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Quick Math Homework Help
Master the 7 pillars of school success that I have learned from 25 years of teaching.
Video covers the characteristics of a quadratic function.
Video Guide
0:21 What is the vertex ?
0:31 What is the axis of symmetry?
1:13 What is the domain and the range?
2:16 Explanation of interval increase and interval decrease
2:44 What is the rate of change?
3:55 What is the max and min?
4:10 How do you find the y intercept and the x intercept?
5:17 What is end behavior?
The graph of a quadratic function is in the shape of a parabola. A parabola can take many shapes but is basically in the shape of a U. See an example below.
Let's break down the equation in order to understand what each part tells us.
• If a is NEGATIVE then the graph reflects across the x-axis.
• If |a| is less than 1, the graph SHRINKS.
• If |a| is greater than 1, the graph STRETCHES.
• h is TRICKY!
• If h is POSITIVE then the graph moves LEFT.
• If h is NEGATIVE then the graph moves RIGHT.
• Axis of Symmetry: x = h
• Special Case
• If k is POSITIVE then the graph moves UP.
• If k is NEGATIVE then the graph moves DOWN.
• Vertex: (h, k)
1. Put the equation in standard form. A quadratic equation has two forms.
2. Identify the values of a, b, &c
3. Find the axis of symmetry:
4. Construct a table of values for x and y
5. Plot the points and connect them with a U- shaped curve and arrows
Several examples of Graphing Quadratic Equations
0:15 Graph y=2(x-2)^2-5
This graph is a change of direction of a parabola.
Remember: "The opposite of H and the same sign as K"
3:08 This example is not in vertex form so you begin by finding your vertex. To convert from standard form to vertex use -b/2a
Graph y=x^2+6x +2
Sample problems changing a quadratic equation from standard form to vertex form.
The two forms are:
Standard equals y= ax^2 + bx +c
Vertex form equals y=a(x-h)^2 +k
0:47 Convert y=-2x^2 +34x -8 (standard form to vertex form)
3:34 Plug the numbers into the vertex formula to finish converting from the standard form to the vertex form.
Given an equation in vertex form how do you switch the equation to standard form?
Video Guide
0:48 Given the equation y=-(x-8)^2 +2 which is in vertex form,switch to standard form. This problem involves distributing a negative which involves changing all of the signs.
Let's look at quadratic equations and what transformations are occurring to the parent function.y=x^2
Video Guide
0:24 y=2(x-9)^2 +3 The skeleton equation equals y=a(x-K)^2 +K
a=stretch h=left right shift,k=up down shift
1:55 What are the transformations of: y=-4/7(x)^2 -6
3:05 What if you are given descriptions and you have to write an equation? Once you write your descriptions you plug these into the vertex equation |
# Newton-Raphson Root Finding
## The Problem
A root of an equation $f$ is any solution $x$ such that $f(x) = 0$. Roots of some equations can be found analytically; e.g. $f(x) = x + 1$, then $x = -1$ is a root which can be proven analytically ($-1 + 1 = 0$.)
However, some functions do not have analytic roots, or the roots are harder to find. Numerical methods have been developed to help find these roots algorithmically.
## Newton-Raphson Method
For a given function $f(x)$, estimate $x_0$ near the desired root. Now consider the Taylor expansion of $f(x)$ around $x_0$.
\eqalign{ f(x) &= f(x_0) + f'(x_0)(x - x_0) + f''(x_0)\frac{(x - x_0)^2}{2} \cdots \cr f(x) &\approx f(x_0) + f'(x_0)(x - x_0) } \label{nr-taylor}
Because we want to find $f(x)=0$, set $x = x_0 - \frac{f(x_0)}{f'(x_0)}$ and plug it into equation \eqref{nr-taylor}:
\eqalign{ f(x) &= f(x_0) + f'(x_0)(x_0 - \frac{f(x_0)}{f'(x_0)} - x_0) \cr &= f(x_0) - f'(x_0)\frac{f(x_0)}{f'(x_0)} \cr &= f(x_0) - f(x_0) \cr &= 0 } \label{nr-taylor-0}
In more general terms,
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. \label{newton-raphson}$$
Graphically, $x_{n+1}$ is equivalent to the $x$-intercept of the tangent to $f(x)$ at $x_n$.
To approximate the root, iteratively find $x_{n+1}$ such that $x_{n+1} - x_n < \epsilon$ for some chosen tolerance value $\epsilon$. |
May 20, 2018
Event A: $\frac{5}{6}$
Event B: $\frac{11}{18}$
#### Explanation:
Event A: Their sum is greater than 7
We can solve through complementary events.
P(sum greater than 7) = 1- P(sum equal to 7)
What are the numbers that equal to 7?
1,6
2,5
3,4
4,3
5,2
6,1
In each case, you have $\frac{1}{36}$ chance of rolling those two numbers. Here's why:
Looking at 1,6
You have $\frac{1}{6}$ chance of rolling a 1 and you also have $\frac{1}{6}$ chance of rolling a 6. Since these two chances are independent events, then you multiply the numbers together. $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$
The $\frac{1}{36}$ applies to all 6 numbers. So $\frac{1}{36} \times 6 = \frac{1}{6}$
Therefore, P(sum equal to 7) = $\frac{1}{6}$
So, P(sum greater than 7) = $1 - \frac{1}{6} = \frac{5}{6}$
Event B: the sum is not divisible by 4 and not divisible by 6
The largest sum you can have between two numbers is 12 ie 6+6 and the smallest number you can have is 2 ie 1+1
So numbers divisible by 4 in that range are: 4, 8, 12
and numbers divisible by 6 in that range are: 6, 12
Therefore, looking at the table below, you can see that there are 14 combinations that you cannot throw. Since each combination ie 1,3 or 6,6 and you have $\frac{1}{36}$ chance of throwing each combination, then P(sum is divisible by 4 and 6) is equal to$\frac{1}{36} \times 14 = \frac{7}{18}$
P(sum not divisible by 4 and 6) = 1-P(sum is divisible by 4 and 6)
P(sum not divisible by 4 and 6) = $1 - \frac{7}{18} = \frac{11}{18}$
May 20, 2018
$P \left(A\right) = \frac{5}{12}$
$P \left(B\right) = \frac{11}{18}$
#### Explanation:
Consider all the possible results of two dice rolls.
$1 + 1 = 2$
$1 + 2 = 3$
$\cdots$
$2 + 1 = 3$
$2 + 2 = 4$
$\cdots$
$6 + 5 = 11$
$6 + 6 = 12$
In total, there are $36$ combinations:
$2 , 3 , 3 , 4 , 4 , 4 , 5 , 5 , 5 , 5 , 6 , 6 , 6 , 6 , 6 , 7 , 7 , 7 , 7 , 7 , 7 , 8 , 8 , 8 , 8 , 8 , 9 , 9 , 9 , 9 , 10 , 10 , 10 , 11 , 11 , 12$
$15$ out of these $36$ are strictly greater than 7.
$22$ out of these $36$ are not divisible by 4 or by 6.
Therefore,
$P \left(A\right) = \frac{15}{36} = \frac{5}{12}$
$P \left(B\right) = \frac{22}{36} = \frac{11}{18}$ |
# How do you find the tension between two vertical objects?
For a system of two masses hanging from a vertical pulley, tension equals 2g(m1)(m2)/(m2+m1), where “g” is the acceleration of gravity, “m1” is the mass of object 1, and “m2” is the mass of object 2.
## How do you solve a tension problem in physics?
1. Step 1: Identify the direction of the tension force.
2. Step 2: Identify any other forces on the same axis as the tension force.
3. Step 3: Identify the acceleration along the tension force axis.
4. Step 4: Using Newton’s second law.
5. Step 5: Check units to be sure they are in NewtonsN
## What are 5 examples of tension force?
• Car Towing.
• Pulling a Bucket of Water from Well.
• Weighing.
• Gym Equipment.
• Crane Machine.
• Whirligig.
• Tug of war.
• Pulling a block with the help of a rope.
## How do you find the tension between two horizontal objects?
1. T = m1(a + μkg)
2. Friction(fk) = μk N = μk*(mtotalg) fk = μk(m1+m2)g.
3. Fnet = F – friction.
4. acceleration(a) = F/Total mass.
5. a = [F-μk(m1+m2)g]/(m1+m2)
6. Fnet = T – friction.
7. T= Fnet + friction.
8. T = μk*m1*g + m1*a ——–(1)
## What is an everyday example of tension?
tension: Two pulling forces, directly opposing each other, that stretch an object and try to pull it apart. For example, pulling on a rope, a car towing another car with a chain – the rope and the chain are in tension or are “being subjected to a tensile load.”
## What is tension force give example?
Tension is defined as the force transmitted through a rope, string or wire when pulled by forces acting from opposite sides. The tension force is directed over the length of the wire and pulls energy equally on the bodies at the ends. Every physical object which is in contact exerts some force on one another.
## What are 10 examples of normal force?
• #1 Girl standing on a flat surface. When a girl is standing on a flat surface, one perpendicular force acts on a girl.
• #2 Book placed on a table.
• #3 Banana crate placed on a floor.
• #5 Flower pot placed on a table.
## Is net force the same as tension?
The normal force and the force of gravity (weight) of the block are balanced, so the net force is all from the tension directed to the right. Hanging mass: On the right, tension pulls the block upwards while gravity pulls it downwards, so the net force must be the difference between them.
## How do you find the tension in a string between two blocks?
The tension between two blocks can be found by knowing the net forces acting on the two blocks attached to the string, we can calculate the tension exerted on the string due to the two blocks.
## Is tug of war a tension force?
Tension is when an object is pulled from one or multiple sides. Since tug of war is a game of a rope being pulled by people (applied force) on both sides (tension force), it is both applied force and tension force.
## How do you find tension in a spring?
F = -kx. The proportional constant k is called the spring constant. It is a measure of the spring’s stiffness. When a spring is stretched or compressed, so that its length changes by an amount x from its equilibrium length, then it exerts a force F = -kx in a direction towards its equilibrium position.
## Is the tension in the rope less than equal to or greater than 50 N?
Is the tension in the rope less than, equal to, or greater than 50 N? Use the parallelogram rule to defend your answer. Less than. 50N would be the resultant of the two vectors combined.
## Is tension equal to weight?
Thus, just as you would expect, the tension equals the weight of the supported mass: T=w=mg.
## What is the tension in newtons in rope 1?
Physicists use a metric unit called the newton to measure force; the tension on a rope suspending a 100-gram weight is roughly 1 newton. Multiply the weight’s mass in kilograms by 9.8, the acceleration in meters per second squared due to gravity.
## How do you find tension when given mass and angle?
Using the formula, T = (Tx2 + Ty2)1/2, the tension is calculated. The component TX provides centripetal force and so Tx = mv2 (m=mass of the object; v=velocity). The component TY corresponds to weight of the object, i.e. TY = mg (m=mass of the object; g= acceleration due to gravity).
## What are 3 examples of applied force?
• Pulling a bucket of water from a well.
• Rotating a doorknob by human hand.
• Lifting an object from the floor.
• Checking the weight of the person (This is a practice way to measure the applied force)
• Pushing a box on a floor.
## What is tension force kids?
Tension is a force that stretches something. Compression is a force that squeezes something together. Materials are only useful if they can withstand forces. Force flows through a material like water flows through a pipe.
## Does tension increase with mass?
With heavier masses the tension may be higher, but it will be higher at all 3 points, so the tension, higher or lower, will always be equal at all 3 points.
## Is the normal always 90 degrees?
Due to Newton’s third law, the normal forces will be equal in magnitude and in opposite direction. The normal force always makes a 90 degree angle with the surface (perpendicular).
## Is tension a contact force?
In order for objects to exert a contact force, the objects must be touching. Common examples of contact forces include tension, the normal force, and friction.
## What type of force would be responsible for holding up a lamp hanging from the ceiling?
Tension. An object that is being stretched experiences a tension force. For example, a cable holding a ceiling lamp. |
Question
The circle drawn with variable chord $x+ay-5=0$($a$being parameter) of the parabola${{y}^{2}}=20x$as diameter will always touch the line(a) $x+5=0$(b) $y+5=0$(c) $x+y+5=0$(d) $x-y+5=0$
Hint: To find the circle with variable chord of the parabola as its diameter, we will
first find the exact equation of the chord and then use the fact that a circle drawn with a
chord to the parabola as its diameter has a tangent which is directrix to the parabola.We
have a variable chord $x+ay-5=0$ of the parabola${{y}^{2}}=20x$.
We want to find the equation of tangent to the circle with the variable chord of the parabola
as its diameter.
We will consider the parabola${{y}^{2}}=4ax$.
By comparing it with our parabola${{y}^{2}}=20x=4\times 5x$, we observe that$a=5$.
Substituting the value of$a$in the equation of the chord$x+ay-5=0$of the parabola, we
have$x+5y-5=0$as the chord of the parabola${{y}^{2}}=20x$.
We know that a circle drawn with a chord to the parabola as its diameter has a tangent
which is directrix to the parabola.
We know that the equation of directrix to the parabola${{y}^{2}}=4ax$is$x+a=0$.
Substituting$a=5$, we get$x+5=0$as the equation of directrix to the parabola${{y}^{2}}=20x$.
Thus, the tangent to the circle drawn with chord of the parabola${{y}^{2}}=20x$ as its diameter is of the form$x+5=0$
Hence, the correct answer is $x+5=0$.
Note: We can also solve the question by finding the exact equation of the circle and then finding the equation of tangent to the circle. To find the equation of the tangent to the
circle, we will find the equation of circle and points of intersection of the chord to the
parabola and the circle. However, it’s not necessary to do as it’s a longer way to solve the
question. |
# Rational input, integer output
Standard
Consider the following polynomial equation [source: Berkeley Problems in Mathematics, problem 6.13.10]:
$f(t) = 3t^3 + 10t^2 - 3t$
Let’s try to figure out the rational values of $t$ for which $f(t)$ is an integer. Clearly, if $t\in\mathbb{Z}$ then $f(t)$ is an integer. So let’s consider the case when $t=m/n$ where $\gcd(m,n)=1$ and $m\neq \pm 1$. Substituting this value of $t$ we get:
$\displaystyle{f\left(\frac{m}{n}\right) = \frac{3m^3}{n^3} + \frac{10m^2}{n^2} - \frac{3m}{n}= \frac{m(3m^2+10mn-3n^2)}{n^3}=k \in \mathbb{Z}}$
Since, $n^3\mid (3m^2+10mn-3n^2)$ we conclude that $n\mid 3$. Also it’s clear that $m\mid k$. Hence, $n=\pm 3$ and we just need to find the possible values of $m$.
For $n=3$ we get:
$\displaystyle{f\left(\frac{m}{3}\right) = \frac{m(m^2+10m-9)}{9}=k \in \mathbb{Z}}$
Hence we have $9\mid (m^2+10m)$. Since $\gcd(m,n)=\gcd(m,3)=1$, we have $9\mid (m+10)$, that is, $m\equiv 8\pmod 9$.
Similarly, for $m=-3$ we get $n\equiv 1 \pmod 9$. Hence we conclude that the non-integer values of $t$ which lead to integer output are:
$\displaystyle{t = 3\ell+ \frac{8}{3}, -3\ell-\frac{1}{3}}$ for all $\ell\in\mathbb{Z}$
# Polynomials and Commutativity
Standard
In high school, I came to know about the statement of the fundamental theorem of algebra:
Every polynomial of degree $n$ with integer coefficients have exactly $n$ complex roots (with appropriate multiplicity).
In high school, a polynomial = a polynomial in one variable. Then last year I learned 3 different proofs of the following statement of the fundamental theorem of algebra [involving, topology, complex analysis and Galois theory]:
Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.
A more general statement about the number of roots of a polynomial in one variable is the Factor Theorem:
Let $R$ be a commutative ring with identity and let $p(x)\in R[x]$ be a polynomial with coefficients in $R$. The element $a\in R$ is a root of $p(x)$ if and only if $(x-a)$ divides $p(x)$.
A corollary of above theorem is that:
A polynomial $f$ of degree $n$ over a field $F$ has at most $n$ roots in $F$.
(In case you know undergraduate level algebra, recall that $R[x]$ is a Principal Ideal Domain if and only if $R$ is a field.)
The key fact that many times go unnoticed regarding the number of roots of a given polynomial (in one variable) is that the coefficients/solutions belong to a commutative ring (and $\mathbb{C}$ is a field hence a commutative ring). The key step in the proof of all above theorems is the fact that the division algorithm holds only in some special commutative rings (like fields). I would like to illustrate my point with the following fact:
The equation $X^2 + X + 1$ has only 2 complex roots, namely $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$. But if we want solutions over 2×2 matrices (non-commutative set) then we have at least 3 solutions (consider 1 as 2×2 identity matrix and 0 as the 2×2 zero matrix.)
$\displaystyle{A=\begin{bmatrix} 0 & -1 \\1 & -1 \end{bmatrix}, B=\begin{bmatrix} \omega & 0 \\0 & \omega^2 \end{bmatrix}, C=\begin{bmatrix} \omega^2 & 0 \\0 & \omega \end{bmatrix}}$
if we allow complex entries. This phenominona can also be illusttrated using a non-commutative number system, like quaternions. For more details refer to this Math.SE discussion.
# Huntington’s Red-Blue Set
Standard
While reading Lillian Lieber’s book on infinity, I came across an astonishing example of infinite set (on pp. 207). Let’s call the property of existence of a rational number between given two rational number to be “beauty” (a random word introduced by me to make arguments clearer).
The set of rational numbers between 0 and 1 are arranged in ascending order of magnitude, and all of them are coloured blue. This is clearly a beautiful set. Then another another set of rational numbers between 0 and 1 is taken and arrange in ascending order of magnitude, but all of them are coloured red. This is also a beautiful set. Now, put these two sets together in such a way that each blue number is immediately followed by the corresponding red number. For example, 1/2 is immediately followed by 1/2 etc. It appears that if we interlace two beautiful sets, the resulting set should be even more beautiful. But since each blue number has an immediate successor, namely the corresponding red number, so that between these two we can’t find even a single other rational number, red or blue, the resulting set is NOT beautiful.
The set created above is called Huntington’s Red-Blue set. It is an ingenious invention, where two beautiful sets combined together lead to loss of beauty. For more details, read the original paper:
Huntington, Edward V. “The Continuum as A Type of Order: An Exposition of the Modern Theory.” Annals of Mathematics, Second Series, 7, no. 1 (1905): 15-43. doi:10.2307/1967192.
# Rationals…
Standard
A few days ago I noticed some fascinating properties of so called rational numbers.
Natural Bias:
Our definition of a number being rational or irrational is very much biased. We implicitly assume our numbers to be in decimals (base 10), and then define rational numbers as those numbers which have terminating or recurring decimal representation.
But it is interesting to note that, for example, √5 is irrational in base-10 (non-terminating, non-repeating decimal representation) but if we consider “golden-ratio base“, √5 = 10.1, has terminated representation, just like rational number!!
Ability to complete themselves:
When we construct numbers following Peano’s Axioms we can “easily” create (set of) natural numbers ($\mathbb{N}$), and from them integers ($\mathbb{Z}$) and rational numbers ($\mathbb{Q}$). Notice that $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q}$.
But it is comparatively difficult to create real numbers ($\mathbb{R}$) from rational numbers ($\mathbb{Q}$) although still we want to create a set from its subset. Notice that unlike previous cases, to create $\mathbb{R}$ we will first have to create so-called irrational numbers ($\overline{\mathbb{Q}}$) from $\mathbb{Q}$. The challenge of creating the complementary set ($\overline{\mathbb{Q}}$) of a given set ($\mathbb{Q}$) using the given set ($\mathbb{Q}$) itself makes it difficult to create $\mathbb{R}$ from $\mathbb{Q}$ . We overcome this difficulty by using specialized techniques like Dedekind cut or Cauchy sequences (the process is called “completion of rational numbers”). |
# Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number. f(x)=frac{1}{sqrt{x}} with a=4, approximate frac{1}{sqrt{3}}
Question
Series
Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}$$ with $$\displaystyle{a}={4}$$, approximate $$\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}$$
2021-02-06
Consider the provided function,
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}$$ with $$\displaystyle{a}={4}$$, approximate $$\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}$$
Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.
The Taylor series as a form of $$\displaystyle{\sum_{{{k}={0}}}^{\infty}}{c}_{{k}}{\left({x}-{a}\right)}^{{k}}$$, where $$\displaystyle{c}_{{k}}={\frac{{{{f}^{{k}}{\left({a}\right)}}}}{{{k}!}}},{k}={0},{1},{2},\ldots$$
We evaluate the derivative of the function at point $$\displaystyle{a}={4}$$
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}\Rightarrow{f{{\left({4}\right)}}}={\frac{{{1}}}{{{2}}}}$$
$$\displaystyle{f}'{\left({x}\right)}=-{\frac{{{1}}}{{{2}{x}^{{{\frac{{{3}}}{{{2}}}}}}}}}\Rightarrow{f}'{\left({4}\right)}=-{\frac{{{1}}}{{{16}}}}$$
$$\displaystyle{f}{''}{\left({x}\right)}={\frac{{{3}}}{{{4}{x}^{{{\frac{{{5}}}{{{2}}}}}}}}}\Rightarrow{f}{''}{\left({4}\right)}={\frac{{{3}}}{{{128}}}}$$
$$\displaystyle{f}{'''}{\left({x}\right)}=-{\frac{{{15}}}{{{8}{x}^{{{\frac{{{7}}}{{{2}}}}}}}}}\Rightarrow{f}{'''}{\left({4}\right)}=-{\frac{{{15}}}{{{1024}}}}$$
Hence, the first four term of the series is shown below.
$$\displaystyle{\frac{{{1}}}{{{2}}}}-{\frac{{{1}}}{{{16}}}}{\left({x}-{4}\right)}+{\frac{{{3}}}{{{128}}}}{\left({x}-{4}\right)}^{{2}}-{\frac{{{15}}}{{{1024}}}}{\left({x}-{4}\right)}^{{3}}$$
So, $$\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}={\frac{{{1}}}{{{2}}}}-{\frac{{{1}}}{{{16}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}+{\frac{{{3}}}{{{128}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}^{{2}}-{\frac{{{15}}}{{{1024}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}^{{3}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}-{\frac{{{1}-{4}\sqrt{{{3}}}}}{{{16}\sqrt{{{3}}}}}}+{\frac{{{49}\sqrt{{{3}}}-{24}}}{{{128}\sqrt{{{3}}}}}}-{\frac{{{5}{\left({145}-{204}\sqrt{{{3}}}\right)}}}{{{1024}\sqrt{{{3}}}}}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}+{\frac{{{1668}-{327}\sqrt{{{3}}}}}{{{1024}}}}$$
Hence.
### Relevant Questions
Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.
$$f(x)=\frac{1}{\sqrt{x}}$$ with $$a=4$$, approximate $$\frac{1}{\sqrt{3}}$$
Any method
a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients.
b. Determine the radius of convergence of the series.
$$f(x)=\cos2x+2\sin x$$
Any method
a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients.
$$f(x)=x^2\cos x^2$$
Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
$$f(x)=\cosh(2x-2),a=1$$
Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
$$f(x)=\tan^{-1}4x,a=0$$
Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
$$\displaystyle{f{{\left({x}\right)}}}={\text{cosh}{{\left({2}{x}-{2}\right)}}},{a}={1}$$
Binomial series
a. Find the first four nonzero terms of the binomial series centered at 0 for the given function.
b. Use the first four terms of the series to approximate the given quantity.
$$f(x)=(1+x)^{\frac{2}{3}}$$, approximate $$(1.02)^{\frac{2}{3}}$$
Taylor series and interval of convergence
a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.
b. Write the power series using summation notation.
c. Determine the interval of convergence of the series.
$$f(x)=\log_3(x+1),a=0$$
$$f(x)=x\ln x-x+1,a=1$$
$$f(x)=2^x,a=1$$ |
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Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
78
Number Lines and the Coordinate Plane
Which should you use to graph?
5.1 What characteristic of an equation dictates whether its solution should be
graphed on a number line or a coordinate plane?
A number line is, like any line, one-dimensional, whereas a coordinate plane
is two-dimensional. Equations with one variable are one-dimensional, so their
solutions must be graphed on a number line. Equations with two different
variables have a two-dimensional graph.
5.2 Graph the values w = 6, x = –5, y = , and z = on the number line in
Figure 5-1.
Figure 5-1: Plot w, x, y, and z on this number line.
To plot w, count six units to the right of 0 and mark the value with a solid dot.
Because x is negative, it should be five units to the left of 0. Plot y two-thirds of a
unit to the right of 0 ( of the distance from 0 to 1 on the number line). It
is easier to plot z if you first convert it from am improper fraction into a mixed
number: . It is located 2.25 units to the left of 0, two full units and
then one-fourth of a unit beyond that. All four values, w, x, y, and z, are
illustrated in Figure 5-2.
Figure 5-2: Negative values, like x and z, are located left of the number 0, and positive
values, like w and y, are located to the right of 0.
5.3 Solve the equation 2x – 3(x + 1) – 5 = –6 and graph the solution.
Use the method described in Problems 4.14.36 to isolate x and solve the
equation. Begin by distributing –3 through the parentheses, then combine like
terms, isolate x on the left side of the equation and eliminate the coefficient of x.
A line is
one-dimensional
because it just
allows horizontal
travel, left and right
along the line. On
a coordinate plane,
however, you can
travel horizontally
and vertically, in two
dimensions. Think of an
ant crawling along a
stick as opposed to
crawling around
on a piece of
paper.
If youre
not sure how
to do this, check
out Problems
2.7–2.8.
Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
79
To plot the solution, mark the value –2 on a number line, as illustrated by
Figure 5-3.
Figure 5-3: Only one real number, x = –2, satisfies the equation 2x – 3(x + 1) – 5 = –6.
5.4 Identify lines k and m and point A on the coordinate plane in Figure 5-4.
Figure 5-4: Lines k and m intersect at point A = (0,0).
A coordinate plane is a two-dimensional plane created by two perpendicular
lines. Typically, the horizontal line (m in Figure 5-4) is called the x-axis and
has equation y = 0. The vertical line (k in Figure 5-4) is called the y-axis and
has equation x = 0. The axes intersect at a point called the origin, which has
coordinates (0,0), point A in Figure 5-4.
A point
on the plane
is represented
by coordinates,
sort of like an
of two numbers in
parentheses and looks
like B = (2,–1). The
rst number in the
parentheses gives
the horizontal location
of the point (+2 means
two units right of the
origin), and the second
number represents
its vertical location
(1 means one unit
below the origin).
You’ll practice
plotting points
in Problem
5.5.
Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
80
Note: Problems 5.5–5.6 refer to the points A = (–1,5); B = (4,4); C = (6,–2); D = ;
E = (0,1); and F = (–6,0).
5.5 Plot the points on the coordinate plane in Figure 5-5.
Figure 5-5: Use the coordinates of points A, B, C, D, E, and F to plot the points on this
coordinate plane.
