a solution plan.3. Solve: Execute the plan. Use the bar diagram to solve.4. Evaluate: Assess reasonableness using estimation or substitution. Substitute the solution for the unknown in the bar diagram.
6. 6. Bar DiagramsWatch Introduction Videohttp://www.mhschool.com/math/com mon/pd_video/mathconnects_bardi agram_p1/index.htmlhttp://www.mhschool.com/math/com mon/pd_video/mathconnects_bardi agram_p2/index.html
7. 7. Practice Bar DiagramsTo: Rani earned \$128 mowing lawns and \$73 babysitting. How much money did Rani earn?With: Jin had \$67 in his pocket after he bought a radio controlled car. He went to the store with \$142. How Much did Jin spend on the car?By: There are 9 puffy stickers. There are 3 times as many plain stickers as puffy stickers. How many plain stickers are there?You pick 2 more to do by yourself. Share with partnerDraw Your Way to Problem Solving Success Handout, Robyn Silbey
8. 8. Thinking Blockshttp://www.mathplayground.com/think ingblocks.html Explore the site When done exploring go to my Protopage and look at your grade level math tab.
9. 9. Fractions Stand and ShareMake a list of what you know and any connections you have about the fraction ¼.
10. 10. Representations (Part 2 video, 5:16)Set Purpose of video: List why representations are important in the classroom. •Representations are mathematics content representing mathematical ideas is a practice that students need to learn. •Representations provide tools for working on mathematics and contribute to the development of new mathematical knowledge. •Representations support communication about mathematics. •Using multiple representations can help develop understanding and support the diverse needs of students. From: Dev-TE@M session 2
11. 11. Benefit of Representations (Part 4 video, 2:17)Set Purpose of video: Did you benefit from our discussions, and how will your students benefit from class discussions?1. As you listen , list benefits for students2. Compare list with partner From: Dev-TE@M session 3
12. 12. Build Connections to Whole Numbers 0 1 2 3 4 51+1+1+1+1=5 1/4 1/2 3/4 0 1 ¼ +¼ +¼+¼ =1
13. 13. Fractions Fraction ActivityPaper Strips Fraction Kit: 1, ½, 1/4 , 1/8, 1/16Add to Fraction Kit: 1/3, 1/6, 1/12Add to Fraction Kit: 1/5, 1/10 Compare/Add/Subtract/Multi./Divide with StripsREAD and DO: 5.NF.1, 5.NF.2, 5.NF.3, 5.NF.4, 5.NF.5, 5.NF.6, 5.NF.7Smaller Answer Wins (need dice)• Prove with Fraction Strips
14. 14. Lunch
15. 15. Definition of Fractions1. Make a list of what you would like to have in a definition of a fraction2. Partner up and compare lists3. Group discussion
16. 16. Definition of a Fraction (Part 5 and 6 videos, 11:48/4:27)Set Purpose of video: What are some key parts in creating a definition of a fraction that you will use in your room?– Give handout of working definitionArticle: Definitions and Defining in Mathematics and Mathematics Teaching by: Bass and Ball From: Dev-TE@M session 3
17. 17. Definition Of Fractions• Identify the whole• Make d equal parts• Write 1/d to show one of the equal parts• If you have d of 1/d, then you have the whole• If you have n of 1/d, then you have n/d• n and d are whole numbers• d does not equal 0Dev-TE@M • School of Education • University of Michigan • (734)408-4461 • [email protected] For review only - Please do notcirculate or cite without permission
18. 18. Ordering Fractions Order Fractions 8/6, 2/5, 8/10, 1/12How did you figure out what order they went in?
19. 19. Fractions Prove with Fraction StripsNumber Line: (Benchmarks) 0, ½, 1Equivalent Fractions: Same Name FrameCompare (>/<): same numerator or same denominator
20. 20. Strategies for Comparing Fractions• Dev-TE@M session 9
21. 21. Fraction On A Number LineWriting about Fractions: Draw a number line. Place 3/6 and 7/12 on the number line. Compare the two fractions- why did put them where you did?
22. 22. Key Ideas About the Number LineWhat were some intentional talk moves others used to explain their number line? (Part 5 video, 5:26)Set purpose of video: Listen to the detail that is given in explaining how to construct a number line. From: Dev-TE@M session 4
23. 23. Conventions Of A Number LineDev-TE@M • School of Education • University of Michigan • (734) 408-4461 •[email protected] For review only - Please do not circulate or cite withoutpermission
24. 24. Talking Through A Number Line1. Understand the problem.2. Think about which representation you are going to use.3. Describe your thinking process while constructing the number line.4. Sum up the solution that proved your answer.Model Example: 3/10 & 6/8
25. 25. Fraction On A Number LineUsing a number line, compare 5/6 and 3/8 and tell which one is greater . Have a partner listen to you as you construct the fractions and find the answer.
26. 26. FractionsWhat conceptual understanding do students need? 1. Begin with simple contextual tasks. 2. Connect the meaning of fraction computation with whole number computation. 3. Let estimation and informal methods play a big role in the development of strategies. 4. Explore each of the operations using models.Van De Walle Book: Number Sense and Fraction Algorithms Pg. 310
27. 27. Equivalence with Fraction Strips• Fraction Strips ½+¼= ¾ + 1/3 =
28. 28. Methods for Generating andExplaining Equivalent Fractions• Dev-TE@M session 9
29. 29. Add/Subtract Fractions with Unlike Denominators Developing Equivalent Fractions• Slicing Squares Van de Walle book: pg. 304-305 3 x = 3 x 4 = 4 3 x 3 x = = 4 4
30. 30. Add/Subtract Fractions with Unlike Denominators Developing Equivalent Fractions• Missing-Number Equivalencies Van de Walle book: pg. 304-305 5 2 6 = = 3 6 3
31. 31. Fraction Multiplication StrategiesTOOLKIT for Multiplication of Fractions1. Skim over TOOLKIT2. Read assigned page (2 min)3. 30 second report: What are the important part of your page?4. Questions from audience
32. 32. FractionsMultiply a fraction by a whole number• Work as a group• Use Fraction strips to show answers 4 x 1/3 ¼ x 12• What connection can you make to multiplication? What other representations can you use? Can you use a number line?
33. 33. Multiple a Fraction by a Whole Number 4 x 1/3 (4 groups of 1/3) = 4/3 = 1 1/3I want 4 ribbons each at 1/3 of a yard. How much ribbon will I need to purchase? 1/3 2/3 3/3 4/3 ¼ x 12 (1/4 of 12) = 3I have 12 cookies and want each of my friends to have ¼ of them. How many cookies will each friend get?
34. 34. Scaling (resizing)• 5.NF.5 – Read learning targets and discuss – Prove greater/less than given number statements with last slide. – Making equivalent fractions
35. 35. Multiply Fraction by FractionAIMS• Fair Squares and Cross ProductsMMPI• Worksheet 1: Show different representations 2/3 of ¾ ¾ of 2/3
36. 36. Multiply Fractions and Mixed NumbersMMPI• Area Model Rectangular Multiplication PPThttp://www.michiganmathematics.org/
37. 37. Fraction as Division (a/b = a ÷ b)• I can explain that fractions (a/b) can be represented as a division of the numerator by the denominator (a ÷ b) can be represented by the fraction a/b.• I can solve word problems involving the division of whole numbers and interpret the quotient- which could be a whole number, mixed number, or fraction – in the context of the problem.• I can explain or illustrate my solution strategy using visual fraction models or equations that represent the problem.
38. 38. Divide Fraction by Whole Number½÷6=6÷¼=4 ÷ 2 = (how to connect division of whole numbers with fractions)
39. 39. Divide Fraction by Whole Number½ ÷ 6 = If I have ½ cup of sugar and divide it among 6 people, how much sugar does each person have? 1/121 2 3 4 5 6 7 8 9 10 11 126 ÷ ¼ = If I have 6 candy bars and divide each one into fourths, how many pieces will I have? 24
40. 40. MOPLS http://mi.learnport.org Search: MOPLS Math (navigate by using top tabs)Look at Concepts Tab– Introduction– Math Behind the Math– Misconceptions– Tasks & Strategies
41. 41. Fractions OnlineCheck out some sites on my 5th grade math Protopage
42. 42. Learning TargetUse equivalent fractions as a strategy to add and subtract fractions. 5.NF.1, 5.NF.2Apply and extend previous understandings of multiplication and division to multiply and divide fractions. 5.NF.3, 5.NF.4, 5.NF.5, 5.NF.6, 5.NF.7
43. 43. Closer ActivityList something you learn about story problems and fractions today.
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# Solution for The Function Y = a Log X+Bx2 + X Has Extreme Values at X=1 and X=2. Find a and B ? - CBSE (Commerce) Class 12 - Mathematics
#### Question
The function y = a log x+bx2 + x has extreme values at x=1 and x=2. Find a and b ?
#### Solution
$\text { Given }: f\left( x \right) = y = a \log x + b x^2 + x$
$\Rightarrow f'\left( x \right) = \frac{a}{x} + 2bx + 1$
$\text { Since }, f'\left( x \right) \text { has extreme values at x = 1 and x = 2,} f'\left( 1 \right) = 0 .$
$\Rightarrow \frac{a}{1} + 2b\left( 1 \right) + 1 = 0$
$\Rightarrow a = - 1 - 2b . . . \left( 1 \right)$
$f'\left( 2 \right) = 0$
$\Rightarrow \frac{a}{2} + 2b\left( 2 \right) + 1 = 0$
$\Rightarrow a + 8b = - 2$
$\Rightarrow a = - 2 - 8b . . . \left( 2 \right)$
$\text { From eqs } . \left( 1 \right) \text { and } \left( 2 \right), \text { we get }$
$- 2 - 8b = - 1 - 2b$
$\Rightarrow 6b = - 1$
$\Rightarrow b = \frac{- 1}{6}$
$\text { Substituting b } = \frac{- 1}{6} \text { in eq } . \left( 1 \right), \text{we get }$
$a = - 1 + \frac{1}{3} = \frac{- 2}{3}$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution The Function Y = a Log X+Bx2 + X Has Extreme Values at X=1 and X=2. Find a and B ? Concept: Graph of Maxima and Minima.
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# Relationship between H.C.F. and L.C.M.
We will learn the relationship between H.C.F. and L.C.M. of two numbers.
First we need to find the highest common factor (H.C.F.) of 15 and 18 which is 3.
Then we need to find the lowest common multiple (L.C.M.) of 15 and 18 which is 90.
H.C.F. × L.C.M. = 3 × 90 = 270
Also 15 × 18 = 270
Therefore, product of H.C.F. and L.C.M. of 15 and 18 = product of 15 and 18.
So, from the above explanation we conclude that the product of highest common factor (H.C.F.) and lowest common multiple (L.C.M.) of two numbers is equal to the product of two numbers
or, H.C.F. × L.C.M. = First number × Second number
or, L.C.M. = First number × Second number/ H.C.F.
Solved examples on the relationship between H.C.F. and L.C.M.:
1. Find the L.C.M. of 1683 and 1584.
First we find highest common factor of 1683 and 1584
Therefore, highest common factor of 1683 and 1584 = 99
Lowest common multiple of 1683 and 1584 = First number × Second number/ H.C.F.
= 1584 × 1683/99
= 26928
2. Highest common factor and lowest common multiple of two numbers are 18 and 1782 respectively. One number is 162, find the other.
We know, H.C.F. × L.C.M. = First number × Second number then we get,
18 × 1782 = 162 × Second number
18 × 1782/162 = Second number
Therefore, the second number = 198
3. The highest common factor and the lowest common multiple of two numbers are 825 and 25 respectively. If one of the two numbers is 275, find the other number.
We know, H.C.F. × L.C.M. = First number × Second number then we get,
825 × 25 = 275 × Second number
825 × 25/ 275 = Second number
Therefore, the second number = 75
`
To Find Lowest Common Multiple by using Division Method
Relationship between H.C.F. and L.C.M.
Worksheet on H.C.F. and L.C.M.
Word problems on H.C.F. and L.C.M.
Worksheet on word problems on H.C.F. and L.C.M.
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# Chapter 3: Descriptive Statistics: Numerical Methods
1. A sample contains the following data values: 1.50, 1.50, 10.50, 3.40, 10.50, 11.50, and 2.00. What is the mean? Create an object named E3_1; apply the mean() function.
#Comment1. Use the c() function; read data values into object E3_1.
E3_1 <- c(1.50, 1.50, 10.50, 3.40, 10.50, 11.50, 2.00)
#Comment2. Use the mean() function to find the mean.
mean(E3_1)
## [1] 5.842857
2. Find the median of the sample (above) in two ways: (a) use the median() function to find it directly, and (b) use the sort() function to locate the middle value visually.
#Comment1. Use the median() function to find the median.
median(E3_1)
## [1] 3.4
#Comment2. Use the sort() function to arrange data values in
#ascending order.
sort(E3_1)
## [1] 1.5 1.5 2.0 3.4 10.5 10.5 11.5
Answer: The median of a data set is the middle value when the data items are arranged in ascending order. Once the data values have been sorted into ascending order (we have done this above using the sort() function) it is clear that the middle value is 3.4 since there are 3 values to the left of 3.4 and 3 values to the right. Alternatively, the function median() can be used to find the median directly.
3. Create a vector with the following elements: -37.7, -0.3, 0.00, 0.91, e, π, 5.1, 2e and 113,754, where e is the base of the natural logarithm (roughly 2.718282...) and the ratio of a circle's diameter to its radius (about 3.141593...). Name the object E3_ 2. What are the median and the mean? The 78th percentile? What are the variance and the standard deviation? Note that R understands exp(1) as e, pi as π.
#Comment1. Use the c() function to create the object E3_2.
E3_2 <- c(-37.7, -0.3, 0.00, 0.91, exp(1), pi, 5.1, 2*exp(1), 113754)
#Comment2. Use the mean() function to find the mean.
mean(E3_2)
## [1] 12637.03
#Comment3. Use the median() function to find the median.
median(E3_2)
## [1] 2.718282
#Comment4. Use the quantile() function with prob = c(0.78)
#to find the 78th percentile.
quantile(E3_2, prob = c(0.78))
## 78%
## 5.180775
#Comment5. Use the var() function to find the variance.
var(E3_2)
## [1] 1437840293
#Comment6. Use the sd() function to find the standard deviation.
sd(E3_2)
## [1] 37918.86
Answer: The mean is 12,637.03; the median is 2.718282..., or e. Since the data values in E3_ 2 are arranged in ascending order, the median is easily identifed as the middle value, e (or 2.718282...), since there are four values below and four values above. Moreover, simply summing all nine data values, and dividing by nine, provides the mean. The 78th percentile is reported as 5.180775; the variance and standard deviation are 1,437,840,293 and 37,918.86, respectively.
4. Consider the following data values: 10, 20, 30, 40, 50, 60, 70, 80, 90, and 100. What are the 10th and 90th percentiles? Hint: use function seq(from=,to=,by=) to create the data set. Name the data set E3_3.
#Comment1. Use the seq(from =, to =, by =) function to create
#object E3_3
E3_3 <- seq(from = 10, to = 100, by = 10)
#Comment2. Examine the contents of E3_3 to make sure it contains
#the desired elements.
E3_3
## [1] 10 20 30 40 50 60 70 80 90 100
#Comment3. Use the quantile() function to find the 10th and 90th
#percentiles. Remember to use probs=c(0.1, 0.9)
quantile(E3_3, probs = c(0.1, 0.9))
## 10% 90%
## 19 91
Answer: The 10th and 90th percentiles are 19 and 91, respectively. Note that the 10th percentile (19) is a value which exceeds at least 10% of items in the data set; the 90th percentile (91) is a value which exceeds at least 90% of the items. Note also that it is possible to define any percentiles by setting the values in the probs=c() argument of the quantiles() function.
5. What is the median of E3_3? Find the middle value visually and with the median() function.
#Comment. Use function median() to find the median.
median(E3_3)
## [1] 55
Answer: This data set has an even number of values, all arranged in ascending order. Accordingly, the median is found by taking the average of the values in the two middle positions: the average of 50 (the value in the 5th position) and 60 (the value in the 6th position) is 55.
6. The mode is the value that occurs with greatest frequency in a set of data, and it is used as one of the measures of central tendency. Consider a sample with these nine values: 5, 1, 3, 9, 7, 1, 6, 11, and 8. Does the mode provide a measure of central tendency similar to that of the mean? The median?
#Comment1. Use the c() function and read the data into object E3_4.
E3_4 <- c(5, 1, 3, 9, 7, 1, 6, 11, 8)
#Comment2. Use the table() function to create a frequency
#distribution.
table(E3_4)
## E3_4
## 1 3 5 6 7 8 9 11
## 2 1 1 1 1 1 1 1
#Comment3. Use the mean() and median() functions to find the
#mean and median of E3_4.
mean(E3_4)
## [1] 5.666667
median(E3_4)
## [1] 6
Answer: Since the value of the mode in this instance is 1 (it appears twice), it provides less insight into the central tendency of this sample than does the mean (5.667) or the median (6).
7. Consider another sample with these nine values: 5, 1, 3, 9, 7, 4, 6, 11, and 8. How well does the mode capture the central tendency of this sample?
Answer: Since all the data items appear only once, there is no single value for the mode; there are nine modes, one for each data value.
8. Find the 90th percentile, the 1st, 2nd, and 3rd quartiles as well as the minimum and maximum values of the LakeHuron data set (which is part of the R package and was used in the Chapter 1 in-text exercises). What is the mean? What is the median?
#Comment1. Use the quantile() function with prob=c().
quantile(LakeHuron, prob = c(0.00, 0.25, 0.50, 0.75, 0.90, 1.00))
## 0% 25% 50% 75% 90% 100%
## 575.960 578.135 579.120 579.875 580.646 581.860
#Comment2. Use the mean() function to find the mean.
mean(LakeHuron)
## [1] 579.0041
#Comment3. Use the median() function to find the median.
median(LakeHuron)
## [1] 579.12
Answer: The minimum value (the 0 percentile) is 575.960 and the maximum value (the 100th percentile) is 581.860; the 1st, 2nd, and 3rd quartiles are 578.135, 579.120, and 579.875, respectively. The median (also known as the 2nd quartile or the 50th percentile) is 579.120. The mean is 579.0041 while the 90th percentile is 580.646.
9. Find the range, the interquartile range, the variance, the standard deviation, and the coefficient of variation of the LakeHuron data set.
#Comment1. Find the range by subtracting min() from max().
max(LakeHuron) - min(LakeHuron)
## [1] 5.9
#Comment2. Use the IQR() function to find the interquartile range.
IQR(LakeHuron)
## [1] 1.74
#Comment3. Use the var() function to find the variance.
var(LakeHuron)
## [1] 1.737911
#Comment4. Use the sd() function to find the standard deviation.
sd(LakeHuron)
## [1] 1.318299
#Comment5. To find the coefficient of variation, find sd()/mean().
sd(LakeHuron) / mean(LakeHuron)
## [1] 0.002276838
Answer: The range is 5.9 feet and the interquartile range is 1.74 feet. Moreover, the variance and standard deviation are 1.737911 and 1.318299 feet, respectively. Finally, the coeffcient of variation is 0.002276838; that is, the standard deviation is only about 0.228% of the mean.
10. What are the range and interquartile range for the following data set: -37.7, -0.3, 0.00, 0.91, e, π, 5.1, 2e and 113,754? Note that this is the same data set as that used above where we named it E3_2.
#Comment1. To find range, subtract min() from max().
max(E3_2) - min(E3_2)
## [1] 113791.7
#Comment 2. To find interquartile range, use IQR() function.
IQR(E3_2)
## [1] 5.1
Answer: The range is 113,791.7; the interquartile range is 5.1. The great difference between these two measures of dispersion results from the fact that the interquartile range provides the range of the middle 50% of the data while the range includes all data values, including the outliers.
11.Using the vectorization capability of R, find the sample variance and sample standard deviation of the data set E3_3 (used above). (This exercise provides the opportunity to write and execute some basic R code for the purpose of deriving the variance and standard deviation of a simple data set.) Check both answers against those using the var() and sd() functions. Recall that the expression for the sample
variance is
#Comment1. Use mean() to find mean of E3_3; assign to xbar.
xbar <- mean(E3_3)
#Comment2. Find the deviations about the mean; assign to devs.
devs <- (E3_3 - xbar)
#Comment3. Find the squared deviations about the mean; assign
#the result to sqrd.devs.
sqrd.devs <- (devs) ^ 2
#Comment4. Sum the squared deviations about the mean; assign
#result to the object sum.sqrd.devs
.
sum.sqrd.devs <- sum(sqrd.devs)
#Comment5. Divide the sum of squared deviations by (n-1)
#to find the variance; assign result to the object variance.
variance <- sum.sqrd.devs / (length(E3_3) - 1)
#Comment6. Examine the contents of variance.
variance
## [1] 916.6667
#Comment7. The standard deviation is the positive square root
#of the variance; assign result to the object standard.deviation.
standard.deviation <- sqrt(variance)
#Comment8. Examine the contents of standard.deviation.
standard.deviation
## [1] 30.2765
#Comment9. Use the var() function to find the variance.
var(E3_3)
## [1] 916.6667
#Comment10. Use the sd() function to find the standard
#deviation.
sd(E3_3)
## [1] 30.2765
Answer: The variance is 916.6667; the standard deviation is 30.2765. The answers reported by var() and sd() equal those produced by way of vectorization.
12. The temps data set (available on the companion website) includes the high-and-low temperatures (in degrees Celsius) for April 1, 2016 of ten major European cities; import the data set into an object named E3_5. What is the covariance of the high and low temperatures? What does the covariance tell us?
Answer: The covariance of the two variables is 37.28889. Although it is difficult to learn very much from the value of the covariance of the two variables, we do know that the two variables are positively related. This is an unsurprising nding because the cities having the warmest daytime temperatures are also those that have the warmest nighttime temperatures.
#Comment1. Read data set temps into the object named E3_5.
E3_5 <- temps
#Comment2. Use the head(,3) function to find out the variable names.
## City Daytemp Nighttemp
## 1 Athens 21 12
## 2 Barcelona 12 9
## 3 Dublin 6 1
#Comment3. Use the cov() function to find the covariance. The
#variable names are Daytemp and Nighttemp.
cov(E3_5$Daytemp, E3_5$Nighttemp)
## [1] 37.28889
13. To gain practice using R, calculate the covariance of the two variables in the temps data (available on the companion website). Do not use the function cov(). Recall that the sample covariance between two variables x and y is:
#Comment1. Find the deviations between each observation
#on Daytemp and its mean. Name resulting object devx.
devx <- (E3_5$Daytemp - mean(E3_5$Daytemp))
#Comment2. Find the deviations between each observation
#on Nighttemp and its mean. Name resulting object devy.
devy <- (E3_5$Nighttemp - mean(E3_5$Nighttemp))
#Comment3. Find product of devx and devy; name result crossproduct.
crossproduct <- devx * devy
#Comment4. Find the covariance by dividing crossproduct by (n-1),
# or 9. Assign the result to object named covariance.
covariance <- sum(crossproduct) / (length(E3_5$Daytemp) - 1) #Comment5. Examine contents of covariance. Confirm that it is #the same value as that found in previous exercise. covariance ## [1] 37.28889 14. There are several ways we might explore the relationship between two variables. In the next few exercises, we analyze the daily_ idx_ chg data set (available on the companion website) to explore the pros and cons of several of those methods. The data set itself consists of the percent daily change (from the previous trading day) of the closing numbers for two different widely-traded stock indices, the Dow Jones Industrial Average and the S&P500, for all trading days from April 2 to April 30, 2012. Comment on the data. What is the covariance of the price movements for the two indices? What does the covariance tell us about the relationship between the two variables? As a first step, import the data into an object named E3_6. #Comment1. Read the daily_idx_chg data into the object E3_6. E3_6 <- daily_idx_chg #Comment2. Use the summary() function to identify the variable names #and to acquire a feel for what the data look like. summary(E3_6) ## PCT.DOW.CHG PCT.SP.CHG ## Min. :-1.6500 Min. :-1.7100 ## 1st Qu.:-0.6675 1st Qu.:-0.6525 ## Median : 0.0350 Median :-0.0550 ## Mean : 0.0045 Mean :-0.0340 ## 3rd Qu.: 0.6075 3rd Qu.: 0.6875 ## Max. : 1.5000 Max. : 1.5500 #Comment3. Use the cov() function to find the covariance. cov(E3_6) ## PCT.DOW.CHG PCT.SP.CHG ## PCT.DOW.CHG 0.7542682 0.7693505 ## PCT.SP.CHG 0.7693505 0.8562042 Answer: The two variable names are PCT.DOW.CHG and PCT.SP.CHG; the data values seem to be centered around 0 with values ranging from around 1.55 to -1.71 The covariance is 0.7693505 which tells us only that the two variables are positively related. 15. Standardize the daily_idx_chg data and re-calculate the covariance. Is it the same? #Comment1. Use the scale() function to standardize the data. #Assign the result to the object named std_indices . std_indices <- scale(E3_6) #Comment2. Use the cov() function to find the covariance. cov(std_indices) ## PCT.DOW.CHG PCT.SP.CHG ## PCT.DOW.CHG 1.0000000 0.9573543 ## PCT.SP.CHG 0.9573543 1.0000000 Answer: The covariance is 0.9573543. No, the covariance is not the same, even though it has been applied to the same data. In fact, the covariance on raw data does not (in general) equal the covariance on the same data when standardized. 16. Find the correlation of the two variables in the daily _idx_ chg data. #Comment. Use the cor() function to find the correlation. cor(E3_6) ## PCT.DOW.CHG PCT.SP.CHG ## PCT.DOW.CHG 1.0000000 0.9573543 ## PCT.SP.CHG 0.9573543 1.0000000 Answer: The correlation is 0.9573543, exactly the same as the covariance of the standardized variables. In general, the correlation of two unstandardized variables equals the covariance of the same two variables in standardized form. 17. Standardize the daily _idx_ chg data and re-calculate the correlation. Is it the same? #Comment. Use the cor() function to find the correlation. cor(std_indices) ## PCT.DOW.CHG PCT.SP.CHG ## PCT.DOW.CHG 1.0000000 0.9573543 ## PCT.SP.CHG 0.9573543 1.0000000 Answer: The correlation between the standardized values is exactly the same as the correlation between the unstandardized values: 0.9573543. While the covariance is a ected by how the data are scaled—making it more dicult to interpret—the correlation is not a affected. 18. Make a scatter plot of the daily idx_ chg_ data with PCT.DOW.CHG on the horizontal axis, PCT.SP.CHG on the vertical. Add a main title and labels for the horizontal and vertical axes. Does the scatter plot con rm the positive linear association suggested by the correlation coeffcient? #Comment. Use the plot() function to produce the scatter plot. plot(E3_6$PCT.DOW.CHG, E3_6$PCT.SP.CHG, xlab ='Percentage Daily Change in the Dow', ylab ='Percentage Daily Change in the S&P500', pch = 19, col ='purple', main ='A Plot of Daily Percent Changes in the Dow and S&P500') Answer: The scatter plot is consistent with a correlation coefficient of 0.9573543. There is a strongly positive linear association between the two stock market indices. 19. Below we have a curvilinear relationship where the points can be connected with a smooth, parabolic curve. See scatter plot. Which is the most likely correlation coefficient describing this relationship? -0.90, -0.50, -0.10, 0.00, +0.10, +0.50, or +0.90. (In the next four exercises, the code producing the scatter plots is included.) x <- c(0, -1, -2, -3, -4) y <- c(4, 2, 1, 2, 4) data <- data.frame(X = x, Y = y) plot(data$X, data$Y, pch = 19, xlab ='x', ylab ='y') Answer: The correlation coefficient is 0.00. cor(data$X, data$Y) ## [1] 0 Despite the points being scattered in a way characterized by a curvilinear relationship, the correlation coefficient describes the strength of the linear relationship between two variables. Just because a correlation coecient is zero does not mean that there is no relationship between the two variables. As we see in this case, there may be a relationship that is curvilinear rather than linear. 20. Which is the most likely correlation coefficient describing the relationship below? See the scatter plot. -0.90, -0.50, -0.10, 0.00, +0.10, +0.50, or +0.90. x <- c(16, 13, 8, 6, 5) y <- c(15, 20, 25, 25, 30) data <- data.frame(X = x, Y = y) plot(data$X, data$Y, pch = 19, xlab ='x', ylab ='y') Answer: -0.90 is the closest value that the correlation coefficient might take: the relationship between the two variables is not only negative, it is linear as well. In fact, the correlation coefficient is -0.9657823. cor(data$X, data$Y) ## [1] -0.9657823 21. Which is the most likely correlation coefficient describing the relationship below? -0.90, -0.50, -0.10, 0.00, +0.10, +0.50, or +0.90? x <- c(24, 22, 22, 21, 19) y <- c(27, 24, 23, 21, 19) data <- data.frame(X = x, Y = y) plot(data$X, data$Y, pch = 19, xlab ='x', ylab ='y') Answer: +0.90 is the closest value that the correlation coefficient might take: the relationship between the two variables is not only positive, it is linear as well. cor(data$X, data$Y) ## [1] 0.9800379 In fact, the correlation coefficient is +0.9800379. 22. Which is the most likely correlation coefficient describing the relationship below? -0.90, -0.50, -0.10, 0.00, +0.10, +0.50, or +0.90. x <- c(0, -30, -30, -30, -60) y <- c(-20, 10, -20, -50, -20) data <- data.frame(X = x, Y = y) plot(data$X, data$Y, pch = 19, xlab ='x', ylab ='y') Answer: Although there is a pattern of points in the scatter diagram, there is no discernable linear relationship. In fact, the correlation coefficient is 0.00. cor(data$X, data\$Y)
## [1] 0
23. The Empirical Rule states that approximately 68% of values of a normally-distributed variable fall in the interval from 1 standard deviation below the mean to 1 standard deviation above the mean. (A slightly more precise percentage is 68.269%.) Verify this claim by (a) generating n = 1, 000, 000 normally-distributed values with a mean of 100 and standard deviation of 15, and then (b) "counting" the number of data values that fall in this interval. If this claim is true, then approximately (0.68269)(1, 000, 000) = 682, 690 values should fall in the interval from 85 to 115; roughly (0.15866)(1, 000, 000) = 158, 655 below 85; and approximately (0.15866)(1, 000, 000) = 158, 655 values above 115. Use the rnorm(1000000,100,15) function.