Each coordinate pair has the form (x,y). The left value, x, indicates a signed
horizontal distance from the y-axis. In other words, positive values of x
correspond with points to the right of the y-axis, and negative x-values produce
points left of the y-axis. Similarly, the right value of each coordinate pair, y,
indicates a signed vertical distance from the x-axis. Coordinates with positive
values of y are located above the x-axis, and negative y-values indicate points
below the x-axis. With this in mind, plot each of the points; the results are
graphed in Figure 5-6.
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# Counting On in Math: Definition & Strategy
Instructor: Shannon Orr
Because mental math is a very useful life skill, there are many different concepts. This lesson is designed to define what counting on means. It will also give strategies for helping students understand and have the ability to use counting on in addition.
## Retrain Your Thinking
When you learn to read, you are taught to start at the beginning of the sentence. When you learn to write, you are taught to start at the beginning of the paper. No surprise that when you are taught how to count, you are told to start at the beginning. There are songs, poems, and pictures that clearly show that counting always starts at the number 1. Now imagine first grade students walking into their classroom, sitting at their desk, and you, the teacher, announcing that they will begin counting from another number. Another number? This can't be real. Can't you just picture the terror on the faces of the little people looking back at you?
As an adult, the idea of counting on may seem pretty clear but to a child who has always started at the number 1, this idea may seem ridiculous and unbearable. Counting on is a mental math skill where, when adding two numbers, you begin counting from the largest number and add the second number to it. Counting on helps us figure out how many more we need of something and how many we will have once more is added.
## Importance
Being able to count starting at any number is important for two main reasons. One reason is that it shows how well a student understands numerical order. For example, when students use rote counting or counting in order, teachers can't really determine how well they understand number order. Every day, Sally may be able to count from 1 to 20 without ever skipping a number. It may appear that Sally understands what number comes first, next, and so on until she reaches 20. The real test comes when the teacher asks Sally to start at number 7 and count to 20. If Sally only knows her numbers starting at 1, she hasn't truly grasped numerical order.
Another reason counting on is important is because it is the beginning steps for teaching students how to add. If Tony understands that after the number 11 there is 12, 13, 14, and so on, he has the foundation to be able to compute 11 + 3 = 14. Until students understand the numerical order, they may have a difficult time understanding how to add.
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# Question: Is A Part Of A Line With Two Endpoints?
## Is a line two dimensional?
We can classify figures on the basis of the dimensions they have.
A circle, triangle, square, rectangle and pentagon are examples of two-dimensional shapes.
A point is zero-dimensional, while a line is one-dimensional, for we can only measure its length..
## Is a line one dimensional?
A line is a straight one-dimensional figure having no thickness and extending infinitely in both directions. A line is sometimes called a straight line or, more archaically, a right line (Casey 1893), to emphasize that it has no “wiggles” anywhere along its length.
## Can a ray sr be called RS?
Ray SR can not be called RS because rays only go in one direction is the ray were RS it would be going the opposite direction or ray SR.
## How many endpoints does a line segment have?
two endpointsA line segment doesn’t go in any direction. It’s just a small piece of a line, with two endpoints.
## What is a part of a line called?
In geometry, a line segment is a part of a line that is bounded by two distinct end points, and contains every point on the line between its endpoints. A closed line segment includes both endpoints, while an open line segment excludes both endpoints; a half-open line segment includes exactly one of the endpoints.
## What is a line with two arrows called?
A line has no beginning point or end point. Imagine it continuing indefinitely in both directions. We can illustrate that by little arrows on both ends. We can name a line using two points on it. This is line EF or line (note the arrowheads).
## What is the difference between a line and a ray?
Student: So, what is the difference between a line and a ray? Mentor: A line goes on to infinity in both directions, but a ray stops on one end. If you cut a line in half, you make two rays. … A line segment is part of a line that has two endpoints.
## Is Ray a part of line?
In geometry, a ray can be defined as a part of a line that has a fixed starting point but no end point. It can extend infinitely in one direction.
## What is the meaning of line?
a continuous extent of length, straight or curved, without breadth or thickness; the trace of a moving point. something arranged along a line, especially a straight line; a row or series: a line of trees. a number of persons standing one behind the other and waiting their turns at or for something; queue.
## Does a line go on forever?
The difference between a line and a line segment is that the line segment has two endpoints and a line goes on forever. A line segment is denoted by its two endpoints, as in . A ray has one endpoint and goes on forever in one direction. … When naming a ray, we always say the endpoint first.
## What does it mean if a line is vertical?
A vertical line is one the goes straight up and down, parallel to the y-axis of the coordinate plane. All points on the line will have the same x-coordinate. … A vertical line has no slope. Or put another way, for a vertical line the slope is undefined.
## What is straight is part of a line and has 2 endpoints?
line segmentA section of a line is a line segment. A line segment has two endpoints. A line segment is a straight line that connects two points. It is the shortest path between the two points. |
Lesson 56 -- Finite and Infinite Sets -- Membership in a Set -- Rearranging Before Graphing
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# Lesson 56 -- Finite and Infinite Sets -- Membership in a Set -- Rearranging Before Graphing - PowerPoint PPT Presentation
Lesson 56 -- Finite and Infinite Sets -- Membership in a Set -- Rearranging Before Graphing. Finite and Infinite Sets. Definitions:. Finite. - Implies the thought of bounded or limited. - can be counted because there is an exact number. Infinite.
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Presentation Transcript
Finite and Infinite Sets
Definitions:
Finite
- Implies the thought of bounded or limited
- can be counted because there is an exact number
Infinite
- Implies the thought of unbounded or unlimited
- cannot be counted because there is a never ending number of things
A set with a finite number of members is called a finite set.
A set with a infinite number of members is called a infinite set.
Membership in a Set
Example 56.1 Represent the following numbers as being members of set K:
Membership in a Set
Example 56.2 Represent the following numbers as being members of set L:
Membership in a Set
Example 56.3 Given the sets A = {0, 1, 3, 5}, B = {0, 4, 6, 7}, and C = {1, 2, 3, 5, 7}, are the following statements true or false?
True
False
True
Membership in a Set
Example 56.4 Given the sets L = {0, 1, 2, 3}, M = {5, 6, 7}, and N = {0, 1}, are the following statements true or false?
False
True
Rearranging Before Graphing
Example 56.5 Graph:
3x + 2y = 4
Rearranging Before Graphing
Example 56.6 Graph:
y – x = 0 |
# Math Snap
## Simplify: $\left(4 m^{4} n^{3}\right)\left(6 m^{2} n^{2}\right)$ [?]m n
#### STEP 1
Assumptions 1. We need to simplify the expression $\left(4 m^{4} n^{3}\right)\left(6 m^{2} n^{2}\right)$. 2. We will use the properties of exponents and multiplication to simplify the expression.
#### STEP 2
First, we will separate the coefficients and the variables with their exponents. $\left(4 m^{4} n^{3}\right)\left(6 m^{2} n^{2}\right) = (4 \cdot 6) \cdot (m^{4} \cdot m^{2}) \cdot (n^{3} \cdot n^{2})$
#### STEP 3
Next, we will multiply the coefficients. $4 \cdot 6 = 24$
#### STEP 4
Now, we will use the property of exponents that states $a^m \cdot a^n = a^{m+n}$ to combine the exponents of $m$. $m^{4} \cdot m^{2} = m^{4+2} = m^{6}$
#### STEP 5
Similarly, we will use the property of exponents to combine the exponents of $n$. $n^{3} \cdot n^{2} = n^{3+2} = n^{5}$
##### SOLUTION
Now, we will combine all the simplified parts together. $24 \cdot m^{6} \cdot n^{5} = 24 m^{6} n^{5}$ The simplified expression is: $24 m^{6} n^{5}$ |
Åk 6–9
English/Русский
4.1 Calculating with Formulas and Functions
It is important to be able to figure out the unknown variable in a function if you know the others. This you do by taking out the unknown variable.
We will look at two examples of formulas where you can do a little different calculation depending on which variables you know:
This is a well known formula which you use to calculate speed.
v: speed (v comes from the word velocity.)
d: distance
t: time
The formula looks like this:
v = d t
If you know the distance and the time, then it is not a problem to calculate the speed.
Example:
d = 240 km
t = 3 h
v = 240 = 80 km/h 3
If you know the speed and the time, then it is not a problem to calculate the distance.
v = 70 km/h
t = 2.5 h
The variable that you don’t know is d (the distance). How do you calculate the distance?
We begin by taking out the distance from the formula:
d · t v · t = We multiply both sides with t. t
d
= v · t
Then we get:
d =
70 · 2.5
d = 175 km
A formula we often use in geometry is the formula for the rectangle’s perimeter.
h: height b: base P: perimeter P = 2h + 2b
If you know the base and the height of the rectangle, it is no problem to calculate the perimeter. But how do you do it if you only know the perimeter and the base and want to calculate the rectangle’s height? We take an example:
P = 38 cm
b = 9 cm
We put these in the formula:
38 = 2h + 2 · 9
38 = 2h + 18
38 - 18 = 2h + 18 - 18
20 = 2h
h = 10 cm
Answer: The rectangle’s height is 10 cm. |
ARCS AND ANGLES - Trigonometric Functions and Their Inverses - Functions - REVIEW OF MAJOR TOPICS - SAT SUBJECT TEST MATH LEVEL 2
## CHAPTER 1Functions
### ARCS AND ANGLES
Although the degree is the chief unit used to measure an angle in elementary mathematics courses, the radian has several advantages in more advanced mathematics. A radian is one radius length. The circle shown in the figure below has radius r. The circumference of this circle is 360°, or 2π radians, so one radian is .
EXAMPLES
1. In each of the following, convert the degrees to radians or the radians to degrees. (If no unit of measurement is indicated, radians are assumed.)
(A) 30°
(B) 270°
(C)
(D)
(E) 24
TIP Although R is used to indicate radians, a radian actually has no units, so the use of R is optional.
SOLUTIONS
(A) To change degrees to radians multiply by , so 30° = 30° .
(B) 270°
(C) To change radians to degrees, multiply by , so
(D)
(E)
In a circle of radius inches with an arc subtended by a central angle of measured in radians, two important formulas can be derived. The length of the arc, s, is equal to r, and the area of the sector, AOB, is equal to .
2. Find the area of the sector and the length of the arc subtended by a central angle of radians in a circle whose radius is 6 inches.
3. In a circle of radius 8 inches, find the area of the sector whose arc length is 6π inches.
4. Find the length of the radius of a circle in which a central angle of 60° subtends an arc of length 8π inches.
The 60° angle must be converted to radians:
Therefore,
EXERCISES
1. An angle of 30 radians is equal to how many degrees?
(A)
(B)
(C)
(D)
(E)
2. If a sector of a circle has an arc length of 2π inches and an area of 6π square inches, what is the length of the radius of the circle?
(A) 1
(B) 2
(C) 3
(D) 6
(E) 12
3. If a circle has a circumference of 16 inches, the area of a sector with a central angle of 4.7 radians is
(A) 10
(B) 12
(C) 15
(D) 25
(E) 48
4. A central angle of 40° in a circle of radius 1 inch intercepts an arc whose length is s. Find s.
(A) 0.7
(B) 1.4
(C) 2.0
(D) 3.0
(E) 40
5. The pendulum on a clock swings through an angle of 25°, and the tip sweeps out an arc of 12 inches. How long is the pendulum?
(A) 1.67 inches
(B) 13.8 inches
(C) 27.5 inches
(D) 43.2 inches
(E) 86.4 inches
|
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Multiplying Polynomials Calculator
Top
Multiplying Polynomials Calculator (polynomial calculator) is an online tool to multiply two polynomials.Multiply polynomials calculator can take any polynomial, such as monomial, binomial, trinomial etc. It is a polynomial multiplier used for any multiplication like, multiplying a polynomial with monomial or binomial or trinomial etc. It is known for multiplying a polynomial by a monomial calculator.
It takes 2 polynomials and calculates their multiplication.
This calculator can also be used as Multiplying Monomials Calculator, Multiplying Trinomials Calculator etc.
Below is given two default polynomials, click "Multiply". It will multiply both the polynomials.
Step by Step Calculation
Step 1 :
First we multiply every term inside the parentheses by the polynomial (By using distributive property rule).
Step 2 :
Example Problems
1. 6x2+7x+8 multiply by 3x?
Step 1 :
Here we have to multiply 6x2+7x+8 by 3x.
So 3x * (6x2+7x+8)
= 3x * (6x2) + 3x * (7x) + 3x * (8)
Step 2 :
Now 3x * (6x2) + 3x * (7x) + 3x * (8)
= 18(x3) + 21(x2) + 24x
18x3+ 21x2+24x
2. 10x2+4x+5 multiply by 2x + 1?
Step 1 :
Here we have to multiply (10x2+4x+5) multiply by (2x + 1)
So (2x + 1) * (10x2+4x+5)
= 2x * (10x2) + 2x * (4x) + 2x * (5) + 1 * (10x2) + 1 * (4x) + 1 * (5)
Step 2 :
So 2x * (10x2) + 2x * (4x) + 2x * (5) + 1 * (10x2) + 1 * (4x) + 1 * (5)
= 20(x3) + 8(x2) + 10x + 10(x2) + 4x + 5
= 20(x3) + 18(x2) + 14x + 5 |
# MUTUALLY EXCLUSIVE AND INDEPENDENT EVENTS
## Presentation on theme: "MUTUALLY EXCLUSIVE AND INDEPENDENT EVENTS"— Presentation transcript:
MUTUALLY EXCLUSIVE AND INDEPENDENT EVENTS
To understand mutually exclusive events To understand independent events To understand when to use the addition and multiplication rules To be able to draw venn diagrams for different types of events
When events have no outcomes in common they are mutually exclusive.
P(A B) = 0 Using the addition rule P(AB) = P(A) + P(B) - P(A B) Since P(A B) = 0 Then P(AB) = P(A) + P(B) (The OR rule) E.G. Getting a 2 or a 3 in one roll of a dice. A B S 1 6 1 6 4 6
When one event has no effect on another they are independent events.
The probability of B happening is the same whether A has happened or not. P(A B) = P(A) x P(B) (The AND rule) E.G. Getting a 5 on a spinner numbered 1,2,3,4,5,5,7,8 and a 3 on a fair unbiased dice. A B S
A=outcome on red die is 3 B=outcome on blue die is 3
A red die and a blue die are rolled and the outcome of each die is recorded. A=outcome on red die is B=outcome on blue die is 3 C=sum of the two dice is D=the outcome on each die is the same Show that A and B are independent P(A)= 6 = 1 P(B)= 6 = 1 P(A B) = 1 x 1 = 1 This is the same as can be seen on the sample space diagram Therefore A and B are independent Red die 1 2 3 4 5 6 Blue die
A=outcome on red die is 3 B=outcome on blue die is 3
A red die and a blue die are rolled and the outcome of each die is recorded. A=outcome on red die is B=outcome on blue die is 3 C=sum of the two dice is D=the outcome on each die is the same Show that C and D are mutually exclusive Red die P(C) has 4 outcomes P(D) has 6 outcomes They have no overlapping outcomes in common. Therefore C and D are mutually exclusive 1 2 3 4 5 6 Blue die
P(one bead is green and the other is blue)
A bag contains 7 green beads and 5 blue beads. A bead is taken from the bag at random, the colour is recorded and the bead is replaced. A second bead is then taken from the bag and its colour is recorded. a) Find the probability that one bead is green and the other is blue b) Show that the event “the first bead is green” and “the second bead is green” are independent P(one bead is green and the other is blue) =P(GB) + P(BG) = 7 x x 7 = = 70 = 35 G B 7 12 5
A bag contains 7 green beads and 5 blue beads
A bag contains 7 green beads and 5 blue beads. A bead is taken from the bag at random, the colour is recorded and the bead is replaced. A second bead is then taken from the bag and its colour is recorded. a) Find the probability that one bead is green and the other is blue b) Show that the event “the first bead is green” and “the second bead is green” are independent P(first is G) = 7 12 P(second is G) =P(GG) + P(BG) = 7 x x 7 = = 84 = 7 P(GG) = 7 x 7 This is the same as multiplying P(G) x P(G) on the tree diagrams, therefore the events are independent. G B 7 12 5
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# A triangle has sides with lengths: 14, 8, and 12. How do you find the area of the triangle using Heron's formula?
Feb 11, 2016
Area of triangle is $47.906$
#### Explanation:
Heron's formula for area of triangle having three sides as $a$, $b$ and $c$ is given by
sqrt(s(s-a)(s-b)(s-c) where $s = \frac{1}{2} \left(a + b + c\right)$
Here three sides are 14, 8 and 12. Thus $s = \frac{1}{2} \left(14 + 8 + 12\right)$ i.e. $s = 17$. Hence $s - a = 3$, $s - b = 9$ and $s = c = 5$.
Hence area of triangle is sqrt(17*3*9*5 or $\sqrt{2295}$ or $47.906$. |
Chapter 3: Probability Topic
3.1 Terminology
• Experiment: is a planned operation carried out under the controlled condition
• Outcome: Result of an experiment
• Sample space: a set of all possible outcomes
• Event: any combination of outcomes that are usually represented by the letters A and B.
• Equally Likely: Each outcome of an experiment has the same probability.
• Probability: the chance of the outcome of an event
• βORβ Event: an outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B.
• "AND" Event: an outcome is in the event A AND B if the outcome is in both A and B at the same time.
• Conditional probability: A given B is written P(A|B). P(A|B) is the probability that event A will occur given that event B has already occurred
• Complement: A given B is written P(A|Bβ)
3.2 Independent and Mutually Exclusive Events
• Independent: One event occurring does not affect the chance that the other event occurs.
• Sampling with replacement: then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
• Sampling without replacement: the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.
• The events are independent if this is true:
• P(A|B) = P(A)
• P(B|A) = P(B)
• P(A AND B) = P(A)P(B)
• Mutually exclusive: If events occur at the same time
• P(A and B) = 0
3.3 Two Basic Rules of Probability
• Multiplication Rule: If A and B are two events defined on a sample space, then: P(A AND B) = P(B)P(A|B). A and B are independent, then P(A|B) = P(A). Then P(A AND B) = P(A|B)P(B) becomes P(A AND B) = P(A)P(B)
• P(A|B) = π(π΄Β ANDΒ π΅) / π(π΅)
• Addition Rule: If A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) - P(A AND B). If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) - P(A AND B) becomes P(A OR B) = P(A) + P(B)
3.4 Contingency Tables
• Contingency table: provides a way of portraying data that can facilitate calculating probabilities.
3.5 Tree and Venn Diagrams
• Tree diagrams: used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities.
• Venn diagrams: a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. |
Free Algebra Tutorials!
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# Solving Linear Inequalities
We know that a linear equation is an equation of the form ax + b = 0. If we replace the equality symbol in a linear equation with an inequality symbol, we have a linear inequality.
Linear Inequality
A linear inequality in one variable x is any inequality of the form ax + b < 0, where a and b are real numbers, with a ≠ 0. In place of < we may use ≤, >, or ≥.
Inequalities that can be rewritten in the form of a linear inequality are also called linear inequalities.
Before we solve linear inequalities, let's examine the results of performing various operations on each side of an inequality. If we start with the inequality 2 < 6 and add 2 to each side, we get the true statement 4 < 8. Examine the results in the following table.
Perform these operations on each side of 2 < 6:
Add 2 Subtract 2 Multiply by 2 Divide by 2 Resultin inequality 4 < 8 0 < 4 4 < 12 1 < 3
All of the resulting inequalities are correct. However, if we perform operations on each side of 2 < 6 using -2, the situation is not as simple. For example, -2 · 2 = -4 and -2 · 6 = -12, but -4 is greater than -12. To get a correct inequality when each side is multiplied or divided by -2, we must reverse the inequality symbol, as shown in the following table.
Perform these operations on each side of 2 < 6:
Add -2 Subtract -2 Multiply by -2 Divide by -2 Resulting inequality 0 < 4 4 < 8 -4 > -12 -1 > -3 |
# How do you solve log_5 9 - log_5 (x-5)=log_5 45?
Jul 30, 2016
$x = \frac{26}{5}$
#### Explanation:
Using the $\textcolor{b l u e}{\text{laws of logarithms}}$
$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\log x - \log y = \log \left(\frac{x}{y}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$
Applies to logarithms to any base.
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\log}_{b} x = {\log}_{b} y \Rightarrow x = y} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(B\right)$
Using (A)
${\log}_{5} 9 - {\log}_{5} \left(x - 5\right) = {\log}_{5} \left(\frac{9}{x - 5}\right)$
Using (B)
${\log}_{5} \left(\frac{9}{x - 5}\right) = {\log}_{5} 45 \Rightarrow \frac{9}{x - 5} = 45$
solve $45 \left(x - 5\right) = 9 \Rightarrow x = \frac{26}{5}$ |
# How do you get the area of a trapezoid?
#### Understand the Problem
The question is asking for the formula or method to calculate the area of a trapezoid, which involves using the lengths of its bases and its height.
You use the formula $$A = \frac{1}{2} \times (a + b) \times h$$.
To find the area of a trapezoid, you use the formula ( A = \frac{1}{2} \times (a + b) \times h ). You need the lengths of the two bases and the height.
#### Steps to Solve
1. Identify the lengths of the bases and the height
To calculate the area of a trapezoid, you need the lengths of the two parallel sides (called bases). Let's denote the lengths of these bases as $a$ and $b$, and the height (the perpendicular distance between the bases) as $h$.
1. Use the area formula for a trapezoid
The formula to find the area of a trapezoid is:
$$A = \frac{1}{2} \times (a + b) \times h$$
Substitute the values of $a$, $b$, and $h$ into this formula.
1. Calculate the area
Perform the arithmetic calculations to find the area.
### Example Calculation:
Let's assume $a = 5$ units, $b = 7$ units, and $h = 4$ units.
Step 1: Identify the lengths of the bases and the height.
Step 2: Substitute the values into the formula:
$$A = \frac{1}{2} \times (5 + 7) \times 4$$
Step 3: Calculate the area:
$$A = \frac{1}{2} \times 12 \times 4 = 24 ext{ square units}$$
So, the area of the trapezoid is 24 square units.
To find the area of a trapezoid, you use the formula ( A = \frac{1}{2} \times (a + b) \times h ). You need the lengths of the two bases and the height. |
Objectives of the lesson:
After this lesson, you will be able to:
Given the vertex of parabola, find an equation of a quadratic function
Given three points of a quadratic function, find the equation that defines the function
Many real world situations that model quadratic functions are data driven. What happens when you are not given the equation of a quadratic function, but instead you need to find one?
In order to obtain the equation of a quadratic function, some information must be given. Significant data points, when plotted, may suggest a quadratic relationship, but must be manipulated algebraically to obtain an equation.
Two forms of a quadratic equation: General Form and Vertex Form
General Form of a Quadratic Equation Vertex Form of a Quadratic Equation
When do I use each form?
When you are given the vertex and at least one point of the parabola, you generally use the vertex form.
When you are given points that lie along the parabola, you generally use the general form.
Vertex Form
Let's use a vertex that you are familiar with: (0,0).
Use the following steps to write the equation of the quadratic function that contains the vertex (0,0) and the point (2,4).
1. Plug in the vertex.
2. Simplify, if necessary.
3. Plug in x & y coordinates of the point given.
4. Solve for "a."
5. Now substitute "a" and the vertex into the vertex form.
Our final equation looks like this:
Now you try.........
Find the equation of a quadratic function with vertex (0,0) and containing the point (4,8).
Final equation:
General Form
Given the following points on a parabola, find the equation of the quadratic function: (1,1); (2,4); (3,9).
By solving a system of three equations with three unknowns, you can obtain values for a, b, and c of the general form.
1. Plug in the coordinates for x and y into the general form. Remember y and f(x) represent the same quantity.
2. Simplify. (Remember the order of operations)
3. Repeat steps 1 & 2 for the other two points.
4. Take two equations at a time and eliminate one variable (c works well)
5. Then repeat using two equations and eliminate the same variable you eliminated in #4.
6. Take the two resulting equations and solve the system (you may use any method).
7. After finding two of the variables, select an equation to substitute the values back into.
8. Find the third variable.
9. Substitute a, b, and c back into the general equation.
1 = a + b + c
4 = 4a + 2b + c
9 = 9a + 3b + c
-1 = - a - b - c -1 = - a - b - c 4 = 4a + 2b + c 9 = 9a + 3b + c 3 = 3a + b 8 = 8a + 2b
3 = 3a + b>>>>>>>>>-6 = -6a - 2b
8 = 8a + 2b>>>>>>>> 8 = 8a + 2b
>>>>>>>>>>>>>>>> 2 = 2a
>>>>>>>>>>>>>>>> 1 = a
If a = 1, then 3 = 3(1) + b, so b = 0.
Now, if a = 1 and b = 0, then 1 = 1 + 0 + c, so c = 0.
By plugging in the values for a, b, and c into the general equation, we obtain the following:
Our final equation looks like:
Now you try.........
Find the equation of a quadratic function with the given points (3,3) ;(6,12); and (9,27).
Conclusion:
You should now be familiar enough with writing quadratic equation that you will be ready for the next activity! It involves graphing using Excel and writing the equations of quadratic functions. |
# Free Basic Arithmetic - 05 Practice Test - CAT
Anil and Sunil can complete a job in 15 days and 20 days respectively. Both of them started it together. Anil left 6 days before the work was completed. Find the time taken (in days) to complete the work.
A. 8
B. 10
C. 12
D. 14
#### SOLUTION
Solution : C
Let total work is LCM of (15, 20) = 60
Work done by Anil in a day = 4 units
Work done by Sunil in a day = 3 units
Anil worked for (t – 6) days and Sunil worked for t days.
4(t – 6) + 3t = 60
t = 12; Option (c).
2nd method:- During last 6 days Sunil alone will work i.e 30% of work will be done by Sunil alone. Remaining 70% of work will be done together. In one day both will do (10015)+(10020)=11.67% , then 70% will complete in 6 days. Total 6+6= 12 days.Option (c).
The average of temperatures at noontime from Tuesday to Saturday is 50; the lowest one is 45, what can be the possible maximum range of the temperatures.
A. 20
B. 25
C. 45
D. 75
#### SOLUTION
Solution : B
The average of the 5 temperatures is: a+b+c+d+e5 = 50
One of these temps is 45: a+b+c+d+455 = 50.
Solving for the variables: a + b + c + d = 205
In order to find the greatest range of temps, we minimize all temps but one. Remember, though, that 45 is the lowest temp possible, so: 45 + 45 + 45 + d = 205 Solving for the variable: d = 70
70 - 45 = 25 Option (b).
2nd method:- For finding maximum range, we will take 4 days minimum out of 5 days i.e 4 days temp. will take 45, then total negative deviations will be 20. Thus for average 50, positive deviation should be also 20 . Thus highest temp will be (50+20)=70 degrees. Then range =70-45 = 25. Option (b).
A shopkeeper buys 50 footballs at Rs.100 per football. He sells a part of the footballs at a profit of 30% and on the remaining; he incurs a loss of 10% . If the overall profit is 10%, then the number of footballs sold at profit is?
A. 25
B. 30
C. 35
D. 40
#### SOLUTION
Solution : A
Total Cost Price = Rs 5000 . Total profit = 10% . Total Selling Price =1.1×5000=5500
Now go from answer options
If the number of footballs sold at profit is 25, then revenue for these 25 balls =1.3×25×100=3250
And the number of footballs sold at loss will be 25; revenue for these balls =0.9×25×100=2250
Sum= 5500 = 10% profit on 5000
Thus, the assumption is right
A crew can row a certain course upstream in 84 minutes; they can row the same course downstream in 9 minutes less than they can row in still water. The time they would take to row down with the stream___
#### SOLUTION
Solution :
As the distance is kept constant, time is inversely proportional to speed and hence the time taken is in HM. Let the time required for rowing in still water be x, then rowing time down with stream will be x-9; x is the HM of 84 and x-9. Therefore x= 72 required time = x – 9 = 63
Three students Ankit, Abhash and Ajay appeared for an exam. If Ankit and Abhash secured 70 marks and 120 marks respectively, the average marks secured by all of them were 60% of the maximum marks. The passing marks in it were 40 less than their average marks and 40% of the maximum marks. The marks secured by Ajay___
#### SOLUTION
Solution :
Let Ajay’s score be x and maximum marks be Y.
Given that, 70+120+x3=60100×Y
190+x3=3Y5
70+120+x340=40100×Y
190+x340=2Y5
3Y540=2Y5
Y = 200 and X = 170
‘A’ borrowed money at 4% per annum simple interest payable yearly and lent it immediately at 6% per annum compound interest payable half-yearly and gained thereby 209 rupees at the end of a year. What sum of money did he borrow?
A. 10,000
B. 1000
C. 20,000
D. None of these
#### SOLUTION
Solution : A
Let amount borrowed be Rs. X
Amount to be paid back = Rs. 1.04X
Amount received due to lending at 6% compound interest payable half -yearly = x(1+6200)2 = 1.0609x
Profit = 1.0609X – 1.04X = 209
0.0209X = 209, X = 10,000; ‘A’ borrowed Rs.10, 000 initially. Option(a)
Alternate Solution
We see that the answer options are far apart and hence the exact calculations are not required. There is an increase of 2% and he is gaining 209 rupees in one year.Hence the amount has to be close to 10000 and hence the answer.
An Aspirant scored 98 in his 19th Mock CSAT and thus increases his average by 4. What is his average after the 19th Mock CSAT.
A. 22
B. 24
C. 26
D. Cannot be determined
#### SOLUTION
Solution : C
Let the average score after 18th Mock CSAT be x.
1st Method:- Weighted Average: 18x+9819=x+4;x=22 answer is x+4=26, option (c).
2nd Method:- Using Alligation
Use the reverse gear approach now.
Put X = 22; we will get 18: 1; answer is x + 4 = 26, option (c).
Next door lives a man with his son. They both work in the same factory. I watch them going to work through my window. The father leaves for work 10 min earlier than his son. One day I asked him about it and he told me he takes 30 min to walk to his factory ,whereas his son is able to cover the distance in only 20 min. I wondered if the father were to leave the house 5 min earlier than his son, how soon would the son catch up with the father?
A. 3.33 mins
B. 8.3 mins
C. 5.76 mins
D. 8.25 mins
#### SOLUTION
Solution : B
Let speed of son be s,and speed of father be f,then
sf=32 (inverse ratio of the time) ( If D is the distance ,then Df=30 and Ds=20 which implies sf=32)
Let speed of dad be 20 and son be 30m/s
Now in 5 mins, the father covers 100 m
And son covers 150 m
The son covers 100 m in 3.33 min,
Hence the son will be able to catch up with the father after 8.3 mins from the start. Option (b).
Pranab is thrice as efficient as Ghulam, and Ghulam is twice as efficient as Krishna. If the trio work together, how long will they take to complete a job which Ghulam completes in 25 days.
A. 509
B. 259
C. 499
D. 249
#### SOLUTION
Solution : A
Let total work be 100 units.
Ghulam’s one day work = 10025 = 4 units.
Pranab’s one day work = 12 units.
Krishna’s one day work = 2 units.
Combined work of the trio = 4 + 12 + 2 = 18 units.
Hence, They will take 10018=509 days. Option(a).
2nd method:- In place of unit, we can take in percentage form also.In one day, G=4% , P=12% , K=2%
Total in one day= 18% , Then 100% in 1008 days i.e 509 days. Option(a).
Minister of Railways, Shri DineshTrivedi travels from Bangalore to Delhi at 50 km/hr and returns at 100 km/hr. What was his average speed?
A. 60
B. 75
C. 66.67
D. Can’t be determined
#### SOLUTION
Solution : C
Average Speed =2×50×10050+100=10,000150=66.67 . Distance is kept constant, hence Speed and Time are inversely proportional, hence, we can use Harmonic Mean. H.M=(2aba+b) Option(c).
Alternative, Average speed= total distance / total time= 66.67. Option (c).
Minister of Power, Shri Sushilkumar Shinde and Minister of New and Renewable Energy, Dr. Farooq working together can do a piece of work in 18 days, Minister of New and Renewable Energy, Dr. Farooq and Minister of Petroleum and Natural Gas, Shri S. Jaipal Reddy in 24 days, Minister of Petroleum and Natural Gas and Minister of Power in 36 days. If all of them work together, how many days will it take?
A. 15
B. 17
C. 18
D. 16
#### SOLUTION
Solution : D
The LCM of 18, 24 and 36 is 72. Let total work is be 72 units.
Sushilkumar Shinde and, Dr. Farooq’s 1 day work =118×72=4 units
Dr. Farooq and S. Jaipal Reddy’s 1 day work =124×72=3 units
S. Jaipal Reddy and Sushilkumar Shinde’s 1 day work =136×72=2 units
Combined work of Ministers =4+3+22=92 units
Hence, they will take 729×2=16 days;
Option (d).
The population of a town was 3600 three years back. It is 4800 right now. . What is the rate of growth of population, if it has been constant over the years and has been compounding annually?
A. 15%
B. 5%
C. 12%
D. 10%
#### SOLUTION
Solution : D
Option (d)
Conventional approach :
Final Value=Initial Value(1+r100)t
Thus, rate =(FVIV)1t1
FVIV=4836=1.33
Going from answer options
If 10 is the answer
1.13=112×11=12×111.33
Therefore r is approximately 10%.
Amit and Francis are on detention and they need to write 65 pages together, Amit writing for an hour extra than Francis. Francis can write 2 pages/hour more than Amit and therefore, he did 5 pages more than Amit.What is the speed per hour of Francis? (pages/hr)
A. 6
B. 5
C. 7
D. 12
#### SOLUTION
Solution : C
Option (c )
Total = 65 pages
Go from answer options
Option (a)F=6
A= 4
Together 1024 in one hour . This multiplied by (65-4=61),
should be an integer which is not true. Hence, answer a is eliminated Similarly, options (b),(d)are eliminated.
Option (c) =3525×60,which is an integer
Reverse gear method 2
If Francis writes 35 pages, then Amit writes 30 pages. Use answer options, to find each person’s individual time the time difference should be one hour. This happens only with option (c)
20% of a larger number is 2.3 less than 30% of a smaller number. The larger number also exceeds the smaller number by 10. What is the value of the larger number?
A. 34
B. 43
C. 62
D. 53
#### SOLUTION
Solution : D
Let the larger number and smaller number is L and S respectively.
0.2L=0.3S2.3 & 2L3S=23
LS=10 & 2L2S=20
Solving the above two equations we get L = 53. Hence option (e)
Ram bought two cars and C.P of 1st car = 150000, C.P of 2nd car = 175000. He sold first car at a profit of 30% and second car at a loss of 10%. What is his profit?___
#### SOLUTION
Solution : 1st car:
C.P = 150000
S.P = 195000
Profit = 45000
2nd car:
C.P = 175000
S.P = 157500
loss = 17500
Overall profit = 45000 - 17500 = 27500 |
Blog
OLD ID: 36
The common errors children make are mixing up the answers 63 and 64, and 56 and 54. If your child has a tendency to confuse the answers to 7x9 and 8x8 or 7x8 and 6x9, you can sometimes help them sort it out by working from the answer back to the factors.
A good starting point is talking about patterns of the answers when you have odd x odd, even x even or odd x even. Start with the 2x table and you’ll notice all the answers are even. Look at the other times tables – it always works. If you’ve got an even number as a factor, the product (answer) will be even.
So how can you get an odd product (answer)? Only if BOTH factors are odd. Hence an answer of 63 could only come from two odd numbers being multiplied – therefore it can’t be the answer to 8x8.
Continuing our perusal of answers, take a close look at the multiples of 9. If you add their digits, all the answers to the 9x table add to nine. eg 18 (1+8=9), 27 (2+7=9), 36 (3+6=9) etc.
It always works, and it works both ways. If you have a number with a digit sum of 9 you know you have a multiple of nine. Eg 67392: 6+7+3+9+2=27, 2+7=9 so 67392 has a digit sum of 9, therefore 9x something must equal 67392 and sure enough, 9x 7488=67392
You can use this interesting fact about the 9x table to instantly recognize 63 as a multiple of 9 (because 6+3=9) and 54 as a multiple of 9 (because 5+4=9).
Taking time to really examine the answers like this gives children more information about the times tables they are trying to rote learn. Simple tricks like this can help clear up 63/64 and 54/56 confusion because more information makes things easier to remember. |
# 1.3 Radicals and rational exponents
Page 1 / 11
In this section students will:
• Evaluate square roots.
• Use the product rule to simplify square roots.
• Use the quotient rule to simplify square roots.
• Add and subtract square roots.
• Rationalize denominators.
• Use rational roots.
A hardware store sells 16-ft ladders and 24-ft ladders. A window is located 12 feet above the ground. A ladder needs to be purchased that will reach the window from a point on the ground 5 feet from the building. To find out the length of ladder needed, we can draw a right triangle as shown in [link] , and use the Pythagorean Theorem.
$\begin{array}{ccc}\hfill {a}^{2}+{b}^{2}& =& {c}^{2}\hfill \\ \hfill {5}^{2}+{12}^{2}& =& {c}^{2}\hfill \\ \hfill 169& =& {c}^{2}\hfill \end{array}$
Now, we need to find out the length that, when squared, is 169, to determine which ladder to choose. In other words, we need to find a square root. In this section, we will investigate methods of finding solutions to problems such as this one.
## Evaluating square roots
When the square root of a number is squared, the result is the original number. Since $\text{\hspace{0.17em}}{4}^{2}=16,$ the square root of $\text{\hspace{0.17em}}16\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}4.\text{\hspace{0.17em}}$ The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root.
In general terms, if $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is a positive real number, then the square root of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is a number that, when multiplied by itself, gives $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ The square root could be positive or negative because multiplying two negative numbers gives a positive number. The principal square root is the nonnegative number that when multiplied by itself equals $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ The square root obtained using a calculator is the principal square root.
The principal square root of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is written as $\text{\hspace{0.17em}}\sqrt{a}.\text{\hspace{0.17em}}$ The symbol is called a radical , the term under the symbol is called the radicand , and the entire expression is called a radical expression .
## Principal square root
The principal square root of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is the nonnegative number that, when multiplied by itself, equals $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ It is written as a radical expression , with a symbol called a radical over the term called the radicand : $\text{\hspace{0.17em}}\sqrt{a}.$
Does $\text{\hspace{0.17em}}\sqrt{25}=±5?$
No. Although both $\text{\hspace{0.17em}}{5}^{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\left(-5\right)}^{2}\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}25,$ the radical symbol implies only a nonnegative root, the principal square root. The principal square root of 25 is $\text{\hspace{0.17em}}\sqrt{25}=5.$
## Evaluating square roots
Evaluate each expression.
1. $\sqrt{100}$
2. $\sqrt{\sqrt{16}}$
3. $\sqrt{25+144}$
4. $\sqrt{49}-\sqrt{81}$
1. $\sqrt{100}=10\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}{10}^{2}=100$
2. $\sqrt{\sqrt{16}}=\sqrt{4}=2\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}{4}^{2}=16\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{2}^{2}=4$
3. $\sqrt{25+144}=\sqrt{169}=13\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}{13}^{2}=169$
4. $\sqrt{49}-\sqrt{81}=7-9=-2\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}{7}^{2}=49\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{9}^{2}=81$
For $\text{\hspace{0.17em}}\sqrt{25+144},$ can we find the square roots before adding?
No. $\text{\hspace{0.17em}}\sqrt{25}+\sqrt{144}=5+12=17.\text{\hspace{0.17em}}$ This is not equivalent to $\text{\hspace{0.17em}}\sqrt{25+144}=13.\text{\hspace{0.17em}}$ The order of operations requires us to add the terms in the radicand before finding the square root.
Evaluate each expression.
1. $\sqrt{225}$
2. $\sqrt{\sqrt{81}}$
3. $\sqrt{25-9}$
4. $\sqrt{36}+\sqrt{121}$
1. $15$
2. $3$
3. $4$
4. $17$
## Using the product rule to simplify square roots
To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions. For instance, we can rewrite $\text{\hspace{0.17em}}\sqrt{15}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}\sqrt{3}\cdot \sqrt{5}.\text{\hspace{0.17em}}$ We can also use the product rule to express the product of multiple radical expressions as a single radical expression.
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 |
# A triangle has an area of 77 square inches. How do you find the length of the base if the base is 3 inches more than the height?
Jan 8, 2016
Solve the quadratic equation to get $b = 14$
#### Explanation:
If the base of a triangle is $b$ and the height is $h$, the area $A = \frac{1}{2} b h$
In this example $h = b - 3$ and $A = 77$
$\therefore 77 = \frac{1}{2} b \left(b - 3\right)$
$154 = {b}^{2} - 3 b$
${b}^{2} - 3 b - 154 = 0$
$154 = 77 \cdot 2 = 7 \cdot 11 \cdot 2 = 11 \cdot 14$
$11 - 14 = - 3$ so these are the factors we need
$\left(b - 14\right) \left(b + 11\right) = 0$
$b$ cannot be equal to $- 11$ as we are dealing with a real entity (the triangle) so $b = 14$ |
# Question #935d4
Apr 9, 2017
$\frac{2 {x}^{3} {y}^{3}}{6 {x}^{-} 4 {y}^{4}} = {x}^{7} / \left(3 y\right)$
#### Explanation:
Given: $\frac{2 {x}^{3} {y}^{3}}{6 {x}^{-} 4 {y}^{4}}$
We can see the integer multipliers: $\frac{2}{6} = \frac{1}{3}$
The $x$ component is ${x}^{3} / {x}^{-} 4 = {x}^{3} / \left(\frac{1}{x} ^ 4\right) = {x}^{7}$ $\to$after invert and multiply
The $y$ component is ${y}^{3} / {y}^{4} = \frac{1}{y}$ $\to$dividing exponents = subtract
Then we can multiply all the components:
$\frac{2 {x}^{3} {y}^{3}}{6 {x}^{-} 4 {y}^{4}} = \left(\frac{1}{3}\right) {x}^{7} \left(\frac{1}{y}\right) = {x}^{7} / \left(3 y\right)$ |
# Lesson 6: Methods for Multiplying Decimals
Let’s look at some ways we can represent multiplication of decimals.
## 6.1: Which One Doesn’t Belong: Products
Which expression doesn’t belong? Explain your reasoning.
A. $2 \boldcdot (0.3)$
B. $2 \boldcdot 3 \boldcdot (0.1)$
C. $6 \boldcdot (0.1)$
D. $(0.1) \boldcdot 6$
## 6.2: Using Properties of Numbers to Reason about Multiplication
1. Elena and Noah used different methods to compute $(0.23) \boldcdot (1.5)$. Both computations were correct.
Analyze the two methods, then discuss these questions with your partner.
• Which method makes more sense to you? Why?
• What might Elena do to compute $(0.16) \boldcdot (0.03)$? What might Noah do to compute $(0.16) \boldcdot (0.03)$? Will the two methods result in the same value?
1. Compute each product using the equation $21 \boldcdot 47 = 987$ and what you know about fractions, decimals, and place value. Explain or show your reasoning.
1. $(2.1) \boldcdot (4.7)$
1. $21 \boldcdot (0.047)$
1. $(0.021) \boldcdot (4.7)$
## 6.3: Using Area Diagrams to Reason about Multiplication
1. In the diagram, the side length of each square is 0.1 unit.
1. Explain why the area of each square is not 0.1 square unit.
1. How can you use the area of each square to find the area of the rectangle? Explain or show your reasoning.
2. Explain how the diagram shows that the equation $(0.4) \boldcdot (0.2) = 0.08$ is true.
1. Label the squares with their side lengths so the area of this rectangle represents $40 \boldcdot 20$.
1. What is the area of each square?
2. Use the squares to help you find $40 \boldcdot 20$. Explain or show your reasoning.
1. Label the squares with their side lengths so the area of this rectangle represents $(0.04) \boldcdot (0.02)$.
Next, use the diagram to help you find $(0.04) \boldcdot (0.02)$. Explain or show your reasoning.
## Summary
Here are three other ways to calculate a product of two decimals such as $(0.04) \boldcdot (0.07)$.
• First, we can multiply each decimal by the same power of 10 to obtain whole-number factors.
Because we multiplied both 0.04 and 0.07 by 100 to get 4 and 7, the product 28 is $(100 \boldcdot 100)$ times the original product, so we need to divide 28 by 10,000.
$$(0.04) \boldcdot 100 = 4$$ $$(0.07) \boldcdot 100 = 7$$ $$4 \boldcdot 7=28$$
$$28\div 10,\!000=0.0028$$
• Second, we can write each decimal as a fraction, $0.04 = \frac{4}{100}$ and $0.07 = \frac{7}{100}$, and multiply them. $$\frac{4}{100} \boldcdot \frac{7}{100} = \frac{28}{10,\!000}=0.0028$$
• Third, we can use an area model. The product $(0.04) \boldcdot (0.07)$ can be thought of as the area of a rectangle with side lengths of 0.04 unit and 0.07 unit.
In this diagram, each small square is 0.01 unit by 0.01 unit. Its area, in square units, is therefore $\left(\frac{1}{100} \boldcdot \frac{1}{100}\right)$, which is $\frac{1}{10,000}$.
Because the rectangle is composed of 28 small squares, its area, in square units, must be: $$28 \boldcdot \frac{1}{10,000} = \frac{28}{10,000}=0.0028$$
All three calculations show that $(0.04) \boldcdot (0.07) = 0.0028$. |
# Chapter 9 - Quadratic Functions and Equations - 9-5 Completing the Square - Practice and Problem-Solving Exercises - Page 579: 30
$v=7,-2$
#### Work Step by Step
$2v^2-10v-20=8$ Re-write the equations as: $v^2-5v-10=4$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=-5$ Therefore, $b^2=4ac$ $\implies$ $c=\dfrac{b^2}{4a}$ Thus, $c=\dfrac{b^2}{4a}=\dfrac{(-5)^2}{4}=\dfrac{25}{4}$ To complete the square, add $\dfrac{25}{4}$ on both sides. $v^2-5v-10+\dfrac{25}{4}=4+\dfrac{25}{4}$ $\implies (v-\dfrac{5}{2})^2=\dfrac{81}{4}$ $\implies (v-\dfrac{5}{2})= \dfrac{9}{2}$ and $\implies (v-\dfrac{5}{2})= -\dfrac{9}{2}$ or, $v=7,-2$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
# 3.1: Inequalities in One Variable
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When learning about domain and range, you learned about inequalities and using set-builder and interval notation to represent them. The process is very similar to solve equations, but instead of the solution being a single value, the solution will be an inequality.
Notice that if an inequality is true, like 2 < 5, then these operations result in a true statement as well, just like with equations:
Adding a number to both sides:
2 + 4 < 5 + 4 6 < 9 True
Subtracting a number on both sides:
2 – 3 < 5 – 3 -1 < 2 True
Multiplying a positive number on both sides:
2(3) < 5(3) 6 < 15 True
Dividing by a positive number on both sides:
2/2 < 5/2 1 < 2.5 True
We can use these operations just like when solving equations.
Example (PageIndex{1})
Solve [3x + 7 geq 1 onumber]
Solution
[3x + 7 geq 1 onumber ]
Subtract 7 from both sides
[3x geq - 6 onumber ]
Divide both sides by 3
[x geq - 2 onumber ]
This inequality represents the solution set. It tells us that all numbers greater than or equal to -2 will satisfy the original inequality. We could also write this solution in interval notation, as ( [ - 2,infty )).
To understand what is happening, we could also consider the problem graphically. If we were to graph the equation (y = 3x + 7 ), then solving (3x + 7 geq 1 ) would correspond with asking “for what values of (x) is (y geq 1)”. Notice that the part of the graph where this is true corresponds to where (x geq - 2).
While most operations in solving inequalities are the same as in solving equations, we run into a problem when multiplying or dividing both sides by a negative number. Notice, for example:
2(-3) < 5(-3) -6 < -15 Not True
To account for this, when multiplying or dividing by a negative number, we must reverse the sign of the inequality.
Rules for Solving Linear Inequalities
1. You may add or subtract a positive or negative number to both sides of the inequality.
2. You can multiply or divide both sides of the inequality by a positive number.
3. You can multiply or divide both sides of the inequality by a negative number, but you must reverse the direction of the inequality.
Example (PageIndex{2})
Solve [12 - 4x < 6 onumber ]
Solution
[12 - 4x < 6 onumber ]
Subtract 12 from both sides
[ - 4x < - 6 onumber ]
Divide both sides by -4, and reverse the inequality direction
[x > frac{ - 6}{ - 4} onumber ]
Simplify
[x > frac{3}{2} onumber ]
Exercise (PageIndex{1})
Solve: [6 + 2x leq 18 + 5x onumber ]
[x geq - 4 onumber ]
Example (PageIndex{3})
A company spends $1200 per day on overhead and labor, and each item they produce costs$5 for materials. If they sell the items for \$15 each, how many items will they need to sell each day for their profits to be positive?
Solution
While we could solve this problem using equations, it also lends itself to inequalities, since we want the profit to be positive: (P > 0).
Costs: (C(q) = 1200 + 5q)
Revenue: (R(q) = 10q)
Profit: (P(q) = 10q – (1200 + 5q) = 5q – 1200)
Solving (P(q) > 0):
[egin{align*} 5q – 1200 &> 0 5q &> 1200 q &> 240 end{align*}]
The company will need to less at least 240 items a day to make a profit.
## Compound Inequalities
Compound inequalities are inequalities that consist of more than one part. The most common type is called a tripartite inequality. The basic version looks like this:
[ - 1 < 3x + 5 < 14 onumber ].
When we write these, it is important that both inequalities point in the same direction and that the “outside” inequality is also true – in this case (-1 < 14) is true, so this is valid. Expressions like (10 < x < 2 ) and (1 < x > 5 ) are not valid notation.
The most universal way to solve a tripartite inequality is to:
1. Break it into two separate inequalities
2. Solve each inequality separately
3. Combine the solutions if possible.
Example (PageIndex{4})
Solve [ - 1 < - 3x + 5 < 14 onumber ]
Solution
First we separate this into two inequalities:
[ - 1 < - 3x + 5 quad ext{and} quad - 3x + 5 < 14 onumber ]
Now we solve each:
[ - 6 < - 3x quad ext{and} quad -3x < 9 onumber ]
[2 > x onumber quad ext{and} quad x > - 3 onumber ]
Now we can combine these solution sets. The numbers where both (2 > x) and (x > - 3 ) are true is the set:
[2 > x > - 3 onumber ]
While this solution is valid and correct, it is more common to write the solution to tripartite inequalities with the smaller number on the left. We could rewrite the solution as:
[ - 3 < x < 2 onumber ]
This also has the advantage of corresponding better with the answer in interval notation: ((-3, 2))
With this particular inequality, it would also be possible to skip the step of breaking it apart, and instead just subtract 5 from all three “parts” of the inequality. This works for simple problems like this, but may fail if the inequality has variables in more than one “part” of the inequality.
Exercise (PageIndex{2})
Solve: [4 leq 2x + 6 < 16 onumber ]
[-1 leq x < 5 onumber ]
In interval notation, this is ( [-1, 5) ).
## Absolute Value
So far in this section we have been looking at inequalities that are linear. We will now turn to absolute value inequalities. The absolute value function is a piecewise-defined function made up of two linear functions.
Absolute Value of a Function
The absolute value function can be defined as
[f(x) = left| x ight| = left{ egin{array}{*{20}{c}} x & ext{ if }& x geq 0 - x& ext{ if }& x < 0end{array} ight. onumber]
The graph of the absolute value looks like a V:
The absolute value function is commonly used to determine the distance between two numbers on the number line. Given two values (a) and (b), then (left| a - b ight|) will give the distance, a positive quantity, between these values, regardless of which value is larger.
Example (PageIndex{5})
Describe all values, (x), within a distance of 4 from the number 5.
Solution
We want the distance between (x) and 5 to be less than or equal to 4. The distance can be represented using the absolute value, giving the expression
[left| x - 5 ight| leq 4 onumber]
Example (PageIndex{6})
A 2010 poll reported 78% of Americans believe that people who are gay should be able to serve in the US military, with a reported margin of error of 3%[1]. The margin of error tells us how far off the actual value could be from the survey value[2]. Express the set of possible values using absolute values.
[1] http://www.pollingreport.com/civil.htm, retrieved August 4, 2010
[2] Technically, margin of error usually means that the surveyors are 95% confident that actual value falls within this range.
Solution
Since we want the size of the difference between the actual percentage, (p), and the reported percentage to be less than 3%,
[left| p - 78 ight| leq 3 onumber]
Exercise (PageIndex{3})
Students who score within 20 points of 80 will pass the test. Write this as a distance from 80 using the absolute value notation.
Using the variable (p), for passing, [left| {p - 80} ight| leq 20 onumber]
## Solving Absolute Value Equations
To solve an equation like (8 = left| 2x - 6 ight|), we can notice that the absolute value will be equal to eight if the quantity inside the absolute value were 8 or -8. This leads to two different equations we can solve independently:
[egin{align*} 2x - 6 &= 8 2x &= 14 x &= 7 end{align*}]
or
[egin{align*} 2x - 6 &= - 8 2x &= - 2 x &= - 1 end{align*}]
Solutions to Absolute Value Equations
An equation of the form (left|A ight| = B), with (B geq 0), will have solutions when
Example (PageIndex{7})
Solve: (0 = left| 4x + 1 ight| - 7)
Solution
[0 = left| {4x + 1} ight| - 7 onumber]
Isolate the absolute value on one side of the equation
[7 = left| {4x + 1} ight| onumber]
Now we can break this into two separate equations:
[ egin{align*} 7 &= 4x + 1 6 &= 4x x &= frac{6}{4} = frac{3}{2} end{align*} ]
or
[ egin{align*} - 7 &= 4x + 1 - 8 &= 4x x &= frac{- 8}{4} = - 2 end{align*} ]
There are two solutions: (x = frac{3}{2}) and (x = -2).