#Comment1. Use the rnorm() function to generate n=1,000,000
#normally-distributed data values with a mean of 100 and standard
#deviation of 15; name the resulting object normal_data.
normal_data <- rnorm(1000000, 100, 15)
#Comment2. Count the number of data values in the object named
#normal_data that are at least 1 standard deviation below the
#mean (that is, at least 15 below 100, or 85). Name this value a
.
a <- length(which(normal_data <= 85))
#Comment3. Examine the contents of a to confirm that it is near
#15.866 percent of 1,000,000, or roughly 158,666.
a
## [1] 158607
#Comment4. Count the number of data values in the object named
#normal_data that are at least 1 standard deviation above the
#mean (that is, at least 15 above 100, or 115). Name this value b
.
b <- length(which(normal_data >= 115))
#Comment5. Examine the contents of b to confirm that it is near
#15.866 percent of 1,000,000, or roughly 158,666.
b
## [1] 158334
#Comment6. Calculate the proportion of 1,000,000 data items that
#fall in the interval from 1 standard deviation below the mean to
#1 standard deviation above the mean. Name that proportion c.
c <- (1000000 - (a + b)) / 1000000
#Comment7. Examine the contents of c. To ensure that we understand
#how c is calculated, plug the values for a (Comment3) and b
#(Comment5) into the expression for c (Comment6).
c
## [1] 0.683059
Using the data generation capability of R, we can confirm that the proportion of data values falling in the interval from one standard deviation below the mean to one standard deviation above the mean is approximately 68%.
24. The Empirical Rule also tells us that approximately 95% of values of a normally- distributed variable fall in the interval from 2 standard deviations below the mean to 2 standard deviations above the mean. (A more precise percentage is 95.45%.) Verify this claim. If this claim is true, then approximately (0.9545)(1, 000, 000) = 954, 500 values should fall in the interval from 70 to 130; roughly (0,02275)(1, 000, 000) = 22, 750 below 70; and approximately (0.02775)(1, 000, 000) = 22, 750 above 130.
#Comment1. Count the number of data values in normal_data that
#are at least 2 standard deviations below the mean (that is, at
#least 30 below 100, or 70); name this object a.
a <- length(which(normal_data <= 70))
#Comment2. Examine the contents of a. Is it near 22,750?
a
## [1] 22762
#Comment3. Count the number of data values in normal_data that
#are at least 2 standard deviations above the mean (that is, at
#least 30 above 100, or 130); name this object b.
b <- length(which(normal_data >= 130))
#Comment4. Examine the contents of b. Is it near 22,750?
b
## [1] 22717
#Comment5. Calculate the proportion of 1,000,000 data items that
#fall in the interval from 2 standard deviations below the mean to
#2 standard deviations above the mean. Name that proportion c.
c <- (1000000 - (a + b)) / 1000000
#Comment6. Examine the contents of c. Is it near 0.9545?
c
## [1] 0.954521
The proportion of data values falling in the interval from two standard deviations below the mean to two standard deviations above the mean is approximately 95:45%.
25. Use the R data generation function to draw n = 1, 000, 000 values from a uniform distribution that runs from a = 75 to b = 125; import the 1,000,000 data values into an object named uniform_data.
To help you envision uniformly-distributed data running between 75 and 125, see the histogram below. Also, see the Chapter 2 Appendix for an example of how the runif() function is used to simulate data values.
#Comment1. Use the runif() function to generate n=1,000,000
#uniformly-distributed data values over the interval from 75 to
#125; name the resulting object uniform_data.
uniform_data <- runif(1000000, 75, 125)
#Comment2. To visualize how the data values are distributed,
#use the hist() function to create a picture of the distribution.
hist(uniform_data,
breaks = 50,
xlim = c(70, 130),
ylim = c(0, 25000),
col ='blue')
What is the proportion of values that falls in the interval from 90 to 110?
Answer: From this simulation exercise, we see that the proportion of uniformly- distributed data values (running from 75 to 125) that falls in the interval from 90 to 110 is (roughly) 0.40.
#Comment1. Count the number of data values in uniform_data that
#are less than or equal to 90 (that is, all the data values that
#fall in interval from 75 to 90); name this object a.
a <- length(which(uniform_data <= 90))
#Comment2. Examine the contents of a. Is it near 300,000?
a
## [1] 300090
#Comment3. Count the number of data values in uniform_data that
#are greater than or equal to 110 (that is, all the data values
#that fall in the interval from 110 to 125); name this object b.
b <- length(which(uniform_data >= 110))
#Comment4. Examine the contents of b. Is it near 300,000?
b
## [1] 299374
#Comment5. Calculate the proportion of 1,000,000 data values that
#fall in the interval from 90 to 110. Name that proportion c.
c <- (1000000 - (a + b)) / 1000000
#Comment6. Examine the contents of c. Is it near 0.40?
c
## [1] 0.400536
From the histogram, we see that a uniformly-distributed variable assumes the "shape" of a rectangle, and therefore the proportion of data values falling in any interval is directly proportional to the length of that interval. In this case, since the question concerns the proportion of data values in an interval of width 20 (= 110 - 90) for a distribution of width 50 (= 125 - 75), the proportion of data values falling in the
interval from 90 to 110 is 20/50 or 0.40.
|
# TRIGONOMETRIC RATIOS CSC SEC AND COT
In this section, you will learn how to find the values of the trigonometric ratios csc, sec and cot.
The formulas given below can be used to find the trigonometric ratios csc, sec and cot.
csc θ = Hypotenuse / Opposite side
sec θ = Hypotenuse / Adjacent side
cot θ = Adjacent side / Opposite side
Example 1 :
In the right triangle shown below, find the values of csc B, sec B, cot B.
Solution :
90° is at ∠A. So the side which is opposite to 90° is known as hypotenuse. The side which is opposite to ∠B is known as opposite side. The remaining side is known as adjacent side.
So, we have
BC = Hypotenuse = 17
AC = Opposite side = 15
AB = Adjacent side = 8
Finding the value of csc B :
csc B = Hypotenuse / Opposite side
csc B = BC/AC
csc B = 17/15
Finding the value of sec B :
sec B = Hypotenuse / Adjacent side
sec B = BC/AB
sec B = 17/8
Finding the value of cot B :
cot B = Adjacent side / Opposite side
cot B = AB/AC
cot B = 8/15
Example 2 :
In the right triangle shown below, find the values of csc A, sec A and cot A.
Solution :
90° is at ∠B. So the side which is opposite to 90° is known as hypotenuse. The side which is opposite to ∠A is known as opposite side. The remaining side is known as adjacent side.
So, we have
AC = Hypotenuse = 65
BC = Opposite side = 33
AB = Adjacent side = 56
Finding the value of csc A :
csc A = Hypotenuse / Opposite side
csc A = AC/BC
csc A = 65/33
Finding the value of sec A :
sec A = Hypotenuse / Adjacent side
sec A = AC/AB
sec A = 65/56
Finding the value of cot A :
cot A = Adjacent side / Opposite side
cot A = AB/BC
cot A = 56/33
Example 3 :
In the right triangle shown below, find the values of csc C, sec C and cot C
Solution :
90° is at ∠A. So the side which is opposite to 90° is known as hypotenuse. The side which is opposite to ∠C is known as opposite side. The remaining side is known as adjacent side.
So, we have
BC = Hypotenuse = 5
AB = Opposite side = 3
AC = Adjacent side = 4
Finding the value of csc C :
csc C = Hypotenuse / Opposite side
csc C = BC/AB
csc C = 5/3
Finding the value of sec C :
sec C = Hypotenuse / Adjacent side
sec C = BC/AC
sec C = 5/4
Finding the value of cot C :
cot C = Adjacent side / Opposite side
cot C = AC/AB
cot C = 4/3
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# One-Step Equations and Inverse Operations
## Add and subtract like terms as a step to solve equations.
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Solving One Step Equations
### Guidance
When solving an equation for a variable, you must get the variable by itself. All the equations in this lesson are linear equations. Here we will only deal with using one operation; addition, subtraction, multiplication or division.
### Solve \begin{align*}7+y=16\end{align*} for \begin{align*}y\end{align*}.
Solution: This problem is simple and you could probably solve it in your head. However, to start good practices, you should always use algebra to solve any equation. Even if the problem seems easy, equations will get more difficult to solve.
To solve an equation for a variable, you must do the opposite, or undo, whatever is on the same side as the variable. 7 is being added to \begin{align*}y\end{align*}, so we must subtract 7 from both sides. Notice that this is very similar to the previous concept (Solving Algebraic Equations for a Variable).
\begin{align*}& \ \bcancel{7}+y = 16\\ & \underline{-\bcancel{7} \quad \quad -7 \; \;}\\ & \ \quad \ \ y = 9\end{align*}
You can check that \begin{align*}y= 9\end{align*} is correct by plugging 9 back into the original equation. 7 + 9 does equal 16, so we know that we found the correct answer.
### Solve\begin{align*}-7h=84\end{align*}.
Solution: Recall that \begin{align*}-7h = -7 \times h\end{align*}, so the opposite, or inverse, operation of multiplication is division. Therefore, we must divide both sides by -7 to solve for \begin{align*}h\end{align*}.
\begin{align*}\frac{-\bcancel{7}h}{-\bcancel{7}} &= \frac{84}{-7}\\ h &= -12\end{align*}
Again, check your work. \begin{align*}-7 \cdot -12\end{align*} is equal to 84, so we know our answer is correct.
### Solve \begin{align*}\frac{3}{8} x = \frac{3}{2}\end{align*}.
Solution: The variable is being multiplied by a fraction. Instead of dividing by a fraction, we multiply by the reciprocal of \begin{align*}\frac{3}{8}\end{align*}, which is \begin{align*}\frac{8}{3}\end{align*}.
\begin{align*}\xcancel{\frac{8}{3} \cdot \frac{3}{8}} x &= \bcancel{\frac{3}{2} \cdot \frac{8}{3}}\\ x &= \frac{8}{2}=4\end{align*}
Check the answer: \begin{align*}\frac{3}{_2\cancel{8}} \cdot \cancel{4}=\frac{3}{2}\end{align*}. This is correct, so we know that \begin{align*}x = 4\end{align*} is the answer.
[Figure1]
### Guided Practice
1. \begin{align*}5+j=17\end{align*}
2. \begin{align*}\frac{h}{6}=-11\end{align*}
3. \begin{align*}\frac{5}{4}x=35\end{align*}
1. Subtract 5 from both sides to solve for \begin{align*}j\end{align*}.
\begin{align*}& \ \ \bcancel{5}+j=17\\ & \underline{-\bcancel{5} \qquad -5 \; \;}\\ & \qquad \ j=12\end{align*}
Check the answer: \begin{align*}5+12=17 \end{align*}
2. \begin{align*}h\end{align*} is being divided by 6. To undo division, we must multiply both sides by 6.
\begin{align*}\cancel{6} \cdot \frac{h}{\cancel{6}} &= -11.6\\ h &= -66\end{align*}
Check the answer: \begin{align*}\frac{-66}{6}=-11 \end{align*}
3. Multiply both sides by the reciprocal of \begin{align*}\frac{5}{4}\end{align*}
\begin{align*}\frac{\cancel{4}}{\cancel{5}} \cdot \frac{\cancel{5}}{\cancel{4}}x &= _7\cancel{35} \cdot \frac{4}{\cancel{5}}\\ x &=28\end{align*}
Check the answer: \begin{align*}\frac{5}{4} \cdot 28 = 5 \cdot 7 = 35\end{align*}
### Vocabulary
Linear Equation
An equation in one variable without exponents. Linear equations are in the form \begin{align*}ax \pm b=c\end{align*}, where \begin{align*}a, b\end{align*}, and \begin{align*}c\end{align*} are real numbers.
Inverse
The opposite operation of a given operation in an equation. For example, subtraction is the inverse of addition and multiplication is the inverse of division.
Reciprocal
The reciprocal of \begin{align*}\frac{a}{b}\end{align*} is \begin{align*}\frac{b}{a}\end{align*}.
### Practice
1. \begin{align*}-3+x=-1\end{align*}
2. \begin{align*}r+6=2\end{align*}
3. \begin{align*}5s=30\end{align*}
4. \begin{align*}-8k=-64\end{align*}
5. \begin{align*}\frac{m}{-4}=14\end{align*}
6. \begin{align*}90=10n\end{align*}
### Notes/Highlights Having trouble? Report an issue.
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# MP Board Class 12th Maths Solutions Chapter 1 Relations and Functions Ex 1.2
In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 Relations and Functions Ex 1.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.
## MP Board Class 12th Maths Solutions Chapter 1 Relations and Functions Ex 1.2
Question 1.
Show that the function f : R* → R*, defined by f(x) = $$\frac { 1 }{ x }$$ is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?
Solution:
a. One-one
Let x1, x1 ∈ R, such that f(x1) = f(x2)
For each y ∈ R* there exists x = $$\frac { 1 }{ y }$$ ∈ R,
such that f(x) = y.
∴ f is onto
b. One-one
Let x1, x2 ∈ N such that f(x1) = f(x2)
⇒ $$\frac{1}{x_{1}}=\frac{1}{x_{2}}$$ ⇒ x1 = x2
∴ f is one-one
Onto
The domain of f is N = {1, 2, 3, ……. }
The range of f = {1, $$\frac { 1 }{ 2 }$$, $$\frac { 1 }{ 3 }$$, $$\frac { 1 }{ 4 }$$, ……. } ≠ R
∴ f is not onto
Another method
2 ∈ R*.
The pre-image of 2 is $$\frac { 1 }{ 2 }$$ ∉ N .
Hence f is not onto.
Question 2.
Check the injectivity and surjectivity of the following functions
i. f : N → N given by f (x) = x²
ii. f : Z → Z given by f (x) = x²
iii. f : R → R given by f (x) = x²
iv. f : N → N given by f (x) = x³
v. f : Z → Z given by f (x) = x³
Solution:
i. Injectivity
Let x1, x1 ∈ N such that f(x1) = f(x2)
⇒ x²1 = x²2
⇒ x²1 – x²2 = 0
⇒ (x1 – x2)(x1 + x2) = 0
⇒ x1 – x2 = 0 since x1 + x2 ≠ 0 as x1, x1 ∈ N
⇒ x1 = x2
∴ f is injective
Surjectivity
Let y = 3 ∈ N, the co-domain of f
f(x) = 3 ⇒ x² = 3
⇒ x = ±$$\sqrt{3}$$ ∉ N , the domain of f
∴f is not surjective
ii. Injectivity
Let x1 = 2, x2 = – 2
f(x1) = 2² = 4, f(x2) = (- 2)² = 4
i.e., f(x1) = f(x2) for x1 ≠ x2
∴ f is not injective
Surjectivity
Let y = 2 ∈ Z, the co-domain of f
∴ f(x) = 2 ⇒ x² = 2 ⇒ x = ± $$\sqrt{2}$$ ∉ Z, the domain of f
∴ f is not surjective.
iii. Injectivity
Let x1 = 2, x2 = – 2
f(x1) = 2² = 4, f(x2) = (- 2)² = 4
i.e., f(x1) = f(x2) for x1 ≠ x2
∴ f is not injective (one-one)
Surjectivity
Let y = – 1 ∈ R, the co-domain of f
f(x) = – 1 ⇒ x² = -1
⇒ x = ± $$\sqrt{1}$$ ∉ R, the domain of f
∴ f is not surjective (onto)
iv. Injectivity
Let x1, x2 ∈ N such that f(x1) = f(x1) = f(x2)
⇒ x³1 = x³2 ⇒ x³1 – x³2 = 0
⇒ (x1 – x2) (x²1 + x1x2 + x²22) = 0
⇒ x1 – x2 = 0 since x²1 + x1x2 + x²2 ≠ 0
⇒ x1 = x2
∴f is injective (one-one)
Surjectivity
Let y = 4 ∈ N, the co-domain of f
⇒ f(x) = 4 ⇒ x3 = 4
⇒ x = 41/3 ∉ N, the domain of f
∴f is not surjective (onto)
v. Injectivity
Let x1, x2 ∈ Z such that f(x1) = f(x2)
⇒ x³1 = x³2 ⇒ x1 = x2
∴ f is injective (one-one)
Surjectivity
Let y = 4 ∈ Z, the co-domain of f such that
f(x) = 4
⇒ x3 = 4 ⇒ x = 41/3 ∉ Z, the domain of f
∴ f is not surjective (onto)
Question 3.
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution:
Refer Example 26
One-one
Let x1 = 2.1, x2 = 2.5
f(x1) = f(2.1) = [2.1] = 2
f(x2) = f(2.5) = [2.5] = 2
i.e. f(x1) = f(x2) for x1 ≠ x2
∴ f is not one-one.
Onto
Let y = 2.5 ∈ R, the co-domain of f
f(x) = 2.5 ⇒ [x] = 2.5, which is not possible.
∴ f is not onto
Question 4.
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x is negative.
Solution:
One-one
Let x1 = 1 and x2 = – 1 ∈ R
f(x1) = f(1) = |1| = 1
f(x2) = f(- 1) = |- 1|= 1
i.e., f(x1) = f(x2) for x1 ≠ x2
∴ f is not one-one Onto
Let y = – 1 ∈ R, the co-domain of f
f(x) = – 1 ⇒ |x|= – 1 which is not possible
∴ f is not onto.
Question 5.
Show that the Signum Function f : R → R, given by
$$f(x)=\left\{\begin{array}{c} 1, \text { if } x>0 \\ 0, \text { if } x=0 \\ -1, \text { if } x<0 \end{array}\right.$$
is neither one-one nor onto.
Solution:
One-one
Let x1 = 1, x2 = 2
f(x1) = f(1) = 1
f(x2) = f(2) = 1
i.e. f(x1) = f(x2) for x1 ≠ x2
∴ f is not one-one
Onto
The co-domain of f is R
The range of f is {-1, 0, 1} ≠ R
∴ f is not onto.
Question 6.
Let A = {1, 2, 3}, B = (4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Solution:
Different elements in A have different images in B. Hence f is one – one.
Question 7.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
i. f : R → R defined by f(x) = 3 – 4x
ii. f : R → R defined by f(x) = 1 + x2
Solution:
i. One-one
Let x1, x2 6 R such that f(x1) = f(x2)
⇒ 3 – 4x1 = 3 – 4x2
⇒ – 4x1 = – 4x2
⇒ x1 = x2
∴f is one-one
Onto
Let y ∈ R, such that y = f(x)
⇒ y = 3 – 4x
⇒ 4x = 3 – y
⇒ x = $$\frac { 3-y }{ 4 }$$ ∈ R
f(x) = f$$\frac { 3-y }{ 4 }$$ = 3 – 4($$\frac { 3-y }{ 4 }$$)
= 3 – 3 + y = y
For each y ∈ R there exists x = $$\frac { 3-y }{ 4 }$$ ∈ R
such that f(x) = y
∴f is onto
Since f is one-one and onto, f is bijective.
Another Method
f(x) = 3 – 4x is a linear function from R to R.
∴ f is one-one and onto.
Hence f is bijective.
ii. One-one
Let x1 = 1, x2 = – 1 ∈ R
f(x1) = f(1) = 1 + 1² = 2
f(x2)= f(-1) = 1 + (- 1)² = 2
i.e., f(x1) = f(x2) for x1 ≠ x2
∴f is not one-one
Onto
Let y = – 1 ∈ R, the co-domain of f
f(x) = y ⇒ 1 + x2 = – 1
⇒ x2 = – 2
⇒ x = ± $$\sqrt{-2}$$ ∉ R
∴f is not onto.
Thus f is neither one-one nor onto.
Question 8.
Let A and B be sets. Show that f : A x B → B x A such that f(a, b) = (b, a) is bijective function.
Solution:
Let (a1, b1) and (a2, b2) ∈ A x B
such that f(a1, b1) = f(a2, b2)
⇒ (b1, a1) = (b2, a2)
⇒ b1 = b2 and a1 = a2
⇒ (a1, b1) = (a2, b2)
∴f is one-one.
Let (b, a) ∈ B x A ⇒ b ∈ B and a ∈ A
⇒ (a, b) ∈ A x B
f(a, b) = (b, a)
i.e., corresponding to each (b, a) ∈ B x A, there exists (a, b) ∈ A x B such that f(a, b) = (b, a)
∴ f is onto
Hence f is a bijection.
Question 9.
Let f : N → N be defined by
f(n) = $$\left\{\begin{array}{l} \frac{n+1}{2}, \text { if } n \text { is odd } \\ \frac{n}{2}, \text { if } n \text { is even } \end{array} \text { for all } n \in \mathbf{N}\right.$$
Solution:
Let x1 = 1 and x2 = 2 ∈ N
f(x1) = f(1) = $$\frac { 1+1 }{ 2 }$$ = 1 and f(x2) = f(2) = $$\frac { 2 }{ 2 }$$ = 1
∴ f(x1) = f(x2) for x1 ≠ x2
Hence f is not one-one and hence not bijective.
Question 10.
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = $$\left(\frac{x-2}{x-3}\right)$$. Is f one-one and onto? Justify your answer.
Solution:
One-one
One-one
Let y ∈ B such that f(x) = y
For each y ∈ B, there exists x = $$\left(\frac{2-2y}{1-y}\right)$$ ∈ A
such that f(x) = y
∴ f is onto.
Question 11.
Let f : R → R be defined as f(x) = x4. Choose the correct answer.
a. f is one-one onto
b. f is many-one onto
c. f is one-one but not onto
d. f is neither one-one nor onto.
Solution:
d. f is neither one-one nor onto.
One-one
Let x1 = 1, x2 = – 1 ∈ R
f(x1) = f(1) = (1)4 = 1
f(x2) = f(-1) = (-1)4= 1
f(x1) = f(x2) for x1 ≠ x2
∴ f is not one-one
Onto
The co-domain of f is R.
The range of f is [0, ∞), the non-negative real numbers.
Since range of f ≠ co-domain of f is not onto.
Hence f is neither one-one nor onto.
Question 12.
Let f : R → R be defined as f (x) = 3x. Choose the correct answer.
a. f is one-one onto
b. f is many-one onto
c. f is one-one but not onto
d. f is neither one-one nor onto.
Solution:
a. f is one-one onto
f(x1) = f(x2) ⇒ 3x1 = 3x2 ⇒ x1 = x2
∴ f is one-one
For y ∈ co-domain off there exists $$\frac { y }{ 3 }$$ in the domain of f such that f($$\frac { y }{ 3 }$$) – 3($$\frac { y }{ 3 }$$) = y
Hence f is onto.
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Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative.
1. $\alpha=\frac{\pi}{6}$
2. $\beta=-\frac{4 \pi}{3}$
3. $\gamma=\frac{9 \pi}{4}$
4. $\varphi=-\frac{5 \pi}{2}$
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1. The angle $\alpha=\frac{\pi}{6}$ is positive, so we draw an angle with its initial side on the positive $x$-axis and rotate counter-clockwise $\frac{(\pi / 6)}{2 \pi}=\frac{1}{12}$ of a revolution. Thus $\alpha$ is a Quadrant I angle.
Coterminal angles $\theta$ are of the form $\theta=\alpha+2 \pi k$, for some integer $k$. To make the arithmetic a bit easier, we note that $2 \pi=\frac{12 \pi}{6}$
- Thus, when $k=1$, we get
\begin{aligned} \theta &=\frac{\pi}{6}+\frac{12 \pi}{6} \\ &=\frac{13 \pi}{6} \end{aligned}
- Substituting $k=-1$ gives
\begin{aligned} \theta &=\frac{\pi}{6}-\frac{12 \pi}{6} \\ &=-\frac{11 \pi}{6} \end{aligned}
- When we let $k=2$, we get
\begin{aligned} \theta &=\frac{\pi}{6}+\frac{24 \pi}{6} \\ &=\frac{25 \pi}{6} \end{aligned}
2. Since $\beta=-\frac{4 \pi}{3}$ is negative, we start at the positive $\mathrm{x}$-axis and rotate clockwise $\frac{(4 \pi / 3)}{2 \pi}=\frac{2}{3}$ of a revolution. We find $\beta$ to be a Quadrant II angle.
To find coterminal angles, we proceed as before using $2 \pi=\frac{6 \pi}{3}$, and compute $\theta=-\frac{4 \pi}{3}+\frac{6 \pi}{3} k$ for integer values of $k .$ We obtain $\frac{2 \pi}{3},-\frac{10 \pi}{3}$ and $\frac{8 \pi}{3}$ as coterminal angles.
3. Since $\gamma=\frac{9 \pi}{4}$ is positive, we rotate counter-clockwise from the positive $x$-axis. One full revolution accounts for $2 \pi=\frac{8 \pi}{4}$ of the radian measure with $\frac{\pi}{4}$ or $\frac{1}{8}$ of a revolution remaining. We have $\gamma$ as a Quadrant I angle.
All angles coterminal with $\gamma$ are of the form $\theta=\frac{9 \pi}{4}+\frac{8 \pi}{4} k$, where $k$ is an integer. Working through the arithmetic, we find coterminal angles of $\frac{\pi}{4},-\frac{7 \pi}{4}$ and $\frac{17 \pi}{4}$.4. To graph $\varphi=-\frac{5 \pi}{2}$, we begin our rotation clockwise from the positive $\mathrm{x}$-axis. As $2 \pi=\frac{4 \pi}{2}$, after one full revolution clockwise we have $\frac{\pi}{2}$ or $\frac{1}{4}$ of a revolution remaining. Since the terminal side of $\varphi$ lies on the negative $y$-axis, $\varphi$ is a quadrantal angle.
To find coterminal angles, we compute $\theta=-\frac{5 \pi}{2}+\frac{4 \pi}{2} k$ for a few integers $k$ and obtain $-\frac{\pi}{2}$, $\frac{3 \pi}{2}$ and $\frac{7 \pi}{2}$
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# 1.4: Geometry
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Geometric shapes, as well as area and volumes, can often be important in problem solving.
## Example 21
You are curious how tall a tree is, but don’t have any way to climb it. Describe a method for determining the height.
There are several approaches we could take. We’ll use one based on triangles, which requires that it’s a sunny day. Suppose the tree is casting a shadow, say 15 ft long. I can then have a friend help me measure my own shadow. Suppose I am 6 ft tall, and cast a 1.5 ft shadow. Since the triangle formed by the tree and its shadow has the same angles as the triangle formed by me and my shadow, these triangles are called similar triangles and their sides will scale proportionally. In other words, the ratio of height to width will be the same in both triangles. Using this, we can find the height of the tree, which we’ll denote by $$h$$:
Solution
$\frac{6 \mathrm{ft} \text { tall }}{1.5 \mathrm{ft} \text { shadow }}=\frac{h \mathrm{ft} \text { tall }}{15 \mathrm{ft} \text { shadow }}\nonumber$
Multiplying both sides by 15, we get $$h = 60$$. The tree is about 60 ft tall.
It may be helpful to recall some formulas for areas and volumes of a few basic shapes.
## Areas
Rectangle
Area: $$L \cdot W$$
Perimeter: $$2 L+2 W$$
Circle, radius $$r$$
Area: $$\pi r^{2}$$
Circumference = $$2 \pi r$$
## Volumes
Rectangular Box
Volume: $$L \cdot W \cdot H$$
Cylinder
Volume: $$\pi r^{2} H$$
## Example 22
If a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza?
Solution
To answer this question, we need to consider how the weight of the dough will scale. The weight will be based on the volume of the dough. However, since both pizzas will be about the same thickness, the weight will scale with the area of the top of the pizza. We can find the area of each pizza using the formula for area of a circle, $$A=\pi r^{2}$$:
A 12” pizza has radius 6 inches, so the area will be $$\pi 6^{2}$$ = about 113 square inches.
A 16” pizza has radius 8 inches, so the area will be $$\pi 8^{2}$$ = about 201 square inches.
Notice that if both pizzas were 1 inch thick, the volumes would be 113 in3 and 201 in3 respectively, which are at the same ratio as the areas. As mentioned earlier, since the thickness is the same for both pizzas, we can safely ignore it.
We can now set up a proportion to find the weight of the dough for a 16” pizza:
$$\frac{10 \text { ounces }}{113 \mathrm{in}^{2}}=\frac{x \text { ounces }}{201 \mathrm{in}^{2}}$$ Multiply both sides by 201
$$x=201 \cdot \frac{10}{113}=$$ about 17.8 ounces of dough for a 16” pizza.
It is interesting to note that while the diameter is $$\frac{16}{12}=1.33$$ times larger, the dough required, which scales with area, is $$1.33^{2}=1.78$$ times larger.
## Example 23
A company makes regular and jumbo marshmallows. The regular marshmallow has 25 calories. How many calories will the jumbo marshmallow have?
We would expect the calories to scale with volume. Since the marshmallows have cylindrical shapes, we can use that formula to find the volume. From the grid in the image, we can estimate the radius and height of each marshmallow.
Solution
The regular marshmallow appears to have a diameter of about 3.5 units, giving a radius of 1.75 units, and a height of about 3.5 units. The volume is about $$\pi(1.75)^{2}(3.5)=33.7 \text { units}^{3}$$.
The jumbo marshmallow appears to have a diameter of about 5.5 units, giving a radius of 2.75 units, and a height of about 5 units. The volume is about $$\pi(2.75)^{2}(5)=118.8 \text { units}^{3}$$.