Example (PageIndex{8})
Solve (1 = 4left| x - 2 ight| + 2)
Solution
Isolating the absolute value on one side the equation,
[egin{align*} 1 &= 4left| x - 2 ight| + 2 -1 &= 4left| x - 2 ight| - frac{1}{4} &= left| x - 2 ight| end{align*}]
At this point, we notice that this equation has no solutions – the absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value.
Exercise (PageIndex{4})
Find the horizontal & vertical intercepts for the function (f(x) = - left| {x + 2} ight| + 3)
Horizontal: ( (1,0) ) and ((-5,0 ))
Vertical: ((0,1))
## Solving Absolute Value Inequalities
When absolute value inequalities are written to describe a set of values, like the inequality (left| x - 5 ight| leq 4) we wrote earlier, it is sometimes desirable to express this set of values without the absolute value, either using inequalities, or using interval notation.
We will explore two approaches to solving absolute value inequalities:
1. Using a graph
2. Using test values
Example (PageIndex{9})
Solve [left| {x - 5} ight| leq 4 onumber]
Solution
With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where (left| {x - 5} ight| = 4). We do this because the absolute value is a nice friendly function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve (left| {x - 5} ight| = 4),
[egin{align*} x - 5 &= 4 x &= 9 end{align*}]
or
[egin{align*} x - 5 = - 4 x = 1 end{align*} ]
To use a graph, we can sketch the function (f(x) = left| {x - 5} ight|). To help us see where the outputs are 4, the line (g(x) = 4) could also be sketched.
On the graph, we can see that indeed the output values of the absolute value are equal to 4 at (x = 1) and (x = 9). Based on the shape of the graph, we can determine the absolute value is less than or equal to 4 between these two points, when (1 leq x leq 9). In interval notation, this would be the interval ([1,9]).
As an alternative to graphing, after determining that the absolute value is equal to 4 at (x = 1) and (x = 9), we know the graph can only change from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals: (x<1, 19). To determine when the function is less than 4, we could pick a value in each interval and see if the output is less than or greater than 4.
[egin{array}{llll}
ext { Interval } & ext { Test } x & f(x) & <4 ext { or }>4 ?
hline x<1 & 0 & |0-5|=5 & ext { greater }
1x>9 & 11 & |11-5|=6 & ext { greater }
end{array}
onumber]
Since (1 leq x leq 9) is the only interval in which the output at the test value is less than 4, we can conclude the solution to (left| {x - 5} ight| leq 4) is (1 leq x leq 9).
Example (PageIndex{10})
Given the function (f(x) = - frac{1}{2}left| {4x - 5} ight| + 3), determine for what (x) values the function values are negative.
Solution
We are trying to determine where (f(x) < 0), which is when (- frac{1}{2}left| {4x - 5} ight| + 3 < 0). We begin by isolating the absolute value:
[- frac{1}{2}left| {4x - 5} ight| < - 3 onumber]
When we multiply both sides by -2, it reverses the inequality,
[left| {4x - 5} ight| > 6 onumber ]
Next we solve for the equality (left| {4x - 5} ight| = 6)
[egin{align*} 4x - 5 &= 6 4x &= 11 x &= frac{11}{4} end{align*} onumber]
or
[egin{align*} 4x - 5 &= - 6 4x &= - 1 x &= frac{- 1}{4} end{align*} ]
We can now either pick test values or sketch a graph of the function to determine on which intervals the original function value are negative. Notice that it is not even really important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at (x = frac{- 1}{4}) and (x = frac{11}{4}), and that the graph has been flipped.
From the graph of the function, we can see the function values are negative to the left of the first horizontal intercept at (x = frac{ - 1}{4}), and negative to the right of the second intercept at (x = frac{11}{4}). This gives us the solution to the inequality:
In interval notation, this would be (left( - infty ,frac{ - 1}{4} ight) cup left( frac{11}{4},infty ight))
Exercise (PageIndex{6})
Solve ( - 2left| k - 4 ight| leq - 6)
(k < 1 ) or (k > 7); in interval notation this would be (left( -infty ,1 ight) cup left( 7,infty ight))
There is a third approach to solving absolute value inequalities that is formulaic. While it works, and you are welcome to use it, it is much more likely that you will remember the other approaches.
Solutions to Absolute Value Inequalities
To solve (left| A ight| < B ), solve: ( - B < A < B)
To solve (left| A ight| > B ), solve: (A > B ) or (A < - B )
Example (PageIndex{11})
Solve (3left| x + 4 ight| - 2 geq 7)
Solution
We need to start by isolating the absolute value:
[3left| {x + 4} ight| - 2 geq 7 onumber ]
[3left| {x + 4} ight| geq 9 onumber ]
Divide both sides by 3
[left| {x + 4} ight| geq 3 onumber ]
Now we can break this apart and solve each piece separately:
[egin{align*} x + 4 &geq 3 onumber x &geq - 1end{align*} ]
or
[egin{align*} x + 4 &leq - 3 x &leq - 7 end{align*} ]
In interval notation, this would be (( -infty , -7] cup [ -1,infty )).
Important Topics of this Section
The properties of the absolute value function
Solving absolute value equations
Finding intercepts
Solving absolute value inequalities
## INEQUALITIES IN ONE VARIABLE
In Sections 1.1 and 2.1, we saw that of two different numbers, the graph of the lesser number lies to the left of the graph of the greater number on a number line. These order relationships can be expressed by using the following symbols:
&le means "is less than or equal to,"
&ge means "is greater than or equal to."
"1 is less than 3" can be written as 1 -5.
"2 is less than or equal to x" can be written as 2 &le x.
"4 is greater than or equal to y" can be written as 4 &ge y.
Statements that involve any of the above symbols are called inequalities. Inequalities such as
are said to be of opposite order or opposite sense because in one case the left-hand member is less than the right-hand member and in the other case the left-hand member is greater than the right-hand member.
PROPERTIES OF INEQUALITIES
In Section 3.1, we saw that a first-degree equation in one variable has only one solution. But a first-degree inequality has an infinite number of solutions. For example, the graphs of the infinite number of integer solutions of the inequality x > 3 are shown in Figure 3.1.
Sometimes it is not possible to determine the solutions of a given inequality simply by inspection. But using the following properties, we can form equivalent inequalities (inequalities with the same solutions) in which the solution is evident by inspection.
1. If the same expression is added to or subtracted from each member of an inequality, the result is an equivalent inequality in the same order.
a Example 1 a. Because 3 Example 2 a. Because 2 0,
5(z) Example 3 a. Because 3 5( -2) or -6 >-10
The three properties above also apply to inequalities of the form a > b, as well as a Example 4 Solve , where x is an integer.
Solution Multiplying each member by 2 (a positive number), we have
Then dividing each member by 3, we get
The graph of this inequality is
In the above example, all the inequalities were in the same order because we only applied Property 2 above. Now consider the following inequality.
Example 5 Solve - 3x + 1 > 7, where x is an integer.
Solution Adding - 1 to each member, we get
Now we apply Property 3 and divide each member by -3. In this case we have to reverse the order of the inequality.
When solving word problems involving inequalities, we follow the six steps outlined on page 115 except the word equation will be replaced by the word inequality.
## 3.1: Inequalities in One Variable
This course is intended for students looking to create a solid algebraic foundation of fundamental mathematical concepts from which to take more advanced courses that use concepts from precalculus, calculus, probability, and statistics. This course will help solidify your computational methods, review algebraic formulas and properties, and apply these concepts model real world situations. This course is for any student who will use algebraic skills in future mathematics courses. Topics include: the real numbers, equalities, inequalities, polynomials, rational expressions and equations, graphs, relations and functions, radicals and exponents, and quadratic equations.
Module 3: Solving Inequalities
The relative position of two points on a coordinate line is used to define an inequality relationship on the set of real numbers. We say that a is less than b, written a<b, when the real number a lies to the left of the real number b on the coordinate line. From this definition, other inequalities naturally follow.
## Linear Equations and Inequalities in One Variable
Note: These PDF files are included to make printing easier. The links are not live in this format. For the most updated version of materials and working links, scroll down to the Big Ideas and open the Google Doc versions, which are updated continuously.
### Unit Resources:
#### Initial Tasks See 2 items Hide 2 items
The Initial Task for a unit is intended to both preview the upcoming mathematics for a student and help teachers see how their students understand the mathematics prior to the unit.
Please comment below with questions, feedback, suggestions, or descriptions of your experience using this resource with students.
#### Linear Equations and Inequalities in One Variable
Please comment below with questions, feedback, suggestions, or descriptions of your experience using this resource with students.
#### Big Idea 1: Equations can be derived from functions. See 2 items Hide 2 items
The resources for Big Idea 1 focus on analyzing quantities from a situation and using them to write an equation or inequality. The distinction between a function and equation is also explored.
Please comment below with questions, feedback, suggestions, or descriptions of your experience using this resource with students.
#### Big Idea 2: A solution set makes an equation or inequality true. See 2 items Hide 2 items
The resources for Big Idea 2 focus on solving equations and inequalities, and using visual models and properties to explain each step in the process involved in solving equations and inequalities.
Please comment below with questions, feedback, suggestions, or descriptions of your experience using this resource with students.
#### Big Idea 3: Equations and inequalities can be represented in multiple, equivalent ways. See 2 items Hide 2 items
The resources for Big Idea 3 focus on analyzing quantities from a situation and using them to write and solve equations or inequalities with the variable on both sides of the equality or inequality sign.
Please comment below with questions, feedback, suggestions, or descriptions of your experience using this resource with students.
#### Formative Assessment Lesson See 1 item Hide 1 item
A Formative Assessment Lesson (also known as a Classroom Challenge) is a carefully designed lesson that both supports teachers in understanding how students make sense of the unit's mathematics and offers students opportunities to revisit and deepen their understanding of that mathematics.
A Classroom Challenge (aka formative assessment lesson) is a classroom-ready lesson that supports formative assessment. The lesson’s approach first allows students to demonstrate their prior understandings and abilities in employing the mathematical practices, and then involves students in resolving their own difficulties and misconceptions through structured discussion.
Please comment below with questions, feedback, suggestions, or descriptions of your experience using this resource with students.
#### Re-engagement See 1 item Hide 1 item
Re-engagement means going back to a familiar problem or task and looking at it again in different ways, with a new lens, or going deeper into the mathematics. This is often done by showing examples of student work and providing prompts to help students think about the mathematical ideas differently. This guide provides more information on how to design re-engagement lessons for your students which you can use at any time during a unit where you think it will be helpful for students to revisit a specific mathematical idea before moving on.
Re-engagement means going back to a familiar problem or task and looking at it again in different ways, with a new lens, or going deeper into the mathematics. This is often done by showing examples of student work and providing prompts to help students think about the mathematical ideas differently. This guide provides more information on how to design re-engagement lessons for your students, which you can use at any time during a unit, where you think it will be helpful for students to revisit a specific mathematical idea before moving on.
Please comment below with questions, feedback, suggestions, or descriptions of your experience using this resource with students.
#### End of Unit Assessment See 3 items Hide 3 items
The End of Unit Assessment is intended to surface how students understand the mathematics in relation to the end of year goal of a Regents examination. To support retention, the end of unit assessments are intentionally designed with spiralled questions from previous units.
After this unit, how prepared are your students for the end-of-course Regents examination? The end of unit assessment is designed to surface how students understand the mathematics in the unit. It includes spiralled multiple choice and constructed response questions, comparable to those on the end-of-course Regents examination. A rich task, that allows for multiple entry points and authentic assessment of student learning, may be available for some units and can be included as part of the end of unit assessment. All elements of the end of unit assessment are aligned to the NYS Mathematics Learning Standards and PARCC Model Frameworks prioritization.
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## Graphing Inequalities in One Variable
The solutions of inequalities can be graphed on the number line as rays. If the inequality is "strict" ( < or > ), we use an open dot to indicate that the endpoint of the ray is not part of the solution. For the other types of inequalities ( &le and &ge ), we use a closed dot .
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Just as with equations, there are a number of Properties of Inequality A set of rules for inequalities that describe how addition, subtraction, multiplication, or division can be applied to both sides of an inequality in order to produce an equivalent inequality. that help us work with these types of relationships.
Addition and Subtraction Properties of Inequality
Let&rsquos start with addition and subtraction and the simple inequality a>b . If we want to add a quantity c to the left side, we also have to add it to the right side in order to keep the inequality true. We can write this property as:
People&rsquos ages serve as a good real-life example to model this property. For instance, imagine that you know two people: Adam and Bernard. You know that Adam is older than Bernard (although you do not know how much older). In a certain number of years from now, will Adam still be older than Bernard? Of course! Adam is older to begin with, and they are aging the same amount. In algebraic fashion, you could represent this inequality as:
then Adam's age + some years > Bernard's age + the same number of years.
The Subtraction Property is similar. If we begin with the inequality a>b again and we subtract c from a , then we also need to subtract c from b in order to maintain the relationship. We can write this property as:
The age example can help you make sense of this relationship as well: if Adam is older than Bernard now, then five years ago Adam was also older than Bernard (because Bernard was also five years younger). You could represent this inequality as:
## Unit 3: Linear Expressions & Single-Variable Equations/Inequalities
Identify properties of operations that result in equivalent linear expressions.
Use properties of equations to analyze and write equivalent equations.
Solve single-variable linear equations using properties of equality.
Solve equations with a variable in the denominator.
Solve for a variable in an equation or formula.
#### Topic B: Modeling with Single-Variable Linear Equations
Write equations using defined variables to represent a contextual situation.
Define variables write and solve equations to represent a contextual situation.
Write and solve equations to represent contextual situations where estimations and unit conversions are required.
Model a contextual situation and make an informed decision based on the model.
#### Topic C: Properties and Solutions of Single-Variable Linear Inequalities
Solve unbounded single-variable inequalities in contextual and non-contextual situations.
Write and graph compound single-variable inequalities to describe the solution to contextual and non-contextual situations.
Solve and graph compound inequalities where algebraic manipulation is necessary in contextual and non-contextual situations.
#### Understanding
Students develop a sense of “order- and solution-preserving” moves that can be applied to inequalities and how these moves relate to the solution set for an inequality.
### What to look for
Students will encounter the fact that multiplying both sides of an inequality by a negative number does not preserve order.
### Sample Assessment
What are all the possible whole numbers that make 8-___>3 true?
a. 0, 1, 2, 3, 4, 5
b. 0, 1, 2, 3, 4
c. 0, 1, 2
d. 5
#### The Big Idea
Finding the solution to an inequality in one variable involves four ideas:
1. The smaller of two number is to the left of the larger
2. if one number is less than another, (a < b), then some positive (c) added to (a) will equal (b) ((a+c = b))
3. the point of equality (boundary point) divides a set of values into those greater than and those less than that point and
4. some operations on both sides of an inequality preserve order and some do not (in particular, multiplying or dividing by a non-zero number).
### What are the students doing?
Students reason about order-preserving and solution- preserving moves while using interactive visuals on a number line.
### What is the teacher doing?
Encourage students to think about the ways they learned to solve equations that apply similarly to solving inequalities.
## Watch the video: Ισότητες και ανισότητες (July 2022).
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In my opinion, this is just the beginning. I suggest you try to search google.com
3. Abdul-Mujib
I am finite, I apologize, but it all doesn’t come close. Are there other variants?
4. Leeroy
There is something in this. Thanks for the help in this matter, I also think that the simpler the better ... |
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# How do you factor $28x – 49$?
Last updated date: 13th Jun 2024
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Hint: First notice the common factor among 28 and 49. Then, we will take the common factor out of both like we do in reverse of distributive property.
We have two terms; one is $28x$ and -49.
We will first find the prime factorization of both 28 and 49.
$\Rightarrow 28 = 2 \times 2 \times 7$
$\Rightarrow 49 = 7 \times 7$
We can see that both the 28 and 49 have 7 common in them. Let us take out this 7 out of both of them, we can write the given expression as following:-
$\Rightarrow$ 28x – 49 = 7(4x) – 7(7)
Now, taking 7 common out of both of them, we will write the given expression as following:-
$\Rightarrow$ 28x – 49 = 7 (4x – 7)
Thus, we have got our factors as 7 and 4x – 7.
Note: Alternate way to do the same question. Let us do that as follows:-
We are given that we need to find the factors of 28x – 49. We can observe that if we put $x = \dfrac{7}{4}$ in the given equation, we will get 0. So, $x - \dfrac{7}{4}$ must be a factor of the equation 28x – 49.
Now, we will divide the given equation by $x - \dfrac{7}{4}$ to obtain:-
$\Rightarrow x - \dfrac{7}{4})\overline {28x - 49}$
Now, we will multiply the divisor by 28 to get:-
$\Rightarrow x - \dfrac{7}{4})\overline {28x - 49} (28$
$\underline {28x - 49}$
0
Thus, we have our factors as 28 and $x - \dfrac{7}{4}$. By multiplying them, we get the required answer.
The students must note that we are basically just doing the reverse of distributive property. In distributive property, we have a (b + c) = ab + ac
So, we just did the reverse of it. Putting a = 7, b = 4x and c = -7, we get the same result as we got above in the solution that 28 x – 49 = 7 (4x – 7). |
# Lesson 8
Expanding and Factoring
### Lesson Narrative
In grade 6, students worked extensively with the distributive property involving both addition and subtraction, but only with positive coefficients. In the previous lesson, students learned to rewrite subtraction as "adding the opposite" to avoid common pitfalls. In this lesson, students practice using the distributive property to write equivalent expressions when there are rational coefficients. Some of the expressions they will work with are in preparation for understanding combining like terms in terms of the distributive property, coming up in the next lesson. (For example, $$17a-13a$$ can be rewritten $$a(17-13)$$ using the distributive property, so it is equivalent to $$a \boldcdot 4$$ or $$4a.$$
### Learning Goals
Teacher Facing
• Apply the distributive property to expand or factor an expression that includes negative coefficients, and explain (orally and using other representations) the reasoning.
• Comprehend the terms “expand” and “factor” (in spoken and written language) in relation to the distributive property.
### Student Facing
Let's use the distributive property to write expressions in different ways.
### Student Facing
• I can organize my work when I use the distributive property.
• I can use the distributive property to rewrite expressions with positive and negative numbers.
• I understand that factoring and expanding are words used to describe using the distributive property to write equivalent expressions.
Building On
Addressing
Building Towards
### Glossary Entries
• expand
To expand an expression, we use the distributive property to rewrite a product as a sum. The new expression is equivalent to the original expression.
For example, we can expand the expression $$5(4x+7)$$ to get the equivalent expression $$20x + 35$$.
• factor (an expression)
To factor an expression, we use the distributive property to rewrite a sum as a product. The new expression is equivalent to the original expression.
For example, we can factor the expression $$20x + 35$$ to get the equivalent expression $$5(4x+7)$$.
### Print Formatted Materials
Teachers with a valid work email address can click here to register or sign in for free access to Cool Down, Teacher Guide, and PowerPoint materials.
Student Task Statements pdf docx Cumulative Practice Problem Set pdf docx Cool Down Log In Teacher Guide Log In Teacher Presentation Materials pdf docx
### Additional Resources
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### Sum of Digits of Multiples of Nine
```Date: 08/12/2004 at 05:19:50
From: Saba
Subject: number theory: multiples of 9
Why is it that when you add the individual digits of any multiple of
nine until a single digit answer is reached the answer is always nine?
Is it possible to prove this?
For example, 99 => 9 + 9 = 18 => 1 + 8 = 9
Why doesn't it work with other numbers between 1-9 either?
```
```
Date: 08/12/2004 at 10:10:23
From: Doctor Luis
Subject: Re: number theory: multiples of 9
Hi Saba,
Good job finding that pattern! The reason is that the sum of the
digits of ANY multiple of 9 is also a multiple of 9. Since you keep
adding the digits (each time getting a new multiple of 9, but a
smaller multiple), eventually you'll end up with a single digit
number. Eventually you'll get to the multiple 9 itself.
Now, how do I know that the sum of the digits is always a multiple of
9? Suppose that a number N has digits a,b,c,d,...(from right to left),
N = a + 10b + 100c + 1000d + ...
= a + (b + 9b) + (c + 99c) + (d + 999d) + ...
= (a + b + c + d + ...) + (9b + 99c + 999d + ...)
= (a + b + c + d + ...) + 9*(b + 11c + 111d + ...)
N = (sum of digits of N) + 9 * (some number)
Now, look at that equation carefully. It means that
(sum of digits of N) = N - 9 * (some number)
Since N is assumed to be a multiple of 9, we can write it in terms of
another integer k, so that N = 9k
(sum of digits of N) = 9 * k - 9 * (some number)
= 9 * (k - (some number))
= 9 * (some other number)
Since we showed that the sum of the digits is 9 times some integer,
then it is also a multiple of 9 itself.
To summarize, starting from a multiple of 9, you keep adding the
digits, each time arriving to a multiple of 9. This establishes a
chain of decreasing multiples of 9, until eventually you reach 9 (from
a two-digit multiple). Does that make sense?
It doesn't work for other integers because the chain is broken. For
example, multiples of 8 such as 56 don't add up to a multiple of 8.
Well, I hope this helped!
Let us know if you have any more questions.
- Doctor Luis, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 08/12/2004 at 12:48:19
From: Saba
Subject: number theory: multiples of 9
best!
```
Associated Topics:
High School Number Theory
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# QUICK SOLUTION to List T Consists of 30 Positive Decimals GMAT
## Here’s the Solution to this Crazy-ass Problem: List T Consists of 30 Positive Decimals…
Now I’m not going to lie and tell you this question is easy. It’s actually a total beast. And for that reason, I suggest that you repeat this question and repeat it again and then repeat it again until it makes good sense.
The only thing to do with a question such as this one is to learn it thoroughly so that even though you’re unlikely to see a question on the actual GMAT that is anywhere near as bad, you will be able to take some sort of learning from it.
## Here’s the question itself — the old “List T consist of 30 positive decimals” GMAT Question itself:
List T consist of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digits is odd is rounded down to the nearest integer. If 1/3 of the decimals in T have a tenths digit that is even, which of the following is a possible value of E – S ?
I. -16
II. 6
III. 10
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
Now the first thing to notice with this question is that you’re not looking to calculate all the numbers. Rather, you’re looking to calculate the specific amount by which you estimate. That is your difference E – S.
Here are the steps:
1. Calculate the maximum amount by which the estimate for the EVENS might be off.
2. Calculate the minimum amount by which the estimate for the EVENS might be off.
3. Treat this as a range, then multiply by the number of EVENS you have.
4. Calculate the maximum amount by which the estimate for the ODDS might be off.
5. Calculate the minimum amount by which the estimate for the ODDS might be off.
6. Treat this as a range, then multiply by the number of ODDS that you have.
7. Add the ranges together to find the total range of possibilities.
Granted, this is a lot to process in two minutes, but if you precisely how to do this question, it is possible to do the actual calculations in in this time. Ideally, if you were presented with another similar question, the mechanism would be similar enough that you’d be able to use the above method to get the answer.
In the end, if you can’t answer this “List T consists of 30 positive decimals” GMAT question, you won’t, most likely, be able to answer any similar question. |
# Into Math Grade 1 Module 12 Lesson 1 Answer Key Represent Adding Tens
We included HMH Into Math Grade 1 Answer Key PDF Module 12 Lesson 1 Represent Adding Tens to make students experts in learning maths.
## HMH Into Math Grade 1 Module 12 Lesson 1 Answer Key Represent Adding Tens
I Can add multiples of ten with multiples of ten.
Spark Your Learning
Start with 2 tens. Toss the number cube and add that many tens. How can you show your work?
First Time
2 tens + _______ tens = _________ tens
Second Time
2 tens + _______ tens = _________ tens
Tell children they will start with 2 tens and then add some more tens. Children toss the number cube and add that many tens. Children should draw to show the addition. Have pairs repeat the activity.
Answer:
First time:
2 tens + 3 tens = 5 tens
Second time:
2 tens + 4 tens = 6 tens.
Build Understanding
Question 1.
The Beach Shop has 30 adult sunglasses and 40 child sunglasses. How many sunglasses does the shop have?
A. How can you show the problem?
________ tens + _________ tens = _________ tens
Answer:
3 tens + 4 tens = 7 tens
B. How can you write an equation to solve the problem?
Equation: ____________
Answer:
3 + 4 = 7 is the equation that solve the problem.
Connect to Vocabulary
equations:
80 – 50 = 30
9 = 6 + 3
90 = 60 + 30
C. The shop has _________ sunglasses.
Answer:
The shop has 70 sunglasses.
Explanation:
The Beach Shop has 30 adult sunglasses and 40 child sunglasses
30 is 3 tens, 40 is 4 tens
So, 30 + 40 =
3 tens + 4 tens = 7 tens
There are 70 sun glasses in the shop.
Turn and Talk Does your answer make sense? How do you know?
Answer:
Question 2.
A store has 30 large rafts and 50 small rafts. How many rafts does the store have?
A. Use tens to show the problem.
Answer:
3 tens + 5 tens = 8 tens
B. Write an equation to solve the problem.
Equation: __________
Answer:
3 + 5 = 8 is the equation that solves the problem.
C. The store has __________ rafts.
Answer:
The store has 80 rafts.
Explanation:
A store has 30 large rafts and 50 small rafts
30 is 3 tens
50 is 5 tens
3 tens + 5 tens = 8 tens
So, 30 + 50 = 80
The store has 80 rafts in all.
Check Understanding
Draw tens to show the problem. Write an equation to solve.
Question 1.
Kara has 20 blue marbles. She has 40 purple marbles. How many marbles does Kara have?
Equation: ___________
Kara has __________ marbles.
Answer:
Explanation:
Kara has 20 blue marbles
She has 40 purple marbles
20 + 40 =
2 tens + 4 tens = 6 tens
So, 20 + 40 = 60
Kara has 60 marbles.
On Your Own
Draw tens to show the problem. Write an equation to solve.
Question 2.
Model with Mathematics Marcus has 20 oranges. Sally has 10 oranges. How many oranges do they have?
Equation: _____________
They have ___________ oranges.
Answer:
Explanation:
Marcus has 20 oranges
Sally has 10 oranges
20 is 2 tens
10 is 1 tens
2 tens + 1 ten = 3 tens
20 + 10 = 30
So, they have 30 oranges
Question 3.
Open-Ended Jodi has 5 packages of 10 pens each. All the pens in each package are either yellow or blue. How many of each could she have?
Equation: __________
Answer:
20 + 30 = 50
Explanation:
Jodi has 5 packages of 10 pens each
All the pens in each package are either yellow or blue
Question 4.
Reason Explain how to show this equation using .
20 + 20 = 40
Answer:
Explanation:
I drew connecting cubes to show the equation 20 + 20 = 40.
I’m in a Learning Mindset!
What can I do to become a better learner?
Answer:
Scroll to Top
Scroll to Top |
### Theory:
Rules for subtracting two integers
• First, change the sign of the subtrahend(second number) into the opposite.
• Then,change the operation from subtraction to addition.
• Finally, apply the addition rule for integers.
Example:
1. Subtract $$2$$ from $$-5$$.
Step 1: The opposite of $$2$$ is $$-2$$. Change $$2$$ into $$-2$$.
$$(-5) - (-2)$$.
Step 2: Now, change the operation subtraction into addition.
$$(-5) + (-2)$$.
Step 3: $$-5$$ and $$-2$$ both the numbers have the same signs. Thus, add the integers and keep the sign same.
$$(-5) + (-2) = -7$$.
Therefore, $$(-5) - 2 = (-5) + (-2) = -7$$.
2. Find the difference of the integers $$(-16)$$ and $$(-14)$$.
Step 1: The opposite of $$-14$$ is $$14$$. Change $$-14$$ into $$14$$.
$$(-16) - (14)$$.
Step 2: Now, change the operation subtraction into addition.
$$(-16) + (14)$$.
Step 3: $$-16$$ and $$14$$ both having different signs.
Now, by the addition rule, subtract the integers and put a larger number sign.
$$(-16) + 14 = -2$$.
Therefore, $$(-16) - (-14) = (-16) + (14) = -2$$. |
# T4T What is Math? (Lesson 1 of 6)
Lesson excerpt:
NC Mathematics Standards:
Extend and recognize patterns in the counting sequence.
NC.1.NBT.7 Read and write numerals, and represent a number of objects with a written numeral, to 100 20.
NC.1.NBT.1 Count to 150, starting at any number less than 150.
Standards for Mathematical Practice:
1. Make sense of problems and persevere in solving them.
4. Model with mathematics.
6. Attend to precision.
Student Outcomes:
● I can work with a partner to solve math problems.
● I can count on from a given number within 100.
● I can explain the process of solving a problem to my teacher and classmates.
Math Language:
· Count, count on, next, ones, ten
Materials:
● Bags (paper), Cube bags activity sheet, Twenty board, Multi-link (pop) cubes, Number cards, Pattern blocks, Two-color counters
● Gather materials
● Prepare 15 cube bags. Each bag should have between 1 and 20 cubes in it.
Launch:
1. Introduce the word mathematician.
· Say: Mathematicians solve problems. During our daily math time, we will be mathematicians and solve problems with objects, shapes, and numbers.
· Explain: Many times mathematicians need to work together to solve problems.
· Read the task to students: There are 4 children in the park. Three more students come to the park. If Sonya gives each student a number, what numbers will Sonya give the 3 students who showed up? Use your twenty board and counters to prove your answer.
· Provide students with 3-5 minutes to work on the task. Students should work in partners to find the next 3 numbers (5, 6, 7).
· Write the following so students can see it: 4, ___, ___, ____
o What numbers go in the blanks?
o How did you find out the next 3 numbers?
· Pose the follow up task. There are 8 children in the cafeteria. Four more students come in. If Stefan gives each student a number what numbers will Stefan give the 4 children who came in? Use your twenty board and counters to prove your answer.
· Provide students with 3-5 minutes to work on the task. Students should work in partners to find the next 4 numbers (9, 10, 11, 12).
· Write the following so students can see it: 8, __, __, __, __
o What numbers go in the blanks?
o How did you find out the next 4 numbers?
o What happened after 8 on the twenty board?
Explore
1. Introduce the Cube Bags to students.
· Have a student come to the front of the class to model the activity with you.
· You and your partner are going to have a cube bag. You are going to look at your bag and estimate how many cubes are in the bag. All of the bags have less than 20 cubes.
· On your recording sheet you are going to write the letter of your bag and your estimate.
· Then we are going to empty the bag and count how many cubes we have.
· Model how to count the cubes and record the number on the recording sheet.
1. Counting the Cube Bags
· Allow students 10-12 minutes to work on the activity.
· Make sure you have established a system for students to return a bag and grab a new bag after they are done estimating and counting.
Discuss:
5. Bring students together to discuss their strategies on the cube bag activity. |
# How many ways can you roll 5 dice?
Contents
## How many different ways can you roll 5 dice?
The last die may have six values. For each of these six values, the second- to-last die may have six values. Thus, we have 6·6 = 36 possible outcomes for the last two dice. By extension, we have a total of 65 = 7776 possible outcomes for all five dice.
## What is the average roll of 5 dice?
The probability is 0. The average of five dice rolls is some integer divided by 5, but 3.5=72 is not any integer divided by 5. However, the average value of a dice roll is 3.5, so over a large number of turns the average of all throws will roughly equal 3.5.
## What is the probability of rolling a 5 on a 6 sided dice?
Two (6-sided) dice roll probability table
Roll a… Probability
5 10/36 (27.778%)
6 15/36 (41.667%)
7 21/36 (58.333%)
8 26/36 (72.222%)
## Is Yahtzee a skill or luck?
As with all games, players typically need elements of both skill and luck. However, some games are known for needing more skill than others. In a 2020 survey conducted by YouGov, 49 percent of respondents in the United States stated that Yahtzee was a game of luck rather than skill.
THIS IS INTERESTING: How do you get free bets on William Hill?
## What are the odds of rolling a 1 or a 5 with two dice?
That means both show a 2, 3, 4, or 6. Thats (4/6)2. Hence the probability that at least one shows a 1 or 5 is 1−(2/3)2=5/9.
## What are the odds of rolling 5 1s in a row?
The probability is 1216 chance, which is approximately a 0.46% chance.
## What is the average of dice rolls?
Another option for finding the average dice roll is to add all of the possible outcomes together then divide by the number of sides the die has. For example, to find the average dice roll of 1d4 you would add 1, 2, 3, and 4 together and divide by 4.
## How many times can you roll a 7?
As the chart shows the closer the total is to 7 the greater is the probability of it being thrown.
Probabilities for the two dice.
Total Number of combinations Probability
6 5 13.89%
7 6 16.67%
8 5 13.89%
9 4 11.11% |
# Understanding Linear Algebra
With our understanding of symmetric matrices and variance in hand, we’ll now explore how to determine the directions in which the variance of a dataset is as large as possible and where it is as small as possible. This is part of a much larger story involving a type of function, called a quadratic form, that we’ll introduce here.
### Preview Activity7.2.1.
Let’s begin by looking at an example. Suppose we have three data points that form the demeaned data matrix
\begin{equation*} A = \begin{bmatrix} 2 \amp 1 \amp -3 \\ 1 \amp 2 \amp -3 \\ \end{bmatrix} \end{equation*}
1. Plot the demeaned data points in Figure 7.2.1. In which direction does the variance appear to be largest and in which does it appear to be smallest?
2. Construct the covariance matrix $$C$$ and determine the variance in the direction of $$\twovec11$$ and the variance in the direction of $$\twovec{-1}1\text{.}$$
3. What is the total variance of this dataset?
4. Generally speaking, if $$C$$ is the covariance matrix of a dataset and $$\uvec$$ is an eigenvector of $$C$$ having unit length and with associated eigenvalue $$\lambda\text{,}$$ what is $$V_{\uvec}\text{?}$$
Given a matrix $$A$$ of $$N$$ demeaned data points, the symmetric covariance matrix $$C=\frac1N AA^T$$ determines the variance in a particular direction
\begin{equation*} V_{\uvec} = \uvec\cdot(C\uvec), \end{equation*}
where $$\uvec$$ is a unit vector defining the direction.
More generally, a symmetric $$m\times m$$ matrix $$A$$ defines a function $$q:\real^m \to \real$$ by
\begin{equation*} q(\xvec) = \xvec\cdot(A\xvec). \end{equation*}
Notice that this expression is similar to the one we use to find the variance $$V_{\uvec}$$ in terms of the covariance matrix $$C\text{.}$$ The only difference is that we allow $$\xvec$$ to be any vector rather than requiring it to be a unit vector.
#### Example7.2.2.
Suppose that $$A=\begin{bmatrix} 1 \amp 2\\ 2 \amp 1 \end{bmatrix} \text{.}$$ If we write $$\xvec=\twovec{x_1}{x_2}\text{,}$$ then we have
\begin{align*} q\left(\twovec {x_1}{x_2}\right) \amp = \twovec {x_1}{x_2} \cdot \left( \begin{bmatrix} 1 \amp 2 \\ 2 \amp 1 \end{bmatrix} \twovec {x_1}{x_2} \right)\\ \amp = \twovec {x_1}{x_2} \cdot \twovec{x_1 + 2x_2}{2x_1 + x_2}\\ \amp = x_1^2 + 2x_1x_2 + 2x_1x_2 + x_2^2\\ \amp = x_1^2 + 4x_1x_2 + x_2^2. \end{align*}
We may evaluate the quadratic form using some input vectors:
\begin{equation*} q\left(\twovec 10\right) = 1, \hspace{24pt} q\left(\twovec 11\right) = 6, \hspace{24pt} q\left(\twovec 24\right) = 52. \end{equation*}
Notice that the value of the quadratic form is a scalar.
#### Definition7.2.3.
If $$A$$ is a symmetric $$m\times m$$ matrix, the quadratic form defined by $$A$$ is the function $$q_A(\xvec) = \xvec\cdot(A\xvec)\text{.}$$
#### Activity7.2.2.
Let’s look at some more examples of quadratic forms.
1. Consider the symmetric matrix $$D = \begin{bmatrix} 3 \amp 0 \\ 0 \amp -1 \\ \end{bmatrix} \text{.}$$ Write the quadratic form $$q_D(\xvec)$$ defined by $$D$$ in terms of the components of $$\xvec=\twovec{x_1}{x_2}\text{.}$$ What is the value of $$q_D\left(\twovec2{-4}\right)\text{?}$$
2. Given the symmetric matrix $$A=\begin{bmatrix} 2 \amp 5 \\ 5 \amp -3 \end{bmatrix} \text{,}$$ write the quadratic form $$q_A(\xvec)$$ defined by $$A$$ and evaluate $$q_A\left(\twovec{2}{-1}\right)\text{.}$$
3. Suppose that $$q\left(\twovec{x_1}{x_2}\right) = 3x_1^2 - 4x_1x_2 + 4x_2^2\text{.}$$ Find a symmetric matrix $$A$$ such that $$q$$ is the quadratic form defined by $$A\text{.}$$
4. Suppose that $$q$$ is a quadratic form and that $$q(\xvec) = 3\text{.}$$ What is $$q(2\xvec)\text{?}$$ $$q(-\xvec)\text{?}$$ $$q(10\xvec)\text{?}$$
5. Suppose that $$A$$ is a symmetric matrix and $$q_A(\xvec)$$ is the quadratic form defined by $$A\text{.}$$ Suppose that $$\xvec$$ is an eigenvector of $$A$$ with associated eigenvalue -4 and with length 7. What is $$q_A(\xvec)\text{?}$$
Linear algebra is principally about things that are linear. However, quadratic forms, as the name implies, have a distinctly non-linear character. First, if $$A=\begin{bmatrix} a \amp b \\ b \amp c \end{bmatrix}\text{,}$$ is a symmetric matrix, then the associated quadratic form is
\begin{equation*} q_A\left(\twovec{x_1}{x_2}\right) = ax_1^2 + 2bx_1x_2 + cx_2^2. \end{equation*}
Notice how the variables $$x_1$$ and $$x_2$$ are multiplied together, which tells us this isn’t a linear function.
This expression assumes an especially simple form when $$D$$ is a diagonal matrix. In particular, if $$D = \begin{bmatrix} a \amp 0 \\ 0 \amp c \\ \end{bmatrix} \text{,}$$ then $$q_D\left(\twovec{x_1}{x_2}\right) = ax_1^2 + cx_2^2\text{.}$$ This is special because there is no cross-term involving $$x_1x_2\text{.}$$
Remember that matrix transformations have the property that $$T(s\xvec) = sT(\xvec)\text{.}$$ Quadratic forms behave differently:
\begin{equation*} q_A(s\xvec) = (s\xvec)\cdot(A(s\xvec)) = s^2\xvec\cdot(A\xvec)= s^2q_A(\xvec). \end{equation*}
For instance, when we multiply $$\xvec$$ by the scalar 2, then $$q_A(2\xvec) = 4q_A(\xvec)\text{.}$$ Also, notice that $$q_A(-\xvec) = q_A(\xvec)$$ since the scalar is squared.
Finally, evaluating a quadratic form on an eigenvector has a particularly simple form. Suppose that $$\xvec$$ is an eigenvector of $$A$$ with associated eigenvalue $$\lambda\text{.}$$ We then have
\begin{equation*} q_A(\xvec) = \xvec\cdot(A\xvec) = \lambda\xvec\cdot\xvec = \lambda \len{\xvec}^2. \end{equation*}
Let’s now return to our motivating question: in which direction $$\uvec$$ is the variance $$V_{\uvec}=\uvec\cdot(C\uvec)$$ of a dataset as large as possible and in which is it as small as possible. Remembering that the vector $$\uvec$$ is a unit vector, we can now state a more general form of this question: If $$q_A(\xvec)$$ is a quadratic form, for which unit vectors $$\uvec$$ is $$q_A(\uvec)=\uvec\cdot(A\uvec)$$ as large as possible and for which is it as small as possible? Since a unit vector specifies a direction, we will often ask for the directions in which the quadratic form $$q(\xvec)$$ is at its maximum or minimum value.
#### Activity7.2.3.
We can gain some intuition about this problem by graphing the quadratic form and paying particular attention to the unit vectors.
1. Evaluating the following cell defines the matrix $$D = \begin{bmatrix} 3 \amp 0 \\ 0 \amp -1 \end{bmatrix}$$ and displays the graph of the associated quadratic form $$q_D(\xvec)\text{.}$$ In addition, the points corresponding to vectors $$\uvec$$ with unit length are displayed as a curve.
Notice that the matrix $$D$$ is diagonal. In which directions does the quadratic form have its maximum and minimum values?
2. Write the quadratic form $$q_D$$ associated to $$D\text{.}$$ What is the value of $$q_D\left(\twovec10\right)\text{?}$$ What is the value of $$q_D\left(\twovec01\right)\text{?}$$
3. Consider a unit vector $$\uvec=\twovec{u_1}{u_2}$$ so that $$u_1^2+u_2^2 = 1\text{,}$$ an expression we can rewrite as $$u_1^2 = 1-u_2^2\text{.}$$ Write the quadratic form $$q_D(\uvec)$$ and replace $$u_1^2$$ by $$1-u_2^2\text{.}$$ Now explain why the maximum of $$q_D(\uvec)$$ is 3. In which direction does the maximum occur? Does this agree with what you observed from the graph above?
4. Write the quadratic form $$q_D(\uvec)$$ and replace $$u_2^2$$ by $$1-u_1^2\text{.}$$ What is the minimum value of $$q_D(\uvec)$$ and in which direction does the minimum occur?
5. Use the previous Sage cell to change the matrix to $$A=\begin{bmatrix} 1 \amp 2 \\ 2 \amp 1 \end{bmatrix}$$ and display the graph of the quadratic form $$q_A(\xvec) = \xvec\cdot(A\xvec)\text{.}$$ Determine the directions in which the maximum and minimum occur?
6. Remember that $$A=\begin{bmatrix} 1 \amp 2 \\ 2 \amp 1 \end{bmatrix}$$ is symmetric so that $$A=QDQ^T$$ where $$D$$ is the diagonal matrix above and $$Q$$ is the orthogonal matrix that rotates vectors by $$45^\circ\text{.}$$ Notice that
\begin{equation*} q_A(\uvec) = \uvec\cdot(A\uvec) = \uvec\cdot(QDQ^T\uvec) = (Q^T\uvec)\cdot(DQ^T\uvec) = q_D(\vvec) \end{equation*}
where $$\vvec=Q^T\uvec\text{.}$$ That is, we have $$q_A(\uvec) = q_D(\vvec)\text{.}$$
Explain why $$\vvec = Q^T\uvec$$ is also a unit vector; that is, explain why
\begin{equation*} |\vvec|^2 = |Q^T\uvec|^2 = (Q^T\uvec)\cdot(Q^T\uvec) = 1. \end{equation*}
7. Using the fact that $$q_A(\uvec) = q_D(\vvec)\text{,}$$ explain how we now know the maximum value of $$q_A(\uvec)$$ is 3 and determine the direction in which it occurs. Also, determine the minimum value of $$q_A(\uvec)$$ and determine the direction in which it occurs.
This activity demonstrates how the eigenvalues of $$A$$ determine the maximum and minimum values of the quadratic form $$q_A(\uvec)$$ when evaluated on unit vectors and how the associated eigenvectors determine the directions in which the maximum and minimum values occur. Let’s look at another example so that this connection is clear.
#### Example7.2.4.
Consider the symmetric matrix $$A=\begin{bmatrix} -7 \amp -6 \\ -6 \amp 2 \\ \end{bmatrix}\text{.}$$ Because $$A$$ is symmetric, we know that it can be orthogonally diagonalized. In fact, we have $$A=QDQ^T$$ where
\begin{equation*} D = \begin{bmatrix} 5 \amp 0 \\ 0 \amp -10 \\ \end{bmatrix},\hspace{24pt} Q = \begin{bmatrix} 1/\sqrt{5} \amp 2/\sqrt{5} \\ -2/\sqrt{5} \amp 1/\sqrt{5} \\ \end{bmatrix}\text{.} \end{equation*}
From this diagonalization, we know that $$\lambda_1=5$$ is the largest eigenvalue of $$A$$ with associated eigenvector $$\uvec_1 = \twovec{1/\sqrt{5}}{-2/\sqrt{5}}$$ and that $$\lambda_2 = -10$$ is the smallest eigenvalue with associated eigenvector $$\uvec_2 = \twovec{2/\sqrt{5}}{1/\sqrt{5}}\text{.}$$
Let’s first study the quadratic form $$q_D(\uvec) = 5u_1^2 - 10u_2^2$$ because the absence of the cross-term makes it comparatively simple. Remembering that $$\uvec$$ is a unit vector, we have $$u_1^2+u_2^2=1\text{,}$$ which means that $$u_1^2 = 1-u_2^2\text{.}$$ Therefore,
\begin{equation*} q_D(\uvec) = 5u_1^2 - 10u_2^2 = 5(1-u_2^2)-10u_2^2 = 5 - 15u_2^2\text{.} \end{equation*}
This tells us that $$q_D(\uvec)$$ has a maximum value of $$5\text{,}$$ which occurs when $$u_2=0$$ or in the direction $$\twovec10\text{.}$$
In the same way, rewriting $$u_2^2 = 1-u_1^2$$ allows us to conclude that the minimum value of $$q_D(\uvec)$$ is $$-10\text{,}$$ which occurs in the direction $$\twovec01\text{.}$$
Let’s now return to the matrix $$A$$ whose quadratic form $$q_A$$ is related to $$q_D$$ because $$A = QDQ^T\text{.}$$ In particular, we have
\begin{equation*} q_A(\uvec) = \uvec\cdot(A\uvec) = \uvec\cdot(QDQ^T\uvec) = (Q^T\uvec)\cdot(DQ^T\uvec) = \vvec\cdot(D\vvec) = q_D(\vvec)\text{.} \end{equation*}
In other words, we have $$q_A(\uvec) = q_D(\vvec)$$ where $$\vvec=Q^T\uvec\text{.}$$ This is quite useful because it allows us to relate the values of $$q_A$$ to those of $$q_D\text{,}$$ which we already understand quite well.
Now it turns out that $$\vvec$$ is also a unit vector because
\begin{equation*} |\vvec|^2 = \vvec\cdot\vvec = (Q^T\uvec)\cdot(Q^T\uvec) = \uvec\cdot(QQ^T\uvec) = \uvec\cdot\uvec = |\uvec|^2 = 1\text. \end{equation*}
Therefore, the maximum value of $$q_A(\uvec)$$ is the same as $$q_D(\vvec)\text{,}$$ which we know to be $$5$$ and which occurs in the direction $$\vvec=\twovec10\text{.}$$ This means that the maximum value of $$q_A(\uvec)$$ is also $$5$$ and that this occurs in the direction $$\uvec = Q\vvec = Q\twovec10 = \twovec{1/\sqrt{5}}{-2/\sqrt{5}}\text{.}$$ We now know that the maximum value of $$q_A(\uvec)$$ is the largest eigenvalue $$\lambda_1=5$$ and that this maximum value occurs in the direction of an associated eigenvector.
In the same way, we see that the minimum value of $$q_A(\uvec)$$ is the smallest eigenvalue $$\lambda_2=-10$$ and that this minimum occurs in the direction of $$\uvec=Q\twovec01 = \twovec{2/\sqrt{5}}{1/\sqrt{5}}\text{,}$$ an associated eigenvector.
More generally, we have
#### Example7.2.6.
Suppose that $$A$$ is the symmetric matrix $$A=\begin{bmatrix} 0 \amp 6 \amp 3 \\ 6 \amp 3 \amp 6 \\ 0 \amp 6 \amp 6 \\ \end{bmatrix}\text{,}$$ which may be orthogonally diagonalized as $$A=QDQ^T$$ where
\begin{equation*} D = \begin{bmatrix} 12 \amp 0 \amp 0 \\ 0 \amp 3 \amp 0 \\ 0 \amp 0 \amp -6 \\ \end{bmatrix}, \hspace{24pt} Q = \begin{bmatrix} 1/3 \amp 2/3 \amp 2/3 \\ 2/3 \amp 1/3 \amp -2/3 \\ 2/3 \amp -2/3 \amp 1/3 \\ \end{bmatrix}\text{.} \end{equation*}
We see that the maximum value of $$q_A(\uvec)$$ is 12, which occurs in the direction $$\threevec{1/3}{2/3}{2/3}\text{,}$$ and the minimum value is -6, which occurs in the direction $$\threevec{2/3}{-2/3}{1/3}\text{.}$$
#### Example7.2.7.
Suppose we have the matrix of demeaned data points $$A = \begin{bmatrix} 2 \amp 1 \amp -3 \\ 1 \amp 2 \amp -3 \\ \end{bmatrix}$$ that we considered in Preview Activity 7.2.1. The data points are shown in Figure 7.2.8.
Constructing the covariance matrix $$C=\frac13~AA^T$$ gives $$C=\begin{bmatrix} 14/3 \amp 13/3 \\ 13/3 \amp 14/3 \end{bmatrix}\text{,}$$ which has eigenvalues $$\lambda_1 = 9\text{,}$$ with associated eigenvector $$\twovec{1/\sqrt{2}}{1/\sqrt{2}}\text{,}$$ and $$\lambda_2=1/3\text{,}$$ with associated eigenvector $$\twovec{-1/\sqrt{2}}{1/\sqrt{2}}\text{.}$$
Remember that the variance in a direction $$\uvec$$ is $$V_{\uvec} = \uvec\cdot(C\uvec) = q_C(\uvec)\text{.}$$ Therefore, the variance attains a maximum value of 9 in the direction $$\twovec{1/\sqrt{2}}{1/\sqrt{2}}$$ and a minimum value of 1/3 in the direction $$\twovec{-1/\sqrt{2}}{1/\sqrt{2}}\text{.}$$ Figure 7.2.9 shows the data projected onto the lines defined by these vectors.
Remember that variance is additive, as stated in Proposition 7.1.16, which tells us that the total variance is $$V = 9 + 1/3 = 28/3\text{.}$$
We’ve been focused on finding the directions in which a quadratic form attains its maximum and minimum values, but there’s another important observation to make after this activity. Recall how we used the fact that a symmetric matrix is orthogonally diagonalizable: if $$A=QDQ^T\text{,}$$ then $$q_A(\uvec) = q_D(\vvec)$$ where $$\vvec = Q^T\uvec\text{.}$$
More generally, if we define $$\yvec = Q^T\xvec\text{,}$$ we have
\begin{equation*} q_A(\xvec) = \xvec\cdot(A\xvec) = \xvec\cdot(QDQ^T\xvec) = (Q^T\xvec)\cdot(DQ^T\xvec) = \yvec\cdot(D\yvec) = q_D(\yvec) \end{equation*}
Remembering that the quadratic form associated to a diagonal form has no cross terms, we obtain
\begin{equation*} q_A(\xvec) = q_D(\yvec) = \lambda_1y_1^2 + \lambda_2y_2^2 + \ldots + \lambda_my_m^2. \end{equation*}
In other words, after a change of coordinates, the quadratic form $$q_A$$ can be written without cross terms. This is known as the Principle Axes Theorem.
We will put this to use in the next section.
### Subsection7.2.2Definite symmetric matrices
While our questions about variance provide some motivation for exploring quadratic forms, these functions appear in a variety of other contexts so it’s worth spending some more time with them. For example, quadratic forms appear in multivariable calculus when describing the behavior of a function of several variables near a critical point and in physics when describing the kinetic energy of a rigid body.
The following definition will be important in this section.
#### Definition7.2.11.
A symmetric matrix $$A$$ is called positive definite if its associated quadratic form satisfies $$q_A(\xvec) \gt 0$$ for any nonzero vector $$\xvec\text{.}$$ If $$q_A(\xvec) \geq 0$$ for all nonzero vectors $$\xvec\text{,}$$ we say that $$A$$ is positive semidefinite.
Likewise, we say that $$A$$ is negative definite if $$q_A(\xvec) \lt 0$$ for all nonzero vectors $$\xvec\text{.}$$
Finally, $$A$$ is called indefinite if $$q_A(\xvec) \gt 0$$ for some $$\xvec$$ and $$q_A(\xvec) \lt 0$$ for others.
#### Activity7.2.4.
This activity explores the relationship between the eigenvalues of a symmetric matrix and its definiteness.
1. Consider the diagonal matrix $$D=\begin{bmatrix} 4 \amp 0 \\ 0 \amp 2 \\ \end{bmatrix}$$ and write its quadratic form $$q_D(\xvec)$$ in terms of the components of $$\xvec=\twovec{x_1}{x_2}\text{.}$$ How does this help you decide whether $$D$$ is positive definite or not?
2. Now consider $$D=\begin{bmatrix} 4 \amp 0 \\ 0 \amp 0 \\ \end{bmatrix}$$ and write its quadratic form $$q_D(\xvec)$$ in terms of $$x_1$$ and $$x_2\text{.}$$ What can you say about the definiteness of $$D\text{?}$$
3. If $$D$$ is a diagonal matrix, what condition on the diagonal entries guarantee that $$D$$ is
1. positive definite?
2. positive semidefinite?
3. negative definite?
4. negative semidefinite?
5. indefinite?