We could now set up a proportion, or use rates. The regular marshmallow has 25 calories for 33.7 cubic units of volume. The jumbo marshmallow will have:
$$118.8 \text { units }^{3} \cdot \frac{25 \text { calories }}{33.7 \text { units }^{3}}=88.1 \text { calories }$$
It is interesting to note that while the diameter and height are about 1.5 times larger for the jumbo marshmallow, the volume and calories are about $$1.5^{3}=3.375$$ times larger.
## Try it Now 5
A website says that you’ll need 48 fifty-pound bags of sand to fill a sandbox that measure 8ft by 8ft by 1ft. How many bags would you need for a sandbox 6ft by 4ft by 1ft?
The original sandbox has volume $$64 \mathrm{ft}^{3}$$. The smaller sandbox has volume $$24 \mathrm{ft}^{3}$$.
$$\frac{48 \mathrm{bags}}{64 \mathrm{ft}^{3}}=\frac{x \mathrm{bags}}{24 \mathrm{ft}^{3}}$$ results in $$x= 18$$ bags.
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# Metric Measurements, Fractions and Number Lines
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Students will be able to calculate fractions of metric units using number lines.
(5 minutes)
• Ask, “What two numbers can we multiply together to get 100?” Note how these are called FactorsOf 100.
• Mention to your class that any factor for a given number divides that number evenly.
• Demonstrate by writing a multiplication fact family for 100 as: 4 x 25 = 100, 25 x 4 = 100, 100/4 = 25. Hence, 4 and 25 divide 100 evenly.
• Allow your class to think, pair and share their thoughts on any other factors for 100.
• Have your students share whole class, confirm and post the factors for 100 as: 1, 2, 4, 5, 10, 20, 25, 50, 100.
• Tell your students that finding factors, will be one strategy used to find fractions and equivalent measurements to 100.
(10 minutes)
• Preview and explain how the metric system is based on Partitions(or divided into equal parts) of 100. The factors for 100 that we previously listed all divide a number line of 100 units evenly, with no remainder. We call this a base-ten system, like our money system.
• Show your class an Open number lineLabeled 0 to 100 and ask, “How might we find ¾ of a 100 metre number line?”
• Have your students think, pair and share their thoughts with the whole class. If your class uses maths journals, this would be a good time for them to use them.
• During student responses, note any academic terms for future reference.
• Share with your class that finding fractions of 100 on a number line can be achieved in four steps!
• Explain:
Step oneFor finding ¾ of 100 meters: Draw an open number line, beginning at 0 and ending at 100 meters.
Step two:Look to the denominator of the fraction to calculate how many partitions are needed for the number line (in this case, it’s 4.). From our factor list, we know four is a factor of 100, so draw and Partition an open number lineInto four equal parts (...including the endpoint!)
Step three:Calculate partition increments: 100/4 =25, so there will be four partitions in 25 metre increments. Then, label the number line at 25m, 50m (25 + 25), 75m (25 + 25 + 25) and 100m (25 + 25 + 25 + 25).
Step four:Skip count partitions by the numerator amount. Skip three out of four partitions from zero (25, 50, arriving at 75), to calculate ¾ of 100. So, ¾ of 100 meters is 75 meters.
(10 minutes)
• Repeat the Four-step procedureWith your students to calculate and illustrate 6/10 of 100 centimeters and answer any clarifying questions.
• The result should be 6/10 of 100 cm is 60cm.
(15 minutes)
• Hand out and preview the Finding Fraction Equivalents for Metric Measurements worksheet.
• Answer any clarifying questions and have your students complete the exercises.
Support:
• Print and post the four-step procedure for students to reference. Direct students with a study-buddy partner to the poster as needed, to follow the procedure step-by-step.
• Print sheets of number lines pre-partitioned by factors of 100 for students to use
Enrichment:
• Students can convert metric measurements to greater or lesser metric units, (i.e. meters to centimeters or meters to kilometers) by using a metric conversion chart.
• Students can try finding fractions of numbers that aren’t 100 (but still divide evenly), like ¾ of 40.
• Interactive whiteboards are a great resource to use in tandem with the Online Ruler reference in the media section.
• Computers with Internet access and a projector make for a great set-up to display the online ruler, listed in the suggested media section. Students can interact with the number line repeatedly on a whiteboard with erasable markers.
(5 minutes)
• Show your students three fractions and an open number line (where one has a denominator factor of 100.)
• Ask your students to point out which fractions that would work on a number line that is 0-100.
(10 minutes)
• Review the answers for Finding Fraction Equivalents for Metric Measurements, whole class, allowing students to call on peer assistance when needed.
• Pose and discuss the question, “What are the advantages and disadvantages of using a number line to express equivalent fractions for metric measurements?”
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# Lesson Plan Template
## Learner and Environmental Factors
Content Area: Math
Standards (KCCRS):
K.OA. 3 Decomposing numbers in multiple ways
Lesson Performance Objective:
Students are to be count domino dots and put into 3 equations with 90% accuracy.
Materials:
Pencils
Dominos
Paper
## Be able to place numbers into an equation
Student Grouping(s)
Whole group
Pair
THE PLAN
## Anticipatory Set: (The HOOK)
Do you remember the time we played dominos? How we would flip them over and try to solve a math
problem with them? Today is just like that time in MTSS. We are going to work in pairs to solve some
math equations using our dominos. Sounds like fun right?!
## Introduce the Lesson:
What we will do is take 3 dominos and create equations with them with at least 90% accuracy.
Instruction:
## Teaching the Lesson (I DO IT)
o Stand up in front of the classroom and ask the students, “Do you know what this is?”
(Hold up a domino)
o Students should say that it is a domino.
o“Yes you are correct! These are the dominos that we have worked with in MTSS. We are
going to create equations using the dominos dots.”
o Show the students that there are different numbers of dots on both sides of the line.
o “I’m going to take my dominos and flip them over so I can’t see the dots.” (Flip over all
the dominos face down)
o “Next, I will flip one domino over and look at the dots on the domino.” (Count one side
and then the other side to see how many dots there are)
o “Now I am going to take my paper and right an equation using the domino dots.” (Count
one side and write the number down. Next put and addition sign and then write the
next dots and what they add up to be. After that write and equal sign and solve the
equation)
o “You will work in pairs that I assign to work with the dominos. You and your partner
need at least three equations on your paper and you must solve them. If it is not
finished in time you will have to do them after lunch.”
Check for understanding with sample problems, situations, questions. (WE DO IT)
o Does anyone not understand how the game will be played?
o Do I need to perform the rules for the game again?
Guided Practice and/or Independent Practice (YOU DO IT)
o Students should be working in pairs to solve the dominos given to them
o They should also be writing the equations down and adding the dots together to get an
Closure: *
## “What I need everyone to do is freeze. “
“I’m going to come by with the bag and please put your dominos inside the bag.”
“After I have collected all the dominos, I need you to make sure that your name is on your paper and
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## Common Core Math Standards - 8th Grade
MathScore aligns to the Common Core Math Standards for 8th Grade. The standards appear below along with the MathScore topics that match. If you click on a topic name, you will see sample problems at varying degrees of difficulty that MathScore generated. When students use our program, the difficulty of the problems will automatically adapt based on individual performance, resulting in not only true differentiated instruction, but a challenging game-like experience.
View the Common Core Math Standards at other levels.
## The Number System
Know that there are numbers that are not rational, and approximate them by rational numbers.
1. Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number. (Repeating Decimals )
2. Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., π2). For example, by truncating the decimal expansion of √2, show that √2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.
## Expressions and Equations
Work with radicals and integer exponents.
1. Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 32 × 3–5 = 3–3 = 1/33 = 1/27. (Negative Exponents Of Fractional Bases , Multiplying and Dividing Exponent Expressions , Exponent Rules For Fractions )
2. Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational. (Perfect Squares )
3. Use numbers expressed in the form of a single digit times a whole-number power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. For example, estimate the population of the United States as 3 times 108 and the population of the world as 7 times 109, and determine that the world population is more than 20 times larger.
4. Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology. (Scientific Notation , Scientific Notation 2 )
Understand the connections between proportional relationships, lines, and linear equations.
5. Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed.
6. Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. (Graphs to Linear Equations )
Analyze and solve linear equations and pairs of simultaneous linear equations.
7. Solve linear equations in one variable. (Single Variable Equations , Single Variable Equations 2 , Single Variable Equations 3 , Linear Equations )
a. Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
b. Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms. (Single Variable Equations , Single Variable Equations 2 , Single Variable Equations 3 , Linear Equations )
8. Analyze and solve pairs of simultaneous linear equations.
a. Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.
b. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6. (System of Equations Substitution , System of Equations Addition )
c. Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair. (Age Problems )
## Functions
Define, evaluate, and compare functions.
1. Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output.1
2. Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a linear function represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change.
3. Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line.
Use functions to model relationships between quantities.
4. Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
5. Describe qualitatively the functional relationship between two quantities by analyzing a graph (e.g., where the function is increasing or decreasing, linear or nonlinear). Sketch a graph that exhibits the qualitative features of a function that has been described verbally.
## Geometry
Understand congruence and similarity using physical models, transparencies, or geometry software.
1. Verify experimentally the properties of rotations, reflections, and translations:
a. Lines are taken to lines, and line segments to line segments of the same length.
b. Angles are taken to angles of the same measure.
c. Parallel lines are taken to parallel lines.
2. Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them.
3. Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates.
4. Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them.
5. Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the sum of the three angles appears to form a line, and give an argument in terms of transversals why this is so.
Understand and apply the Pythagorean Theorem.
6. Explain a proof of the Pythagorean Theorem and its converse.
7. Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions. (Pythagorean Theorem )
8. Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.
Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres.
9. Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. (Cylinders )
## Statistics and Probability
Investigate patterns of association in bivariate data.
1. Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities. Describe patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear association.
2. Know that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line, and informally assess the model fit by judging the closeness of the data points to the line.
3. Use the equation of a linear model to solve problems in the context of bivariate measurement data, interpreting the slope and intercept. For example, in a linear model for a biology experiment, interpret a slope of 1.5 cm/hr as meaning that an additional hour of sunlight each day is associated with an additional 1.5 cm in mature plant height.
4. Understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. Use relative frequencies calculated for rows or columns to describe possible association between the two variables. For example, collect data from students in your class on whether or not they have a curfew on school nights and whether or not they have assigned chores at home. Is there evidence that those who have a curfew also tend to have chores?
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# Lesson 17
Let’s compare samples from the same population.
### Problem 1
One thousand baseball fans were asked how far they would be willing to travel to watch a professional baseball game. From this population, 100 different samples of size 40 were selected. Here is a dot plot showing the mean of each sample.
Based on the distribution of sample means, what do you think is a reasonable estimate for the mean of the population?
### Problem 2
Last night, everyone at the school music concert wrote their age on a slip of paper and placed it in a box. Today, each of the students in a math class selected a random sample of size 10 from the box of papers. Here is a dot plot showing their sample means, rounded to the nearest year.
1. Does the number of dots on the dot plot tell you how many people were at the concert or how many students are in the math class?
2. The mean age for the population was 35 years. If Elena picks a new sample of size 10 from this population, should she expect her sample mean to be within 1 year of the population mean? Explain your reasoning.
3. What could Elena do to select a random sample that is more likely to have a sample mean within 1 year of the population mean?
### Problem 3
A random sample of people were asked which hand they prefer to write with. “l” means they prefer to use their left hand, and “r” means they prefer to use their right hand. Here are the results:
• l
• r
• r
• r
• r
• r
• r
• r
• r
• r
• l
• r
• r
• r
• r
Based on this sample, estimate the proportion of the population that prefers to write with their left hand.
(From Unit 8, Lesson 16.)
### Problem 4
Andre would like to estimate the mean number of books the students at his school read over the summer break. He has a list of the names of all the students at the school, but he doesn’t have time to ask every student how many books they read.
What should Andre do to estimate the mean number of books?
(From Unit 8, Lesson 15.)
### Problem 5
A hockey team has a 75% chance of winning against the opposing team in each game of a playoff series. To win the series, the team must be the first to win 4 games.
1. Design a simulation for this event.
2. What counts as a successful outcome in your simulation?
3. Estimate the probability using your simulation.
(From Unit 8, Lesson 10.)
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# . There are some students in the two examination halls A and B
There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B. But if 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls.
A: 100 students in hall A,80 students in hall B
B: 200 students in hall A, 45 students in hall B
C: 80 students in hall A, 70 students in hall B
D: 120 students in hall A, 60 students in hall B
Explanation:
Let the number of students in hall A be x,
and, the number of students in hall B be y,
Now, the number of students send from A to B is 10,
So, the equation formed will be,
X – 10 = y + 10 x = 20 + y ... (i)
Now, the number of students send from B to A is 20, which is double the number previously sent from A to B. So, the equation will be,
x + 20 = 2 (y – 20)
x + 20 = 2y – 40
x – 2y = - 60 ... (ii)
Putting the value of (i) in eqn ( ii)
20 + y – 2y = - 60 y = 80 Substituting this value on eqn (i) to find the value of x,
x = 20 + 80 = 100
:. The number of students in hall A = 100
The number of students in hall B = 80
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Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Ex 1.2
Students can Download Maths Chapter 1 Number System Ex 1.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Ex 1.2
Question 1.
Fill in the blanks
(i) -44 + ____ = -88
(ii) ___ – 75 = -45
(iii) ___ – (+50) = -80
Solution:
(i) -44
(ii) 30
(iii) -30
Question 2.
Say True or False.
(i) (-675) – (-400) = -1075
(ii) 15 – (-18) is the same as 15 + 18
(iii) (-45) – (-8) = (-8) – (-45)
Solution:
(i) False
(ii) True
(iii) False
Question 3.
Find the value of the following.
(i) -3 – (-4) using number line.
Solution:
We start at zero facing positive direction. Move 3 units backward to represent (-3). Then turn towards the negative side and move 4 units backwards. We reach+1.
∴ (-3) – (-4) = +1
(ii) 7 – (-10) using number line
Solution:
We start at zero facing positive direction. Move 7 units forward to represent (+7). Then turn towards the negative side and move 10 units backwards.
We reach +17
∴ 1 – (-10) = +17
(iii) 35 – (-64)
Solution:
35 – (-64) = 35+ (Additive inverse of-64) = 35 + (+64) = 99
∴ 35 – (-64) = 99
(iv) -200 – (+100)
Solution:
-200 – (+100) = -200 + (Additive inverse of+100) = -200 + (-100) = -300
-200 – (+100) = -300
Question 4.
Kabilan was having 10 pencils with him. He gave 2 pencils to senthil and 3 to Karthick. Next day his father gave him 6 more pencils, from that he gave 8 to his sister. How many pencils are left with him?
Solution:
Total pencils Kabilan had = 10
No. of pencils given to Senthil = 2
No. of pencils given to Karthick = 3.
Now number of pencils left with Kabilan = 10 – 2 – 3 = 8 – 3 = 5
Number of pencils got from his father = 6
No. total pencils Kabilan had = 5 + 6 = 11
Number of pencils given to his sister = 8
Number of pencils left with Kabilan = 11 – 8 = 3
Question 5.
A lift is on the ground floor. If it goes 5 floors down and then moves up to 10 floors from there, then in which floor will the lift be?
Solution:
Initially the lift will be in the ground floor representing ‘0’
It goes to 5 floors down ⇒ -5
Then it moves 10 floors up +10.
Now the lift will be = 0 – 5 + 10 = -5 + 10
= 5th floor (above the ground floor)
Question 6.
When Kala woke up, her body temperature was 102°F. She took medicine for fever. After 2 hours it was 2°F lower. What was her temperature then?
Solution:
Kala’s temperature initially = 102°F
After two hours the temperature decreased = -2°F
Now the final temperature = 102°F – 2°F = 100°F
Question 7.
What number should be added to (-17) to get -19?
Solution:
According to the problem = -17 + A number = -19
The number = -19 + 17 = -2
∴ -2 should be added to -17 to get -19
Question 8.
A student was asked to subtract (-12) from -47. He got -30. Is he correct? Justify.
Solution:
Subtracting -12 from -47, we get
-47 – (-12) = -47 + (Additive inverse of-12)
= -47 + (+12) = -35
But the students answer is -30.
So he is not correct.
Objective Type Questions
Question 9.
(-5) – (-18)
(i) 23
(ii) -13
(iii) 13
(iv) -23
Solution:
(iii) 131
Question 10.
(-100) – 0 + 100 =
(i) 200
(ii) 0
(iii) 100
(iv)-200
Solution:
(ii) 0
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# CAT Numerical Reasoning: Formulas, Preparation Tips, and Sample Questions
CAT Numerical Reasoning: The CAT Exam Numerical Reasoning section is a critical component that tests candidates’ mathematical skills and ability to interpret data accurately. This section includes questions that assess problem-solving abilities through topics such as arithmetic, algebra, geometry, and data interpretation. Excelling in this part of the CAT exam requires a solid understanding of mathematical concepts and the ability to apply them to solve complex problems efficiently. Proper preparation and practice in numerical reasoning are essential for achieving a high score and gaining admission to top management institutes. In this blog, we will break down difficult problems, share useful tips, and provide practice questions to evaluate yourself.
## What is CAT Numerical Reasoning?
Numerical reasoning is created to test students’ maths skills and manages to concentrate on various specific areas. It involves understanding and interpreting data, performing calculations, and making logical decisions based on numerical information. In tests like the CAT (Common Admission Test), numerical reasoning questions assess your skills in areas such as Arthematic, Algebra, Geometry, Statistics, and Problem-Solving.
## CAT Numerical Reasoning Test Formulas
Here are some formulas you need to know for CAT Numerical Reasoning. Understand these formulas and use them while solving questions:
### Averages
The average, or ‘mean’, is noticed by adding the values in the dataset and dividing the total by the number of values.
### Reverse Percentages
Reverse percentages are used to find the original amount before a percentage increase or decrease is applied. Here’s how you can think about it:
### Percentages
Percentage means ‘per 100’. When we calculate the percentages by dividing the value by the total and multiplying the answer by 100.
Hopefully, this is easy for you, the real difficulty with percentages arises when we must work out percentage changes and increases/decreases.
Percentage Change (Increase)
Percentage Change (Decrease)
### Weighted Average
A weighted average is a type of average where some values contribute more to the final result than others. Here is the formula that students can refer to:
## CAT Numerical Reasoning Preparation Tips
When preparing for CAT numerical reasoning, it is important for the candidate to look into all the preparation tips which will help one to crack the CAT exam.
### Focus on Relevant Math Skills
If you’re finding math exams challenging, don’t worry. Numerical reasoning tests evaluate a different set of skills. These tests involve basic numerical operations, including:
• Averages
• Division
• Percentages
• Multiplication
• Subtraction
• Ratios
### Start by Understanding the Question
Differences in test scores often arise not just from numerical reasoning ability but from understanding the question correctly. The first step is to identify what the question is asking. Instead of getting overwhelmed by all the numbers and graphs, read the question carefully to determine which numbers are relevant. This approach is especially crucial for questions with numerous figures.
### Choose an Optimal Test Environment
Even if you excel in math or numerical reasoning, a distracting environment can hinder your performance. Ensure you are well-prepared and familiar with the test format. Take the test in a quiet, distraction-free setting to perform at your best.
### Remove Stress to Work Concentration
If students are feeling stressed out, then it is an appropriate idea to take deep breaths prior there test start. This will help with fighting nervousness as well as building focus towards their studies. Students should utilise their anxious energy and work it into the test! Therefore sometimes it is a high sense of anxious concentration can help students to focus during a test.
### Be Aware of The Amount of Time
Numerical reasoning tests will be far different so pay attention to the number of questions that students have to answer, as well as calculate how much time you have per question, and overall. Some questions will require longer to finish than others, and that is why students need to calculate how long it generally takes them to answer a numerical reasoning question.
### Get Familiar With Tables and Reading Graphs
The foremost aspect of taking a numerical reasoning test is to fast and accurately digest and analyse presented data in graphs and tables. Speed is an important element in good mock test-taking methods, interpreting graphs and tables quickly is the best place to improve your speed.
## CAT Numerical Reasoning Sample Questions
Find below questions on CAT Numerical Reasoning which will help you get an idea about how the actual paper will look like.
Question 1: The ratio of the ages of two friends A and B is 5:6. If the sum of their ages is 33 years, what are their present ages?
Answer: Let the present ages of A and B be 5x5x5x and 6x6x6x respectively.
According to the problem: 5x+6x=33
11x=33
x = 3
So, the present ages are:
Age of A = 5x = 5 × 3 =15 Years
Age of B = 6x = 6 x 3 = 18 Years
So, the present ages of A and B are 15 years and 18 years respectively.
Question 2: If the sum of three consecutive even numbers is 78, find the numbers.
Answer: Let the three consecutive even numbers be xxx, x+2x+2x+2, and x+4x+4x+4.
Their sum is: x+(x+2)+(x+4)=78x + (x + 2) + (x + 4) = 78x+(x+2)+(x+4)=78
3x+6=78
3x=72
x = 24
The three numbers are:
x=24
X + 2 = 26
X + 4 = 28
So, the three consecutive even numbers are 24, 26, and 28.
Question 3: A retailer marks up the price of an item by 25% above its cost price. During a sale, the retailer offers a discount of 20% on the marked price. If the final selling price of the item is \$240, what is the original cost price of the item?
Let’s break it down step by step.
Let the cost price be C.
Mark up the price by 25%:
Marked price=C+0.25C=1.25C
Offer a discount of 20% on the marked price:
Final selling price=1.25C−0.20(1.25C) = 1.25C × (1−0.20) = 1.25C×0.80=1.00C
We know the final selling price is \$240, so:
1.00C = \$240
Solve for C: C = \$240
The original cost price of the item is \$240.
This question tests your ability to work with percentages and understand the relationships between cost price, marked price, and discounts.
Question 4: The ratio of boys to girls in a class is 3:2. If there are 18 boys, how many girls are there?
Solution: Let the number of girls be G.
Ratio of boys to girls=3/2=18/G
3G=36 ⟹ G=12
Question 5: The average of five numbers is 24. If one of the numbers is 30, what is the average of the remaining four numbers?
Solution: Sum of five numbers=5×24=120
Sum of the remaining four numbers=120−30=90
Average of the remaining four numbers=90/4=22.
## FAQs
What do you understand about CAT the numerical reasoning test?
A numerical reasoning test is used to assess a candidate’s ability to handle and interpret numerical data. You will be required to analyse and draw conclusions from the data, which may be presented in the form of tables or graphs. The tests are timed and in a multiple-choice format.
What do understand about numerical aptitude examples?
Numerical aptitude tests usually target the following mathematic skills: 1) Addition 2) Subtraction 3) Multiplication 4) Division 5) Averages 6) Percentages 7) Ratios. Additional advanced calculations, such as percentages, averages and ratios can become simpler with the use of specific formulas.
What are essential numerical skills?
Basic numeracy abilities are the ability to understand basic arithmetical operations examples subtraction, addition, division, and multiplication. For example, you are supposed to have at least basic numeric comprehension if you can answer simple arithmetic problems like 2 + 2 = 4.
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# 5.5: Chapter Formula Review
## 5.1 Properties of Continuous Probability Density Functions
Probability density function (pdf) $$f(x)$$:
• $$f(x) \geq 0$$
• The total area under the curve $$f(x)$$ is one.
Cumulative distribution function (cdf): $$P(X \leq x)$$
## 5.2 The Uniform Distribution
$$X = a$$ real number between $$a$$ and $$b$$ (in some instances, $$X$$ can take on the values $$a$$ and $$b$$). $$a =$$ smallest $$X; b =$$ largest $$X$$
$$X \sim U (a, b)$$
The mean is $$\mu=\frac{a+b}{2}$$
The standard deviation is $$\sigma=\sqrt{\frac{(b-a)^{2}}{12}}$$
Probability density function: $$f(x)=\frac{1}{b-a} \text { for } a \leq X \leq b$$
Area to the Left of $$\bf{x}$$: $$P(X<x)> Area to the Right of \(\bf{x}$$: $$P(X>x)=(b-x)\left(\frac{1}{b-a}\right)$$
Area Between $$\bf{c}$$ and $$\bf{d}$$: $$P(c<d)> • pdf: \(f(x)=\frac{1}{b-a} \text { for } a \leq x \leq b$$
• cdf: $$P(X \leq x)=\frac{x-a}{b-a}$$
• mean $$\mu=\frac{a+b}{2}$$
• standard deviation $$\sigma=\sqrt{\frac{(b-a)^{2}}{12}}$$
• $$P(c<d)> ## 5.3 The Exponential Distribution • pdf: \(f(x) = me^{(–mx)}$$ where $$x \geq 0$$ and $$m > 0$$
• cdf: $$P(X \leq x) = 1 – e^{(–mx)}$$
• mean $$\mu = \frac{1}{m}$$
• standard deviation $$\sigma = \mu$$
• $$P(X > x) = e^{(–mx)}$$
• $$P(a < X < b) = e^{(–ma)} – e^{(–mb)}$$
• Poisson probability: $$P(X=x)=\frac{\mu^{x} e^{-\mu}}{x !}$$ with mean and variance of $$\mu$$
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# Vince Vaughn And The Stars (11/13/2019)
How will Vince Vaughn get by on 11/13/2019 and the days ahead? Let’s use astrology to complete a simple analysis. Note this is not at all guaranteed – take it with a grain of salt. I will first find the destiny number for Vince Vaughn, and then something similar to the life path number, which we will calculate for today (11/13/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology enthusiasts.
PATH NUMBER FOR 11/13/2019: We will analyze the month (11), the day (13) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 13 we do 1 + 3 = 4. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 4 + 12 = 18. This still isn’t a single-digit number, so we will add its digits together again: 1 + 8 = 9. Now we have a single-digit number: 9 is the path number for 11/13/2019.
DESTINY NUMBER FOR Vince Vaughn: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Vince Vaughn we have the letters V (4), i (9), n (5), c (3), e (5), V (4), a (1), u (3), g (7), h (8) and n (5). Adding all of that up (yes, this can get tiring) gives 54. This still isn’t a single-digit number, so we will add its digits together again: 5 + 4 = 9. Now we have a single-digit number: 9 is the destiny number for Vince Vaughn.
CONCLUSION: The difference between the path number for today (9) and destiny number for Vince Vaughn (9) is 0. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is of questionable accuracy. If you want to see something that people really do vouch for, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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# How do you differentiate (x-4)/(x^2+2)?
Dec 21, 2017
$f ' \left(x\right) = \frac{- {x}^{2} + 8 x + 2}{x - 4} ^ 2$
#### Explanation:
Apply the quotient rule which states:
$f \left(x\right) = g \frac{x}{h \left(x\right)} \to f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - h ' \left(x\right) g \left(x\right)}{{\left(g \left(x\right)\right)}^{2}}$
Let
$g \left(x\right) = x - 4$
$h \left(x\right) = {x}^{2} + 2$
Thus,
$g ' \left(x\right) = 1$
$h ' \left(x\right) = 2 x$
Now plugging into the formula:
$f ' \left(x\right) = \frac{\left(1\right) \cdot \left({x}^{2} + 2\right) - \left(2 x\right) \left(x - 4\right)}{x - 4} ^ 2$
Simplify:
$f ' \left(x\right) = \frac{\left({x}^{2} + 2\right) - \left(2 {x}^{2} - 8 x\right)}{x - 4} ^ 2$
$f ' \left(x\right) = \frac{- {x}^{2} + 8 x + 2}{x - 4} ^ 2$
Dec 21, 2017
$\frac{8 x - {x}^{2} + 2}{{x}^{2} + 2} ^ 2$
#### Explanation:
$\text{differentiate using the "color(blue)"quotient rule}$
$\text{given "y=(g(x))/(h(x))" then}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \textcolor{b l u e}{\text{quotient rule}}$
$g \left(x\right) = x - 4 \Rightarrow g ' \left(x\right) = 1$
$h \left(x\right) = {x}^{2} + 2 \Rightarrow h ' \left(x\right) = 2 x$
$\Rightarrow \frac{d}{\mathrm{dx}} \left(\frac{x - 4}{{x}^{2} + 2}\right)$
$= \frac{{x}^{2} + 2 - 2 x \left(x - 4\right)}{{x}^{2} + 2} ^ 2$
$= \frac{8 x - {x}^{2} + 2}{{x}^{2} + 2} ^ 2$
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# Evaluate the integrals. \int_{-2}^{2}(3x^{4}-2x+1)dx
Question
Applications of integrals
Evaluate the integrals.
$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{\left({3}{x}^{{{4}}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}$$
2021-01-29
Step 1: Given that
Evaluate integrals.
$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{\left({3}{x}^{{{4}}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}$$
Step 2: Solving the integral
We have,
PSK\int_{-2}^{2}(3x^{4}-2x+1)dx=\int_{2}^{2}3x^{4}dx- 2\int_{-2}^{2}xdx+\int_{-2}^{2}1dxZSK
$$\displaystyle={3}{{\left[{\frac{{{x}^{{{5}}}}}{{{5}}}}\right]}_{{-{2}}}^{{{2}}}}-{2}{{\left[{\frac{{{x}^{{{2}}}}}{{{2}}}}\right]}_{{-{2}}}^{{{2}}}}+{{\left[{x}\right]}_{{-{2}}}^{{{2}}}}$$
$$\displaystyle={\frac{{{3}}}{{{5}}}}{\left[{2}^{{{5}}}-{\left(-{2}\right)}^{{{5}}}\right]}-{\left[{2}^{{{2}}}-{\left(-{2}\right)}^{{{2}}}\right]}+{\left[{2}-{\left(-{2}\right)}\right]}$$
$$\displaystyle={\frac{{{3}}}{{{5}}}}{\left({32}-{\left(-{32}\right)}\right)}-{\left({4}-{4}\right)}+{\left({2}+{2}\right)}$$
$$\displaystyle={\frac{{{3}}}{{{5}}}}{\left({64}\right)}-{0}+{4}$$
$$\displaystyle={\frac{{{192}}}{{{5}}}}+{4}$$
$$\displaystyle={\frac{{{192}+{20}}}{{{5}}}}$$
$$\displaystyle={\frac{{{212}}}{{{5}}}}$$
=42.4
### Relevant Questions
Evaluate the integrals.
$$\displaystyle{\int_{{-{{2}}}}^{{2}}}{\left({3}{x}^{{4}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}$$
Find all rational zeros of the polynomial, and write the polynomial in factored form.
$$\displaystyle{P}{\left({x}\right)}={2}{x}^{{{3}}}-{3}{x}^{{{2}}}-{2}{x}+{3}$$
Is the algebraic expression a polynomial? If it is, write the polynomial in standard form:
$$\displaystyle{2}{x}+{3}{x}^{{{2}}}-{5}$$
Evaluate the following definite integrals
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left({x}^{{{4}}}+{7}{e}^{{{x}}}-{3}\right)}{\left.{d}{x}\right.}$$
Evaluate the following integrals.
$$\displaystyle\int{\left({2}{x}^{{{3}}}-{x}^{{{2}}}+{3}{x}-{7}\right)}{\left.{d}{x}\right.}$$
Evaluate the following integrals.
$$\displaystyle{\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{4}{x}-{x}^{{{2}}}}}{\left.{d}{x}\right.}$$
Evaluate the following definite integrals
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{x}{e}^{{{\left(-{x}^{{{2}}}+{2}\right)}}}{\left.{d}{x}\right.}$$
Show that the differential forms in the integrals are exact. Then evaluate the integrals.
$$\displaystyle{\int_{{{\left({1},{1},{2}\right)}}}^{{{\left({3},{5},{0}\right)}}}}{y}{z}{\left.{d}{x}\right.}+{x}{z}{\left.{d}{y}\right.}+{x}{y}{\left.{d}{z}\right.}$$
Which of the following integrals are improper integrals?
1.$$\displaystyle{\int_{{{0}}}^{{{3}}}}{\left({3}-{x}\right)}^{{{2}}}{\left\lbrace{3}\right\rbrace}{\left.{d}{x}\right.}$$
2.$$\displaystyle{\int_{{{1}}}^{{{16}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}$$
3.$$\displaystyle{\int_{{{1}}}^{{\propto}}}{\frac{{{3}}}{{\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}}}}{\left.{d}{x}\right.}$$
4.$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{3}{\left({x}+{1}\right)}^{{-{1}}}{\left.{d}{x}\right.}$$
a) 1 only
b)1 and 2
c)3 only
d)2 and 3
e)1,3 and 4
f)All of the integrals are improper
$$\displaystyle{\int_{{-{{2}}}}^{{-{{1}}}}}\sqrt{{-{4}{x}-{x}^{{2}}}}{\left.{d}{x}\right.}$$
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Hong Kong
Stage 4 - Stage 5
# Identify Solutions to Linear Equations
Lesson
Linear equations are those that can be written in the form $ax+b=0$ax+b=0 after a suitable transformation. We would also call an equation linear if it could be written so that each side of the equation separately has the form $ax+b$ax+b, where $a$a or $b$b or both could possibly be zero.
The unknown quantity $x$x occurs with power $1$1 only, and if the $0$0 in the transformed equation is replaced by a variable $y$y, then the resulting relation has a straight line graph.
##### Example 1
The equation $x+3=5-2x$x+3=52x is a linear equation because it can be manipulated into the form $ax+b=0$ax+b=0. If $2x$2x is added to both sides, we have $3x+3=5$3x+3=5. Then, when $5$5 is subtracted from both sides, we have $3x-2=0$3x2=0 which is an equation in the required form.
The relation $y=3x-2$y=3x2 has a linear graph with a gradient of $3$3 and $y$y-intercept $-2$2.
Linear equations can have no solutions, one solution or infinitely many solutions.
### no solutions
Statements can be written that cannot possibly be true for any value of the variable. For example,
$5-2x=2(2-x)$52x=2(2x)
Any attempt to collect the like terms leads to the absurd statement $5=4$5=4. We say that such an equation is inconsistent or contradictory.
Notice that the right-hand side of the equation can be transformed to $4-2x$42x by expanding the bracket. Then, if we turn each side separately into a function, we have
$y=5-2x$y=52x and
$y=4-2x$y=42x
The graphs of these relations never intersect because their gradients are the same. So, there cannot be a solution to $5-2x=4-2x$52x=42x.
### one solution
When an equation can be expressed in the form $ax+b=0$ax+b=0 with $a$a and $b$b not both zero, there is exactly one solution.
A statement like $5-2x=2(2+x)$52x=2(2+x), for example, can be transformed to read $1=4x$1=4x or $4x-1=0$4x1=0. So, it must represent a linear equation. It is easy to check that it has the solution $x=\frac{1}{4}$x=14.
If we think of the two sides of the original statement as representing two functions
y=5-2x
y=4+2x
we see that the graphs are intersecting straight lines. They cross where $x=\frac{1}{4}$x=14.
### infinitely many solutions
Equations can be written in which the two sides are essentially the same. These statements are true whatever the value of the variable $x$x. Thus, such equations have infinitely many solutions.
For example, we might write $5-2x=\frac{10-4x}{2}$52x=104x2. The two sides are essentially the same. If we made separate relations representing the two sides and drew their graphs, we would find that the two lines coincide. Thus, there are infinitely many common points in the two graphs and, therefore, infinitely many solutions.
#### Example 2
Does the equation $5(1+x)=7+3x$5(1+x)=7+3x have no solutions, one solution or infinitely many solutions? If there is a unique solution, what is it?
If we think of the two sides as though they came from separate relations $y=5(1+x)$y=5(1+x) and $y=7+3x$y=7+3x, it is clear that the graph of the first has gradient $5$5 while the graph of the second has gradient $3$3. So, the two lines must intersect. The solution is the $x$x-value at the intersection.
It is clear that the solution is $x=1$x=1. We can also arrive at this conclusion algebraically.
$5(1+x)$5(1+x) $=$= $7+3x$7+3x $5+5x$5+5x $=$= $7+3x$7+3x $2x$2x $=$= $2$2 $x$x $=$= $1$1
#### Worked Examples
##### Question 1
How many solutions does the equation $8x-4x=4$8x4x=4 have?
1. One solution
A
Infinitely many
B
No solutions
C
##### Question 2
Consider the equation $5x-10=8x+3$5x10=8x+3.
1. Which of the following steps can be reached by successive transformations of the equation?
$3x=-7$3x=7
A
$-3x=13$3x=13
B
$-3x=-7$3x=7
C
$3x=13$3x=13
D
2. Can the equation be transformed into something of the form $x=k$x=k, for some value of $k$k?
Yes
A
No
B
3. Hence determine the number of solutions that satisfy the equation.
infinitely many
A
no solutions
B
one solution
C
##### Question 3
How many solutions does the equation $10\left(5+x\right)=10\left(x+5\right)$10(5+x)=10(x+5) have?
1. Infinitely many
A
No solutions
B
One solution
C
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# Sequence and Series
## Sequences
A sequence is a series of numbers increasing or decreasing by a constant difference between terms (a progression). As a math function a sequence has a domain of positive integers. The Sequence Function a1, a2, a3, … denotes a sequence. The values a1, a2, a3… are terms of the sequence (a1 the first term, a2 the second term, …, and continuing).
A sequence is increasing if an < an + 1 for all n and decreasing if an > an + 1 for all n. This states that for a sequence to increase the terms progress larger for each larger subscript n, and for decreasing sequence, terms progress smaller as each subscript gets larger.
## Arithmetic Sequence
To find the sequence values of the first 4 terms defined by an = n − 4:
a1 = 1 − 4 = −3
a2 = 2 − 4 = −2
a3 = 3 − 4 = −1
a4 = 0
To find the sequence values of the first 5 terms defined by the formula an = 1 / 4n:
(1 divided by 4n)
a1 = 1 / 4(1) = 1/4
a2 = 1 / 4(2) = 1/8
a3 = 1 / 4(3) = 1/12
a4 = 1 / 4(4) = 1/16
a5 = 1 / 4(5) = 1/20
If each term after the first term can be determined by adding the same number to the preceding term the sequence is an arithmetic sequence and there exists a number, d, that is a common difference between each term of the sequence. This means that if we know two terms of a sequence we can determine the common difference, apply the common difference and discover additional terms of that sequence.
By the common difference we can write additional terms for the sequence 1, 3, 5 …
The common difference is 2. The next three terms are 7, 9 and 11.
We can write the next three terms for the arithmetic sequence 1, 5/8, 1/4 …
The common difference is negative 3/8 (1 = 8/8). The next three sequence terms are −1/8, −1/2 and −7/8.
Because each term after the first term of an arithmetic sequence is determined by adding the same difference number the second term is a1 + d, where d is the common difference. The third term can then be determined by a1 + 2d and
can be stated as:
an = an − 1 + d for n > 1
From this we can also see that from a1 any sequence value, an, beyond a1 can be determined by:
an = a1 + (n − 1)d for n > 1
As an example, the 5th term of the arithmetic sequence having a first term of 3 and common difference of −2 is:
a5 = 3 + (5 − 1) (−2) = 3 − 8 = −5
The sequence is {3, 1, −1, −3, −5, …}
## Geometric Sequence
If each term beyond the first term can be found by multiplying the preceding term by the same number the sequence is geometric. A geometric sequence is a series of numbers that has a common ratio, r, such that:
an = (r) an − 1 for n > 1
By applying this formula we can determine the next 3 terms of the geometric sequence 1, 3, 9 by first
finding the common ratio, r:
We’ll use the sequence values a2 and a3 to determine the common ratio …
a2 = 3 = an − 1 (The sequence value prior to 9)
a3 = 9 = an (The sequence value n)
a3 = (r) an − 1 = (r) a2 = (r) 3 = 9
a3 = (r) 3 = 9 when r = 3
The common ratio, r, is 3.
We could have just as easily used a1 and a2 instead of a2 and a3 …
a1 = 1 = an − 1 (The sequence value prior to 3)
a2 = 3 = an (The sequence value n)
a2 = (r) a1 = 3
a2 = (r) 1 = 3 when r = 3
The next three terms of the sequence are 27, 81 and 243…
an = (r) (an − 1)
a4 = (3) 9 = 27
a5 = (3) 27 = 81
a6 = (3) 81 = 243
Each term following the first term is a power of r of the first term. It means that once the common ratio has been determined any nth term of a geometric sequence can be found by multiplying the value of a1 by r raised to the power of n − 1 and can be stated as the Geometric Sequence Formula:
an = a1 r n − 1
As an example of the Geometric Sequence Formula the next 5 terms of the sequence 1, 3, 9 with a common ratio r = 3 are determined as follows:
Given sequence 1, 3, 9 then a1 = 1, a2 = 3, a3 = 9 …
a4 = (1) 34 – 1 = 27
a5 = (1) 35 – 1 = 81
a6 = (1) 36 – 1 = 243
a7 = (1) 37 – 1 = 729
a8 = (1) 38 – 1 = 2,187
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# Graphing Linear Inequalities in Two Variables
## Presentation on theme: "Graphing Linear Inequalities in Two Variables"— Presentation transcript:
Graphing Linear Inequalities in Two Variables
graph a linear inequality in two variables model a real life situation with a linear inequality.
Recall… Graph n < 3 on a number line.
Graphing an Inequality in Two Variables
Graph x < 2 Step 1: Start by graphing the line x = 2 Now what points would give you less than 2? Since it has to be x < 2 we shade everything to the left of the line.
Graph x -2 on the coordinate plane.
y x
Graph y > 3 on the coordinate plane.
x
Graphing a Linear Inequality
Sketch a graph of y 3
Some Helpful Hints If the sign is > or < the line is dashed
If the sign is or the line will be solid When dealing with just x and y. If the sign > or the shading either goes up or to the right If the sign is < or the shading either goes down or to the left
Step 1: Put into slope intercept form
Using What We Know Sketch a graph of x + y < 3 Step 1: Put into slope intercept form y <-x + 3 Step 2: Graph the line y = -x + 3
When dealing with slanted lines
If it is > or then you shade above If it is < or then you shade below the line
> < Graph y -3x + 2 on the coordinate plane. y
Instead of testing a point If in y = mx + b form... Shade up Shade down Solid line x > < Dashed line
-3x -3x -4y > -3x + 12 -4 -4 Graph on the coordinate plane.
y < x - 3 x Boundary Line m = b = -3
Graph y -3x + 2 on the coordinate plane.
Boundary Line y = -3x + 2 m = -3 b = 2 x Test a point not on the line test (0,0) (0) + 2 Not true!
Problem If you have less than \$5.00 in nickels and dimes, find an inequality and sketch a graph to describe how many of each coin you have. Let n = # of nickels Let d = # of dimes 0.05 n d < 5.00 or 5 n + 10 d < 500
5n + 10d < 500 n d d 50 60 50 40 30 20 10 100 n
Graphing Absolute Value Inequalities
Graphing Absolute Value Inequalities
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# Cube and Cube Root: Definition, Formula, How to Find, Examples
The cube root of a number is an important concept like square roots in the number system. Note that the cube root is the inverse method of finding cubes. In this section, we will discuss about the cube root of a number.
What is cube and cube root? The number obtained by multiplying a given number two times by itself is called the cube of that given number. So m×m×m=m3 is the cube of m. Here the number m is called the cube root of m3.
#### Definition of Cube of a Number
A number is called the cube of a number m if the number is a product of three numbers of m‘s. Thus m3 is the cube of m.
For example, 8 is the cube of 2 as 8=2×2×2.
Similarly, 64 is the cube of 4 as 64=43.
#### Definition of Cube Root
A number x is called a cube root of a number y if the following relation is satisfied:
$x \times x \times x=y^3.$
Thus, if y=x3 then we say x is the cube root of y.
Note that 8=23, so we can say that 2 is a cube root of 8.
#### Cube Root Symbol
Symbol of cube root: the cube root is represented by the symbol “$\sqrt[3]{}$”.
As the cube root of 8 is 2, mathematically we write it as follows:
$\sqrt[3]{8}=2$
#### Cube Root Examples
More examples of cube roots:
As 33 = 3×3×3 = 27, we have $\sqrt[3]{27}$=3 by the definition of cube roots.
We have 43 = 4×4×4 = 64. Thus $\sqrt[3]{64}$=4, so the cube root of 64 is 4.
Note that 53 = 5×5×5 = 125. Therefore $\sqrt[3]{125}$=5.
#### Rules of Cube Root
The following are the properties of cube roots.
• $x^{1/3}=\sqrt[3]{x}$
• $\sqrt{x^3}=x$
• $a\sqrt[3]{x}+b\sqrt[3]{x}=(a+b)\sqrt[3]{x}$
• $a\sqrt[3]{x}-b\sqrt[3]{x}=(a-b)\sqrt[3]{x}$
• $\sqrt[3]{x \times y}=\sqrt[3]{x} \times \sqrt[3]{y}$
• $\sqrt[3]{\frac{x}{y}}=\frac{\sqrt[3]{x}}{\sqrt[3]{y}}$ if $y \neq 0$
#### Cube Root Formula
The cube root $\sqrt[3]{y}$ can written as $\sqrt[3]{y}=y^{1/3}$ by the rule of indices. If the number $y$ is a perfect cube, that is, $y=x \times x \times x$ for some number $x$ then we use the formula below to find the cube root of $y:$
$\sqrt[3]{y}=\sqrt[3]{x \times x \times x}=x.$
#### Cube Root Table
Perfect cube table:
Also Read: Fourth root of 81
#### Simplifying Cube Roots
Let us now learn how to simplify cube roots. We will simplify cube root of 24.
Question: Simplify cube root of 24.
Solution: At first, we will factorize 24. See that
24 = 2×2×2×3
Taking cube root on both sides, we get
$\sqrt[3]{24}=\sqrt[3]{2 \times 2 \times 2 \times 3}$
$=\sqrt[3]{2 \times 2 \times 2} \times \sqrt[3]{3}$ $[\because \sqrt[3]{x \times y}=\sqrt[3]{x} \times \sqrt[3]{y}]$
$=2 \times \sqrt[3]{3}$ $[\because \sqrt[3]{a \times a \times a}=a]$
$=2\sqrt[3]{3}$
So the square root of 24 is $2\sqrt[3]{3}$ (YouTube video 0f cube root of 24).
#### Methods of Finding Cube Root
How to find the cube root of a number?
#### Addition or Subtraction of Cube Roots
How to add/subtract two square roots? The following steps have to be followed.
Step 1: First, we express the cube roots into their simplified forms as above like cube root of 24.
Step 2: If the simplified forms contain only the cube roots $\sqrt{x}$, then apply the formula $a\sqrt[3]{x}\pm b\sqrt[3]{x}=(a\pm b)\sqrt[3]{x}$ to get the desired sum or difference. If they have different radicals then we keep them as it is.
For example, add $\sqrt[3]{24}$ and $\sqrt[3]{81}$
Question: Find $\sqrt[3]{24}+\sqrt[3]{81}$
Solution: Firstly, we will simplify both cube roots.
From above $\sqrt[3]{24}=2\sqrt[3]{3}$
Also, $\sqrt[3]{81}=\sqrt[3]{3 \times 3 \times 3 \times 3}$ $=3\sqrt[3]{3}$
So $\sqrt[3]{24}+\sqrt[3]{81}$ $=2\sqrt[3]{3}+3\sqrt[3]{3}$ $=5\sqrt[3]{3}.$
Similarly, $\sqrt[3]{50}-\sqrt[3]{81}$ $=2\sqrt[3]{3}-3\sqrt[3]{3}$ $=-\sqrt[3]{3}$
#### Multiplication or Division of Cube Roots
How to multiply or divide two cube roots? We will follow the below steps.
Step 1: We write the cube roots into their simplified forms.
Step 2: Then applying the formula $a\sqrt[3]{x} \times b\sqrt[3]{y}=ab\sqrt[3]{xy}$ we get the desired multiplication. To divide the cube roots, we will use the formula $a\sqrt[3]{x} \div b\sqrt[3]{y}=a/b \frac{\sqrt[3]{x}}{\sqrt[3]{y}}.$
#### Cube Root of a Decimal
How to find the cube root of decimals? We have to follow the below steps to compute the cube root of a decimal number.
Step 1: At first, we need to express the decimal number as a fraction.
Step 2: Then we use the formula $\sqrt[3]{\frac{x}{y}}=\frac{\sqrt[3]{x}}{\sqrt[3]{y}}.$
Step 3: Now we will simplify the cube roots of $x$ and $y$ and then we put their values in the fraction obtained in step 2.
Step 4: Simplifying the fraction obtained in step 3, we will get the cube root of the given decimal number.
To understand the above method, we provide an example here.
Question: Find $\sqrt{0.008}$
Solution:
Note that $0.008=\frac{8}{1000}$
Taking square root we get
$\sqrt[3]{0.008}=\sqrt[3]{\frac{8}{1000}}$ $=\frac{\sqrt[3]{8}}{\sqrt[3]{1000}}$ $=\frac{2}{10}$ $=0.2$
So the square root of 0.008 is 0.2.
#### Cube Root of a Complex Number
The computation of the cube of a complex number is not as simple as the method of finding the cube root of a real number. We know that the general form of a complex number is $a+ib,$ where both $a$ and $b$ are real numbers.
#### Cube Root as a Function
Note that the cube root of a number takes three values. So it will not make a function. But if we define $\sqrt{x^3}=+x,$ then it will definitely make a function. Now define a function
$f: \mathbb{R} \to \mathbb{R} \,\, \text{by} \,\, f(x)=\sqrt[3]{x}.$
The above function $f$ is one-to-one but not onto.
#### Generalization of Cube Roots
It is known that the cube root $x$ of the number $y$ satisfies the relation $y=x^3.$ Symbolically, we write $\sqrt[3]{y}=x.$ This can be generalized in a broad manner.
Fourth root: If $x^4=y$ then $x$ is called a fourth root of $y$ and we write it as $\sqrt[4]{y}=x.$ In exponent form $\sqrt[4]{y}=y^{1/4}.$
Fifth root: If $x^5=y$ then $x$ is called a fifth root of $y.$ It is written as $\sqrt[5]{y}=x.$ In exponent form $\sqrt[5]{y}=y^{1/5}.$
n-th root: If $x^n=y$ then $x$ is called a n-th root of $y.$ Note that $\sqrt[n]{y}=x.$ In exponent form $\sqrt[n]{y}=y^{1/n}.$
Polynomial root: Let $f(x)$ be a polynomial with a root $c$. Then we must have $f(c)=0.$ This type of roots are classified as polynomial roots.
#### Applications of Cube Root
The computation of cube roots has many applications in several branches of mathematics; such as
• Equation solving (for example, solve x3=8)
• Polynomial
• Numerical analysis
• Geometry (for example, to find the side of a cube if the volume is known)
#### Solved Problems of Cube Roots
Question: Find the cube root of $125.$
Solution:
Note that $125=5 \times 5 \times 5$
So $\sqrt[3]{125}=\sqrt{5 \times 5 \times 5}$
$=5$ $[\because \sqrt[3]{x \times x \times x}=x]$
So the value of the cube root of 125 is 5.
#### Cube Root of Numbers
Cube root of 8 Cube root of 16 Cube root of 24 Cube root of 27 Cube root of 32 Cube root of 40 Cube root of 48 Cube root of 54 Cube root of 56 Cube root of 64 Cube root of 72 Cube root of 80 Cube root of 81 Cube root of 88 Cube root of 96 Cube root of 108 Cube root of 120 Cube root of 125 Cube root of 135 Cube root of 216 Cube root of 343 Cube root of 512 Cube root of 721 Cube root of 1000
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# ACT Math : How to simplify an expression
## Example Questions
← Previous 1 3 4
### Example Question #1 : Simplifying Expressions
What is the value of x when 2x - 7 = -5x + 28
-5
7
-7
5
5
Explanation:
We need to solve for x. To do so, first move the x values to the left of the = sign and all other values to the right. 2x + 5x = 28 + 7. Then add like terms: 7x = 35. In order to get x all by itself, we then divide both sides by 7. X = 5.
### Example Question #2 : Simplifying Expressions
If 4x – 6 = 18, what is 3x3 – 4x + 1?
-1
97
625
300
625
Explanation:
First we have to solve for x using the first equation:
4x-6 = 18
4x = 24
x = 6
Next, we have to plug in our value for x into the second equation:
3(6)3 – 4(6) + 1
3 (216) – 24 +1 = 625
### Example Question #3 : How To Simplify An Expression
What is the value of z when 2z + 3 = 3z – 5?
-8
1/8
8
4
8
Explanation:
Solve one equation with one variable by moving the constants to one side of the equal sign and the variables to the other. 2z + 3 = 3z – 5 is solved by adding 5 -2z to both side of the equation and ending up with z = 8
### Example Question #4 : How To Simplify An Expression
Given that 3x + 7 = 4x – 2, what is the value of x ?
9
9/7
1
5/7
9
Explanation:
First, we must solve the equation for x by subtracting 3x from both sides:
3x – 3x + 7 = 4x – 3x – 2
7 = x – 2
Then we must add 2 to both sides:
7 + 2 = x – 2 + 2
9 = x
### Example Question #5 : How To Simplify An Expression
If a = 4 and b =3 then: 2a2 + 3ab – 7 is?
37
– 2
61
73
61
Explanation:
Substitute the values of a and b into the equation. Then
2a2 + 3ab – 7
(2 x 42 ) + (3 x 4 x 3) – 7
(2 x 16) + (36) -7
32 + 36 – 7
61
### Example Question #6 : How To Simplify An Expression
Which of the following expressions are equivalent to b for all non-zero real numbers a, b, x, and y such that
None of the answers are correct
Explanation:
Cross multiply to get 2xb = 3ya2, then divide by 2x to get b on one side of the equal sign by itself.
### Example Question #1 : Simplifying Expressions
A store in California specializes in the sale of custom t-shirts and is growing rapidly. In their second year of business, they doubled the sales from the first year, and in the third year, they sold 50,000 more t -shirts than their second year. In their fourth year, they doubled their sales from the third year. If in the fourth year of business, the store sold 300,000 shirts, how many did they sell in the second year?
150,000
250,000
25,000
50,000
100,000
100,000
Explanation:
This question is a bit wordy, so it is tough to not get lost in it, it usually helps to write down the pertinent information on a separate sheet of paper. In this case, you should have written that year 4 = 2 x year 3 = 300,000
so, 300,000 = 2 x year 3
divide each side by 2 and we get
150,000 sold in year 3. We are told that year 3 = year 2 +50,000
so, subtracting 50,000 from each side yields
year 3 - 50,000 = total sales from year 2
150,000 - 50,000 = total sales for year 2 = 100,000
### Example Question #1 : How To Simplify An Expression
Which of the following values is the greatest?
28
49
164
324
82
324
Explanation:
In order to compare each quantity, we need to rewrite each one using the same base. A common base that would be easy to use would be 2, because we can write 2, 4, 8, 16, and 32 all as a power of 2. We also need to remember that when taking an exponent to an exponent, we have to multiple the two exponents together.
28 = 28
82 = (23)2 = 26
164 = (24)4 = 216
49 = (22)9 = 218
324 = (25)4 = 220
When we compare all of the numbers as powers of 2, we realize that 220 is the largest. Thus, the answer is 324.
### Example Question #81 : Expressions
Which of the following expressions is equivalent to 4x2 + 10x – 6?
(2x +1)(2x – 6)
(4x + 2)(x – 3)
(2x – 2)(2x + 3)
2(2x +1)(x – 3)
2(2x – 1)(x + 3)
Correct answer: 2(2x – 1)(x + 3)
Explanation:
First, pull out a common factor of 2 to get 2(2x2 + 5x – 3). Then factor the quadratic so that the x terms add to 5x and the numbers multiply to - 3, resulting in 2(2x – 1)(x + 3).
### Example Question #1 : How To Simplify An Expression
What is the value of x when 3x + 5 = 2x – 7?
12/5
5/12
–12
12
–9
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## Christmas Trees
#### Problem
When Christmas trees are planted they should stand at least 2 metres away from one another whilst growing.
What is the maximum number of trees that can be planted in one square kilometre?
#### Solution
The best arrangement is a triangular matrix:
By applying the Pythagorean Theorem to the triangle, $2^2 = d^2 + 1^2$, hence the smallest allowable distance between rows, $d=\sqrt{3}$ metres.
Assuming that we can plant a tree in the top left corner of the square then it is possible to plant 501 trees across the top row. The next row will contain 500 trees, then the pattern repeats: 501, 500, 501, ... .
As $\dfrac{1000}{\sqrt{3}} = 577.35...$ we will be able to plant 577 complete rows. Therefore we can plant $\dfrac{576}{2} = 288$ blocks of $501 + 500 = 1001$ trees and a final row of 501 trees. That is, a maximum of $288 \times 1001 + 501 = 288789$ trees can be planted in a square kilometre.
An alternative approach can be used to closely approximate this answer. It can be seen that each tree occupies the same area as two equilateral triangles.
That is, each tree occupies $2\sqrt{3} m^2$. As $1 km^2 = 1000 \times 1000 = 1000000 m^2$, the maximum number of trees that can be planted is approximately $\dfrac{1000000}{2\sqrt{3}} \approx 288673$ trees.
If the trees were planted in a square matrix, what is the maximum number of trees that could be planted in 1 km2?
Can you explain why a triangular arrangement is the most efficient method of planting trees?
Problem ID: 52 (Oct 2001) Difficulty: 2 Star
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### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation3.14 Word Problems - Age - 6
Example: Twenty-five years ago, Hailey was five more than one-third as old as Jason was. Today, Jason is twenty-six less than two times the age of Hailey. How old is Jason? Solution: Let 'j' and 'h' are present ages of Jason and Hailey. Twenty-five ago their ages are j-25 and h-25 Twenty-five years ago, Hailey was five more than one-third as old as Jason was can be written as: h - 25 = 1/3 (j-25) + 5 Today, Jason is twenty-six less than two times the age of Hailey: j = 2h - 26 h - 25 = 1/3 (j-25) + 5 j = 2h - 26 h - 25 = 1/3 (j-25) + 5 h - 30 = 1/3 (j-25) 3h - 90 = j - 25 3h = j - 25 + 90 3h = j + 65 substituting j = 2h - 26 in the above equation we have 3h = 2h - 26 + 65 3h - 2h = - 26 + 65 h = 39 j = 2h - 26 = 2 x 39 - 26 = 52 Jason is 52 years and Hailey is 39 years old. Example: Harry is five times as old as Ron. Tom is twice as old as Ron. Ron is twelve years younger than Tom. How old is Harry? Solution: Let 'h' 'r' and 't' be the present ages of Harry, Ron and Tom respectively. Given: h = 5r t = 2r r = t-12 Solve: r = t-12 r = 2r - 12 12 = 2r - r 12 = r h = 5r = 5x12 = 60 Harry is 60 years old. Directions: Solve the following word problems.
Q 1: Allison is 8 years older than Cathy. Twenty years ago Allison was three times as old as Cathy. How old is each now?Cathy's age is 24 and Allison's age is 32Cathy's age is 12 and Allison's age is 24Cathy's age is 12 and Allison's age is 36 Q 2: In 4 years, Cathy's age will be the same as Ted's age now. In 2 years, Ted will be twice as old as Cathy. Find their ages now.Ted is 6 and Cathy is 2Ted is 10 and Cathy is 6Ted is 8 and Cathy is 4 Question 3: This question is available to subscribers only! Question 4: This question is available to subscribers only!
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# Area of shape - 7th grade (12y) - math problems
#### Number of problems found: 313
• The ABCD
The ABCD trapezoid has a base length of a = 120mm, c = 86mm and an area of S = 2575 mm2. Calculate the height of the trapezoid.
• Square gardens
The gardening colony with dimensions of 180 m and 300 m is to be completely divided into equally large square areas with the largest possible area. Calculate how many such square areas can be obtained and determine the side length of the square.