4. Suppose that $$A$$ is a symmetric matrix with eigenvalues 4 and 2 so that $$A=QDQ^T$$ where $$D=\begin{bmatrix}4 \amp 0 \\ 0 \amp 2 \end{bmatrix}\text{.}$$ If $$\yvec = Q^T\xvec\text{,}$$ then we have $$q_A(\xvec) = q_D(\yvec)\text{.}$$ Explain why this tells us that $$A$$ is positive definite.
5. Suppose that $$A$$ is a symmetric matrix with eigenvalues 4 and 0. What can you say about the definiteness of $$A$$ in this case?
6. What condition on the eigenvalues of a symmetric matrix $$A$$ guarantees that $$A$$ is
1. positive definite?
2. positive semidefinite?
3. negative definite?
4. negative semidefinite?
5. indefinite?
As seen in this activity, it is straightforward to determine the definiteness of a diagonal matrix. For instance, if $$D=\begin{bmatrix} 7 \amp 0 \\ 0 \amp 5 \end{bmatrix}\text{,}$$ then
\begin{equation*} q_D(\xvec) = 7x_1^2 + 5x_2^2. \end{equation*}
This shows that $$q_D(\xvec) \gt 0$$ when either $$x_1$$ or $$x_2$$ is not zero so we conclude that $$D$$ is positive definite. In the same way, we see that $$D$$ is positive semidefinite if all the diagonal entries are nonnegative.
Understanding this behavior for diagonal matrices enables us to understand more general symmetric matrices. As we saw previously, the quadratic form for a symmetric matrix $$A=QDQ^T$$ agrees with the quadratic form for the diagonal matrix $$D$$ after a change of coordinates. In particular,
\begin{equation*} q_A(\xvec) = q_D(\yvec) \end{equation*}
where $$\yvec=Q^T\xvec\text{.}$$ Now the diagonal entries of $$D$$ are the eigenvalues of $$A$$ from which we conclude that $$q_A(\xvec) \gt 0$$ if all the eigenvalues of $$A$$ are positive. Likewise, $$q_A(\xvec)\geq 0$$ if all the eigenvalues are nonnegative.
We will now apply what we’ve learned about quadratic forms to study the nature of critical points in multivariable calculus. The rest of this section assumes that the reader is familiar with ideas from multivariable calculus and can be skipped by others.
First, suppose that $$f(x,y)$$ is a differentiable function. We will use $$f_x$$ and $$f_y$$ to denote the partial derivatives of $$f$$ with respect to $$x$$ and $$y\text{.}$$ Similarly, $$f_{xx}\text{,}$$ $$f_{xy}\text{,}$$ $$f_{yx}$$ and $$f_{yy}$$ denote the second partial derivatives. You may recall that the mixed partials, $$f_{xy}$$ and $$f_{yx}$$ are equal under a mild assumption on the function $$f\text{.}$$ A typical question in calculus is to determine where this function has its maximum and minimum values.
Any local maximum or minimum of $$f$$ appears at a critical point $$(x_0,y_0)$$ where
\begin{equation*} f_x(x_0,y_0) = 0,\hspace{24pt} f_y(x_0,y_0) = 0. \end{equation*}
Near a critical point, the quadratic approximation of $$f$$ tells us that
\begin{align*} f(x,y)\approx f(x_0,y_0) \amp + \frac12 f_{xx}(x_0,y_0)(x-x_0)^2\\ \amp + f_{xy}(x_0,y_0)(x-x_0)(y-y_0) + \frac12 f_{yy}(x_0,y_0)(y-y_0)^2. \end{align*}
#### Activity7.2.5.
Let’s explore how our understanding of quadratic forms helps us determine the behavior of a function $$f$$ near a critical point.
1. Consider the function $$f(x,y) = 2x^3 - 6xy + 3y^2\text{.}$$ Find the partial derivatives $$f_{x}$$ and $$f_y$$ and use these expressions to determine the critical points of $$f\text{.}$$
2. Evaluate the second partial derivatives $$f_{xx}\text{,}$$ $$f_{xy}\text{,}$$ and $$f_{yy}\text{.}$$
3. Let’s first consider the critical point $$(1,1)\text{.}$$ Use the quadratic approximation as written above to find an expression approximating $$f$$ near the critical point.
4. Using the vector $$\wvec = \twovec{x-1}{y-1}\text{,}$$ rewrite your approximation as
\begin{equation*} f(x,y) \approx f(1,1) + q_A(\wvec) \end{equation*}
for some matrix $$A\text{.}$$ What is the matrix $$A$$ in this case?
5. Find the eigenvalues of $$A\text{.}$$ What can you conclude about the definiteness of $$A\text{?}$$
6. Recall that $$(x_0,y_0)$$ is a local minimum for $$f$$ if $$f(x,y) \gt f(x_0,y_0)$$ for nearby points $$(x,y)\text{.}$$ Explain why our understanding of the eigenvalues of $$A$$ shows that $$(1,1)$$ is a local minimum for $$f\text{.}$$
Near a critical point $$(x_0,y_0)$$ of a function $$f(x,y)\text{,}$$ we can write
\begin{equation*} f(x,y) \approx f(x_0, y_0) + q_A(\wvec) \end{equation*}
where $$\wvec = \twovec{x-x_0}{y-y_0}$$ and $$A = \frac12 \begin{bmatrix} f_{xx}(x_0,y_0) \amp f_{xy}(x_0,y_0) \\ f_{yx}(x_0,y_0) \amp f_{yy}(x_0,y_0) \end{bmatrix}\text{.}$$ If $$A$$ is positive definite, then $$q_A(\wvec) \gt 0\text{,}$$ which tells us that
\begin{equation*} f(x,y) \approx f(x_0,y_0) + q_A(\wvec) \gt f(x_0,y_0) \end{equation*}
and that the critical point $$(x_0,y_0)$$ is therefore a local minimum.
The matrix
\begin{equation*} H = \begin{bmatrix} f_{xx}(x_0,y_0) \amp f_{xy}(x_0,y_0) \\ f_{yx}(x_0,y_0) \amp f_{yy}(x_0,y_0) \end{bmatrix} \end{equation*}
is called the Hessian of $$f\text{,}$$ and we see now that the eigenvalues of this symmetric matrix determine the nature of the critical point $$(x_0,y_0)\text{.}$$ In particular, if the eigenvalues are both positive, then $$q_H$$ is positive definite, and the critical point is a local minimum.
This observation leads to the Second Derivative Test for multivariable functions.
Most multivariable calculus texts assume that the reader is not familiar with linear algebra and so write the second derivative test for functions of two variables in terms of $$D=\det(H)\text{.}$$ If
• $$D \gt 0$$ and $$f_{xx}(x_0,y_0)) \gt 0\text{,}$$ then $$(x_0, y_0)$$ is a local minimum.
• $$D \gt 0$$ and $$f_{xx}(x_0,y_0)) \lt 0\text{,}$$ then $$(x_0, y_0)$$ is a local maximum.
• $$D \lt 0\text{,}$$ then $$(x_0,y_0)$$ is neither a local maximum nor minimum.
The conditions in this version of the second derivative test are simply algebraic criteria that tell us about the definiteness of the Hessian matrix $$H\text{.}$$
### Subsection7.2.3Summary
This section explored quadratic forms, functions that are defined by symmetric matrices.
• If $$A$$ is a symmetric matrix, then the quadratic form defined by $$A$$ is the function $$q_A(\xvec) = \xvec\cdot(A\xvec)\text{.}$$ Quadratic forms appear when studying the variance of a dataset. If $$C$$ is the covariance matrix, then the variance in the direction defined by a unit vector $$\uvec$$ is $$q_C(\uvec) = \uvec\cdot(C\uvec)=V_{\uvec}\text{.}$$
Similarly, quadratic forms appear in multivariable calculus when analyzing the behavior of a function of several variables near a critical point.
• If $$\lambda_1$$ is the largest eigenvalue of a symmetric matrix $$A$$ and $$\lambda_m$$ the smallest, then the maximum value of $$q_A(\uvec)$$ among unit vectors $$\uvec\text{,}$$ is $$\lambda_1\text{,}$$ and this maximum value occurs in the direction of $$\uvec_1\text{,}$$ a unit eigenvector associated to $$\lambda_1\text{.}$$
Similarly, the minimum value of $$q_A(\uvec)$$ is $$\lambda_m\text{,}$$ which appears in the direction of $$\uvec_m\text{,}$$ an eigenvector associated to $$\lambda_m\text{.}$$
• A symmetric matrix is positive definite if its eigenvalues are all positive, positive semidefinite if its eigenvalues are all nonnegative, and indefinite if it has both positive and negative eigenvalues.
• If the Hessian $$H$$ of a multivariable function $$f$$ is positive definite at a critical point, then the critical point is a local minimum. Likewise, if the Hessian is negative definite, the critical point is a local maximum.
### Exercises7.2.4Exercises
#### 1.
Suppose that $$A = \begin{bmatrix} 4 \amp 2 \\ 2 \amp 7 \end{bmatrix}\text{.}$$
1. Find an orthogonal diagonalization of $$A\text{.}$$
2. Evaluate the quadratic form $$q_A\left(\twovec11\right)\text{.}$$
3. Find the unit vector $$\uvec$$ for which $$q_A(\uvec)$$ is as large as possible. What is the value of $$q_A(\uvec)$$ in this direction?
4. Find the unit vector $$\uvec$$ for which $$q_A(\uvec)$$ is as small as possible. What is the value of $$q_A(\uvec)$$ in this direction?
#### 2.
Consider the quadratic form
\begin{equation*} q\left(\twovec{x_1}{x_2}\right) = 3x_1^2 - 4x_1x_2 + 6x_2^2. \end{equation*}
1. Find a matrix $$A$$ such that $$q(\xvec) = \xvec^TA\xvec\text{.}$$
2. Find the maximum and minimum values of $$q(\uvec)$$ among all unit vectors $$\uvec$$ and describe the directions in which they occur.
#### 3.
Suppose that $$A$$ is a demeaned data matrix:
\begin{equation*} A = \begin{bmatrix} 1 \amp -2 \amp 0 \amp 1 \\ 1 \amp -1 \amp -1 \amp 1 \\ \end{bmatrix}. \end{equation*}
1. Find the covariance matrix $$C\text{.}$$
2. What is the variance of the data projected onto the line defined by $$\uvec=\twovec{1/\sqrt{2}}{1/\sqrt{2}}\text{.}$$
3. What is the total variance?
4. In which direction is the variance greatest and what is the variance in this direction?
#### 4.
Consider the matrix $$A = \begin{bmatrix} 4 \amp -3 \amp -3 \\ -3 \amp 4 \amp -3 \\ -3 \amp -3 \amp 4 \\ \end{bmatrix} \text{.}$$
1. Find $$Q$$ and $$D$$ such that $$A=QDQ^T\text{.}$$
2. Find the maximum and minimum values of $$q(\uvec) = \xvec^TA\xvec$$ among all unit vectors $$\uvec\text{.}$$
3. Describe the direction in which the minimum value occurs. What can you say about the direction in which the maximum occurs?
#### 5.
Consider the matrix $$B = \begin{bmatrix} -2 \amp 1 \\ 4 \amp -2 \\ 2 \amp -1 \\ \end{bmatrix}\text{.}$$
1. Find the matrix $$A$$ so that $$q\left(\twovec{x_1}{x_2}\right) = \len{B\xvec}^2=q_A(\xvec)\text{.}$$
2. Find the maximum and minimum values of $$q(\uvec)$$ among all unit vectors $$\uvec$$ and describe the directions in which they occur.
3. What does the minimum value of $$q(\uvec)$$ tell you about the matrix $$B\text{?}$$
#### 6.
Consider the quadratic form
\begin{equation*} q\left(\threevec{x_1}{x_2}{x_3}\right) = 7x_1^2 + 4x_2^2 + 7x_3^2 - 2x_1x_2 -4x_1x_3-2x_2x_3. \end{equation*}
1. What can you say about the definiteness of the matrix $$A$$ that defines the quadratic form?
2. Find a matrix $$Q$$ so that the change of coordinates $$\yvec = Q^T\xvec$$ transforms the quadratic form into one that has no cross terms. Write the quadratic form in terms of $$\yvec\text{.}$$
3. What are the maximum and minimum values for $$q(\uvec)$$ among all unit vectors $$\uvec\text{?}$$
#### 7.
Explain why the following statements are true.
1. Given any matrix $$B\text{,}$$ the matrix $$B^TB$$ is a symmetric, positive semidefinite matrix.
2. If both $$A$$ and $$B$$ are symmetric, positive definite matrices, then $$A+B$$ is a symmetric, positive definite matrix.
3. If $$A$$ is a symmetric, invertible, positive definite matrix, then $$A^{-1}$$ is also.
#### 8.
Determine whether the following statements are true or false and explain your reasoning.
1. If $$A$$ is an indefinite matrix, we can’t know whether it is positive definite or not.
2. If the smallest eigenvalue of $$A$$ is 3, then $$A$$ is positive definite.
3. If $$C$$ is the covariance matrix associated with a dataset, then $$C$$ is positive semidefinite.
4. If $$A$$ is a symmetric $$2\times2$$ matrix and the maximum and minimum values of $$q_A(\uvec)$$ occur at $$\twovec10$$ and $$\twovec01\text{,}$$ then $$A$$ is diagonal.
5. If $$A$$ is negative definite and $$Q$$ is an orthogonal matrix with $$B = QAQ^T\text{,}$$ then $$B$$ is negative definite.
#### 9.
Determine the critical points for each of the following functions. At each critical point, determine the Hessian $$H\text{,}$$ describe the definiteness of $$H\text{,}$$ and determine whether the critical point is a local maximum or minimum.
1. $$f(x,y) = xy + \frac2x + \frac2y\text{.}$$
2. $$f(x,y) = x^4 + y^4 - 4xy\text{.}$$
#### 10.
Consider the function $$f(x,y,z) = x^4 + y^4 +z^4 - 4xyz\text{.}$$
1. Show that $$f$$ has a critical point at $$(-1,1,-1)$$ and construct the Hessian $$H$$ at that point.
2. Find the eigenvalues of $$H\text{.}$$ Is this a definite matrix of some kind?
3. What does this imply about whether $$(-1,1,-1)$$ is a local maximum or minimum? |
# Indian Army Clerk Maths Online Test – Important Question
Indian Army Clerk Maths Online Test – Important Question(Army Clerk Exam Ki Taiyari): The most important section in the clerk written exam in which most of the candidates struggle a lot is Maths. Since the syllabus is vast there is a lot to study. The maths section carries 10 questions out of 25 that are asked in the first part. All those candidates who are preparing for the clerk post should know that it is important to practice maths daily along with the previous year paper and sample paper. With regular practice and hard work, you can achieve anything.
Here we are with some important questions of maths for Indian army clerk with full the explanation that has been collected from previous year paper. Now, you can give this online test of Maths and prepare yourself for the written exam. All the questions that are mentioned below are selected from exam point of view.
Along with maths you can also practice for- Indian Army Clerk General Knowledge(GK) Online Test
## Indian Army Clerk Maths Online Test – Important Question(क्लर्क एग्जाम की तैयारी)
Q 1.The product of two numbers is 216 and their HCF is 3, what is their LCM?
1. 648
2. 72
3. 219
4. 231
2. We have given, product of two numbers= 216. HCF of two numbers= 3. We know that, LCM×HCF = Product of two numbers
LCM×3 = 216
LCM = 216/3
LCM = 72
Q 2.A man buys a toy for Rs.100 and sells it for Rs.150. Then his gain percent is
1. 50%
2. 60%
3. 30%
4. 40%
2. Cost price of toy(CP) = Rs 100
Selling price of toy(SP) = Rs 150
Profit % = (SP-CP/CP)×100
= (150-100/100)×100
= 50%
Q 3.In how many years will Rs.7500 double at 8% simple interest?
1. 12
2. 8
3. 12.5
4. 10
2. Principal = Rs 7500
Rate% = 8
Let the total amount be = A (where A = P+I)
Let in X years A = 2P
Now, 2P-P = P×R×T/100
P = P×R×X/100
X = 100/R
X = 100/8
X = 12.5
Q 4.Express 63 km/h into m/s
1. 16.5
2. 17.5
3. 18.5
4. 19.5
2. 1 Km = 100 m
1 hr = 3600 sec
Hence, 63 km/hr = (63×1000/3600)m/sec
= 63×5/18
= 17.5 m/sec
Q 5.A fruit seller had some apples.He sells 40% apples and still has 420 apples.How many apples he had originally?
1. 600
2. 650
3. 700
4. 750
2. Let the actual number of apples be = a
Now, he sold 40% of apples and are left with 420.
a-(40/100)×a = 420
100a-40a/100 = 420
60a = 42000
a = 42000/60
a= 700
Q 6.If 2994/14.5 = 172, then 29.94/1.45 =?
1. 172
2. 17.2
3. 1.72
4. 0.172
2. We have given that, 2994/14.5 = 172
(2994/14.5 × 1/10) = 172/10
(29.94/1.45) = 17.2
Q 7.The angle of elevation of a ladder leaning against a wall is〖60〗^0 and the foot of the ladder is 4.6m away from the wall. The length of the ladder is
1. 4.6
2. 9.2
3. 10
4. 11.2
2. Angle of elevation = 60°
Distance between wall and foot off ladder = 4.6
Let the length of ladder be = H
cos 60° = 4.6/H
1/2 = 4.6/H
H = 4.6×2
H = 9.2 m
Q 8.The average of 20 values is 18. If 3 is subtracted from each of the value, then the new average will be
1. 21
2. 15
3. 16
4. 17
2. Let the sum of 20 values be = a
then, a/20 = 18
a= 20×18
a= 360
Now, 3 is subtracted from each values. So, total values subtracted from the sum is = 20×3= 60
Remaining value= 360-60 => 300
New average = 300/20 => 15
Q 9.The length of a tunnel, which a train 90 m long, travelling at 48 km/h, can cross in 21 s is
1. 180
2. 280
3. 190
4. 290
2. Length of train = 90 m
Speed of train = 48 km/hr
= 48×1000/3600
= 240/18 m/sec
Time taken by the train = 21 sec
Let length of platform = A
Length of platform+length of train = speed of train×time taken
A+90 = (240/18)×21
90+A = 280
A = 190 m
Q 10.What percentage is 390 g of 1 kg 500 g?
1. 26
2. 20
3. 22
4. 25 |
# Crash Course, Mathematics Permutations and Combinations Part 1
## Permutations and Combinations
Permutations and Combinations is a topic which involves basic counting techniques. We would learn how to calculate number of ways of arranging and selecting a group of objects without listing down all the possibilities.
Example problem:
Find the total number of possible outcomes when 3 coins are tossed at the same time?
## What else you can do inside qs leap ?
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Listing down all the possibilities: {H, H, H}, {H, H, T}, {H, T, H}, {H, T, T}, {T, H, H}, {T, H, T}, {T, T, H}, {T, T, T}
Where each set represents a possible outcome.
{T, H, T} means 1st coin has Tails, 2nd coin has Heads and 3rd coin has Tails and so on.
Total number of possibilities is 8
Knowledge of Permutations and Combinations allows us to come to the conclusion without making the list. Most of the numbers we work in this chapter are large so it becomes impractical to list down all the possibilities.
This technique is widely used in statistics where we have to deal with data. So it is used everywhere is science and technology.
### Principle of Multiplication (the rule of AND)
Let’s assume a situation in which Luffy the pirate wants to attack Marines headquarters. He has 3 ships AND 2 robots. In how many ways he can choose 1 ship and 1 robot to make the attack.
There are 3 ways in which a ship can be chosen because there are 3 ships. For every choice of ship Luffy has 2 robots.
So total number of ways he can choose 1 ship and 1 robot is 3 x 2 = 6
Let’s call ships as S¬1, S2 and S3 and robots R1 and R2. Our choices are
S1 R1
S1 R2
S2 R1
S2 R2
S3 R1
S3 R2
Total 6 ways to select.
In fundamentals of counting this is called the multiplication principle.
It states that “If there are two events, one of them can occur in p ways while the other can occur in q ways, the total number of ways both 1st and 2nd event can occur is p x q.”
### In general
“If n events, x1, x2, x3, . . . xn can occur in a1, a2, a3 . . . an ways. The total number of ways x1 and x2 and x3 . . . and xn can happen is a1 x a2 x a3 . . . x an .”
### Principle of addition (the rule of OR):
Namrata is going to draw one card from a deck of cards. In how many ways she can draw a king or a queen.
There are 4 kings and 4 queens in a deck. Namrata has to choose either king OR queen she can’t choose both (unlike the previous case where Luffy had to choose one ship AND one robot).
So number of ways of choosing = 4 + 4 = 8
Situation 2:
Namrata has to choose either a Queen or a hearts
We know that there are 4 queens, 13 hearts and there is 1 queen of hearts. So total number of choices would be 4 + 13 – 1 = 16
“In general, if n events x1, x2, x3, . . . xn can occur in a1, a2, a3 . . . an ways respectively and there is no way of two events happening at once the number of ways x1 or x2 or x3 . . . or xn can occur is a1 + a2 + a3 + . . . + an .”
[Similarly, "If there are two events x1 and x2 and they can occur in a1 and a2 ways and there are n ways in which they can occur together. Total number of ways x1 or x2 can occur is a1 + a2 – n.” Warning this second rule of addition should not be extrapolated for n objects!]
## What else you can do inside qs leap ?
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## Permutations:
Permutations simply mean the number of ways of arranging a certain number of objects.
To understand permutations let’s see what are the different types of problems we can encounter.
### Problem type 1. n distinct objects arranged in n different places in a row
How many 7 letter words can be formed by using the letters of the word PRODUCT where repetition is not allowed.
Solution:
We have 7 empty spots to fill with 7 letters.
1st spot can have 7 different letters
2nd spot cannot have the letter we placed on the first spot because repetition is not allowed. So 2nd spot has 6 choices of letters.
Similarly, 3rd spot has 5 choices, 4th spot 4 choices, 5th spot 3 choices, 6th spot 2 choices and 7th spot is left with 1 choice.
Since we have to fill 1st spot AND 2nd spot AND 3rd spot and so on
Multiplication rule will apply (the rule of AND).
Total number of words that can be made using the letters of the word product = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 7!
This solution can be generalised for arrangement of n-objects.
N! is the product of all the integers from 1 through N.
N! = 1 x 2 x 3 x . . . x (N-1) x N
In general number of ways of arranging n DISTINCT objects in n distinct places in a row is n!
Important to remember 0! = 1
Reason why 0! =1? Because 0! Is number of ways of arranging 0 objects. And number of ways of arranging 0 objects is 1; i.e. you don’t arrange them. (this is a crude explanation. Actual proof involves use of what we call a gamma function in statistics. It is one of fundamentals in statistics but is not in the syllabus. Gamma function defines factorials for non-whole numbers i.e. it is generalisation of factorial. If you have any inquiries related to the gamma function you can discuss on official quant group on LEAP. I would be happy to reply.)
### Problem type 2: n distinct objects arranged in r distinct places in a row.
How many ways are there to make 4 letter word with or without meaning from the letters of the word PRODUCE without repetition? (i.e. each letter can appear only once in the word)
We have 4 spots to create a 4 letter word __ __ __ __
Word PRODUCE has 7 distinct characters.
1st spot has 7 choices
2nd spot has 6 choices
3rd spot has 5 choices
4th spot has 4 choices
So total number of words that can be made using 4 letters of the word PRODUCE is = 7 x 6 x 5 x 4
Multiplying and dividing this number by 3 x 2 x 1
= (7 x 6 x 5 x 4 x 3 x 2 x 1) / 3 x 2 x 1
= 7! / 3! = 7! / (7 – 4)!
The term 7! / (7 – 4)! Is called 7P4
In general, nPr is the number of ways of arranging n different objects in r different places in a row where nPr = n! / (n-r)!
### Problem type 3: How many 4 letter words can be formed by letters of the word PRODUCT with Repetition allowed?
We have four spots to create a 4 letter word __. __. __. __.
1st place has 7 options
Since repetition is allowed second place also has 7 options
Similarly, 3rd and 4th place have 7 options.
So total number of permutations (arrangement of letters or words) =
7 x 7 x 7 x 7 = 74
### Problem type 4 (very important):
We have noticed that in all the above problems the objects we have to arrange are distinct. But what would happen if we have arranged a set of objects in which there are identical sets of objects?
We took the word PRODUCT for arrangements. Notice that each and every letter of the word PRODUCT is different. What would happen if rather than choosing the word PRODUCT we chose the word CAREERS? Both have 7 letters the only difference is in the word CAREERS the letters ‘R’ and ‘E’ are repeated twice.
Let’s compare the two situations with a very basic example:
Situation 1: number of ways of arranging letters of the word AB;
We know that there are 2. AB and BA.
Situation 2: number of ways of arranging letters of the word AA;
There is only 1 way because even if we switch the letters we will get the same word AA.
So if we have non-distinct objects, number of arrangements is reduced because we have to remove similar arrangements.
The problem statement:
Find the number of ways of arranging the letters of the word TREE.
To solve this let’s begin with assuming the word has all distinct letters.
Total number of ways of arranging a 4 letter word is 4! (problem type 1)
But there are two E’s (let’s call them E1 and E2)
The permutation TRE1E2 will be same as TRE2E1 and should be counted as 1
So from 4! We have remove all the possible arrangements of E and E2 that are counted more than once.
Number of ways of arranging E1 and E2 = 2!
Therefore, total number of permutations = 4! / 2!
Alternate Clarification
If E1 and E2 were distinct If E1 and E2 are same TRE1E2 TREE TRE2E1 TE1RE2 TERE TE2RE1 E1TRE2 ETRE E2TRE1
And so on . . .
We can see that for every 2 arrangements there is only one UNIQUE arrangement that should be counted. Hence actual number of unique permutations = total number of permutations / number of ways E1 and E2 can be arranged among themselves = 4! / 2!
Example 2: Same problem type
Find number of ways to arrange letters of the word CHEESE?
Solution:
Here we see E appears thrice.
Let’s call the three E’s E1, E2 and E3
Total number of ways of arranging the words of letter CHEESE considering E1, E2 and E3 to be different = 6! (problem type 1)
Total number of ways of arranging E1, E2 and E3 = 3!
So number of unique arrangements for the word CHEESE = 6! / 3!
Visualisation:
CHE1E2SE3, CHE1E3SE2, CHE2E1SE3, CHE2E3SE1, CHE3E1SE2, CHE3E2SE1
Are all considered same as CHEESE. We have kept C, H and S at the same spot and arranged E1, E2 and E3. And number of ways of arranging 3 different objects is 3! (just like in the case of previous TREE example every 3! Objects have to be considered as 1)
Generalisation:
Problem: Number of ways of arranging n objects where r1 objects are of 1 kind, r2 objects are of second kind, r3 objects are of 3rd kind and so on . . . then
Total number of unique arrangements possible = n! / (r1! r2! r3! . . .)
Example:
How many ways are there to arrange the letters of the word MISSISSIPPI?