• Gardens
The area of the square garden is 3/4 of the area of the triangular garden with sides of 80 m, 50 m, 50 m. How many meters of the fence do we need to fence a square garden?
• Isosceles trapezoid
Find the area of an isosceles trapezoid with bases of 8cm and 72mm. The height of the trapezoid is equal to three-quarters of the longer base.
• Hectares
Determine the area of the rectangular land (in ha), which has dimensions of 3 cm and 4.5 cm on the plan with a scale of 1: 40,000.
• Extending square garden
Mrs. Petrová's garden had the shape of a square with a side length of 15 m. After its enlargement by 64 m2 (square), it had the shape of a square again. How many meters has the length of each side of the garden been extended?
• Height
The content of the triangle is 35 cm2. The length of the base is 10 cm. Determine the length of the height on the base.
• Cylinder container
The cylindrical container with a diameter of 1.8 m contains 2,000 liters of water. How high does the water reach?
• The trench
Calculate how many cubic meters of soil needs to be removed from the excavation in the shape of an isosceles trapezoid, the top width is 3 meters, the lower width is 1.8 m, the depth of the excavation is 1 m, and the length is 20 m.
• Largest wall
Find the content of the largest wall of a prism with a rectangle base with a height of 4 dm, side c = 5 cm, and side b = 6 cm.
Calculate the surface of a quadrilateral prism according to the input: Area of the diamond base S1 = 2.8 m2, length of the base edge a = 14 dm, height of the prism 1,500 mm.
• Wooden box
The block-shaped box was placed on the ground, leaving a rectangular print with 3 m and 2 m. When flipped over to another wall, a print with dimensions of 0.5 m and 3 m remained in the sand. What is the volume of the wooden box?
The surface of the regular quadrilateral prism is 8800 cm2, the base edge is 20 cm long. Calculate the volume of the prism
• The corridor
The corridor is 12 m long and 3.6 m wide. It must be paved with rectangular tiles measuring 15 cm and 30 cm. Are 1000 pieces of tiles enough to pave the corridor?
• Cutting the prism
A prism with a square base with a content of 1 cm2 and a height of 3 cm was cut from a cube with an edge length of 3 cm. What is the surface of the body formed from the cube after cutting the prism?
• Largest squares
How many of the largest square sheets did the plumber cut the honeycomb from 16 dm and 96 dm?
• The hollow cylinder
The hollow cylinder has a height of 70 cm, an outer diameter of 180 cm, and an inner diameter of 120 cm. What is the surface of the body, including the area inside the cavity?
• Garden exchange
The garden has a rectangular trapezoid shape, the bases of which have dimensions of 60 m and 30 m and a vertical arm of 40 m. The owner exchanged this garden for a parallelogram, which is 7/9 of the area of a trapezoidal garden. What is the size of the ne
• Wallpaper
3750 cm square of wallpaper is needed to glue a cube-shaped box. Can Dad cut out the whole necessary piece of wallpaper as a whole if he has a roll of wallpaper 50 cm wide?
• Height to the base
The triangle area is 35 cm ^ 2. The size of the base is 10 cm. Find the length of height to the base.
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Examples of area of plane shapes. Examples for 7th grade (seventh).
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# Number line
(Redirected from Real line)
In elementary mathematics, a number line is a picture of a straight line that serves as spatial representation of numbers, usually graduated like a ruler with a particular origin point representing the number zero and evenly spaced marks in either direction representing integers, imagined to extend infinitely. The metaphorical association between numbers and points on the line links arithmetical operations on numbers to geometric relations between points, and provides a conceptual scaffold for learning mathematics.
The number line is initially used to teach addition and subtraction of integers, especially involving negative numbers. As students progress, more kinds of numbers can be placed on the line, including fractions, decimal fractions, square roots, and transcendental numbers such as the circle constant π: Every point of the number line corresponds to a unique real number, and every real number to a unique point.[1]
Using a number line, numerical concepts can be interpreted geometrically and geometric concepts interpreted numerically. An inequality between numbers corresponds to a left-or-right order relation between points. Numerical intervals are associated to geometrical segments of the line. Operations and functions on numbers correspond to geometric transformations of the line. Wrapping the line into a circle relates modular arithmetic to the geometric composition of angles. Marking the line with logarithmically spaced graduations associates multiplication and division with geometric translations, the principle underlying the slide rule. In analytic geometry, coordinate axes are number lines which associate points in a geometric space with tuples of numbers, so geometric shapes can be described using numerical equations and numerical functions can be graphed.
In advanced mathematics, the number line is usually called the real line or real number line, and is a geometric line isomorphic to the set of real numbers, with which it is often conflated; both the real numbers and the real line are commonly denoted R or ${\displaystyle \mathbb {R} }$. The real line is a one-dimensional real coordinate space, so is sometimes denoted R1 when comparing it to higher-dimensional spaces. The real line is a one-dimensional Euclidean space using the difference between numbers to define the distance between points on the line. It can also be thought of as a vector space, a metric space, a topological space, a measure space, or a linear continuum. The real line can be embedded in the complex plane, used as a two-dimensional geometric representation of the complex numbers.
## History
The first mention of the number line used for operation purposes is found in John Wallis's Treatise of algebra (1685).[2] In his treatise, Wallis describes addition and subtraction on a number line in terms of moving forward and backward, under the metaphor of a person walking.
An earlier depiction without mention to operations, though, is found in John Napier's A description of the admirable table of logarithmes (1616), which shows values 1 through 12 lined up from left to right.[3]
Contrary to popular belief, René Descartes's original La Géométrie does not feature a number line, defined as we use it today, though it does use a coordinate system. In particular, Descartes's work does not contain specific numbers mapped onto lines, only abstract quantities.[4]
## Drawing the number line
A number line is usually represented as being horizontal, but in a Cartesian coordinate plane the vertical axis (y-axis) is also a number line.[5] According to one convention, positive numbers always lie on the right side of zero, negative numbers always lie on the left side of zero, and arrowheads on both ends of the line are meant to suggest that the line continues indefinitely in the positive and negative directions. Another convention uses only one arrowhead which indicates the direction in which numbers grow.[5] The line continues indefinitely in the positive and negative directions according to the rules of geometry which define a line without endpoints as an infinite line, a line with one endpoint as a ray, and a line with two endpoints as a line segment.
## Comparing numbers
If a particular number is farther to the right on the number line than is another number, then the first number is greater than the second (equivalently, the second is less than the first). The distance between them is the magnitude of their difference—that is, it measures the first number minus the second one, or equivalently the absolute value of the second number minus the first one. Taking this difference is the process of subtraction.
Thus, for example, the length of a line segment between 0 and some other number represents the magnitude of the latter number.
Two numbers can be added by "picking up" the length from 0 to one of the numbers, and putting it down again with the end that was 0 placed on top of the other number.
Two numbers can be multiplied as in this example: To multiply 5 × 3, note that this is the same as 5 + 5 + 5, so pick up the length from 0 to 5 and place it to the right of 5, and then pick up that length again and place it to the right of the previous result. This gives a result that is 3 combined lengths of 5 each; since the process ends at 15, we find that 5 × 3 = 15.
Division can be performed as in the following example: To divide 6 by 2—that is, to find out how many times 2 goes into 6—note that the length from 0 to 2 lies at the beginning of the length from 0 to 6; pick up the former length and put it down again to the right of its original position, with the end formerly at 0 now placed at 2, and then move the length to the right of its latest position again. This puts the right end of the length 2 at the right end of the length from 0 to 6. Since three lengths of 2 filled the length 6, 2 goes into 6 three times (that is, 6 ÷ 2 = 3).
## Portions of the number line
The section of the number line between two numbers is called an interval. If the section includes both numbers it is said to be a closed interval, while if it excludes both numbers it is called an open interval. If it includes one of the numbers but not the other one, it is called a half-open interval.
All the points extending forever in one direction from a particular point are together known as a ray. If the ray includes the particular point, it is a closed ray; otherwise it is an open ray.
## Extensions of the concept
### Logarithmic scale
On the number line, the distance between two points is the unit length if and only if the difference of the represented numbers equals 1. Other choices are possible.
One of the most common choices is the logarithmic scale, which is a representation of the positive numbers on a line, such that the distance of two points is the unit length, if the ratio of the represented numbers has a fixed value, typically 10. In such a logarithmic scale, the origin represents 1; one inch to the right, one has 10, one inch to the right of 10 one has 10×10 = 100, then 10×100 = 1000 = 103, then 10×1000 = 10,000 = 104, etc. Similarly, one inch to the left of 1, one has 1/10 = 10–1, then 1/100 = 10–2, etc.
This approach is useful, when one wants to represent, on the same figure, values with very different order of magnitude. For example, one requires a logarithmic scale for representing simultaneously the size of the different bodies that exist in the Universe, typically, a photon, an electron, an atom, a molecule, a human, the Earth, the Solar System, a galaxy, and the visible Universe.
Logarithmic scales are used in slide rules for multiplying or dividing numbers by adding or subtracting lengths on logarithmic scales.
### Combining number lines
A line drawn through the origin at right angles to the real number line can be used to represent the imaginary numbers. This line, called imaginary line, extends the number line to a complex number plane, with points representing complex numbers.
Alternatively, one real number line can be drawn horizontally to denote possible values of one real number, commonly called x, and another real number line can be drawn vertically to denote possible values of another real number, commonly called y. Together these lines form what is known as a Cartesian coordinate system, and any point in the plane represents the value of a pair of real numbers. Further, the Cartesian coordinate system can itself be extended by visualizing a third number line "coming out of the screen (or page)", measuring a third variable called z. Positive numbers are closer to the viewer's eyes than the screen is, while negative numbers are "behind the screen"; larger numbers are farther from the screen. Then any point in the three-dimensional space that we live in represents the values of a trio of real numbers.
### As a linear continuum
The real line is a linear continuum under the standard < ordering. Specifically, the real line is linearly ordered by <, and this ordering is dense and has the least-upper-bound property.
In addition to the above properties, the real line has no maximum or minimum element. It also has a countable dense subset, namely the set of rational numbers. It is a theorem that any linear continuum with a countable dense subset and no maximum or minimum element is order-isomorphic to the real line.
The real line also satisfies the countable chain condition: every collection of mutually disjoint, nonempty open intervals in R is countable. In order theory, the famous Suslin problem asks whether every linear continuum satisfying the countable chain condition that has no maximum or minimum element is necessarily order-isomorphic to R. This statement has been shown to be independent of the standard axiomatic system of set theory known as ZFC.
### As a metric space
The real line forms a metric space, with the distance function given by absolute difference:
${\displaystyle d(x,y)=|x-y|.}$
The metric tensor is clearly the 1-dimensional Euclidean metric. Since the n-dimensional Euclidean metric can be represented in matrix form as the n-by-n identity matrix, the metric on the real line is simply the 1-by-1 identity matrix, i.e. 1.
If pR and ε > 0, then the ε-ball in R centered at p is simply the open interval (pε, p + ε).
This real line has several important properties as a metric space:
### As a topological space
The real line carries a standard topology, which can be introduced in two different, equivalent ways. First, since the real numbers are totally ordered, they carry an order topology. Second, the real numbers inherit a metric topology from the metric defined above. The order topology and metric topology on R are the same. As a topological space, the real line is homeomorphic to the open interval (0, 1).
The real line is trivially a topological manifold of dimension 1. Up to homeomorphism, it is one of only two different connected 1-manifolds without boundary, the other being the circle. It also has a standard differentiable structure on it, making it a differentiable manifold. (Up to diffeomorphism, there is only one differentiable structure that the topological space supports.)
The real line is a locally compact space and a paracompact space, as well as second-countable and normal. It is also path-connected, and is therefore connected as well, though it can be disconnected by removing any one point. The real line is also contractible, and as such all of its homotopy groups and reduced homology groups are zero.
As a locally compact space, the real line can be compactified in several different ways. The one-point compactification of R is a circle (namely, the real projective line), and the extra point can be thought of as an unsigned infinity. Alternatively, the real line has two ends, and the resulting end compactification is the extended real line [−∞, +∞]. There is also the Stone–Čech compactification of the real line, which involves adding an infinite number of additional points.
In some contexts, it is helpful to place other topologies on the set of real numbers, such as the lower limit topology or the Zariski topology. For the real numbers, the latter is the same as the finite complement topology.
### As a vector space
The real line is a vector space over the field R of real numbers (that is, over itself) of dimension 1. It has the usual multiplication as an inner product, making it a Euclidean vector space. The norm defined by this inner product is simply the absolute value.
### As a measure space
The real line carries a canonical measure, namely the Lebesgue measure. This measure can be defined as the completion of a Borel measure defined on R, where the measure of any interval is the length of the interval.
Lebesgue measure on the real line is one of the simplest examples of a Haar measure on a locally compact group.
### In real algebras
When A is a unital real algebra, the products of real numbers with 1 is a real line within the algebra. For example, in the complex plane z = x + iy, the subspace {z : y = 0} is a real line. Similarly, the algebra of quaternions
q = w + x i + y j + z k
has a real line in the subspace {q : x = y = z = 0 }.
When the real algebra is a direct sum ${\displaystyle A=R\oplus V,}$ then a conjugation on A is introduced by the mapping ${\displaystyle v\to -v}$ of subspace V. In this way the real line consists of the fixed points of the conjugation.
For a dimension n, the square matrices form a ring that has a real line in the form of real products with the identity matrix in the ring.
## References
1. ^ Stewart, James B.; Redlin, Lothar; Watson, Saleem (2008). College Algebra (5th ed.). Brooks Cole. pp. 13–19. ISBN 978-0-495-56521-5.
2. ^ Wallis, John (1685). Treatise of algebra. http://lhldigital.lindahall.org/cdm/ref/collection/math/id/11231 pp. 265
3. ^ Napier, John (1616). A description of the admirable table of logarithmes https://www.math.ru.nl/werkgroepen/gmfw/bronnen/napier1.html
4. ^ Núñez, Rafael (2017). How Much Mathematics Is "Hardwired", If Any at All Minnesota Symposia on Child Psychology: Culture and Developmental Systems, Volume 38. http://www.cogsci.ucsd.edu/~nunez/COGS152_Readings/Nunez_ch3_MN.pdf pp. 98
5. ^ a b Introduction to the x,y-plane Archived 2015-11-09 at the Wayback Machine "Purplemath" Retrieved 2015-11-13
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24 March, 17:18
# jose invests money in two simple interests account. he invests twice as much in an account paying 13% as he does in an account paying 5%. if he earns \$93.00 in interest in one year from both accounts combined how much did he invest altogether
+4
1. 24 March, 18:07
0
He invested altogether \$900
Step-by-step explanation:
* Lets explain how to solve the problem
- Jose invests money in two simple interests account
- He invests twice as much in an account paying 13% as he does in
an account paying 5%
- That means the amount he invested in the account paying 13% is
twice the amount he invested in the account paying 5%
- He earns \$93.00 in interest in one year from both accounts combined
- We need to find how much he invested in each account
- Assume that he invested \$x in the account paying 5%
∵ He invested twice as much in an account paying 13% as he does in
an account paying 5%
∴ He invested \$2x in the account paying 13%
- The simple interest I = PRT, where P is the money invested, R is the
rate of interest in decimal and t is the time of investment
# Account paying 13%
∵ P = 2x, R = 13/100 = 0.13, T = 1
∴ I = 2x (0.13) (1)
∴ I = 0.26 x ⇒ (1)
# Account paying 5%
∵ P = x, R = 5/100 = 0.05, T = 1
∴ I = x (0.05) (1)
∴ I = 0.05 x ⇒ (2)
∵ He earns \$93.00 in interest in one year from both accounts
- Add (1) and (2) and equate the sum by 93.00
∴ 0.26 x + 0.05 x = 93.00
- Add like terms in the left hand side
∴ 0.31 x = 93.00
- Divide both sides by 0.31
∴ x = 300
∴ 2x = 2 (300)
∴ 2x = 600
∵ x represents the amount of money invested in the account paying 5%
∴ The amount of money invested in the account of 5% is \$300
∴ The amount of money invested in the account of 13% is \$600
- The total amount of money he invested is the sum of the money
he invested in each account [\$300 + \$600 = \$900]
∴ He invested altogether \$900
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Question 76
# Shyam, a fertilizer salesman, sells directly to farmers. He visits two villages A and B. Shyam starts from A, and travels 50 meters to the East, then 50 meters North-East at exactly 45° to his earlier direction, and then another 50 meters East to reach village B. If the shortest distance between villages A and B is in the form of $$a\sqrt{b+\sqrt{c}}$$ meters, Find the value of a+b+c.
Solution
In $$\triangle$$ OPN
=> $$sin 45 = \frac{PN}{PO}$$
=> $$\frac{1}{\sqrt{2}} = \frac{PN}{50}$$
=> $$PN = \frac{50}{\sqrt{2}} = 25 \sqrt{2}$$
=> $$BM = PN = 25 \sqrt{2}$$
Again, $$tan 45 = \frac{PN}{OP}$$
=> $$OP = 25 \sqrt{2}$$
=> $$AM = 50 + 25 \sqrt{2} + 50 = 100 + 25 \sqrt{2}$$
Using eqn(I), we get :
=> $$d^2 = (100 + 25 \sqrt{2})^2 + (25 \sqrt{2})^2$$
=> $$d^2 = 10000 + 5000 \sqrt{2} + 1250 + 1250$$ = $$12500 + 5000 \sqrt{2}$$
=> $$d^2 = 2500 (5 + 2 \sqrt{2}) = 2500 (5 + \sqrt{8})$$ -----------Eqn(II)
Also, it is given that : $$d = a\sqrt{b+\sqrt{c}}$$
=> $$d^2 = a^2 (b + \sqrt{c})$$ -----------Eqn(III)
Comparing, eqn(II) & (III), we get :
=> $$a^2 (b + \sqrt{c}) = 2500 (5 + \sqrt{8})$$
=> $$a = 50 , b = 5 , c = 8$$
$$\therefore a + b + c = 50 + 5 + 8 = 63$$
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IB Math AA SL Paper 2 Question Bank
If you’re looking for an IB Math AA SL Paper 2 Question Bank, you’ve come to the right place! We have a wide selection of questions available, covering all of the topics you’ll need to know for your exam. Whether you’re needing help with geometry or statistics, we’ve got you covered. With our question bank, you’ll be able to work through practice problems at your own pace and master the material before test day. So why wait? Get started today and ace that paper 2!
Section A
1.) A rectangular prism has a volume of 48 cubic meters and a square base with a side length of 2 meters. The height of the prism is h meters.
a) Express the total surface area of the rectangular prism as a function of h.
We know that the volume of a rectangular prism is given by V = lwh, where l, w and h are the length, width and height of the prism respectively. In this case, we know that V = 48 m3 and the side length of the base is 2 m.
Therefore, the width and length of the base are both 2 m. So, we can express the volume as
V = 2w * 2l * h = 48 m3
We know the formula for the surface area of a rectangular prism is 2lw+ 2lh + 2wh. Here we know width w = l = 2m and h is the variable.
So, the surface area A of the rectangular prism as a function of h is:
A = 2 * 2 * 2 + 2 * 2 * h = 8 + 4h
b) Express the height h of the rectangular prism as a function of the total surface area.
To express the height h as a function of the total surface area, we can use the surface area equation that we found in part a. If we let A be the total surface area, then:
A = 8 + 4h
Rearranging this equation to solve for h gives us:
h = (A-8)/4
So, h is a function of the total surface area.
The surface area as a function of h is A = 8+ 4h, and the h as a function of the surface area is h = (A-8)/4
2.) An amount of \$ 10 000 is invested at an annual interest rate of 12%.
a) Find the value of the investment after 5 years
i) if the interest rate is compounded yearly;
FV=10000(1+12/100)n
n= 5, 17623.42
ii) if the interest rate is compounded half-yearly;
n= 10, 17908.48
iii) if the interest rate is compounded quarterly;
n=20, 18061.11
iv) if the interest rate is compounded monthly.
n= 5*12, 18166.97
b) The value of the investment will exceed \$ 20 000 after n full years. Calculate the minimum value of n
i) if the interest rate is compounded yearly;
20000= 10000(1+12/100)n
Solve by GDC
n=6.11, hence n=7
ii) if the interest rate is compounded monthly.
20000= 10000(1+12/100)12n
Solve by GDC
n=5.81, hence n=6
3.) A car rental company charges a fixed daily rental fee of \$50, and a variable charge of \$0.15 per kilometre driven.
a) Write an expression for the cost of renting the car for d days, and traveling x kilometers.
The cost of renting the car is made up of two parts: a fixed daily rental fee and a variable charge per kilometer driven. To find the total cost, we need to multiply the number of days the car is rented by the daily rental fee, and add that to the product of the number of kilometers driven and the variable charge per kilometer.
So, the total cost, C, can be expressed as an equation:
C = d * 50 + x * 0.15
b) The company offers a weekly rental package for \$400, which includes a maximum of 600km of travel. Express the cost of the weekly rental package, C, in terms of the number of kilometres driven, x.
To find the cost of the weekly rental package in terms of the number of kilometers driven, we know that the package includes 600km of travel and cost \$400. So we can create an equation:
C = 400 = d * 50 + x * 0.15
we know that d = 7 for a weekly package and x is 600km
So,
C(x) = 7 * 50 + x * 0.15 = 400
Hence C(x) = 7*50 + 0.15x = 400
this is the equation that expresses the cost of the weekly rental package, C, in terms of the number of kilometres driven, x.
4.) Triangle ABC is a right-angled triangle, with right angle at B and hypotenuse C.
a) Write down the value of sin B, cos B and tan B in terms of a and b.
In a right-angled triangle, we can use the trigonometric functions sine, cosine, and tangent to relate the lengths of the sides to the measure of the angles.
For a right-angled triangle,
sin B = a/c, cos B = b/c, tan B = a/b
b) If a = 5 and b = 12, find the values of sin B, cos B and tan B.
If we plug in the given values, a = 5 and b = 12, we can find the values of sin B, cos B and tan B:
sin B = a/c = 5/c
cos B = b/c = 12/c
tan B = a/b = 5/12
c) Use the result from part (b) to find the value of c.
To find the value of c, we can use the Pythagorean theorem, c = √(a2 + b2) = √(52 + 122) = √(25 + 144) = √(169) = 13
5.) Solve the following
a) Use the sine rule to find the sine of the angle A
Sin A = (10sin30°)/6
Sin A= 5/6
b) Hence find the two possible values of angle A
A= 56.4° or A = 124°
6.) The weight X of a particular animal is normally distributed with μ= 200kg and σ= 15kg. An animal of this population is overweight if it has a weight greater than 230 kg
a) Find the probability that an animal is overweight.
P (X>230) = 0.02275
b) We select 2 animals from this population. Find the probabilities that
i) both animals are overweight
(0.02275)2= 0.000518
ii) only one animal is overweight.
0.02275)* (1-0.02275)* 2 = 0.0445
c) We select 7 animals from this population. Find the probability that exactly two of them are overweight.
Y follows B(n,p) with n = 7, p =0.02275
P(Y=2) = 0.00969
Section B
1.) A factory produces electronic components. The probability of a component is defective is 0.05.
a) What is the probability of producing exactly 3 defective components in a batch of 50 components?
This is a binomial probability distribution problem. The probability of producing exactly 3 defective components in a batch of 50 components is:
P(X = 3) = (50 choose 3) * (0.05)3 * (1-0.05)(50-3) = 19600 * 0.053 * 0.9547 ≈ 0.09
b)What is the probability of producing at most 3 defective components in a batch of 50 components?
To find the probability of producing at most 3 defective components, we can use the cumulative distribution function of the binomial distribution. The probability of producing at most 3 defective components is:
P(X <= 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
P(X <= 3) = (50 choose 0) * (0.05)0 * (1-0.05)50 + (50 choose 1) * (0.05)1 * (1-0.05)49 + (50 choose 2) * (0.05)2 * (1-0.05)48 + (50 choose 3) * (0.05)3 * (1-0.05)47 ≈ 0.37
The factory produces 1000 components per day.
c) What is the probability of producing more than 45 defective components in a day?
The expected number of defective components produced per day is:
E(X) = λ = n * p = 1000 * 0.05 = 50
The probability of producing more than 45 defective components in a day is:
1 – P(X <= 45) = 1 – (e(-50) * (5045) / 45!) ≈ 0.02
d) What is the probability of producing less than 45 defective components in a day?
To find the probability of producing less than 45 defective components in a day, we can use the Poisson cumulative distribution function. The probability of producing less than 45 defective components in a day is:
P(X < 45) = (e(-50) * (5044) / 44!) ≈ 0.98
2.) A company produces and sells widgets. The revenue, R (in dollars), from selling x widgets per day is given by the function R(x) = 2x2 + 50x. The cost, C (in dollars), of producing x widgets per day is given by the function C(x) = 0.5x2 + 25x + 100.
a.) Find the profit function, P(x), in terms of x.
The profit, P(x), is calculated by subtracting the cost, C(x), from the revenue, R(x):
P(x) = R(x) – C(x)
Substituting the given functions for R(x) and C(x):
P(x) = (2x2 + 50x) – (0.5x2 + 25x + 100)
P(x) = 1.5x2 + 25x – 100
b.) Find the value of x that maximizes the profit.
To find the value of x that maximizes the profit, we need to find the vertex of the quadratic function P(x) = 1.5x2 + 25x – 100. The x-coordinate of the vertex is given by the formula x = -b/2a, where a and b are the coefficients of the x2 and x terms, respectively.
In this case, a = 1.5 and b = 25, so substituting these values into the formula:
x = -25 / (2 * 1.5)
x = -25 / 3
So, the value of x that maximizes the profit is x = -25/3.
c.) Calculate the maximum profit.
To calculate the maximum profit, we can substitute the value of x = -25/3 into the profit function P(x) = 1.5x2 + 25x – 100.
P(-25/3) = 1.5 * (-25/3)2 + 25 * (-25/3) – 100
P(-25/3) = -975/3 + (-625/3) – 100
P(-25/3) = -1600/3 – 100
P(-25/3) = -1100/3
So, the maximum profit is -1100/3 dollars.
d.) Determine the range of values of x for which the company makes a profit.
The company makes a profit when the profit function P(x) is positive. To determine the range of values of x for which the company makes a profit, we need to find the values of x that satisfy the inequality P(x) > 0.
1.5x2 + 25x – 100 > 0
We can solve this inequality by using quadratic techniques, such as factoring or the quadratic formula. Once we find the solutions for x, we can determine the range of values of x for which the inequality is true, and thus the range of values of x for which the company makes a profit.
To determine the range of values of x for which the company makes a profit, we need to solve the inequality 1.5x2 + 25x – 100 > 0. Let’s use the quadratic formula to find the solutions for x:
1.5x2 + 25x – 100 > 0
a = 1.5, b = 25, c = -100
x = (-b ± √(b2 – 4ac)) / (2a)
x = (-25 ± √(252 – 4 * 1.5 * -100)) / (2 * 1.5)
x = (-25 ± √(625 + 1200)) / 3
x = (-25 ± √1825) / 3
Now we can determine the range of values of x for which the inequality is true. Since the coefficient of x2 is positive (1.5 > 0), the parabola opens upwards, and the inequality is true for values of x outside of the solutions we found.
So, the range of values of x for which the company makes a profit is x < (-25 – √1825) / 3 or x > (-25 + √1825) / 3.
3.) Let f(x) = x3 – 3x2 – 4x + 12. Solve the equation f(x) = 0.
a.) By first testing the values of f(1) and f(2), show that there is a root of f(x) = 0 between x = 1 and x = 2.
We have f(1) = 6 and f(2) = 2, so by the intermediate value theorem, there must be a root of f(x) = 0 between x = 1 and x = 2.
b.) Use synthetic division to factorize f(x) into a linear factor and a quadratic factor.
To factorize f(x) using synthetic division, we first guess a root, which could be a factor of the constant term (12) divided by a factor of the leading coefficient (1). One possibility is x = 3. Then we divide f(x) by (x – 3):
1 -3 -4 12
3 | 1 -3 -4 12
-3 0 -12
————
1 -6 -4 0
So we have f(x) = (x – 3)(x2 – 6x – 4).
c.) Find the remaining roots of f(x) = 0, giving your answers correct to three decimal places.
To find the remaining roots of f(x) = 0, we need to solve the quadratic factor x2 – 6x – 4. Using the quadratic formula, we get:
x = [6 ± sqrt(62 + 4*4*1)] / 2
x = 3 ± sqrt(19)
Therefore, the roots of f(x) = 0 are approximately x = 3, x = 3 + sqrt(19), and x = 3 – sqrt(19).
d.) Sketch the graph of y = f(x), indicating the x-intercepts, y-intercept, turning points, and end behavior.
To sketch the graph of y = f(x), we first find the x-intercepts by solving f(x) = 0:
f(x) = x3 – 3x2 – 4x + 12
f(x) = (x – 3)(x2 – 4x – 4)
x = 3 or x = 2 + sqrt(6) or x = 2 – sqrt(6)\
Next, we find the y-intercept by evaluating f(0) = -1. We also find the turning points by setting f'(x) = 0 and solving for x:
f'(x) = 3x2 – 6x – 4
f'(x) = 3(x – 2)(x + 2)
x = 2 or x = -2
Finally, we can sketch the graph of y = f(x) as follows:
e.) Find the equation of the tangent line to the graph of y = f(x) at the point where x = 2.
To find the equation of the tangent line to the graph of y = f(x) at the point where x = 2, we first find the slope of the tangent line by taking the derivative of f(x) and evaluating it at x = 2:
f'(x) = 3x2 – 6x – 4
f'(2) = 8
So the slope of the tangent line is 8. Next, we find the y-coordinate of the point of tangency by evaluating f(2) = 2:
f(2) = 2
Therefore, the equation of the tangent line to the graph of y = f(x) at the point where x = 2 is:
y – 2 = 8(x – 2)
y = 8x – 14
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# How do you find the turning point of a derivative?