In the word MISSISSIPPI
Total number of letters = 11
“M” occurs 1 time, “I” occurs 4 times, “S” occurs 4 times, “P” occurs 2 times
So total number of unique arrangements = 11! / (1! X 4! X 4! X 2!)
We usually don’t write 1! Because it is equal to 1.
### Important Example:
A binary number x contains a total of n digits and r 1’s. What is the number of ways of arranging the digits of x?
Solution:
A binary number consists of 0s and 1s
Total number of digits in x = n
Total number of 1s in x = r
Hence total number of 0s in x = n – r
So x has r 1s and (n – r) 0s
From previous example we can say that Number of ways of arranging digits in x = n! / r! x (n – r)!
I will be referring back to this example in the article relating to Combinations.
Coming up next week Circular Permutations.
Will post the link once it is up.
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389 minus 1 percent
This is where you will learn how to calculate three hundred eighty-nine minus one percent (389 minus 1 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 389 minus 1 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 389 of something.
389
(100%)
1 percent means 1 per hundred, so for each hundred in 389, you want to subtract 1. Thus, you divide 389 by 100 and then multiply the quotient by 1 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(389 ÷ 100) × 1
= 3.89
We made a pink square that we put on top of the image shown above to illustrate how much 1 percent is of the total 389:
The dark blue not covered up by the pink is 389 minus 1 percent. Thus, we simply subtract the 3.89 from 389 to get the answer:
389 - 3.89
= 385.11
The explanation and illustrations above are the educational way of calculating 389 minus 1 percent. You can also, of course, use formulas to calculate 389 minus 1%.
Below we show you two formulas that you can use to calculate 389 minus 1 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
389 - ((389 × 1/100))
389 - 3.89
= 385.11
Formula 2
Number × (1 - (Percent/100))
389 × (1 - (1/100))
389 × 0.99
= 385.11
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
390 minus 1 percent
Here is the next percent tutorial on our list that may be of interest. |
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# The sum of squares of two parts of a number 100 is minimum, then two parts are:$\left( a \right)50,50 \\ \left( b \right)25,75 \\ \left( c \right)40,60 \\ \left( d \right)30,70 \\$
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Hint: Use application of derivative to find maxima and minima .For maximum and minimum point derivative of function $f'\left( x \right) = \frac{{df}}{{dx}} = 0$ and for check maxima and minima use second derivative test, $f''\left( x \right) > 0$ minima point and $f''\left( x \right) < 0$ maxima point.
Let $x$ and $y$ be two parts of 100.
So, we can write as $x + y = 100$
$\Rightarrow y = 100 - x$
So, $x$ and $100 - x$ are two parts of 100 .
Now, according to question
$f\left( x \right) = {\left( x \right)^2} + {\left( {100 - x} \right)^2}............\left( 1 \right)$
For maxima and minima, $f'\left( x \right) = \frac{{df}}{{dx}} = 0$ .
So, Differentiate (1) equation with respect to x .
$f'\left( x \right) = \frac{d}{{dx}}\left( {{{\left( x \right)}^2} + {{\left( {100 - x} \right)}^2}} \right) \\ \Rightarrow f'\left( x \right) = 2x + 2\left( {100 - x} \right)\left( { - 1} \right) \\ \Rightarrow f'\left( x \right) = 4x - 200..........\left( 2 \right) \\ f'\left( x \right) = 0 \\ \Rightarrow 4x - 200 = 0 \\ \Rightarrow 4x = 200 \\ \Rightarrow x = 50 \\$
Now, use the second derivative test for check x=50 is a maxima or minima point .
So, Differentiate (2) equation with respect to x .
$f''\left( x \right) = \frac{d}{{dx}}\left( {4x - 200} \right) \\ \Rightarrow f''\left( x \right) = 4 \\$
$f''\left( x \right) > 0$ for all value of x .
Now, $f''\left( x \right) > 0$ for x=50
So, x=50 is a minimum point.
Hence the function $f\left( x \right) = {\left( x \right)^2} + {\left( {100 - x} \right)^2}$ minimum at x=50 .
So, the required parts are 50 and 50 .
So, the correct option is (a).
Note: Whenever we face such types of problems we use some important points. First we assume the parts of a number and make a function in one variable according to the question then differentiate the function for maxima and minima then use a second derivative test to confirm the point is maxima or minima. |
United States of AmericaPA
High School Core Standards - Algebra I Assessment Anchors
# 1.07 Units resulting from a formula or graph
Lesson
## Determine the units from the attributes in a formula
We have learned how to make all kinds of mathematical calculations. However, if I told you the answer is $42$42, your initial response may be, "$42$42 what?" It could be $42$42 ounces or it could be $42$42 goats. That's why units of measurement are really important!
We can use units of measurement to define any physical phenomenon, such as quantities, weights, lengths, areas, volumes, and rates. Let's look at some different units of measurement now.
### Measurements of length
Length or distance is a measurement in one dimension. For example, we could measure from one point to another
Units of measurements for length include millimeters ($mm$mm), centimeters ($cm$cm), meters ($m$m), and kilometers ($km$km) or inches ($in$in), feet ($ft$ft), yards ($yd$yd), and miles ($mi$mi).
### Measurements of area
Area is a measure of two dimensions, usually called length and width, but sometimes called base and height. Area is measured in square units.
Units of measurements for area include square millimeters ($mm^2$mm2), square centimeters ($cm^2$cm2), square feet ($ft^2$ft2), and square miles($mi^2$mi2) just to name a few.
### Measurements of volume
Volume reflects the amount of space an object takes up and is a measure of three dimensions usually called length, width, and height and is measured in cubic units.
Units of measurements for volume include cubic millimeters ($mm^3$mm3), cubic inches ($in^3$in3), or cubic feet ($ft^3$ft3) to name a few.
### Measurements of mass
Units of measurement for mass include grams ($g$g), kilograms ($kg$kg), ounces ($oz$oz), pounds ($lbs$lbs) and tons ($t$t), among others. Mass is a measure of how heavy something is. We often refer to the mass of an object as its weight.
### Measurements of capacity
Capacity is similar to volume but is a measure of how much something holds, such as how much liquid will fit in a bottle.
Measures of capacity include gallons ($gal$gal), milliliters ($ml$ml), and liters ($l$l).
### Measurements in rates
A rate is a ratio between two measurements with different units. There are any number of combinations of measurement units for rates, such as miles per gallon ($mi$mi/$gal$gal), kilometers per hour ($km$km/$hr$hr) and so on.
### Units from a formula
Any formula that is made up of measurements will have units attached to each of the variables.
Here is a formula we are familiar with. The speed of an object is a measure of the distance traveled over time.
$S=\frac{d}{t}$S=dt
The units for Speed in the formula are derived from the units used for distance and time. So if the distance is measured in kilometers and time is measured in hours, then the
$\text{Speed (unit) }=\frac{\text{distance (units) }}{\text{time (unit) }}$Speed (unit) =distance (units) time (unit)
$\text{Speed (unit) }=\frac{\text{kilometers }}{\text{hour }}$Speed (unit) =kilometers hour
$\text{Speed (unit) }=\frac{\text{kilometers }}{\text{hour }}$Speed (unit) =kilometers hour
#### Worked Examples
##### Question 1
The area of a rectangle is given by $A=l\times w$A=l×w, where $l$l is the length and $w$w is the width. Both length and width must be of the same units when performing the multiplication to find the area.
1. What would the unit for area be if the length and width are in millimeters?
m2
A
cm-km
B
mm2
C
mm
D
m2
A
cm-km
B
mm2
C
mm
D
##### Question 2
Adam plotted a point to represent material purchases ($x$x) and the costs involved ($y$y). When Adam bought $140$140cm of material, it cost $\$2.202.20.
1. What unit is the $x$x-axis using?
cost in dollars of material purchased per meter
A
cost in dollars of material purchased per sale
B
meters of material purchased
C
centimeters of material purchased
D
cost in dollars of material purchased per meter
A
cost in dollars of material purchased per sale
B
meters of material purchased
C
centimeters of material purchased
D
2. What unit is the $y$y-axis using?
cost in dollars of material purchased
A
centimeters of material purchased
B
cost in cents of material purchased
C
meters of material purchased
D
cost in dollars of material purchased
A
centimeters of material purchased
B
cost in cents of material purchased
C
meters of material purchased
D
## Dimensional analysis (construction of units)
Remember!
Length has $1$1 dimension, therefore the power for the units is $1$1 such as $in$in$m$m, or $ft$ft.
Area has $2$2 dimensions, therefore the power for the units is $2$2 such as $in^2$in2$m^2$m2, or $ft^2$ft2.
Volume has $3$3 dimensions, therefore the power for the units is $3$3 such as $in^3$in3$m^3$m3, or $ft^3$ft3.
#### Worked examples
##### question 3
If $F=3q+r$F=3q+r, where both $q$q and $r$rrepresent lengths, what does $F$F represent?
Think: Consider each part of this equation. The power of each part is just $1$1 at the moment. Does the addition of $3q$3q and $r$r change the power?
Do: As you are adding, there is no change to the power. Therefore this is still a measurement of length.
##### question 4
If $F=\frac{s^3}{p}$F=s3p where both s and p represent lengths, what does $F$F represent?
Think: $s$s has a power of $3$3, so it must be a volume. However, it is then divided by a length ($F$F). What happens to the power in this case?
Do: If a volume (power of $3$3) is divided by a length (power of $1$1), then the resulting power is $2$2 which represents an area.
#### Practice questions
##### question 5
If $\text{work}=\text{force}\times\text{distance}$work=force×distance, what are the dimensions of work?
1. $\text{time}\times\text{length}^2\times\text{mass}^2$time×length2×mass2
A
$\frac{\text{mass}^2\times\text{time}}{\text{length}^2}$mass2×timelength2
B
$\frac{\text{mass}\times\text{length}}{\text{time}}$mass×lengthtime
C
$\frac{\text{mass}\times\text{length}^2}{\text{time}^2}$mass×length2time2
D
$\text{time}\times\text{length}^2\times\text{mass}^2$time×length2×mass2
A
$\frac{\text{mass}^2\times\text{time}}{\text{length}^2}$mass2×timelength2
B
$\frac{\text{mass}\times\text{length}}{\text{time}}$mass×lengthtime
C
$\frac{\text{mass}\times\text{length}^2}{\text{time}^2}$mass×length2time2
D
##### Question 6
This graph's slope has a unit of km/h.
1. What is the unit on the $x$x-axis?
km/h
A
hours
B
kilometers
C
km/h
A
hours
B
kilometers
C
2. What is the unit on the $y$y-axis?
hours
A
km/h
B
kilometers
C
hours
A
km/h
B
kilometers
C
### Outcomes
#### CC.2.1.HS.F.3
Apply quantitative reasoning to choose and Interpret units and scales in formulas, graphs and data displays.
#### CC.2.1.HS.F.4
Use units as a way to understand problems and to guide the solution of multi-step problems. |
Complex Counting
Home > Lessons > Complex Counting Search | Updated February 26th, 2017
Introduction
In this section, you will learn how to use the formulas for combinations and permutations for various problems. Here are the sections within this lesson:
Factorials When counting complex problems, we will use something called a 'factorial.' A factorial looks like an exclamation point; like this: It means this: In general, it means this: In the next section, we will learn about two problems that can use factorials as a means for solving them. Factorial Problems Let's look at two problems that involve factorials. Problem 1: If there are 8 different books on a shelf, how many ways are there to arrange those books? Imagine placing all the books on the floor. One would have to choose between 8 different books to be placed first on the shelf. Once picked, there would be 7 books remaining that could also be picked for the second place on the shelf. Once that book is placed on a shelf, there would be 6 books to choose from, and so on. Using the fundamental counting principal, the math would look like this. There are 40,320 ways to arrange these 8 books. Problem 2: If there are 4 people who need to be seated at a circular table, how many different ways can this be done? It appears this problem is the same as problem #1 above but it has less objects to arrange. We could list all the possibilities. We will list the people as person A, person B, person C, and person D. The problem with this list, which lists all combinations of A, B, C, and D, it does not take in to consideration that there is a circular table. However, it is a great list for placing people in a line. The difference is, there are situations listed there that are the same. Here are arrangements that are exactly the same, because of the circular situation. Take any letter. Look to its left and right and you will see identical adjacent letters in each set. This means the only true unique situations we can gain is by looking down one of the columns from our large 24-set list above. So, there really are only 6 unique arrangements. This means if we take one less than our number of people, we can simply use factorials. There are 6 ways to arrange 4 people at a circular table. Note: This means if we started with 8 people, we could calculate the total possible ways to arrange them around a table by calculating 7!, which is 5,040. Introduction to Combinations If one were to award prizes to runners in a race, it is interesting to calculate how many ways prizes can be awarded. Imagine having two \$20 prizes to give to the first two winners of the race. If there were 5 runners in the race, we can see how many ways we can get two winners. We can list the runners as numbers, as if we gave them tags to wear. Here are all the possible winning pairs: Note that we cannot list {3,1} as a possible pair in addition to {1,3} because doing so would count the same pair of runners twice. Therefore, there are 10 possible outcomes. We will see in the next section that there is an easier way to calculate these situations because listing all subsets if there were 200 runners would be very impractical. Combination Formula When there are 'm' runners and we are to give out 'n' awards, there is a formula to use. Sometimes, this formula is referred using several methods, such as these: Note: This formula is used when there is no significance to order of outcomes. Combination Problems Now, we will use the combination formula from the last section for two problems. Problem 1: If 300 tickets were sold for a raffle, how many ways could 5 \$20 awards be given? There are 300 tickets and we will choose 5 winners, like so. The work above demonstrates how one can simplify the problem using the combination formula. Most people do not do the work above and instead use the built in combination functionality of a calculator to get the final answer. The answer is roughly 19,582,837,560 combinations or over 19 billion different ways winners can be awarded. ideo: Tickets Problems uiz: Tickets Problems Problem 2: If 5 cards are dealt from a standard deck of cards, how many ways are there to do this if order is not important? We simply need to use the formula for a standard 52-card deck. We will use "52 choose 5." Simply plug this into a scientific calculator (or a better calculator) to get 2,598,960, which is over 2.5 million possible outcomes. Permutation Formula Permutations describe counting subsets when order matters. When referring to racing horses and coded locks, we know that order does matters. When we calculate permutations, like how many ways can 8 horses place first, second, and third place, we would use this formula. The problems in the next section will reveal how to use this formula. Permutation Problems The problems in this section will indicate how to use the permutation formula, which was described in the previous section. Problem 1: How many ways can 8 horses place first, second, and third place? For our horse race situation, we would take 8 horses and place 3, like this. There are 336 ways to have 8 horses place for 1st, 2nd, and 3rd place. The work above demonstrates how one can simplify the problem using the permutation formula. Most people do not do the work above and instead use the built in permutation functionality of a calculator to get the final answer. Problem 2: At a certain school, there are 850 seniors. From this group, a president and vice-president will be chosen. How many ways can this be done? This is another problem that involves order. When we choose two people from the large group, the first person will be the president and the second person will be the vice-president. So, we will take 850 and order 2, like so. This means there are 721,650 possible ways to pick these two positions. The work above demonstrates how one can simplify the problem using the permutation formula. Most people do not do the work above and instead use the built in permutation functionality of a calculator to get the final answer. Instructional Videos Try our instructional videos to determine if you understand the lessons above. Interactive Quizmasters Try our interactive quizzes to determine if you understand the lessons above. Activities Examine our activities related to the lessons above. ctivity: Combinations ctivity: Permutations Related Lesson Try these lessons, which are closely related to the lesson above. esson: Basic Counting esson: Placement Counting esson: Probability Basics esson: The Monty Hall Problem |
# What Is 7/9 as a Decimal + Solution With Free Steps
The fraction 7/9 as a decimal is equal to 0.777.
In mathematics, we employ a Division operation to split a number into equal parts. Division frequently appears to be more challenging than other mathematical processes. However, there is a way to simplify this extremely complex process. Long Division is the method employed here to solve the given fraction.
The mathematical operation of dividing large numbers into smaller and more manageable units or groups is known as Long Division. It helps to simplify complex and difficult problems.
Here, we will simplify the fraction 7/9 by the Long Division method and will find its equivalent decimal number.
## Solution
Separating a division problem into its constituent parts according to each one’s functionality is the first and most significant step in solving the problem. The Dividend is the number that is divided, and Divisor is the number that is used to divide the dividend. The following problem has a dividend equal to 7 but a divisor equal to 9.
In the presented problem, we have:
Dividend = 7
Divisor = 9
Quotient and Remainder are the other two important terms associated with division. A fraction can be divided completely to produce a quotient, which is the result of the division. But a partial division produces a remainder, which expresses the value that remains after an incomplete division.
Quotient = Dividend $\div$ Divisor = 7 $\div$ 9
Here, the decimal value of 7/9 is determined by using the Long Division method.
Figure 1
## 7/9 Long Division Method
The detailed steps to simplify the given fraction are shown below.
We possess:
7 $\div$ 9
It can be seen that the dividend is a smaller number than the divisor. So, to solve the supplied fraction, we need a Decimal Point. If we place a zero to the right of the remainder, this is possible. As a result, we receive 70, which must now be divided by 9. Below is a description of the division’s steps:
70 $\div$ 9 $\approx$ 7
Where:
9 x 7 = 63
A Remainder is produced and is given as:
70 – 63 = 7
As we get a non-zero remainder, we once again add a zero to the right of the remainder, but this time without the insertion of a decimal point. So, the remainder of 7 becomes 70 again and we have to divide it by 9.
70 $\div$ 9 $\approx$ 7
Where:
9 x 7 = 63
To calculate the remainder, we proceed as follows:
70 – 63 = 7
We observe that both the Remainder and Quotient are the same as that obtained in the previous step. This illustrates that a given fraction is a non-terminating and recurring fraction. Thus, we don’t need to calculate further, and our final answer is 0.777 with a remainder of 7.
]Images/mathematical drawings are created with GeoGebra. |
# How to Falsely Prove That Pi Equals 3
Author Info
Updated: July 14, 2019 | References
Are you burdened by the evil of memorizing a sequence of random and intangible numbers simply to perform a simple operation like calculating the area of a circle? Do you feel oppressed by the regime that has drilled into our minds the inevitable infinitude of numbers that make up the irrational constant π? Today is the day to shake off the shackles! Read on, and you, too, will discover how to unravel the hidden truth that pi actually equals 3.
## Steps
1. 1
Take out a sheet of paper. If you are really going to stick it to the man, you might want to use something other than the everyday 8-1/2" x 11" sheets of paper.
2. 2
Set up your proof. At the top of the paper, write that a=b.
• Square both sides of the equation.
a2 = b2
• Rewrite this in another form. This will be your starting point for the proof.
a2 = ab
3. 3
Set up a secondary equation. Simply multiply both sides of your most recent equation by 3. You will need to incorporate this into your proof later.
• 3a2 = 3ab
4. 4
Perform the following operations:
• Multiply both sides of the starting point for your proof, a2 = ab, by π.
πa2 = πab
• Subtract one (equal) half of your secondary equation, 3a2 = 3ab, from each side.
πa2 - 3ab = πab - 3b2
• Add 3ab and subtract πab on both sides.
πa2 - πab = 3ab - 3b2
• Add ab and subtract b2 on both sides.
πa2 - πab + ab - b2 = 4ab - 4b2
• Factor out common terms.
πa(a-b) + b(a-b) = 4b(a-b)
• Remove common terms.
πa + b = 4b
• Subtract b from both sides.
πa = 3b
• Substitute a for b (since a = b).
πb = 3b
• Remove common terms.
π = 3
5. 5
Let out a gasp of incredulity! How many years of unnecessary pain did you endure in math, calculating the area of a circle with the clearly fabricated 3.14, or, even worse, 3.1416?
6. 6
Take a moment to relish your new-found freedom by calculating areas and volumes with the new value of π: 3. Now, it's perfectly easy: Given a circle of radius 10 units, the area is π*radius2, or 3*102 = 300 units2. Ah, the power!
7. 7
Why stop at π? You may have been one of the poor, enslaved students who also used 2.718 for e or 1.414 for the square root of 2. Be free of all of them!
## Community Q&A
200 characters left
## Tips
• Discuss. Show this to a person to share your opinion with, like a math teacher, and ask if it is mathematically correct.
• Use this as a fun trick to confuse your friends, parents, or math teacher.
• Can you spot the mistake? Hint: it’s in πa(a-b) + b(a-b) = 4b(a-b). That’s because a = b, so a-b = 0.
• If you can prove that π is equal to 3, then you can also prove that π is also equal to any other number and that &pi is actually an infinite series of numbers.
## Warnings
• Do not use this in your math class or in real problems.
• While the article is amusing, the math is incorrect. PI = (Circumference of a circle)/(Diameter of a Circle). Using similar math as the article, I can prove that Pi is equal to 10, 20 or any other number (even 1!).
• Many will try to tell you that you've got it all wrong and that π is an irrational number approximately equal to 3.1415926535897932384626433... They may even try to show "proofs" of this using phrases like "infinite summation" and "convergence". Maybe they will try to "show you the light" by drawing circles and showing how to estimate π. Well, it's their argument that is circular! Just remember that the people that fabricated those phrases and arguments are the same that fashioned these vicious lies about π! Imagine: A number whose digits are random and go on forever! Inconceivable!
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 34 people, some anonymous, worked to edit and improve it over time. This article has also been viewed 42,541 times.
26 votes - 44%
Co-authors: 34
Updated: July 14, 2019
Views: 42,541
Categories: Numbers Magic | Mathematics |
# Class 12 Maths NCERT Solutions for Chapter 4 Determinants Exercise 4.1
### Determinants Exercise 4.1 Solutions
1. Evaluate the determinants in Exercises 1 and 2.
Solution
= 2(-1) - 4(-5) = -2 +20 = 18
2. Evaluate the determinants in Exercises 1 and 2.
Solution
(i)
= (cos θ)(cos θ) − (−sin θ)(sin θ) = cos2 θ+ sin2 θ = 1
(ii)
= (x2 − x + 1)(x + 1) − (x − 1)(x + 1)
x3 − x2 + x + x2 − x + 1 − (x2 − 1)
x3 + 1 − x2 + 1
x3 − x2 + 2
3. If A = then show that |2A| = 4|A|
Solution
The given matrix is A =
4. If A = then show that |3A| = 27|A| .
Solution
The given matrix is A =
It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.
Hence, the given result is proved.
5. Evaluate the determinants
(i)
(ii)
(iii)
(iv)
Solution
(i) Let A =
It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.
= (-15 + 3) = -12
(ii) Let A =
By expanding along the first row, we have :
|A| =
= 3(1 + 6) + 4(1 + 4) + 5(3 - 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 +5
= 46
(iii) Let A =
= 0 - 1(0 - 6) + 2(-3-0)
= -1(-6) + 2(-3)
= 6 -6 = 0
(iv) Let A =
By expanding along the first column, we have :
= 2(0 - 5) - 0 + 3(1 + 4)
= -10 + 15 = 5
6. If A = , Find |A|
Solution
Let A =
By expanding along the first row, we have :
= 1(-9 + 12) - 1(-18 + 15) - 2(8 - 5)
= 1(3) - 1(-3) - 2(3)
= 3 + 3 - 6
= 6 - 6 = 0
7. Find values of x, if
Solution
(i)
⇒ 2× 1 - 5× 4 = 2x × x - 6×4
⇒ 2 - 20 = 2x2 – 24
⇒ 2x2 = 6
⇒ x2 = 3
⇒ x = ±√3
(ii)
⇒ 2× 5 - 3× 4 = x × 5 - 3 × 2x
⇒ 10 - 12 = 5x - 6x
⇒ -2 = -x
⇒ x = 2
8. If , then x is equal to
(A) 6
(B) ± 6
(C) -6
(D) 0
Solution
⇒ x2 - 36 = 36 - 36
⇒ x2 - 36 = 0
⇒ x2 = 36
⇒ x = ± 6
Hence, the correct answer is B. |
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# Thousandth Place in Decimals
Last updated date: 16th Apr 2024
Total views: 152.4k
Views today: 1.52k
## Introduction
Thousand, I hope you kids have heard of this word. You might have heard it from your parents, or teacher or while on your way to school or the park. It might have been used for a thousand rupees note or quantity many more. The fourth position to the left before the decimal point is at the thousands place value.
Whereas today we will learn about the thousandth place. The third position right after the decimal point is at the thousandth place value.
## What is Place Value in Decimal?
A digit has a value called its place value depending where on its position in the number. A decimal point divides a decimal number into a whole number and a fractional part.
From these two statements, we can conclude that the digit's position in a number after the decimal point can be called the place value of a digit in decimal.
Place Value in Decimal
## Thousandth Place in Decimal:
The thousandth place in decimals is the third numerical after the decimal.
One-thousandth is one part of the thousand parts or pieces. It does not start with ones. Instead, the number towards the right of the decimal is Tenths, the second digit after the decimal is Hundredths, and the third digit is called Thousandths.
For example, in 6.489 (SIX POINT FOUR EIGHT NINE),
4 (FOUR) is in Tenth place.
8 (EIGHT) is in the Hundredth place.
9 (NINE) is at the Thousandth place.
The Thousandth Place of 0.345
## How to Convert the Decimal Fraction 0.825 into a Percentage
To convert a decimal number 0.825 into a percentage, follow the steps:
• The first step is to multiply the decimal by 100.
Let's take the number 0.825
$= \dfrac{825}{1000} \times 100 = 82.5$
The second step is to add the % (PERCENTAGE) sign to the value.
$= 82.5\%$
## Tenth, Hundredth, Thousandth - Chart
INTEGER PART Thousands 1000 Hundreds 100 Tens 10 Ones 1 DECIMAL PART Tenths $\dfrac{1}{{10}}$ Hundredths $\dfrac{1}{{100}}$ Thousandths $\dfrac{1}{{1000}}$
## 3 Decimal Places Example
The Thousandths place in place value means $\dfrac{1}{{1000}}$ (ONE UPON THOUSAND). The Decimal number is a particular portion of the thousand. Let's take a look at some Thousandths place in decimals examples:
Example 1:
0.033 (ZERO POINT ZERO THREE THREE) is also represented as $\dfrac{{33}}{{1000}}$ (THIRTY-THREE UPON THOUSAND) in fraction. It signifies that it is 33 over one thousand.
Example 2:
0.087 (ZERO POINT ZERO EIGHT SEVEN) is also represented as $\dfrac{{87}}{{1000}}$ (EIGHTY-SEVEN UPON THOUSAND) in fraction. It signifies that it is 87 over one thousand.
Example 3:
0.002 (ZERO POINT ZERO ZERO TWO) is also represented as $\dfrac{2}{{1000}}$ UPON THOUSAND) in fraction. It signifies that it is 2 over one thousand.
## Conclusion
The value that a digit in a number represents depending upon its position is known as place value. The third position to the decimal's right is the thousandth place.