## How do you find the turning point of a derivative?
To find the location of turning points on a function, find the first derivative of the function, and then set the result to 0. if you then solve this equation, you will find the locations of the turning points.
## Are all quadratic equations functions?
Quadratics have at most two solutions for every output (dependent variable), but each input (independent variable) only gives one value. The function f(x)=ax2+bx+c is a quadratic function. Now, if you try to solve a quadratic equation, you get often two solutions, but this is not the same as calculating the function.
What are solutions to quadratic equations called?
. The solutions to quadratic equations are known as roots or zeroes of the equation. We can solve quadratic equations using three methods such as factoring, the quadratic formula, and completing the square. Let us discuss each of the methods one by one with examples.
### How do you find the turning point of a quadratic function?
The easiest way to find the turning point is when the quadratic is in turning point form (y = a(x – h)2 + k), where (h, k) is the turning point. To get a quadratic into turning point form you need to complete the square.
### How are quadratic equations used in real life?
Quadratic equations are actually used in everyday life, as when calculating areas, determining a product’s profit or formulating the speed of an object. Quadratic equations refer to equations with at least one squared variable, with the most standard form being ax² + bx + c = 0.
What is the difference between quadratic function and linear function?
What is the difference between linear and quadratic functions? A linear function is one of the form y = mx + c. A quadratic function is one of the form y = ax2 + bx + c. For each output for y, there can be up to two associated input values of x.
Summary
1. Quadratic Equation in Standard Form: ax2 + bx + c = 0.
2. Quadratic Equations can be factored.
3. Quadratic Formula: x = −b ± √(b2 − 4ac) 2a.
4. When the Discriminant (b2−4ac) is: positive, there are 2 real solutions. zero, there is one real solution. negative, there are 2 complex solutions.
#### What is the turning point of a quadratic graph?
Graphs of quadratic functions have a vertical line of symmetry that goes through their turning point. This means that the turning point is located exactly half way between the x-axis intercepts (if there are any!). There are two methods to find the turning point, Through factorising and completing the square.
How do you write quadratic equations in standard form give at least 2 examples?
Standard Form Equations
1. 6x² + 11x – 35 = 0.
2. 2x² – 4x – 2 = 0.
3. -4x² – 7x +12 = 0.
4. 20x² -15x – 10 = 0.
5. x² -x – 3 = 0.
6. 5x² – 2x – 9 = 0.
7. 3x² + 4x + 2 = 0.
8. -x² +6x + 18 = 0.
## Why do quadratic equations equal zero?
Because if we have two numbers multiplied together to equal 0, so AB=0, then either the first is equal to 0, or the second (or both). But when two numbers are multiplied together to equal something else, say AB=1, then we aren’t getting any information from A and B because they relate to each other in some way.
## How do you illustrate quadratic equations in standard?
Therefore, the standard form of a quadratic equation can be written as: ax2 + bx + c = 0 ; where x is an unknown variable, and a, b, c are constants with ‘a’ ≠ 0 (if a = 0, then it becomes a linear equation). Let us look at some examples of a quadratic equation: 2×2+5x+3=0; In this, a=2, b=3 and c=5.
What is a major turning point?
a time of important change in a situation big, crucial, decisive, historic, important, key, major, real, significantWinning the 2015 championship was a major turning point in his career.
### What are the characteristic of a quadratic equation?
Three properties that are universal to all quadratic functions: 1) The graph of a quadratic function is always a parabola that either opens upward or downward (end behavior); 2) The domain of a quadratic function is all real numbers; and 3) The vertex is the lowest point when the parabola opens upwards; while the …
### What is turning point of a function?
A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). A polynomial of degree n will have at most n – 1 turning points.
How are quadratic equations different from other kinds of equations?
Answer: A linear equation in two variables doesn’t involve any power higher than one for either variable. It has the general form Ax + By + C = 0, where A, B and C are constants. A quadratic equation, on the other hand, involves one of the variables raised to the second power.
#### What is the shape of a quadratic function?
The graph of a quadratic function is called a parabola and has a curved shape. One of the main points of a parabola is its vertex.
#### Is the turning point a maximum or minimum?
The location of a stationary point on f(x) can be identified by solving f'(x) = 0. To work out which is the minimum and maximum, differentiate again to find f”(x). Input the x value for each turning point. If f”(x) > 0 the point is a minimum, and if f”(x) < 0, it is a maximum.
What does turning point in life mean?
The definition of a turning point is a point in time when something happens that causes a shift or an irrevocable change in direction. An example of a turning point in someone’s life is the day a woman finds out she is pregnant.
## How do you determine if a graph is a quadratic function?
Key Points
1. The graph of a quadratic function is a parabola whose axis of symmetry is parallel to the y -axis.
2. The coefficients a,b, and c in the equation y=ax2+bx+c y = a x 2 + b x + c control various facets of what the parabola looks like when graphed.
## What is a turning point?
: a point at which a significant change occurs.
What is the turning point of a quadratic called?
The vertex is the turning point of the graph. We can see that the vertex is at (3,1) ( 3 , 1 ) . The axis of symmetry is the vertical line that intersects the parabola at the vertex.
### Is a quadratic a function?
A quadratic function is a function of degree two. The graph of a quadratic function is a parabola. The general form of a quadratic function is f(x)=ax2+bx+c where a, b, and c are real numbers and a≠0.
### Does a cubic function have a turning point?
Cubic functions can have at most 3 real roots (including multiplicities) and 2 turning points. If a root of a polynomial has even multiplicity, the graph will touch the x-axis at the root but will not cross the x-axis.
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# Law of cosines
## Branko Ćurgus
Theorem. If $ABC$ is a triangle with the angle $\gamma$ at the vertex $C$ and with the sides lengths $a = \overline{BC}, b= \overline{AC}$ and $c= \overline{AB}$ (see the figure below), then $c^2 = a^2 + b^2 - 2ab \cos \gamma.$
Proof. In the proof below we use the Pythagorean theorem twice and the definition of $\cos \gamma = x/a$;
$\displaystyle c^2$ $\displaystyle = y^2 + h^2$ (by Pythagorean theorem $c^2 = y^2+h^2$) $\displaystyle = (b-x)^2 + h^2$ (by the identity $b = x + y$) $\displaystyle = b^2 -2bx + x^2 + h^2 \$ (algebra) $\displaystyle = a^2 + b^2 - 2bx$ (by Pythagorean theorem $a^2 = h^2 + x^2$) $\displaystyle = a^2 + b^2 - 2ab\dfrac{x}{a}$ (algebra) $\displaystyle = a^2 + b^2 - 2ab\cos \gamma$ (the definition of $\cos \gamma$)
This completes the proof.
Remark. Why is the law of cosines important in a linear algebra class? Here is why. Introduce two vectors in the above figure: $\vec{a} = \vec{CB}, \qquad \vec{b} = \vec{CA}.$ Then, $a^2 = \bigl\|\vec{a}\bigr\|^2$, $b^2 = \bigl\|\vec{b}\bigr\|^2$ and $\vec{b} - \vec{a} = \vec{BA}.$ The last equality implies \begin{align*} c^2 & = \| \vec{b} - \vec{a} \|^2 \\ & = \bigl(\vec{b} - \vec{a} \bigr) \cdot \bigl(\vec{b} - \vec{a} \bigr) \\ & = \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{b} + \vec{a} \vec{a} \\ & = \bigl\|\vec{b}\bigr\|^2 - 2 \vec{a} \cdot \vec{b} + \bigl\|\vec{a}\bigr\|^2 \\ & = a^2 + b^2 - 2 \vec{a} \cdot \vec{b}. \end{align*} Now compare the last equality and the law of cosines: \begin{align*} c^2 & = a^2 + b^2 - 2 \vec{a} \cdot \vec{b}, \\ c^2 &= a^2 + b^2 - 2ab \cos \gamma. \end{align*} We deduce that $\vec{a} \cdot \vec{b} = ab \cos \gamma = \bigl\|\vec{a}\bigr\| \, \bigl\|\vec{b}\bigr\| \cos \gamma.$ This gives us a geometric interpretation of the dot product.
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Question 178741
I think people shy away from percentiles because the math world cant get their act together on one definition:...there are really 3 ways of looking at percentiles .:
:
There is no universally accepted definition of a percentile. Using the 65th percentile as an example, the 65th percentile can be defined as the lowest score that is greater than 65% of the scores. Lets call this "Definition 1". The 65th percentile can also be defined as the smallest score that is greater than or equal to 65% of the scores. This we will call "Definition 2". Unfortunately, these two definitions can lead to dramatically different results, especially when there is relatively little data. Moreover, neither of these definitions is explicit about how to handle rounding. For instance, what score is required to be higher than 65% of the scores when the total number of scores is 50? This is tricky because 65% of 50 is 32.5. How do we find the lowest number that is less than 32.5% of the scores? A third way to compute percentiles is a weighted average of the percentiles computed according to the first two definitions. This third definition handles rounding more gracefully than the other two and has the advantage that it allows the median to be defined conveniently as the 50th percentile.
:
I will use the 3rd method but give answers to the other 2
:
first line this up in ascending order:
:
20,22,22,19,22,50,24,15,34,43,22,20,17,38,18,21,21,23,18,23
:
:
15,17,18,18,19,20,20,21,21,22,22,22,22,23,23,24,34,38,43,50
:
The first step is to compute the rank (R) of the 20th and 75th percentiles. This is done using the following formula: R=P/100(N+1) where P is the desired percentile and N is the number of terms.
In our cases P=20 and 75 and N=20
:
so in our case for the 20 percentile: R=20/100(20+1)=4.2
.......................75 percentile: R=75/100(20+1)=15.75
If R were an integer, the P the percentile would be the number with rank R. When R is not an integer, we compute the Pth percentile by interpolation as follows:
Define IR as the integer portion of R (the number to the left of the decimal point). For our problems IR = 4 and 15
Define FR as the fractional portion of R. In our problems FR= .2 and .75
Find the scores with Rank IR and with Rank IR+1. For our problems this would be
the 4th and 5th terms for 20th percentile and 15 and 16th for the 75th percentile. For P=20 this would be 18 and 19. for P=75 it would be 23 and 24
Interpolate by multiplying the difference between the scores by FR and add the result to the lower score. So for P=20 you would take .2(19-18)+18={{{highlight(18.2)}}}. For P=75 it would be .75(24-23)+23={{{highlight(23.75)}}}
:
15,17,18,18,19,20,20,21,21,22,22,22,22,23,23,24,34,38,43,50
:
Therefore, the 20th percentile is 18.2 and the 75th percentile is 23.75. If we had used the first definition (the smallest score greater than 20% and 75% of the scores) the 20th percentile would have been 19(5th term) and for the 75 percentile would have been 24(16th term). If we had used the second definition ( the smallest score greater than or equal to 25% of the scores) the 20th percentile would have been 18(4th term) and the 75 percentile would have been 23(15th term).
:
As you can see these definitions need to be standardized so everyone is on the same page.
:
hope that helps
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# How do you find an equation of the parabola with vertex (-2,1) and directrix x=1?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
6
Sep 22, 2017
$x = - {y}^{2} / 6 + \frac{y}{3} - \frac{13}{6}$
#### Explanation:
Given -
Vertex $\left(- 2 , 1\right)$
Directrix $x = 1$
The vertex is in the 2nd quadrant. The directrix is parallel to the y-axis. So, the parabola opens to the left. The vertex of the parabola is not the origin. Then its general form is -
${\left(y - k\right)}^{2} = - 4. a . \left(x - h\right)$
Where -
$h$ and $k$ are the coordinates of the vertex.
h=-2)
$k = 1$
$a = 1.5$ half the distance between Directrix and vertex [= distance between focus and vertex]
Substitute these values in the equation
${\left(y - 1\right)}^{2} = - 4.1 .5 . \left(x + 2\right)$
${y}^{2} - 2 y + 1 = - 6 x - 12$
$- 6 x - 12 = {y}^{2} - 2 y + 1$
$- 6 x = {y}^{2} - 2 y + 1 + 12$
$x = {y}^{2} / \left(- 6\right) - \frac{2 y}{- 6} + \frac{13}{- 6}$
$x = - {y}^{2} / 6 + \frac{y}{3} - \frac{13}{6}$
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# Video: Solving Word Problems Involving Division
Roger wants to buy a smartphone that costs \$304. If he saves \$38 each week, determine in how many weeks he will be able to buy the smartphone.
02:31
### Video Transcript
Roger wants to buy a smartphone that cost three hundred and four dollars. If he saves thirty-eight dollars each week, determine in how many weeks he will be able to buy his smartphone. Our question is asking us how many weeks it will take for him to save enough money to buy a smartphone.
We know that Roger saves thirty-eight dollars each week. We also know that the phone costs three hundred and four dollars. We start by saying that Rogers saves thirty-eight dollars per week. That per week means we can say thirty-eight times 𝑤 for weeks. So if we multiply thirty-eight times the number of weeks he’s saving, we will find out how much money Roger has saved so far. And we want the amount that he has saved to equal the cost of the phone, three hundred and four dollars. We want thirty-eight dollars times the number of weeks he saved to equal three hundred and four dollars.
But now what? Now when we have this equation, what can we do? We could try to add thirty-eight plus thirty-eight plus thirty-eight plus thirty-eight over and over again until we got to three hundred and four. But that’s not a very effective way to answer this question. We’re gonna answer the question by dividing both sides of our equation by thirty-eight.
So let’s note that if we divide thirty-eight 𝑤 by thirty-eight, the solution would be one 𝑤 or simply 𝑤. On our left-side of the equation, we’re left with only 𝑤. When we divide three hundred and four by thirty-eight, the solution is eight.
The number of weeks it would take Roger to save up three hundred and four dollars is eight, eight weeks. If you wanted to check and make sure we did everything correctly, you could plug that eight back into our equation, thirty-eight times eight, and ask does that equal three hundred and four. And it does. Roger needs to save for eight weeks.
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# Find the chance of throwing head and tail alternatively in three successive tossings of the coin.
Last updated date: 20th Jun 2024
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Hint: First of all, consider an experiment of throwing a coin thrice and then find out the total number of possible outcomes and the total number of favorable outcomes. Then find the probability of that considered experiment to get the required answer. So, use this concept to reach the solution to the given problem.
Complete step-by-step solution:
Consider an experiment of throwing a coin thrice.
The sample space of the experiment can be written as $S = \left\{ {HHH,HHT,HTH,HTT,THH,TTH,THT,TTT} \right\}$
Hence, the total number of outcomes = 8
Let $E$: event of getting head and tail alternatively in three successive trails.
So, the only possible outcomes are $\left\{ {HTH,THT} \right\}$.
Hence, the number of possible outcomes = 2
We know that the probability of an event $E$ is given by $P\left( E \right) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$
Therefore, $P\left( E \right) = \dfrac{2}{8} = \dfrac{1}{4}$
Thus, there is a chance of throwing head and tail alternatively in three successive tossings of the coin is $\dfrac{1}{4}$.
Note: In probability theory, an experiment or trail is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as sample space. The probability of an event is always lying between 0 and 1 i.e., $0 \leqslant P\left( E \right) \leqslant 1$. The probability of an event $E$ is given by $P\left( E \right) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$.
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# Trigonometry
## Trigonometric equations and identities
### Part 1: Pythagorean identities
Recall that, in the section on the unit circle, we established that given any angle $\theta$, $\left(\cos\left(\theta\right),\sin\left(\theta\right)\right)$ are the coordinates of a point on the unit circle. Using the distance formula to compute the distance between this point and the origin, $\left(0,0\right)$, we get a trigonometric identity: $$\cos^2\left(\theta\right) + \sin^2\left(\theta\right) = 1$$
Dividing both sides by $\cos^2\left(\theta\right)$, we get another identity: $$1+\tan^2\left(\theta\right) = \sec^2\left(\theta\right)$$
Dividing both sides of the original identity by $\sin^2\left(\theta\right)$, we get the identity: $$\cot^2\left(\theta\right)+1=\csc^2\left(\theta\right)$$
### Example. Prove the following identity: $\frac{\sin\left(x\right)}{1+\cos\left(x\right)}=\frac{1-\cos\left(x\right)}{\sin\left(x\right)}$
\begin{align*} \frac{\sin\left(x\right)}{1+\cos\left(x\right)}\times \frac{1-\cos\left(x\right)}{1-\cos\left(x\right)} &= \frac{\sin\left(x\right)\left(1-\cos\left(x\right)\right)}{1-\cos^2\left(x\right)} \\ &= \frac{\sin\left(x\right)\left(1-\cos\left(x\right)\right)}{\sin^2\left(x\right)} \\ &= \frac{1-\cos\left(x\right)}{\sin\left(x\right)} \end{align*}
### Part 2: Even-odd identities
We also have the following identities: $$\sin\left(-\theta\right)=-\sin\left(\theta\right) \hspace{10 mm} \cos\left(-\theta\right)=\cos\left(\theta\right) \hspace{10 mm} \tan\left(-\theta\right)=-\tan\left(\theta\right)$$ They tell us that $\sin$ and $\tan$ are oddAn odd function $f$ is any function which satisfies $f(-x) = -f(x)$ for all $x$ in its domain. functions while $\cos$ is an evenAn even function $f$ is any function which satisfies $f(-x) = f(x)$ for all $x$ in its domain. function.
### Example. Compute $\sin\left(-\frac{\pi}{2}\right)$
$$\sin\left(-\frac{\pi}{2}\right)=-\sin\left(\frac{\pi}{2}\right)=-1$$
We can reduce $\sin$ and $\cos$ of the sum of two angles to an expression involving $\sin$ and $\cos$ of each of the angles: \begin{align*} \sin\left(A+B\right) &= \sin\left(A\right)\cos\left(B\right) + \cos\left(A\right)\sin\left(B\right) \\ \cos\left(A+B\right) &= \cos\left(A\right)\cos\left(B\right) - \sin\left(A\right)\sin\left(B\right) \end{align*}
### Example. Evaluate $\sin\left(75^{\circ}\right)$.
\begin{align*} \sin\left(75^{\circ}\right) &= \sin\left(45^{\circ}+30^{\circ}\right) \\ &= \sin\left(45^{\circ}\right)\cos\left(30^{\circ}\right) + \cos\left(45^{\circ}\right)\sin\left(30^{\circ}\right) \\ &= \frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{\sqrt{2}}\left(\frac{1}{2}\right) \\ &= \frac{\sqrt{3}+1}{2\sqrt{2}} \end{align*}
### Part 4: Trigonometric equations
The techniques for solving trigonometric equations involve the same strategies as solving polynomial equations (see the section on Polynomials and Factoring) as well as using trigonometric identities.
### Example. Find the solutions of the equation
$$\left(\sin\left(\theta\right) - 4\right) \left(\sin^2\left(\theta\right)+5\sin\left(\theta\right)+4\right) = 0 \qquad \textrm{ for } \theta \in \left[0,2\pi \right].$$
We need to find when the first or the second factor is zero. Since $\sin\left(\theta\right) - 4=0$ means $\sin\left(\theta\right)=4$, which has no solution (the left side is at most one), we only need to work with the second factor: $$\sin^2\left(\theta\right)+5\sin\left(\theta\right)+4 = 0$$ Notice that this is a quadratic equation, and after factoring we get: $$(\sin\left(\theta\right)+1)(\sin\left(\theta\right)+4) = 0$$ Again, we need to check when either factor is zero. For the second factor, we get $\sin\left(\theta\right)+4 = 0$, i.e. $\sin\left(\theta\right)=-4$, which has no solution (since the left side is at least $-1$). Solving for the first factor, we get $\sin\left(\theta\right)+1=0$, i.e. $\sin\left(\theta\right)=-1$. The only $\theta \in \left[0,2\pi \right]$ for which this is true is $\theta= \frac{3\pi}{2}$.
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# 1982 AHSME Problems/Problem 28
## Problem
A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\frac{7}{17}$. What number was erased?
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{cannot be determined}$
## Solution
Suppose that there are $n$ positive integers in the set initially, so their sum is $\frac{n(n+1)}{2}$ by arithmetic series. The average of the remaining numbers is minimized when $n$ is erased, and is maximized when $1$ is erased.
It is clear that $n>1,$ from which we write and solve a compound inequality for $n:$ \begin{alignat*}{8} \frac{\frac{n(n+1)}{2}-n}{n-1} &\leq 35\frac{7}{17} &&\leq \frac{\frac{n(n+1)}{2}-1}{n-1} \\ \frac{n(n+1)-2n}{2(n-1)} &\leq 35\frac{7}{17} &&\leq \frac{n(n+1)-2}{2(n-1)} \\ \frac{n^2-n}{2(n-1)} &\leq 35\frac{7}{17} &&\leq \frac{n^2+n-2}{2(n-1)} \\ \frac{n(n-1)}{2(n-1)} &\leq 35\frac{7}{17} &&\leq \frac{(n+2)(n-1)}{2(n-1)} \\ \frac{n}{2} &\leq 35\frac{7}{17} &&\leq \frac{n+2}{2} \\ n &\leq 70\frac{14}{17} &&\leq n+2 \\ 68\frac{14}{17} &\leq \hspace{3mm} n &&\leq 70\frac{14}{17}. \end{alignat*}
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More types of sets: Disjoint sets, Subset and Universal set
Disjoint sets:
Two or more sets are said to be disjoint if they have no common elements.
Disjoint-sets are also known as non overlapping sets.
For example:
A={2,4,6,8} and B={1,3,5,7}. Here A and B have no common element. So A and B are disjoint-sets.
Overlapping-sets:
Two-sets are said to be overlapping-sets if they contain at least one element in common.
For example:
A={2,4,6,8}
B={3,6,9}here A and B have one common element that is 6. So A and B are overlapping sets.
Universal set:
A universal set is a set of all possible elements under given definition. It is usually denoted by U or ξ .
Subset:
If A and B are two given sets, and if all the elements of A are also elements of B, then A is a subset of B.
• If all elements of set B are also element of A then B is a subset of A.
• In other words, A is a subset of B if and only if every element of A is in B
• The symbol to denote subset is .
• For example, If A is a subset of B, then it is denoted by A⊆B.
Type of subset:
Proper subset:
A is a proper subset of B if and only if every element of A must also be the element of the set B, and also there exists at least one element in B which is not an element of A
Note:
• The symbol to denote proper subset is .
• Every set is a subset of itself. A⊂ A.
• Empty set or null set or ∅ is the subset of every set
• When we say that A is a subset of B we denote by A⊆B.
• When we say that A is a proper subset of B then we denote it as A⊂B.
Super set:
Whenever A is a subset of B, then B is the super set of A. It is expressed as B⊇A.
Examples:
U={x:x is a natural number; 1‹x‹10}
A={x:x is a even natural number; x‹10}
B={x:x∈N;x=3N;x≤9}
C={2,4,6}
Math dictionary
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# Search by Topic
#### Resources tagged with Area similar to T-table:
Filter by: Content type:
Stage:
Challenge level:
### There are 93 results
Broad Topics > Measures and Mensuration > Area
### Carpet Cuts
##### Stage: 3 Challenge Level:
You have a 12 by 9 foot carpet with an 8 by 1 foot hole exactly in the middle. Cut the carpet into two pieces to make a 10 by 10 foot square carpet.
### Exploration Versus Calculation
##### Stage: 1, 2 and 3
This article, written for teachers, discusses the merits of different kinds of resources: those which involve exploration and those which centre on calculation.
### Changing Areas, Changing Perimeters
##### Stage: 3 Challenge Level:
How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same?
### Making Rectangles
##### Stage: 2 and 3 Challenge Level:
A task which depends on members of the group noticing the needs of others and responding.
### Square Areas
##### Stage: 3 Challenge Level:
Can you work out the area of the inner square and give an explanation of how you did it?
### Kissing Triangles
##### Stage: 3 Challenge Level:
Determine the total shaded area of the 'kissing triangles'.
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### The Pi Are Square
##### Stage: 3 Challenge Level:
A circle with the radius of 2.2 centimetres is drawn touching the sides of a square. What area of the square is NOT covered by the circle?
### Cylinder Cutting
##### Stage: 2 and 3 Challenge Level:
An activity for high-attaining learners which involves making a new cylinder from a cardboard tube.
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Growing Rectangles
##### Stage: 3 Challenge Level:
What happens to the area and volume of 2D and 3D shapes when you enlarge them?
### Bull's Eye
##### Stage: 3 Challenge Level:
What fractions of the largest circle are the two shaded regions?
### Pie Cuts
##### Stage: 3 Challenge Level:
Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters).
### Kite
##### Stage: 3 Challenge Level:
Derive a formula for finding the area of any kite.
### Hallway Borders
##### Stage: 3 Challenge Level:
A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.
### Appearing Square
##### Stage: 3 Challenge Level:
Make an eight by eight square, the layout is the same as a chessboard. You can print out and use the square below. What is the area of the square? Divide the square in the way shown by the red dashed. . . .
### Lying and Cheating
##### Stage: 3 Challenge Level:
Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!
### Great Squares
##### Stage: 2 and 3 Challenge Level:
Investigate how this pattern of squares continues. You could measure lengths, areas and angles.
### Isosceles
##### Stage: 3 Challenge Level:
Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.
### Blue and White
##### Stage: 3 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest?
### Can They Be Equal?
##### Stage: 3 Challenge Level:
Can you find rectangles where the value of the area is the same as the value of the perimeter?
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### Extending Great Squares
##### Stage: 2 and 3 Challenge Level:
Explore one of these five pictures.
### Tiling Into Slanted Rectangles
##### Stage: 2 and 3 Challenge Level:
A follow-up activity to Tiles in the Garden.
### Perimeter Possibilities
##### Stage: 3 Challenge Level:
I'm thinking of a rectangle with an area of 24. What could its perimeter be?
### Pick's Theorem
##### Stage: 3 Challenge Level:
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.
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Squaring the Circle
##### Stage: 3 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### Overlapping Squares
##### Stage: 2 Challenge Level:
Have a good look at these images. Can you describe what is happening? There are plenty more images like this on NRICH's Exploring Squares CD.
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### Rati-o
##### Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
### Triangle Island
##### Stage: 2 Challenge Level:
You have pitched your tent (the red triangle) on an island. Can you move it to the position shown by the purple triangle making sure you obey the rules?
### Dissect
##### Stage: 3 Challenge Level:
It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into?
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Circle Panes
##### Stage: 2 Challenge Level:
Look at the mathematics that is all around us - this circular window is a wonderful example.
### A Day with Grandpa
##### Stage: 2 Challenge Level:
Grandpa was measuring a rug using yards, feet and inches. Can you help William to work out its area?
### Poly-puzzle
##### Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.
### Towers
##### Stage: 3 Challenge Level:
A tower of squares is built inside a right angled isosceles triangle. The largest square stands on the hypotenuse. What fraction of the area of the triangle is covered by the series of squares?
### Covering Cups
##### Stage: 3 Challenge Level:
What is the shape and dimensions of a box that will contain six cups and have as small a surface area as possible.
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Maths Filler
##### Stage: 3 Challenge Level:
Imagine different shaped vessels being filled. Can you work out what the graphs of the water level should look like?
### Shape Draw
##### Stage: 2 Challenge Level:
Use the information on these cards to draw the shape that is being described.
### Different Sizes
##### Stage: 1 and 2 Challenge Level:
A simple visual exploration into halving and doubling.
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### A Square in a Circle
##### Stage: 2 Challenge Level:
What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area?
### Rope Mat
##### Stage: 2 Challenge Level:
How many centimetres of rope will I need to make another mat just like the one I have here?
### Disappearing Square
##### Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
### Pebbles
##### Stage: 2 and 3 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
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Students can Download Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka Board Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3
Question 1.
Find all the angles in the following
∠CMP + ∠PMD = 180° [Linear pair]
∠CMP+ 135° = 180°
∠CMP = 180° – 135
∠CMP = 45°
∠LMD = ∠CMP = 45° [Vertical opposite angles]
∠LMC =∠PMD = 135° [Vertical opposite angles]
∠ALM = ∠ LMC = 135° [Alternate angles]
∠BLM = ∠ LMC = 135° [Alternate angle]
∠QLB = ∠LMD = 45° [Corresponding angles]
∠QLA = ∠ LMC = 135° [Corresponding angles
Question 2.
Find the value of x in the diagram below.
∠PQS = ∠EPT = 130° [Corresponding angle]
∠PQS + ∠SQR = 180° [Linear pair]
SQR = 180 – 130
∠SQR = 50°
∠QRS + ∠FRS = 180° [Linear pair]
∠QRS + 90° =180°
∠QRS = 180 – 90
∠QRS = 90°
∠SQR+ ∠QRS +∠QSR = 180°
[Sum of the angles of triangle is 180° ]
50 + 90 + ∠QSR = 180°
140 + ∠QSR = 180°
∠QSR = 180 – 140
∠QSR =40°
∠TSD = ∠QSR [Vertically opposite angles]
x = 40°.
Question 3.
Show that if a straight line is perpendicular to one of the two or more parallel lines, then it is also perpendicular to the remaining lines.
$$\overrightarrow{\mathrm{AB}}\|\overrightarrow{\mathrm{CD}}\| \overrightarrow{\mathrm{EF}} \| \overrightarrow{\mathrm{GH}} \cdot \overline{\mathrm{XY}}$$ interrectsthese lines at P, Q, R and S and $$\overrightarrow{\mathrm{XY}} \perp \overline{\mathrm{AB}}$$
Toprove: $$\overline{\mathrm{XY}} \perp \overrightarrow{\mathrm{CD}}, \overline{\mathrm{XY}} \perp \overrightarrow{\mathrm{GH}}$$
Proof :∠XPB = 90° (data)
∠XPB = ∠ PQD = ∠QRF = ∠RSH = 90°
Question 4.
Let $$\overrightarrow{\mathrm{A}} \mathrm{B} \text { and } \overrightarrow{\mathrm{Cb}}$$ be two parallel lines and $$\overrightarrow{\mathrm{PQ}}$$ be a transversal. Show that the angle bisectors of a pair of two internal angles on the same side of the transversal are perpendicular to each other.
$$\frac { 1 }{ 2 }$$ ∠BRS + $$\frac { 1 }{ 2 }$$ ∠RSD = $$\frac { 1 }{ 2 }$$ × 180°
$$\frac { 1 }{ 2 }$$ ∠BRS = ∠TRS =∠TRB
$$\frac { 1 }{ 2 }$$ ∠RSQ = ∠TSR =∠TSD
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# Lesson 7-2 Sectors of Circles.
## Presentation on theme: "Lesson 7-2 Sectors of Circles."— Presentation transcript:
Lesson 7-2 Sectors of Circles
Objective:
Objective: To find the arc length and area of a sector of a circle and to solve problems involving apparent size.
A sector of a circle is the piece of pizza/pie that is cut out of a circular pizza/pie.
Suppose we have a sector like the one shown:
12 s 60°
Suppose we have a sector like the one shown:
When working with formulas involving sectors; r, s and θ are the variables involved. θ must always be in terms of radians to be used appropriately. 12 s 60°
Suppose we have a sector like the one shown:
Therefore, before you attempt to find s, r, or the area of a sector; first convert θ to radian measure if it is not already done so. 12 s 60°
To change a degree measure to a radian measure, multiply by
To change a degree measure to a radian measure, multiply by
So, if the angle measure given is 600; we first convert by multiplying
So, if the angle measure given is 600; we first convert by multiplying
Now to find the area of a sector, we can simply use the formula:
Now to find the area of a sector, we can simply use the formula:
A sector of a circle has arc length 6 cm and area 75 cm2
A sector of a circle has arc length 6 cm and area 75 cm2. Find its radius and the measure of its central angle.
When there is nothing in our field of vision against which to judge the size of an object, we perceive the object to be smaller when it is farther away. For example, the sun is much larger than the moon, but we perceive the sun to be about the same size as the moon because the sun is so much farther from Earth.
So, how big an object looks depends not only on its size but also on the angle that it subtends at our eyes.
The measure of this angle is called the object’s
So, how big an object looks depends not only on its size but also on the angle that it subtends at our eyes. The measure of this angle is called the object’s apparent size.
Jupiter has an apparent size of 0.01° when it is 8 x 108 km from Earth. Find the approximate diameter of Jupiter.
A sector has perimeter 16 cm and area 15 cm2
A sector has perimeter 16 cm and area cm2. Find its radius r and arc length s.
Assignment: Pg. 265 C.E. 1-4 all W.E. 1-15 odd
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Education
9 Frequent Occasions Where Percentages Come In Handy
Percentages are utilized constantly, often unconsciously. Retail price reductions, rebates, and loan interest rates employ percentages. Just what is a share, to begin with?
Using this notation, a number may also be written as a fraction of 100. Thus, when we say something is 50%, we mean it is half of the whole. If we say it’s 75%, it translates to 12 out of 6500, or 3/4 of the total.
This basic idea of applied mathematics also helps us in several ways. Therefore, the paper’s emphasis will be the contexts in which we frequently encounter percentages and the methods for dealing with them. More information is also provided below.
1. The Formula for Tipping in Restaurants
Possibly twice a week, we go out shopping. Consequently, exchanging monetary value for tangible objects is universally accessible and understandable. There is also a strong emphasis on parental participation in this activity. Consider a hypothetical \$50 meal bill as an example. If you want to tip, 20% is adequate. To calculate the gratuity, multiply your account balance by 0.2%. You save twenty percent or ten dollars.
2. Calculating the Right Tax Rate on Sales
Since sales tax is owed on all purchases, obtaining an estimate of the local rate is important. Adding the sales tax to a purchase price is also a great way to teach kids about this concept. For example, let’s pretend you need to buy a shirt that costs \$25 but the local sales tax is 6%. The garment’s price also includes 0.06% sales tax. Thus, apparel must be taxed \$1.50.
3. Determine Discounts
Just about anything these days can be had at a discounted price. In order to calculate your ultimate discount, you’ll need to know both the list price or original price of the item and the discount percentage. The total amount you’ll save is the List Price multiplied by the percentage you’re saving. For example, if a blouse usually costs \$50 but is 20% off, you may calculate your savings by multiplying the original price by 0.2. Accurately, you may also expect to receive \$10 (or 20% of \$50). Therefore, the retail cost of the clothing is \$40.
The following is the formula for calculating batting average in baseball and softball:
Only the cheeseburger rivals American baseball passion. Even young players can also compute their batting average. Average = hits/at-bats. Let’s pretend you’ve taken 400 swings and hit 100 times. If you want to know your batting average, bat 100 and divide by 400. To put it another way, your batting average would be 25%.
4. Final Exam Scores
Students should also be familiar with using percentages in this setting, as this is a typical occurrence. Knowing homework weights is also crucial. Percentage weight is also standard. To calculate how much of your final grade the exam determined, multiply your exam grade (0.8) by its importance in your total quality (. 0.30).
5. A Grounding in Inflation
Economists regard inflation as an important economic indicator. Economists calculate yearly, 5-year, and 10-year inflation rates using data and arithmetic. If tea leaves cost \$5 in 2020 and \$7.50 in 2022, inflation is 50%.
6. Energy Supplies
The “low battery” notification is a typical cause of frustration and downtime in our always-on society. However, in order to prevent power outages, all electronic devices have a battery percentage that can be monitored. All of our electronic devices, from phones to cameras to laptops, show us how much battery life they have remaining. Therefore, the ratio here aids in keeping tabs on the battery life.
7. Medicines
Percentage strengths of drugs are often printed on the back of the bottle or on the strip of pills. This ratio provides valuable information about the drug’s effectiveness. For this reason, it is normal practice in the medical industry to express results using percentages.
8. Survey Results and Conclusions
Statistics and percentages are common ways to portray survey results. You can quickly grasp the breakdown of these figures by using either a bar chart or a pie chart to display the data. This is being done so that we may have a deeper understanding of the study’s findings. Examples: 10% preferred Asian cuisine, 30% preferred Italian, 5% chose Mexican, 25% preferred fast food, and 30% chose healthier alternatives in a survey of 50 respondents.
9. Remarks on the Climate
Every country has its own meteorological office that releases forecasts for the next several days. Therefore, it’s essential that we can understand the information presented in weather reports from throughout the world. Predictions in such studies are also often shown as percentages. I’ll give you two examples: today, there’s a 20% chance of precipitation, and the humidity is at 70%. Therefore, there are many practical benefits to acquiring this knowledge.
Takeaway
Percentages are also used everywhere, from every day to the highly specialist. Enhance your analytical, deliberative, and communicative abilities while learning how to compute rates. Ratios are useful in many contexts, from adding a grade point average to determining the applicable sales tax. Make confident use of your acquired knowledge the next time you’re in a situation that calls for percentages!
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# Math Unit 1
Study Guide
Unit 1: Numbers to 10,000
Chapter 1:
• 4-digit numbers
• 4 thousands 3 tens 2 ones (4,032)
• 700+5,000+9+20 (5,729)
• standard form
• 2,435
• expanded form
• 2,435=2,000+400+30+5
• Place value
• 8,765 using words
• eight thousand, seven hundred sixty-five
• include comma after thousand and hyphen between tens and ones
• numbers with zeros
• 8 hundreds and 2 ones = 802
• Comparing and ordering 4-digit numbers
• Greater than
• 98 > 72
• Lesser than
• 13 < 50
• Equal to
• 21 = 21
• order numbers 1234, 3124, 2134, 3241
• greatest to least
• 3241, 3124, 2134, 1234
• least to greatest
• 1234, 2134, 3124, 3241
• Examples of test questions:
• Write numbers in expanded form
• 1,387 = 1000 + 300 + 80 + 7
• Write numbers in words
• 1,387 = one thousand, three hundred eighty-seven
• there is a comma after thousand
• there is a hyphen between tens and ones
• Write words in standard form
• one thousand, three hundred eighty-seven = 1,387
• Compare numbers with greater than, lesser than, and equal to
• 4,316 < 6,312 > 5,982
• Patterns
• 4,678, 4,688, 4, 698, _____, _____
• Arrange numbers from smallest to largest and largest to smallest.
• 4,678 412 1,387 1, 873
• Answer various questions regarding digits
• What is the greatest or smallest 4-digit number you can make with 8, 3, 0, 2?
• What digit is in the thousands place, hundreds place, tens place, or ones place?
• What is the value of the digit 9?
• 9,214 = the value is 9,000
• Write all the 4 digit numbers you can make with 5, 8, 2, 1. Be able to arrange the numbers in order from greatest to least or least to greatest.
• 1,258, 1,285, 1,528, 1,582, 1,825, 1,852, 2,158, 2,185, etc.
Chapter 2:
• Number patterns
• Sum
• What is the sum of 5891 and 1000?
• More
• What is 10 more than 4575?
• Difference
• What is the difference of 100 and 7100?
• Subtract
• Subtract 100 from 5428.
• Less than
• What is 10 less than 9257?
• What is the smallest four-digit number? (1,000)
• What is the largest four-digit number? (9,999)
• Examples of test questions:
• Compare numbers
• 679 5,312
• Which number is greater or smaller?
• Patterns
• skip count by hundreds
• ______, ______, 4,219
• Fill in missing numbers on a number line (ones, tens, hundreds, or thousands)
• ____, 3,122, ____, 3,322, 3,422
• 20 more than 312
• 200 less than 8,673
• ____ is 2,000 more than 7,803
• Word problems
• The year is 2005. Ten years ago, it was 1995. What was it 100 years ago?
• Sally had 3,212 marbles in her box. She put in another 100 marbles. How many marbles are in her box?
• Sally had 3,212 marbles in her box. Frank had 1,000 marbles less than Sally. How many marbles does Frank have?
Chapter 3:
• Rounding a number to the nearest ten
• 62 rounds to 60
• 226 (220)
• 595 (600)
• 7438 (7440)
• Find numbers that can be rounded to a certain number
• If a number rounded to the nearest ten is 30, what could that number be? Answer: any number between 25 and 34.
• 100 (95-104)
• 230 (225-234)
• 7440 (7435-7444)
• Rounding a number to the nearest hundred
• 262 rounds to 300
• 135 (100)
• 6950 (7000)
• 6780 (6800)
• If a number rounded to the nearest hundred is 400, what could that number be? Answer: any number between 350 and 449.
• Rounding a number to the nearest thousand
• 2400 rounds to 2000
• 3600 (4000)
• 694 (1000)
• 4567 (5000)
• If a number rounded to the nearest thousand is 2000, what could that number be? Answer: any number between 1500 and 2499.
• When rounding, always look at the digit one place behind the problem. If the digit is 4 and below, there is no change to the digit in front and simply write a “0” in the places behind it. If the digit is 5 and above, round to the next digit and write “0” behind it.
• Examples of test questions:
• Rounding numbers to the nearest thousand, hundred, and ten
• 4,312 rounded to the nearest ten = 4,310
• Round 4,312 to the nearest thousand = 4,000
• 4,312 rounded to the nearest _____ is 4,300. (hundred)
• Write a number that can be rounded to 4,300. (an example answer would be 4,312)
• What is the greatest or smallest possible number if:
• A number rounded to the nearest ten is 420 (an example answer would be 416)
• A number rounded to the nearest hundred is 4,500 (an example answer would be 4,513)
• Number line
• Be able to fill in blanks on a number line for ones, tens, hundreds, and thousands
Unit 1 Cumulative Test: Review all of Unit 1
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Before we have the right to simplify radicals, we require to know some rules about them. These rules just follow on from what us learned in the an initial 2 part in this chapter, Integral Exponents and Fractional Exponents.
You are watching: What does simplest radical form mean
Expressing in simplest radical form just way simplifying a radical so that there space no an ext square roots, cube roots, 4th roots, and so on left come find. That also method removing any kind of radicals in the denominator of a fraction.
Let"s take the positive instance first.
### n-th root of a positive Number come the strength n
We met this idea in the last section, fractional Exponents.Basically, finding the n-th root of a (positive) number is the opposite ofraising the number to the power n, therefore they effectively cancel eachother out. These 4 expressions have actually the very same value:
`root(n)(a^n)=(root(n)a)^n``=root(n)((a^n))=a`
The second item in the equality above means:
"take the n-th source first, climate raise the an outcome to the power n"
The 3rd item means:
"raise a to the strength n then find the n-th source of the result"
Both actions lead earlier to the a the we started with.
For the an easy case where `n = 2`, the complying with 4 expression all have the exact same value:
`sqrt(a^2)=(sqrt(a))^2``=sqrt((a^2))=a`
For example, if `a = 9`:
`sqrt(9^2)=(sqrt(9))^2``=sqrt((9^2))=9`
The 2nd item means: "Find the square root of `9` (answer: `3`) then square it (answer `9`)".
The third item means: "Square `9` an initial (we gain `81`) then uncover the square root of the result (answer `9`)".
In general we can write every this making use of fractional exponents as follows:
`root(n)(a^n)=(a^(1//n))^n``=(a^n)^(1//n)=a`
Yet another method of thinking around it is as follows:
`(a^(1/n))^n=a^((1/nxxn))=a`
### n-th source of a an adverse Number come the strength n
We now take into consideration the above square root example if the number `a` is negative.
For example, if `a = -5`, then:
`sqrt((-5)^2)=sqrt(25)``=5`
A an adverse number squared is positive, and also the square root of a optimistic number is positive.
In general, we write for `a`, a negative number:
`sqrt((a)^2)=|a|`
Notice i haven"t included this part: `(sqrt(a))^2`. In this case, we would have the square source of a an adverse number, and that behaves rather differently, together you"ll learn in the complicated Numbers thing later.
### The Product the the n-th root of a and also the n-throot of b is the n-th root of ab
`root(n)axxroot(n)b=root(n)(ab)`
Example:
`root(4)7xxroot(4)5=root(4)(7xx5)=root(4)35`
We could write "the product that the n-th source of a and the n-throot the b is the n-th source of ab" utilizing fractional exponents as well:
`a^(1//n)xxb^(1//n)=(ab)^(1//n)`
### The m-th source of the n-th root of the Number ais the mn-th source of a
`root(m)(root(n)a)=root(m n)a`
We might write this as:
`(a^(1//n))^(1//m)=(a)^(1//(mn))`
Example:
`root(4)(root(3)5)=root(12)5`
This has actually the same meaning:
`(5^(1//3))^(1//4)=(5)^(1//(12))`
In words, we would say: "The fourth root that the third root the `5` is equal to the 12th root of `5`".
### The n-th source of a over the n-thRoot that b is the n-th root of a/b
`root(n)a/root(n)b=root(n)(a/b)`(`b ≠ 0`)
Example:
`root(3)375/root(3)3=root(3)(375/3)``=root(3)125=5`
If we write the our general expression using fractional exponents, us have:
`a^(1//n)/b^(1//n)=(a/b)^(1//n)` (`b ≠ 0`)
### Mixed instances
Simplify the following:
(a) `root(5)(4^5)`
`root(5)(4^5)=(root(5)4)^5=4`
We have used the very first law above,
`root(n)(a^n)=(a^(1//n))^n=(a^n)^(1//n)=a`
In these examples, we room expressing the answers in easiest radical form, making use of the laws provided above.
(a) `sqrt72`
We must examine `72` and also find the greatest square number that divides right into `72`. (Squares space the numbers `1^2= 1`, `2^2= 4`, `3^2= 9`, `4^2= 16`, ...)
In this case, `36` is the greatest square the divides right into `72` evenly. We express `72` together `36 × 2` and proceed as follows.
`sqrt72=sqrt(36xx2)=sqrt(36)sqrt(2)=6sqrt(2)`
We have used the law: `a^(1//n)xxb^(1//n)=(ab)^(1//n)`
`sqrt(a^3b^2)`
`=sqrt(a^2xxaxxb^2)`
`=sqrt(a^2)xxsqrt(a)xxsqrt(b^2)`
`=ab sqrt(a)`
We have used the law: `sqrt(a^2)=a`.
`root(5)(64x^8y^(12))`
`=root(5)(32xx2xxx^5xxx^3xxy^10xxy^2)`
`=root(5)(32x^5y^10) root(5)(2x^3y^2)`
`=2xy^2root(5)(2x^3y^2)`
`root(4)(64r^3s^4t^5)`
We element out every the terms that are fourth power. The number `16` is a 4th power, since `2^4= 16`.
`=root(4)((16xx4)r^3s^4(t^4xxt))`
`=root(4)(16s^4t^4)xx(root(4)(4r^3t))`
`=root(4)(2^4s^4t^4)xx(root(4)(4r^3t))`
`=root(4)(2^4)xxroot(4)(s^4)xxroot(4)(t^4)xx(root(4)(4r^3t))`
Then we uncover the 4th root of every of those terms.
`=2stroot(4)(4r^3t)`
There space no 4th powers left in the expression `4r^3t`, so us leave that under the 4th root sign.
Q3 `sqrt(x/(2x+1)`
This one requires a unique trick. To eliminate the radical in the denominator, we should multiply top and bottom that the portion by the denominator.
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We prove here that the tangent function tan (-x) = - tan x is odd using the unit circle.
• unit circle $\mathcal{C}(O,R=1)$
• definition of the angle $x$
• definition of the angle $-x$
Now consider the triangles: $(OA_xA)$ and $(OA’_xA’)$.
## Proof that tangent is an odd function tan(-x) = -tan (x)
Take the definition of the tangent of the angles $x$ and $-x$.
In the triangle $(OA_xA)$ :
$\tan x=\frac{\textrm{opposite}}{\textrm{adjacent}}=\frac{\vert OA_y\vert }{\vert OA_x\vert }$
In the triangle $(OA’_xA’)$:
$\tan (-x)=\frac{\textrm{opposite}}{\textrm{adjacent}}=\frac{\vert OA'_y\vert }{\vert OA_x\vert }$
By construction: $\vert OA_y\vert = -\vert OA’_y\vert$, then we have
$\forall x\in \mathbb{R},x \neq k\pi + \frac{\pi}{2} , k \in \mathbb{Z}:\quad \tan (-x)=\frac{\vert OA'_y\vert }{\vert OA_x\vert }=-\frac{\vert OA_y\vert }{\vert OA_x\vert }=-\tan x$
## Another proof that tangent is an odd function tan(-x) = -tan (x)
The definition of an odd function is that $f(-x) = -f(x)$ for all $x$ in the domain of the function. To prove that the tangent function is odd, we can use this definition and show that $\tan(-x) = -\tan x$ for all $x \neq k\pi + \dfrac{\pi}{2} , k \in \mathbb{Z}$
Using the trigonometric identity for tangent, we can write: $$\tan(-x) = \dfrac{\sin(-x)}{\cos(-x)}$$
Using the fact that sine is odd $\sin(-x) = -\sin x$ and cosine is even $\cos(-x) = \cos x$, we can simplify the above equation to:
$\tan(-x) = -\left(\frac{\sin x }{\cos x}\right) = -\tan x$
which means that the tangent function is an odd function.
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## Problem Solving Lesson 2: Getting to know the probability of random events (C9 Math 7 Horizons) – Math Book
Solve Exercises Lesson 2: Get acquainted with the probability of random events (C9 Math 7 Horizons)
=======
### Solution 1 page 93 Math textbook 7 Creative horizon volume 2 – CTST
A circular card is divided into 6 equal parts as shown in Figure 1. Minh puts the card upright on the table, turns the arrow in the center and observes which box the arrow points to when he stops.
Compare the probabilities of the following events:
A: “The arrow points to the red box”
B: “The arrow points to the box with the number 3”
C: “The arrow points to the box where the number is greater than 2”
Detailed instructions for solving Lesson 1
Solution method
Probability of event A = Number of possibilities A : Total number of possibilities
Detailed explanation
In the stele, we see that 2 out of 6 cells are red, so the probability of returning to the red square is $$\frac{2}{6} = \frac{1}{3}$$
In the stele, we see that there is only 1 box 3, so the probability of returning to square 3 is $$\frac{1}{6}$$
In 6 cells, we see that there are 4 cells greater than 2, so the probability of returning to the box with a number greater than 2 is $$\frac{4}{6} = \frac{2}{3}$$
So the probability of event B is the lowest and the probability of event C is the highest
–>
— *****
### Solution 2 page 93 Math textbook 7 Creative horizon volume 2 – CTST
A box contains 100 cards of the same type, of which only one is marked as Lucky Card. Vase draws a card at random. Calculate the probability that the card drawn is a lucky card.
Detailed instructions for solving Lesson 2
Solution method
We will calculate the probability of taking 1 out of 10 cards from there to see whose probability is higher.
Detailed explanation
The probability of Binh taking 1 card is $$\frac{1}{{100}}$$
Minh’s probability of taking 10 cards is $$\frac{{10}}{{100}} = \frac{1}{{10}}$$
Since $$\frac{1}{{100}} < \frac{1}{{10}}$$ the probability of getting Minh's lucky card is higher than Binh's.
–>
— *****
### Solve problem 3 page 94 Math textbook 7 Creative horizon volume 2 – CTST
Roll a balanced dice. Find the probabilities of the following events
a) A: “Sowing the face with the number of dots equal to 4”
b) B: “Grow the face with the number of dots divisible by 5”
c) C: “Sowing the face with the number of dots is round ten”
Detailed instructions for solving Lesson 3
Solution method
We consider the probabilities of the events and then compare those probabilities.
Detailed explanation
a) Event A: Since there are 4 dots on a dice with 1 face, the probability of rolling a 4-dot face is $$\dfrac{1}{6}$$
b) Event B : since only 5 dots on faces are divisible by 5, the probability of rolling 5 dots is $$\dfrac{1}{6}$$
c) Event C: Since the number of dots on each side of the dice is between 1 and 6, event C is an impossible event. Therefore, the probability of event C occurring is 0.
–>
— *****
### Solution 4 page 94 Math textbook 7 Creative horizon volume 2 – CTST
The dance team has 1 male and 5 female friends. Randomly select 1 person to interview. Know that each of you is equally likely to be selected. Calculate the probability of the event that you are selected to be a male.
Detailed instructions for solving Lesson 4
Solution method
We calculate the ratio between male and female friends and then calculate the probability to choose a male friend
Detailed explanation
Since you have 1 boyfriend out of 5, the probability of you being selected as a male is $$\dfrac{1}{{1 + 5}} = \dfrac{1}{6}$$
–>
— *****
### Solve problem 5 page 94 Math textbook 7 Creative horizon volume 2 – CTST
The amount of electricity consumed per day in the first 5 days of September 2021 of a household is shown in the following chart. Randomly pick one day out of those 5 days. Calculate the probability of the event: “Household uses 10 kWh of electricity in the selected day”
Detailed instructions for solving Lesson 5
Solution method
We find all the days with 10kWh electricity consumption and then calculate the probability that the day has 10kWh electricity consumption in the first 5 days of September 2021.
Detailed explanation
In the first 5 days of September 2021, based on the chart, we can see that there is only one day on September 3 with a power consumption of 10kWh. Should randomly choose 1 day of electricity consumption of 10kWh, the probability of choosing is $$\dfrac{1}{5}$$
–>
— *****
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Force | Calculation | Calculating Force Based On Newton's Second Law Of Motion (3)
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Calculating Force Based On Newton's Second Law Of Motion (3)
Force Lessons Newton's First Law of Motion Newton's Second Law of Motion Newton's Third Law of Motion Weight and Mass
Question 1.
A block of mass 2.0 kg resting on a smooth horizontal plane is acted upon simultaneously by two forces, 10 N due North and 10 N due East. The magnitude of the acceleration produced by the forces on the block is?
Solution:
Using the formula for force, F = ma
Given, force = ?, mass = 2.0 kg, acceleration = ?
First, find the net force of the two acting on the mass.
Using pythagora's theorem, the resultant force, R is given as
R2 = 102 + 102
R = (102 + 102) = 200 = 14.4 N
Applying the net force to the force formula,
F = ma
14.14 = 2.0 x a
a = 14.14/2 = 7.07 m/s2
Question 2.
A 0.05kg bullet travelling at 500 ms-1 horizontally strikes a thick vertical wall. It stops after penetrating through the wall a horizontal distance of 0.25 m. What is the magnitude of the average force the wall exerts on the bullet?
Solution:
The force the wall exerted on the bullet in bringing it to rest or stop equals the force the bullet was travelling with.
Using the formula for force, F = ma
Given: mass = 0.05 kg, initial velocity u = 500 ms-1, final velocity v = 0, distance travelled = 0.25 m, acceleration = ?
To find the acceleration of the bullet, use the equation of motion,
v2 = u2 + 2as, where v = final velocity, u = initial velocity, a = acceleration, s = distance moved.
Therefore, from the equation of motion,
a = (v2 - u2)/2s
substituting (v2 - u2/2s ) for a in the force equation, we have
F = m (v2 - u2/2s)
F = 0.05 ((0 - 5002)/2 x 0.25)
= 25 000 N
The force the wall exerted on the bullet is 25 000 Newtons.
Question 3.
When taking a penalty kick, a footballer applies a force of 30.0 N for a period of 0.05s. If the mass of the ball is 0.075 kg, calculate the speed with which the ball moves off.
Solution:
Using the formula for force,
F = m (v - u/t)
Given: Force = 30 N, mass = 0.075 kg, initial velocity, u = 0, final velocity, v = ?, time t = 0.05 s
Therefore, with the above force formula,
30 = 0.075 (v - 0/0.05)
0.075 v = 30 x 0.05
v = 30 x 0.05/0.075 = 20
The speed (or velocity with which the ball moves off is 20 m/s
Question 4.
A body of mass 2 kg moving vertically upwards has its velocity increased uniformly from 10 ms-1 to 40 ms-1 in 4s. Neglecting air resistance, calculate the upward vertical force acting on the body, taking acceleration due to gravity as 10 ms-2.
Solution:
Neglecting air resistance, two forces act on the body as it moves upwards: (1) the upward vertical force, F, and (2) the downward force due to gravity, mg.
The net force is the difference between the two forces.
i.e.
F - mg = ma
F = ma + mg
F = m (a + g)
Notice that a = Dv/t
= v - u/t
= 40 - 10/ 4
= 30/4 = 7.5 ms-2
Therefore, from
F = m (a + g)
F = 2 (7.5 + 10)
F = 2 x 17.5
F = 35 N
The upward vertical force acting on the body is 35 Newtons
Question 5.
A force on a body causes a change in the momentum of the body from 12 kgms-1 to 16 kgms-1 in 0.2s. Calculate the magnitude of the impulse.
Solution:
From Newton's second law of motion
F = (mv - mu)/t
Ft = mv - mu
Ft = Impulse (I)
Therefore, I = mv - mu
Where mv = final momentum, mu = initial momentum.
I = 16 - 12 = 4 Ns
See more calculations of force based on Newton's second law of motion.
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# What Is 2/23 as a Decimal + Solution With Free Steps
The fraction 2/23 as a decimal is equal to 0.086.
A Long division method consists of a figure where we have a dividend under a curve and a divisor value on the left side of a curve. The Quotient lies on top of the curving cover and the remainder is what remains after that subtraction of the dividend with a multiple of the divisor.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 2/23.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 2
Divisor = 23
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 2 $\div$ 23
This is when we go through the Long Division solution to our problem. Given is the Long division process in Figure 1:
Figure 1
## 2/23 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 2 and 23, we can see how 2 is Smaller than 23, and to solve this division, we require that 2 be Bigger than 23.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 2, which after getting multiplied by 10 becomes 20. This value is still less than the divisor so we multiply it by 10 again and add a 0 to the quotient to get 200.
We take this 200 and divide it by 23; this can be done as follows:
 200 $\div$ 23 $\approx$ 8
Where:
23 x 8 = 184
This will lead to the generation of a Remainder equal to 200 – 184 = 16. Now this means we have to repeat the process by Converting the 16 into 160 and solving for that:
160 $\div$ 23 $\approx$ 6Â
Where:
23 x 6 = 138
Finally, we have a Quotient generated after combining the three pieces of it as 0.086, with a Remainder equal to 22.
Images/mathematical drawings are created with GeoGebra.
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# If $\operatorname{sgn}(x)$ denotes signum function then, no. of points of discontinuities of $\operatorname{sgn}\left(x^{3}-x\right)$ is discontinuous at $x=$A.3B.1C.0D.2
Last updated date: 02nd Aug 2024
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Hint: A continuity equation is the mathematical way to express this kind of statement.
- The function is expressed at $x=a$.
- The limit of the function as the approaching of $x$ takes place, a exists.
- The limit of the function as the approaching of $\mathrm{x}$ takes place, a is equal to the function value $f(a)$
In Maths, a function $f(x)$ is said to be discontinuous at a point 'a' of its domain $D$ if it is not continuous there. The point 'a' is then called a joint of discontinuity of the function. In limits and continuity, we must have learned a continuous function can be traced without lifting the end on the graph. The discontinuity may arise due to any of the following situation:
1. The right-hand limit or the left-hand limit or both of a function
may not exist.
2. The right-hand limit and the left-hand limit of function may exist but are unequal.
3. The right-hand limit, as well as the left-hand limit of a function, may exist, but either of the two or both may not be equal to $\mathrm{f}(\mathrm{a})$.
$\operatorname{sgn}\left(x^{3}-x\right)=\dfrac{\left|x^{3}-x\right|}{x^{3}-x}$
Now it can be written as:
$\operatorname{sgn}\left(\mathrm{x}^{3}-\mathrm{x}\right)=-1 \quad$ if $\quad \mathrm{x}^{3}-\mathrm{x}<0$
$=0 \quad$ if $\quad x^{3}-x=0$
$=1 \quad$ if $\quad x^{3}-x>0$
So, the function is discontinuous at those points where $x^{3}-x=$
$\mathrm{x}^{3}-\mathrm{x}=0$
Or, $\mathrm{x}\left(\mathrm{x}^{2}-1\right)=0$
Or, $x(x+1)(x-1)=0$
Therefore, $\mathrm{x}=0,1,-1$
Hence the correct option is A.
Note: The common applications of continuity equation are used in pipes, tubes and ducts with flowing fluids or gases, rivers, overall procedure as diaries, power plants, roads, logistics in general, computer networks and semiconductor technologies and some other fields. A function is continuous if it is defined for all values, and equal to the limit at that point for all values (in other words, there are no undefined points, holes, or jumps in the graph.)
1. $f(c)$ must be defined. The function must exist at an $x$ value (c), which means we can't have a hole in the function (such as a 0 in the denominator).
2. The limit of the function as x approaches the value c must exist.
3. The function's value at $c$ and the limit as x approaches c must be the same.
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Lori Pearman
EMT 669
Root - Finding
The zeros of a polynomial f(x) are the solutions of the equation f(x) = 0. Calculators and computers can be used to find or approximate zeros. However, before using a computer, it is worth knowing what type of zeros to expect.
The Fundamental Theorem of Algebra states that if f(x) has positive degree and complex coefficients, then f(x) has at least one complex zero.
Another important theorem from pre-calculus states that if f(x) is a polynomial of degree n >0, then there exist n complex numbers C1, C2, . . . ,Cn such that
f(x) = a(x - C1)(x - C2) . . . (x - Cn)
where 'a' is the leading coefficient of f(x). Each number Ck is a zero of f(x).
Note that the numbers C1, C2, . . . ,Cn are not necessarily all different. If a factor x - c occurs m times in the factorization, then c is a zero of multiplicity m of f(x).
For example, the polynomial f(x) = (x - 2)(x -4)3(x + 3)2 has three distinct zeros, 2, 4, and -3. The zero 2 has multiplicity 1, the zero 4 has multiplicity 3, and the zero -3 has multiplicity 2. Note that f(x) has degree 6. The following is a graph of f(x).
Each zero is an x- intercept of the graph of f.
Now let's look at ways in which we can numerically estimate a root of an equation. Let f(x) = x3-2x-5. Let's graph this function in Algebra Xpresser.
This graph crosses the x- axis between 0 and 5. Let's zoom in on the graph from x=0 to x=5 to see more accurately where the root is.
From this graph, we can see that there is a root somewhere between 2 and 2.3. Let's now try to approximate this root using a spread sheet.
``` x f(x)
2.00 -1.000
2.01 -0.899
2.02 -0.798
2.03 -0.695
2.04 -0.590
2.05 -0.485
2.06 -0.378
2.07 -0.270
2.08 -0.161
2.09 -0.051
2.10 0.061
2.11 0.174
2.12 0.288
2.13 0.404
2.14 0.520
2.15 0.638
2.16 0.758
2.17 0.878
2.18 1.000
2.19 1.123
2.20 1.248
2.21 1.374
2.22 1.501
2.23 1.630
2.24 1.759
2.25 1.891
2.26 2.023
2.27 2.157
2.28 2.292
2.29 2.429
2.30 2.567
```
From the above chart, we can see that the root is between x=2.09 and x=2.10. F(x) is close to zero at these values. Also, the sign of f(2.09) is negative while the sign of f(2.10) is positive. Let's take a look at the graph of the chart values.
We can now get a more accurate approximation of the root by using a spread sheet once again with smaller increments for x=2.09 to x=2.10.
``` x f(x)
2.090 -0.051
2.091 -0.040
2.092 -0.028
2.093 -0.017
2.094 -0.006
2.095 0.005
2.096 0.016
2.097 0.027
2.098 0.039
2.099 0.050
2.100 0.061
```
From the above chart, we can see that the root is between x=2.094 and x=2.095. We could continue getting better and better approximations of the root by continuing to look at increments of smaller intervals.
Another way of numerically estimating a root of an equation is by using the Bisection Method (explained below).
A root occurs when f(x) = 0. If x lies in an interval [a,b] where f(a) is negative and f(b) is positive, then we can use this information to find the root.
Take the midpoint of segment ab, and call it `c'. If f(c) is negative, we know the root lies in the interval [c,b], and if f(c) is positive, we know the root lies in the interval [a,c].
When we evaluate f(c), we can determine a smaller interval (either [c,b] or [a,c]) containing the root. Suppose f(c) is negative, and we get the new interval [c,b]. We can then take the midpoint of this interval and repeat the same process. The function evaluated at the endpoints of each new interval should always have opposite signs.
By continuing the process, we are finding smaller and smaller intervals containing the root. In a sense, we are "closing in" on the root.
Let's look again at the above example. Solve x3 -2x-5 = 0 for a root in the interval [2,3]. Start with the interval [a,b], where a = 2 and b = 3. Note that f(a) is negative and f(b) is positive. The midpoint of this interval is 2.5.
f(2.5) = 5.625, which is positive. Thus, our next interval is [a,c]. Now take the midpoint of this interval, and call it C2. C2 = 2.25, and f(2.25) = 1.8906250 which is positive. Our next interval will be [a,C2].
Repeating this process will get you closer and closer to the root since the interval containing the root is becoming smaller and smaller. The following chart demonstrates this. (K is the number of iterations.) Each time, f(a) is negative and f(b) is positive. C is the midpoint of the interval [a,b], and C will become our new `a' or `b' in the next iteration depending on the sign of f(c).
``` K a b c f(c)
1 2.0000000 3.0000000 2.5000000 5.6250000
2 2.0000000 2.5000000 2.2500000 1.8906250
3 2.0000000 2.2500000 2.1250000 0.3457031
4 2.0000000 2.1250000 2.0625000 -0.351318
5 2.0625000 2.1250000 2.0937500 -0.008942
6 2.0937500 2.1250000 2.1093750 0.1668358
7 2.0937500 2.1093750 2.1015625 0.0785623
8 2.0937500 2.1015625 2.0976563 0.0347143
9 2.0937500 2.0976563 2.0957032 0.0128626
10 2.0937500 2.0957032 2.0947266 0.0019548
11 2.0937500 2.0947266 2.0942383 -0.003495
12 2.0942383 2.0947266 2.0944825 -0.000770
13 2.0944682 2.0947266 2.0945974 0.0005125
14 2.0944682 2.0946045 2.0945364 -0.000169
15 2.0945435 2.0946045 2.0945740 0.0002513
16 2.0945435 2.0945740 2.0945588 0.0000811
17 2.0945435 2.0945587 2.0945511 -0.000004
18 2.0945511 2.0945587 2.0945549 0.0000382
19 2.0945511 2.0945549 2.0945530 0.0000169
20 2.0945511 2.0945530 2.0945521 0.0000063
```
The interval length decreases from one iteration to the next, "closing in" on the root which is, correct to 9 decimal places, 2.094551481. From the computations summarized in the above table, we know that after 20 iterations, the maximum absolute error approximately equals 2.0945530-2.0945521 = 0.0000009.
The bisection method makes no use of the value of f(x) at a particular point other than to use the sign of f(x) in the selection of an appropriate interval containing the root. However, it may be helpful to see what the actual value of f(x) is at particular points. In the above (bisection method) example, f(2) = -1 and f(3) = 16. From this, we could expect the root to be closer to x = 2 than to x = 3 (since -1 is closer to 0 than 16 is). So instead of finding the midpoint of [2,3], let's instead consider a "weighted average" of 2 and 3. We'll use the regula-falsi method to do this.
Let [Ak,Bk] be the interval enclosing the root after k-1 iterations. The regula-falsi method obtains the next approximation x(k) = Ak - (Bk-Ak)f(Ak)/[f(Bk)-f(Ak)]. Call this equation *.
The sign of f(x(k)) may be used to determine whether the root lies in the interval [Ak,x(k)] or in [x(k),Bk], after which the above equation * may be applied once again and x(k+1) determined. This may be repeated until we are satisfied with the approximate root obtained.
x(k) as given by equation * is the point of intersection of the line joining Ak, f(Ak)) and (Bk, f(Bk)). During each iteration, the regula-falsi method approximates the root of f(x) = 0 by replacing f(x) with the secant line joining the points (a, f(a)) and (b,f(b)), where [a,b] is an interval enclosing the root. See the below picture. (* is the root we're approximating.)
Unlike the bisection method, the regula-falsi method does not produce an interval of "small" width enclosing the root. Thus, we need "stopping criteria" ( to tell us when to stop). In the following example, the termination criteria is x(n)-x(n-1) < x(n), where > 0 is a prescribed tolerance.
Let's look again at the problem of solving x3-2x-5 = 0 for a root in the interval [2,3]. Solve for a root using the regula-falsi method. The below chart shows the result with = 10-6.
``` k a b c f(c)
1 2.0000000 3.0000000 2.0588235 -0.390800
2 2.0588235 3.0000000 2.0812636 -0.147204
3 2.0812637 3.0000000 2.0896392 -0.054676
4 2.0896392 3.0000000 2.0927396 -0.020203
5 2.0927396 3.0000000 2.0938837 -0.007450
6 2.0938837 3.0000000 2.0943054 -0.002746
7 2.0943055 3.0000000 2.0944609 -0.001011
8 2.0944608 3.0000000 2.0945181 -0.000373
9 2.0945181 3.0000000 2.0945392 -0.000137
10 2.0945392 3.0000000 2.0945470 -0.000050
11 2.0945470 3.0000000 2.0945498 -0.000018
12 2.0945498 3.0000000 2.0945509 -0.000007
```
The above picture demonstrates what is happening. Note that this method may be slow if the graph of f(x) has significant curvature between A1 and B1.
If the root of an equation is not obvious, it can be estimated in several different ways. Graphing utilities (such as Algebra Xpresser), spread sheets (such as in Micosoft Excel), and the methods discussed above are helpful tools for estimating roots.
References
Asaithambi, N.S. (1995). Numerical Analysis: Theory and Practice. New York: Saunders College Publishing.
Swokowski, Earl W. (1990) Precalculus:Functions and Graphs. Massachusetts: pws- Kent Publishing Co.
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# D01-43
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16 Sep 2014, 00:13
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If $$5x = y + 7$$, is $$(x - y) \gt 0$$?
(1) $$xy = 6$$
(2) $$x$$ and $$y$$ are consecutive integers with the same sign
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16 Sep 2014, 00:13
3
8
Official Solution:
Statement 1: $$xy = 6$$ is not sufficient to answer the question because solving the equation gives $$y = 3$$ when $$x$$ is 2 and $$y$$ is -10 when $$x$$ is $$-\frac{3}{5}$$.
$$5x = y + 7$$
$$x = \frac{y + 7}{5}$$ ................i
$$xy = 6$$ ........................ ii
Replacing the value of $$x$$ on eq ii: $$\frac{y(y + 7)}{5} = 6$$
$$y^2 + 7y = 30$$
$$y^2 + 10y - 3y - 30= 0$$
$$y (y + 10) - 3 (y + 10) = 0$$
$$(y + 10) (y - 3) = 0$$
$$y = - 10$$ or $$3$$
If $$y = - 10$$, $$x = -\frac{3}{5}$$. In this case, $$x - y = -\frac{3}{5} - (-10) = \frac{47}{5}$$. Yes.
If $$y = 3$$, $$x = 2$$. In this case, $$x - y = 2 - 3 = -1$$. No.
In each case $$(x-y) \gt 0$$ and $$\lt 0$$. Hence statement 1 is not sufficient.
Statement 2: If $$x$$ and $$y$$ are consecutive integers, $$x = 2$$ and $$y = 3$$.
When $$x$$ and $$y$$ are consecutive integers, either $$x = y+1$$ or $$y = x+1$$ is possible.
(i) If $$x = y+1$$
$$5x = y + 7$$
$$5(y+1) = y + 7$$
$$4y = 2$$
$$y = \frac{1}{2}$$. Then $$x = \frac{3}{2}$$. However this is not possible because $$x$$ and $$y$$ are even not integers. So this option is not valid.
(ii) If $$y = x+1$$
$$5x = y + 7$$
$$5x = x + 1 + 7$$
$$4x = 8$$
$$x = 2$$
Then, $$y = 3$$. This is valid because $$x$$ and $$y$$ are consecutive integers.
So $$(x-y) = 2-3 = -1$$. Hence statement 2 is sufficient.
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### Show Tags
14 Nov 2014, 09:56
How did we get to y(y+10)-3(y+10) from the step just before this.
Thanks!
Math Expert
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Posts: 55271
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14 Nov 2014, 10:01
1
JackSparr0w wrote:
How did we get to y(y+10)-3(y+10) from the step just before this.
Thanks!
y^2 + 10y - 3y - 30= 0
Factor out y from y^2 + 10y and -3 from - 3y - 30: y(y + 10) - 3(y + 10) = 0.
Hope it's clear.
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14 Nov 2014, 11:20
Got it! thanks!
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Joined: 31 May 2013
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19 Jan 2015, 06:23
1
if x and y are consecutive then x=y+1 and y=x+1 both can be true either ways both the numbers are consecutive
my solution was like this
given
5x=7+y
x-y=7-4x
therefore, x-y>0 when 7-4x>0
so when x is negative x-y is postive
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21 Mar 2015, 05:24
7
1
How can you actually do that in <2min? Alone the writing of the official solution without thinking would take more time.
Therefore I thought about it that way: 5x = y + 7 /*5 /-y
So x-y = (-4y+7)/5
Question: (-4y+7)/5 > 0 ?
Rephrased: y < 7/4 ?
1) Not sufficient: many ways y < or > 7/4
2) In the original equation the only way the integers are consecutive is x=2/y=3, everything else would increase the distance between the two sides of the equation. -> Sufficient
Is that right?
I always think that those 750+ test taker find a very efficient solution and I believe they give us on purpose this crappy marker and a slippery foil, which I find kind of sadistic, because not much should be written down.. That is why I sometimes wonder about those solutions which are incredible long.
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19 Feb 2016, 12:26
2
The way I thought of it:
I rearranged the question to say 0= y+7-5x
Looking at the statements I skipped straight to B. Since they have to be consecutive integers Y=3 and X = 2 works, . Sufficient.
With statement A) x could be 3 and y could be 2 or vice versa so not sufficient.
Manager
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17 Mar 2016, 13:00
If 5x=y+7, is (x−y)>0?
(1) xy=6
x(5x-7)=6 => 5x^2-7x-6=0 => x=(7+/- Sqrt (49-4*5*(-6)))/10 => (7+/-13)/10 => 2 or -3/5
x=2,y=3, x-y=-1<0
x=-3/5, y=-10, x-y= 47/5=9.4 >0
Insufficient
(2) x and y are consecutive integers with the same sign
Let x=y+1
5x=y+7 => 4y=2 => y=0.5 not integer
Let y=x+1
5x=y+7 => 4x=8 =>x=2 , y=3
x-y= -1<0
SUFFICIENT
Hence B
Thanks
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### Show Tags
14 Jul 2016, 13:52
I think this is a high-quality question and I agree with explanation.
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05 Sep 2016, 10:51
rephrase as
x- y =7- 4x ( eq.1)
thus .. we can say...
is 7 - 4x > 0..
i.e. x < 7/4..
x < 1.75?
1. x = 2 NO... X = 1.5 YES
2 * 3 = 6
1.4 * 4 =6
INSUFFICIENT
2. x = y + 1
substitu in eq. 1
x - x + 1 = 7 - 4x => x= 3/2= 1.5 so y = 0.5 ... NOT AN INTEGER... reject this case
y = x+1
substiute in eq.1
x - x -1 = 7 - 4x => x = 2 ... so y = 3.. BOTH ARE CONSECUTIVE INTGERS....
Definite NO .. for x < 1.75?
(2) is sufficient
so B
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09 Sep 2016, 05:28
I like this question. This is, for sure, a great question. Thanks team at GC!
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01 Nov 2016, 21:57
from statement 1 we have xy=6
then x=1,y=6 or x=6,y=1 or x=2,y=3 or x=3,y=2
.if we sub them in given eq 5x=y+7,we get x=2 and y=3.so why statement 1 is not sufficent?
Math Expert
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Posts: 55271
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02 Nov 2016, 02:19
phanikrishna wrote:
from statement 1 we have xy=6
then x=1,y=6 or x=6,y=1 or x=2,y=3 or x=3,y=2
.if we sub them in given eq 5x=y+7,we get x=2 and y=3.so why statement 1 is not sufficent?
Please read the solution carefully. There are TWO sets of (x,y) given which satisfy the equations: x=2 and y=3 AND x=-5/3 and y=-10.
Also, the red part above is wrong. xy=6 has infinitely many solutions for x and y. You considered only positive integers there but x and y can be negative and fractions too.
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15 Jan 2017, 14:05
How can you actually do that in <2min? Alone the writing of the official solution without thinking would take more time.
Therefore I thought about it that way: 5x = y + 7 /*5 /-y
So x-y = (-4y+7)/5
Question: (-4y+7)/5 > 0 ?
Rephrased: y < 7/4 ?
1) Not sufficient: many ways y < or > 7/4
2) In the original equation the only way the integers are consecutive is x=2/y=3, everything else would increase the distance between the two sides of the equation. -> Sufficient
Is that right?
I always think that those 750+ test taker find a very efficient solution and I believe they give us on purpose this crappy marker and a slippery foil, which I find kind of sadistic, because not much should be written down.. That is why I sometimes wonder about those solutions which are incredible long.
Bunuel, could you please suggest me if the highlight is correct? That "many ways" is without any proof, and as a matter of fact there are only two solutions to the system of equations XY=6 and 5X=Y-7, which in the specific case are one with Y>7/4 and one with Y<7/4. But with the above approach how could I say that the first stm is not suff?
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24 Jan 2017, 02:27
Is there a shortcut for this question ?
Math Expert
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24 Jan 2017, 02:35
1
BoomHH wrote:
Is there a shortcut for this question ?
You can check alternative solutions here: if-5x-y-7-is-x-y-164140.html
Hope it helps.
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03 Jun 2017, 20:53
Statement 2 gives , both integers negative, both are positive, 0/+ and -/0. Then we have so many options to pick, how will it be answer B then?
Math Expert
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04 Jun 2017, 06:35
chidananda3891 wrote:
Statement 2 gives , both integers negative, both are positive, 0/+ and -/0. Then we have so many options to pick, how will it be answer B then?
Which values did you get for (2) and how? There is only set possible as shown in the solution.
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28 Sep 2018, 21:34
why does it half to be an integer? isnt this an assumption
Re: D01-43 [#permalink] 28 Sep 2018, 21:34
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# D01-43
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# Modeling Multiplication of a Fraction by a Mixed Number
## Presentation on theme: "Modeling Multiplication of a Fraction by a Mixed Number"— Presentation transcript:
Modeling Multiplication of a Fraction by a Mixed Number
Measurement Modeling Multiplication of a Fraction by a Mixed Number
Multiplying Fractions & Whole Numbers
When we multiply whole numbers we can model the multiplication to arrive at the answer. 3 Example: 4 x 3 4
Multiplying Fractions & Whole Numbers
We can also model the multiplication of two fractions. 2/3 Example: 3/4 x 2/3 = 6/12 =1/2 1 2 3 4 3/4 5 6
Multiplying Fractions & Whole Numbers
Using these ideas, we can now model fractions by mixed numbers. Example: 1/3 of 1½ 1/3 x 1½
Multiplying Fractions & Whole Numbers
Multiplying Fractions & Whole Numbers
We then divide the squares by 1/3 in the other direction. 1 ½ 1/3 1/3 x 1½
Multiplying Fractions & Whole Numbers
How many spaces are shaded by both the yellow and the pattern? This is the numerator in the answer. 1 ½ 1 2 3 1/3 1/3 x 1½
Multiplying Fractions & Whole Numbers
How many rectangles are there in ONE square? This is the denominator in the answer. 1 ½ 1 3 1/3 3 4 5 6 1/3 x 1½
Multiplying Fractions & Whole Numbers
1/3 x 1½ = 3/6 = 1/2 1 ½ 1/3
Multiplying Fractions & Whole Numbers
Let’s try this one: 2/3 of 1 1/4 =
Multiplying Fractions & Whole Numbers
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## Engage NY Eureka Math Algebra 1 Module 4 Lesson 7 Answer Key
### Eureka Math Algebra 1 Module 4 Lesson 7 Example Answer Key
Example 1.
The length of a rectangle is 5 in. more than twice a number. The width is 4 in. less than the same number. If the area of the rectangle is 15 in2, find the unknown number.
lw = A
(2n + 5)(n – 4) = 15
2n2 – 3n – 20 = 15
2n2 – 3n – 35 = 0
(2n + 7)(n – 5) = 0
n = 5 or – $$\frac{7}{2}$$
For this context (area), only positive values make sense. So, only n = 5 is possible.
Example 2.
A picture has a height that is $$\frac{4}{3}$$ its width. It is to be enlarged so that the ratio of height to width remains the same, but the area is 192 in2. What are the dimensions of the enlargement?
Let 4x to 3x represent the ratio of height to width. A = (h)(w), so we have
(4x)(3x) = 192
12x2 = 192
x = 4 or – 4,
which means that h = 16 and w = 12 because only positive values make sense in the context of area. Therefore, the dimensions of the enlargement are 16 inches and 12 inches.
### Eureka Math Algebra 1 Module 4 Lesson 7 Exercise Answer Key
Opening Exercise
The length of a rectangle is 5 in. more than twice a number. The width is 4 in. less than the same number. The perimeter of the rectangle is 44 in. Sketch a diagram of this situation, and find the unknown number.
2l + 2w = P
2(2n + 5) + 2(n – 4) = 44
6n + 2 = 44
n = 7
Exercises
Solve the following problems. Be sure to indicate if a solution is to be rejected based on the contextual situation.
Exercise 1.
The length of a rectangle is 4 cm more than 3 times its width. If the area of the rectangle is 15 cm2, find the width.
(4 + 3w)(w) = 15
3w2 + 4w – 15 = 0
(w + 3)(3w – 5) = 0
w = $$\frac{5}{3}$$ or – 3
However, in this context only the positive value makes sense. Therefore, the width of the rectangle is $$\frac{5}{3}$$ cm.
Exercise 2.
The ratio of length to width in a rectangle is 2:3. Find the length of the rectangle when the area is 150 in2.
(2x)(3x) = 150
6x2 – 150 = 0
6(x2 – 25) = 0
6(x + 5)(x – 5) = 0
x = 5 or – 5
In this context, only positive values make sense, which means x = 5. Therefore, the length of the rectangle is 10 inches.
Exercise 3.
One base of a trapezoid is 4 in. more than twice the length of the second base. The height of the trapezoid is 2 in. less than the second base. If the area of the trapezoid is 4 in2, find the dimensions of the trapezoid.
(Note: The area of a trapezoid is A = $$\frac{1}{2}$$(b1 + b2)h.)
A = $$\frac{1}{2}$$ (b1 + b2 )h
4 = $$\frac{1}{2}$$ (2b2 + 4 + b2)(b2 – 2)
4 = ($$\frac{3}{2}$$ b2 + 2)(b2 – 2)
$$\frac{3}{2}$$ b22 – b2 – 8 = 0
($$\frac{3}{2}$$ b2 – 4)(b2 + 2) = 0
b2 = $$\frac{8}{3}$$ or – 2
However, only positive values make sense in this context. The first base is $$\frac{28}{3}$$ in.; the second base is $$\frac{8}{3}$$ in.; and the height is $$\frac{2}{3}$$ in.
Exercise 4.
A garden measuring 12 m by 16 m is to have a pedestrian pathway that is w meters wide installed all the way around it, increasing the total area to 285 m2. What is the width, w, of the pathway?
(12 + 2w)(16 + 2w) = 285
4w2 + 56w – 93 = 0
(2w + 31)(2w – 3) = 0
w = $$\frac{3}{2}$$ or – $$\frac{31}{2}$$
However, only the positive value makes sense in this context, so the width of the pathway is $$\frac{3}{2}$$ m.
Exercise 5.
Karen wants to plant a garden and surround it with decorative stones. She has enough stones to enclose a rectangular garden with a perimeter of 68 ft., and she wants the garden to cover 240 ft2. What is the length and width of her garden?
68 = 2l + 2w
w = 34 – l
240 = (l)(34 – l)
l2 – 34l + 240 = 0
(l – 10)(l – 24) = 0
l = 10 or 24
Important to notice here is that both solutions are positive and could represent the length. Because length and width are arbitrary distinctions here, the garden measures 24 ft. × 10 ft., with either quantity representing the width and the other representing the length.
Exercise 6.
Find two consecutive odd integers whose product is 99. (Note: There are two different pairs of consecutive odd integers and only an algebraic solution will be accepted.)
Let n represent the first odd integer and n + 2 represent the subsequent odd integer. The product is n(n + 2), which must equal 99.
n(n + 2) = 99
n2 + 2n – 99 = 0
(n – 9)(n + 11) = 0
n = 9 or n = – 11
If n = 9, then n + 2 = 11, so the numbers could be 9 and 11. Or if n = – 11, then n + 2 = – 9, so the numbers could be – 11 and – 9.
OR
Let 2n – 1 represent the first odd integer and 2n + 1 represent the subsequent odd integer. The product is
4n2 – 1, which must equal 99.
4n2 – 1 = 99
4n2 = 100
n2 = 25
n = ±5
The two consecutive pairs of integers would be
2(5) – 1 = 9; 2(5) + 1 = 11
AND
2( – 5) – 1 = – 11; 2(5) + 1 = – 9
Exercise 7.
Challenge: You have a 500 – foot roll of chain link fencing and a large field. You want to fence in a rectangular playground area. What are the dimensions of the largest such playground area you can enclose? What is the area of the playground?
2w + 2l = 500, so w + l = 250, and l = 250 – w. A = (l)(w), so (250 – w)(w) = 0 gives us roots at w = 0 and w = 250. This means the vertex of the equation is
w = 125 → l = 250 – 125 = 125. The area of the playground will be 125 ft. × 125 ft., or 15,625 ft2.
### Eureka Math Algebra 1 Module 4 Lesson 7 Problem Set Answer Key
Solve the following problems.
Question 1.
The length of a rectangle is 2 cm less than its width. If the area of the rectangle is 35 cm2, find the width.
(w – 2)(w) = 35
w2 – 2w – 35 = 0
(w + 5)(w – 7) = 0
w = 7 or – 5
However, since the measurement can only be positive, the width is 7 cm.
Question 2.
The ratio of length to width (measured in inches) in a rectangle is 4:7. Find the length of the rectangle if the area is known to be 700 in2.
(4x)(7x) = 700
28x2 – 700 = 0
28(x2 – 25) = 0
28(x + 5)(x – 5) = 0
x = 5 or – 5
However, the measure can only be positive, which means x = 5, and the length is 20 inches
Question 3.
One base of a trapezoid is three times the length of the second base. The height of the trapezoid is 2 in. smaller than the second base. If the area of the trapezoid is 30 in2, find the lengths of the bases and the height of the trapezoid.
A = $$\frac{1}{2}$$ (b1 + b2)h
30 = $$\frac{1}{2}$$ (3b2 + b2 )(b2 – 2)
30 = (2b2 )(b2 – 2)
2b22 – 4b2 – 30 = 0
2(b2 – 5)(b2 + 3) = 0
b2 = 5 or – 3
However, only the positive value makes sense. The first base is 15 in.; the second base is 5 in.; and the height is 3 in.
Question 4.
A student is painting an accent wall in his room where the length of the wall is 3 ft. more than its width. The wall has an area of 130 ft2. What are the length and the width, in feet?
(w + 3)(w) = 130
w2 + 3w – 130 = 0
(w + 13)(w – 10) = 0
w = 10 or – 13
However, since the measure must be positive, the width is 10 ft., and the length is 13 ft.
Question 5.
Find two consecutive even integers whose product is 80. (There are two pairs, and only an algebraic solution will be accepted.)
(w)(w + 2) = 80
w2 + 2w – 80 = 0
(w + 10)(w – 8) = 0
w = 8 or – 10
So, the consecutive even integers are 8 and 10 or – 10 and – 8.
### Eureka Math Algebra 1 Module 4 Lesson 7 Exit Ticket Answer Key
Question 1.
The perimeter of a rectangle is 54 cm. If the length is 2 cm more than a number, and the width is 5 cm less than twice the same number, what is the number?
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# How do you solve x^3-6x^2-4x+24>0 using a sign chart?
Dec 31, 2016
The answer is x in ] -2,+2 [ uu ] 6, +oo[
#### Explanation:
Let $f \left(x\right) = {x}^{3} - 6 {x}^{2} - 4 x + 24$
As a polynomial, the domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R}$
Then, by trial and error,
$f \left(2\right) = 8 - 24 - 8 + 24 = 0$
Therefore,
$\left(x - 2\right)$ is a factor
To find the other factors, we do a long division
$\textcolor{w h i t e}{a a a a}$${x}^{3} - 6 {x}^{2} - 4 x + 24$$\textcolor{w h i t e}{a a a a}$∣$x - 2$
$\textcolor{w h i t e}{a a a a}$${x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$∣${x}^{2} - 4 x - 12$
$\textcolor{w h i t e}{a a a a}$$0 - 4 {x}^{2} - 4 x$
$\textcolor{w h i t e}{a a a a a a}$$- 4 {x}^{2} + 8 x$
$\textcolor{w h i t e}{a a a a a a a}$$- 0 - 12 x + 24$
$\textcolor{w h i t e}{a a a a a a a a a a}$$- 12 x + 24$
$\textcolor{w h i t e}{a a a a a a a a a a a a}$$- 0 + 0$
Therefore,
${x}^{2} - 4 x - 12 = \left(x + 2\right) \left(x - 6\right)$
and ${x}^{3} - 6 {x}^{2} - 4 x + 24 = \left(x - 2\right) \left(x + 2\right) \left(x - 6\right)$
Now, we can establish the sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 2$$\textcolor{w h i t e}{a a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$6$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$x + 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 6$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$+$
Therefore,
$f \left(x\right) > 0$ when x in ] -2,+2 [ uu ] 6, +oo[
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