Now, that you guys know what we really mean by a thousandth. Then we can easily understand that if there are 1000 squares equal on a grid and if we spilt one of the squares into 10 equal squares, then one of such small squares will represent the thousandth, which is 1 over one thousand.
## FAQs on Thousandth Place in Decimals
1. What is the significance of place value?
The foundation for regrouping, multiple-digit multiplication, and other operations in the decimal system is set out through place value, which also serves as a starting point for learning other base systems.
2. How may place value be used in daily life?
Place value helps in decision-making for everyday issues including pricing, weight, distances, and timing. Our system of numbers is based on the Base Ten system. Base ten represents the ten base of our number system.
3. What are some place value facts?
Place value is the value that a digit in a number represents based on where it is placed in the number. For example, the place value of 7 in 9,712 is 700, which is 7 hundred. The place value of 7 in 7,689 is, however, 7 thousand, or 7,000. |
In Mathematics, numbers are the building blocks and their fundamental properties. One such property is factors of numbers. In this Math tutorial, we will study the factors of 2, how to calculate the factors of 2, and the prime factors of 2.
As we know, 2 is an even number and a prime number. Which means, it is divisible by itself.
All the even numbers are multiple and divisible by 2.
The number 2 is the smallest number and smallest prime number.
Factors of 2: 1 and 2
Prime Factors of 2: 2
## What are the factors of 2?
When we talk about the factors of a number, we are referring to the whole numbers that can be multiplied together to give that number as a product. In the case of the number 2, its factors are simply 1 and 2 itself. Let’s understand it using an example:
Example 1: Let’s take the number as 4 and 4 is a multiple of 2. Its factors are 1, 2, and 4. In this case, 2 is a prime factor of 4
Example 2: Let’s take the number as 8 and 8 is yet another multiple of 2. Its factors are 1, 2, 4, and 8. In this case, 2 is both a factor and a prime factor, as it cannot be divided further into prime numbers.
Note :
1. All the numbers except for 1 have at least 2 factors.
2. The Prime factor of a prime number would always be the number itself.
3. Even numbers will always have 2 as their factors.
### What Is a Factor of 2? | What Is the Prime Factorization of 2?
A factor of 2 is any whole number that can be multiplied by another whole number to yield 2 as the result.
In the case of 2, the factors are 1 and 2, because 1 x 2 = 2.
We have only taken the positive integers in the article.
Let’s see the factors of other prime numbers:
• The factor of 3: 1 and 3
• Factors of 5: 1 and 5
• The factor of 7: 1 and 7
So, from the above factors of prime numbers, we have understood that the prime number has only two factors i.e. 1 and the number itself.
Likewise, we will cover a very important topic on prime factors. Prime factorization involves expressing a number as a product of prime numbers.
However, 2 is a prime number itself, which means it cannot be further broken down into a product of smaller prime numbers.
Therefore, the prime factorization of 2 is simply 2.
• Prime factor of 2: 2
• Prime factor of 3: 3
• Prime factor of 6: 2,3
Hence, the prime factor of the prime number would be the number itself.
## Why is it important to factorize a number?
Factors are one of the most important topics in Mathematics for various reasons :
1. Divisibility: If you understand factors then here is what it does. It helps determine whether one number is divisible by another. For example, if a number is divisible by 2, it means 2 is one of its factors.
2. Prime Numbers: It is important to identify prime factors as it is essential for working with prime numbers and solving problems related to prime factorization. For example, if a number is 3 then, the prime factor for 3 would be 3.
3. Simplifying Fractions: Usually, factors are used to simplify fractions in Math. When we know the common factors of the numerator and denominator it will help us to reduce fractions to their simplest form.
## How to calculate Factors of a Number?
Understanding factors helps determine whether one number is divisible by another.
For instance, if a number is divisible by 2, it means 2 is one of its factors. Also, to calculate the factor of a number we need to see the divisibility of that given number with the other numbers.
The factors of any number exist in Pairs. So, the Pair factor of 2 would be (1,2 ).
Let’s find out the factors of a few numbers and their pair factor.
Number Pair factor Factors of the number 2 (1, 2) 1,2 3 (1, 3) 1,3 4 (1, 2), (1, 4), (2, 4) 1,2,4 5 (1, 5) 1,5 12 (1, 12), (2, 6), (3, 4) 1,2,3,4,6,12 15 (1, 15), (3, 5) 1, 3, 5, 15 28 (1, 28), (2, 14), (4, 7) 1, 2, 4, 7, 14, 28 37 (1, 37) 1, 37 56 (1, 56), (2, 28), (4, 14), (7, 8) 1, 2, 4, 7, 8, 14, 28, 56 99 (1, 99), (3, 33), (9, 11) 1, 3, 9, 11, 33, 99 100 (1, 100), (2, 50), (4, 25), (5, 20), (10, 10) 1, 2, 4, 5, 10, 20, 25, 50, 100
### What is the prime factor of 2?
The prime factor of 2 is the number itself i.e. 2.
### What is the factor of 2?
The factor of 2 is 1 and 2.
### Can I get assistance with my Math homework ?
Yes, you can hire our expert for your Maths homework help from Letstacle. |
# How do you find the scale factor of similar figures?
## How do you find the scale factor of similar figures?
Suppose you have two similar figures , one larger than the other. The scale factor is the ratio of the length of a side of one figure to the length of the corresponding side of the other figure. Here, XYUV=123=4 . So, the scale factor is 4 .
How do you calculate scale factor?
The basic formula to find the scale factor of a figure is: Scale factor = Dimensions of the new shape ÷ Dimensions of the original shape. This can also be used to calculate the dimensions of the new figure or the original figure by simply substituting the values in the same formula.
What is the scale factor of 2 triangles?
When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangles. In Figure 1 , Δ ABC∼ Δ DEF. Figure 1 Similar triangles whose scale factor is 2 : 1. The ratios of corresponding sides are 6/3, 8/4, 10/5.
### What is a scale factor in math in 7th grade?
7th Grade Math – Scale Drawings Lesson. 1 of 5 – view full lesson. Similar figures are the same shape, but not necessarily the same size. Scale factor is the ratio of corresponding side lengths. (new image to old image) of similar figures.
What is the area of two similar triangles?
The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles. For example, for any two similar triangles ΔABC and ΔDEF, Area of ΔABC/Area of ΔDEF = (AB)2/(DE)2 = (BC)2/(EF)2 = (AC)2(DF)2.
What is the scale factor in math mean?
VOCABULARY. ● Scale Factor: The ratio of any two corresponding lengths in two similar. geometric figures.
#### How do you find the scale factor of a triangle?
Divide one of the sides in the bigger triangle by its corresponding side in the smaller triangle to determine the scale factor for the smaller triangle to the bigger triangle. In the example, if you divided 40 by 20 you would get a scale factor of 2.
How to find the scale factor of a triangle?
Set up ratios of the corresponding sides of the triangles. For example, the ratio of small to big triangle sides in two triangles is 5/10, 10/20 and 20/40. Divide both numbers in one of the ratios by their highest common factor. This will give you the scale factor of the bigger triangle to the smaller triangle.
How do you determine scale factor for two similar figures?
In two similar geometric figures, the ratio of their corresponding sides is called the scale factor. To find the scale factor, locate two corresponding sides, one on each figure. Write the ratio of one length to the other to find the scale factor from one figure to the other.
## What is a scale factor on a triangle?
Scale Factor of Triangle The triangles which are similar have same shape and measure of three angles are also same. The only thing which varies is their sides. However, the ratio of the sides of one triangle is equivalent to the ratio of sides of another triangle , which is called here the scale factor . |
# Solve this linear equation using the elimination method. 1/2x+2y=27,x+1/3y=10 X=? and Y=?
Solving linear systems using the elimination method is also known as using the addition/subtraction method. The goal is to "eliminate" one of the variables in order to solve the system of equations.
`1/2x+2y=27`
`x+1/3y =10`
Simplifying we get:
`x/2 +2y=27`
`x+y/3=10`
Now let's look at the first equation. Multiply each term in the equation by 2.
`2 (x/2 +2y) = 2(27)`
`x+4y = 54`
Now let's go back to the second equation. Multiply each term in the equation by 3.
`3(x+ y/3) =3(10)`
`3x+y =30`
Now the two equations will be easier to work with.
`x+4y=54`
`3x+y=30`
Let's eliminate the x variable to solve for y first. To do this multiply the first equation by -3.
`-3(x+4y=54)` >>> `-3x-12y=-162`
`3x+y=30`
Now we have:
`-3x-12y=-162`
`3x +y =30`
The 3x and -3x cancel out. Combine -12y with 1y which is -11y. Finally -162 combined with 30 is -132.
`-11y = -132`
Divide both sides by -11 to get y alone.
`y=12`
Almost finished. Since you now know what y is, simply plug this value into one of the original equations to solve for x.
`1/2 x +2(12) =27`
`x/2 +24 =27`
`x/2 =3`
`2 (x/2) = 2 (3)`
`x=6`
The solution to the system of equations is
x=6, y=12 |
# GMAT Simple interest
One of the topics commonly tested in the GMAT Quant section is that of Simple Interest. While the formulas and concepts applied herein are simple enough, a lot of test takers get caught up in the terminology surrounding the concepts and therefore get their answers on this topic wrong. Here, then, are the concepts on Simple Interest that you can get tested on when you take the GMAT, including the terminology:
Simple Interest = P x R x T/100
Where P: Principal (or the amount originally lent out)
R: Rate of interest (expressed in terms of %)
T: Time period for which the loan is made
Common mistakes made by test takers when dealing with Simple Interest problems
1. Taking the rate of interest in decimals, instead of in percentage terms: Remember, the 100 in the denominator of the formula is there because it assumes that the rate of interest (R) will be taken in percentage terms, not in decimals.
Example: If P = \$1000, R = 4% per annum, and T = 3 years,
SI = 1000 x 4 x 3 / 100 = 120
Do NOT take SI = 1000 x 0.04 x 3 / 100 = 1.2
2. Interpreting SI to mean the amount to be paid back: The SI formula calculates just the interest to be paid back at the end of the loan period. It does NOT include the amount.
Example: If \$1000 is loaned for 3 years at 4% annual rate of interest, the amount to be paid back at the end of the 3 years = 1000 x 3 x 4 /100 + 1000 = 1120
Remember to add the principal when calculating the amount due.
3. Adjusting both the time and the rate of interest when dealing with periods of loan that are not full years: Remember, if the calculation is to be made on a semi-annual basis, then adjust either the time period or the rate of interest, not both.
Example: If \$1000 is loaned for 6 months at 4% annual rate of interest then
SI = 1000 x 4 x (6/12)/100 = 20 [Dividing the time period by 12 to convert it into half an year or 6 months]
OR
SI = 1000 x (4/2) x 6 /100 = 20 [Dividing the annual rate of interest by 2 to convert it into a monthly rate of interest]
DO NOT make the mistake of using the formula as:
SI = 1000 x (4/2) x (6/12) / 100 = 10 [Here the rate of interest has been converted to a monthly rate but the time period has been converted from months (6) to years (6/12), which IS WRONG]
Author :
GyanOne is an educational services firm focused on premium GMAT coaching, application advisory (with a focus on ISB applications), and interview preparation for B-school admissions. GyanOne operates in the New Delhi region of India and has top global B-school alumni with a minimum score of 770 on the GMAT, as instructors. All GyanOne counsellors are professionally trained top B-school alumni with an experience of 100+ MBA applications behind them.
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# Practice Test on Framing the Formula | Framing the Formula Questions and Answers
Practice Test on Framing the Formula let you know about the different problems available on the Framing the Formula. Most of the below questions appear on the exam which helps you to get good marks in the exam. Practice all the questions available below and have a perfect grip on Framing the Formula Problems.
All the concepts such as formulas, framing a formula, find the value of the variable using a change of subject of formula or an equation, change the subject of the formula, and method of substitution are included in the below article. Learn Tricks to solve Formula and Framing the Formula problems easily.
1. Write the formulas for the following statements.
(a) Area of the square is equal to the square of its side (b)
(b) Area B of the rhombus is equal to half the product of its diagonals (s₁, s₂).
(c) Perimeter (p) of a parallelogram is thrice the sum of its adjacent sides.
(d) Area of four walls (D) of a room is the product of two times the sum of length (l), breadth (b) and height (h).
(e) Profit (P) is calculated by taking the difference of cost price (C) and selling price (S).
Solution:
(a) Given that Area of the square is equal to the square of its side (b)
Area of the square = A
A = b²
(b) Area B of the rhombus is equal to half the product of its diagonals (s₁, s₂).
B = s₁ × s₂
(c) Perimeter (p) of a parallelogram is thrice the sum of its adjacent sides.
sum of its adjacent sides = x
p = 3(x)
(d) Area of four walls (D) of a room is the product of two times the sum of length (l), breadth (b) and height (h).
D = 2 (l + b + h)
(e) Profit (P) is calculated by taking the difference of cost price (C) and selling price (S).
P = S – C
2. Mention the formulas for the following statements.
(a) A side of a chessboard (D) is 3.14 times the side of a board (e).
(b) The difference between the two different variables is 36.
(c) The sum (X) of all the interior angles of a regular polygon of m sides is 2 less than m times 180°.
(d) Subtracting 2/5 from a number and multiplying this difference by 4 gives two times the same number.
(e) Fourteen years from now Sam’s age will be three times her present age.
Solution:
(a) Given that A side of a chessboard (D) is 3.14 times the side of a board (e).
D = 3.14e
(b) The difference between the two different variables is 36.
Two different variables are a and b
a – b = 36
(c) The sum (X) of all the interior angles of a regular polygon of m sides is 2 less than m times 180°.
X = (m – 2) × 180°
(d) Subtracting 2/5 from a number and multiplying this difference by 4 gives two times the same number.
Let the number is X.
X = 4(X – 2/5)
(e) Fourteen years from now Sam’s age will be three times her present age.
Sam’s age is C
C + 14 = 3C
3. Change the subject of the formula and find the value of the new subject.
(a) X = 2YZ make the subject Z. X = 20, Y = 5
(b) l² = r² + h², l is the height of the cone, h is the height and r is the radius. Make the subject h when l = 12 and r = 6.
(c) M = 1/2 × b × h where M is the area of a triangle with base b and height h. Make h the subject, find h when M = 50, b = 24.
(d) D = d × q + r where D is the dividend, q is the quotient, d the divisor, and r is the remainder. Make the subject r, when D = x² – x, d = x – 2, q = x + 3.
(e) M × N = c × d, Make N as the subject where M = 2, c = 4, d = 10.
Solution:
(a) X = 2YZ make the subject Z. X = 20, Y = 5
Given that X = 2YZ
Divide both sides with 2Y
X/2Y = 2YZ/2Y
X/2Y = Z
Therefore, Z = X/2Y
Substitute the given values X = 20, Y = 5
Z = 20/2(5) = 20/10 = 2
Z = 2.
The final answer is Z = 2.
(b) l² = r² + h², l is the height of the cone, h is the height and r is the radius. Make the subject h when l = 12 and r = 6.
Given that l² = r² + h², l is the height of the cone, h is the height and r is the radius.
l² = r² + h²
Subtract r² on both sides
l² – r² = r² – r² + h²
l² – r² = h²
h = √(l² – r²)
Substitute the given values l = 12 and r = 6
h = √((12)² – (6)²) = √144 – 36 = √108
The final answer is h = √108
(c) M = 1/2 × b × h where M is the area of a triangle with base b and height h. Make h the subject, find h when M = 50, b = 24.
Given that M = 1/2 × b × h where M is the area of a triangle with base b and height h. Make h the subject, find h when M = 50, b = 24.
M = 1/2 × b × h
Divide 1/2 × b on both sides.
M/(1/2 × b) = (1/2 × b × h)/(1/2 × b)
M/(1/2 × b) = h
Therefore, h = M/(1/2 × b)
Substitute the given values M = 50, b = 24.
h = 50/(1/2 × 24)
h = 50/12
The final answer is h = 50/12
(d) D = (d × q) + r where D is the dividend, q is the quotient, d the divisor, and r is the remainder. Make the subject r, when D = x² – x, d = x – 2, q = x + 3.
Given that D = d × q + r where D is the dividend, q is the quotient, d the divisor, and r is the remainder. Make the subject r, when D = x² – x, d = x – 2, q = x + 3.
D = (d × q) + r
Subtract (d × q) on both sides
D – (d × q) = (d × q) – (d × q) + r
D – (d × q) = r
Therefore, r = D – (d × q)
Substitute the given values D = x² – x, d = x – 2, q = x + 3.
r = (x² – x) – ((x – 2)(x + 3))
r = (x² – x) – (x² + 3x – 2x – 6)
r = x² – x – x² – x + 6
r = -2x + 6
The final answer is r = -2x + 6
(e) M × N = c × d, Make N as the subject where M = 2, c = 4, d = 10.
Given that M × N = c × d
Divide M on both sides
MN/M = cd/M
N = cd/M
Substitute the given values M = 2, c = 4, d = 10
N = (4 × 10)/2
N = 40/2
N = 20
The final answer is N = 20.
3. If the base of the triangle is 3/2 times its height, then find the area of the triangle.
Solution:
Given that the base of the triangle is 3/2 times its height, then find the area of the triangle.
The base of the triangle = b
Height of the triangle = h
The area of the triangle = A
A = 1/2 × b × h
Base of the triangle is 3/2 times its height
b = 3/2 × h
A = 1/2 × 3/2 × h × h
A = 3/4 × h²
4. If ‘S’ is equal to the 3/4 of the r, then find r.
Solution:
Given that If ‘S’ is equal to the 3/4 of the r, then find r.
S = 3/4 r
Divide 3/4 on both sides.
S/(3/4) = (3/4)/(3/4) . r
4S/3 = r
The final answer is r = 4S/3
5. If x, y, z are in continued proportion, then find the value of z.
Solution:
Given that If x, y, z are in continued proportion, then find the value of z.
x/y = y/z
x . z = y . y
xz = y²
y² = xz
y = √xz
The final answer is y = √xz
6. If y workers can build a wall in 24 days, in how many days will 16 workers take to build a same wall.
Solution: Given that y workers can build a wall in 24 days
So, find the time taken to build a wall by one worker.
y workers = 24 days
1 worker = 24y days
Now, the time taken by 16 workers to build a wall.
1 worker = 24y days
16 worker = 24y/16 = 3/4y
Therefore, The time taken to build a wall by 16 workers is 3/4y.
7. A shirt is marked $y and the shopkeeper allows 100 rupees off as a discount on it. What is its selling price? Solution: Given that a shirt is marked$ y and the shopkeeper allows $100 off as a discount on it. Subtract$100 from $y to get the selling price.$y – $100 The selling price is$y – $100. 8. A weighs 3 kg more than B and B weighs 6 kg less than C. If the weights of A, B, C is n, find the weights of A, B, C separately. Solution: Given that A weighs 3 kg more than B A = 3 + B B weighs 6 kg less than C B = C – 6; C = B + 6 A + B + C = n 3 + B + B + B + 6 = n 3 + 3B + 6 = n 3B + 9 = n 3B = n – 9 B = (n – 9)/3 A = 3 + B A = 3 + (n – 9)/3 C = B + 6 C = (n – 9)/3 + 6 The values of A, B, and C is A = 3 + (n – 9)/3, B = (n – 9)/3, and C = (n – 9)/3 + 6. 9. Half of a herd of deer are grazing in the field and 2/3 of the remaining are playing nearby. The rest 8 are drinking water from the pond. Find the number of deer in the herd? Solution: Let the total no.of deer = y Half of a herd = y/2 2/3 of remaining half herd = (y/2)(2/3) = y/3 remaining deer = 8 From the given information, y = (y/2) + (y/3) + 8 y = (3y + 2y)/6 + 8 y = (5y + 48)/6 Multiply 6 on both sides 6y = (5y + 48)/6 × 6 6y = 5y + 48 Move 5y to the left side 6y – 5y = 48 y = 48. The total number of deers is 48. 10. Arun is twice as old as Sam six years ago his age was four times Shriya’s age. Find their present ages. Solution: Let Arun age is x and Sam age is y Arun age is twice as old as Sam x = 2y if it is five years ago x – 5 = 4(y – 5) x – 5 = 4y – 20 x – 4y + 15 = 0 Substitute x = 2y in x – 4y + 15 = 0 2y – 4y + 15 = 0 -2y + 15 = 0 2y – 15 = 0 2y = 15 y = 15/2 = 7.5 x = 2y = 2 (15/2) = 15 Arun’s age is 15 and Sam’s age is 7.5 11. A car travels 12 km at the speed of x km/hr. Find the time taken by the car to reach the destination. Solution: Given that a car travels 12 km at the speed of x km/hr. We know that Speed = Distance/Time Time = Distance/Speed Time = 12km/x km/hr. The time = 12 hrs. The time taken by the car to reach the destination is 12 hrs. 12. Ram had$192 with him. He purchased x kg potatoes for $40 a kg and y kg tomatoes for$45 a kg and z kg onions at $43 a kg. Find the money left with him? Solution: Given that Ram had$192 with him.
He purchased x kg potatoes for $40 a kg and y kg tomatoes for$45 a kg and z kg onions at $43 a kg. Total cost of potatoes = 40x The total cost of tomatoes = 45y Total cost of onions = 43z Total amount spent = 40x + 45y + 43z. Money left with Ram is$192 – (40x + 45y + 43z) |
# Ncert Solutions for Mensurations Class 8 Chapter 11 CBSE Part 4
In this page we have Ncert Solutions for Mensurations Class 8 Chapter 11 for EXERCISE 4 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
Given a cylindrical tank, in which situation will you find 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.
(a) We need to calculate the volume to find the capacity
(b) As plastering will cover the surface so we need surface area to know this
(c) Volume will give the capacity and that can be compared with capacity of smaller tanks
Question 2
Diameter of cylinder A is 7 cm, and the height is 14 cm. 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 the cylinders. Check whether the cylinder with greater volume also has greater surface area?
As cylinder A’s radius is half of radius of cylinder B so its volume will be lesser than that of cylinder B. Although Cylinder B’s height is half of height of cylinder A but as you know while calculating the volume we need to square the radius so halving the radius has a greater impact than halving the height.
Volume = πr2H
Volume of Cylinder A= (22/7)(7/2)2 14=539
Volume of Cylinder B= (22/7)(7)2 7=1078
So Volume Of Cylinder B is greater than Volume of cylinder A
Total Surface Area of Cylinder is given by
= 2πr(r+H)
Total Surface Area of Cylinder A= 385 cm2
Total Surface Area of Cylinder B= 616 cm2
So Cylinder will greatest volume has greater surface area
Question 3
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?
Volume of cuboid is given by
=LBH
= Base Area x Height
Now given here
V=900cm3 and Base Area=180cm2
So
900=180H
Or H=5cm
Question 4
A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Number of small cubes will be given by
= (Volume of Big Cube)/(Volume of small cube)
=60×54×30/6×6×6
=450
Question 5
Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm ?
Volume of cylinder is given by
=πr2h
Here V=1.54 m3
r= 70 cm=.7 m
So
(22/7) (.7)2 H= 1.54
H=1 m
Question 6
A milk tank is in the form of 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?
Volume of Milk Tank= πr2H
=49.5 m3
Now As we know, 1 cubic metre = 1000 litres
So, 49.5 m3 = 49500 litres
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?
Answer: Let a be the side of the cube
Surface Area of Cube= 3a2
Volume of Cube=a3
(i) So Whenever sides are doubled in any structure then area becomes 4 times the original structure
(ii) Volume becomes 8 times of the original volume if sides are doubled in any structure
Question 8
Water is pouring into a cuboidal 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.
Answer: 108 cubic meter = 108000 liter
So, time = Volume Rate per minute
=108000/60 minutes
= 108000/(60 ×60)
=30 hours |
# Additive and multiplicative notation of a binary operation in a group, examples
The most common notations of a binary operation in a group are additive: + and multiplicative: •.
## Additive notation of a binary operation
If in a group the binary operation is an addition operation, then this group is called an additive one and is denoted as .
• In additive group, the result of applying the addition operation to elements is called the sum and is denoted as .
• A identity element eG in an additive group is called zero and is denoted as 0.
• The inverse element of the g element is denoted as -g and is called the opposite element.
1. Any ring or field is a group with an addition operation (the addition operation is taken from the ring) and is called an additive group of the ring or field.
• - is the additive group of the ring of integers. The set of integers is commutative associative ring with 1 with the familiar addition and multiplication operations.
• - is the additive group of the field of real numbers. The set of real numbers is a field with the familiar operations of addition and multiplication of real numbers.
• - is the additive group of the field of complex numbers. The set of complex numbers is a field with the familiar operations of addition and multiplication of complex numbers.
• - is the additive group of the field of rational numbers. The set of rational numbers is a field with the familiar operations of addition and multiplication of fractions.
2. - is the additive group of all geometric vectors in space.
## Multiplicative notation of a binary operation
If in a group the binary operation is an multiplicative operation, then this group is called a multiplicative one and is denoted as .
• In multiplicative group, the result of applying the multiplication operation to elements is called the product and is denoted as or as .
• An identity element eG in a multiplicative group is called unit and is denoted as 1.
• The inverse element to the element g is denoted as g-1 and is also called the inverse element.
## Note
There are also multiplicative groups of rings or fields. To consider examples of such groups, first consider the definitions of some terms.
Definition of an inverse element in a ring. Let - is an associative ring with 1, then the element is called invertible, if there is an element such, that: , and b = a−1.
Notation. The set of all invertible elements of a ring is denoted as .
Corollary. In the , all elements except 0 are invertible elements. That is .
## Examples of multiplicative groups
1. The set of all elements except 0 (the identity element for addition) of the field is a group with a multiplication operation (the multiplication operation is taken from the field). This group is called the multiplicative group of the field and is denoted as .
Consider the examples of multiplicative groups of fields:
• - is the multiplicative group of the field of real numbers.
• - is the multiplicative group of the field of complex numbers.
• - is the multiplicative group of the field of rational numbers.
2. The set of all invertible elements of an associative ring with 1 - is the group with a multiplication operation (the multiplication operation is taken from the ring). This group is called the multiplicative group of the ring and is denoted by .
Consider the examples of multiplicative groups of rings:
• - is the multiplicative group of the ring of integers. Note that in a ring of integers, only the numbers 1 and -1 are invertible numbers, because the invertible elements for integers are fractions, but fractions are not integers. For example, for the integer 5, the inverse element is the inverse fraction . This fraction is not an integer. Therefore there is no inverse element in integers for the integer 5. However, for the numbers 1 and −1 there are inverse elements - these are inverse fractions: and . This fractions are integers. So, the invertible elements for integers 1 and −1 are themselves.
3. - is the group of all invertible (nondegenerate) square matrices of size n×n with the matrix multiplication operation.
This group is called the general linear group and is defined as: . The matrix is called non-degenerate or invertible if its determinant is not equal to 0. - in parentheses means that the elements of the matrix X are real numbers. This group will be discussed in more details in the following articles. |
Lesson Objectives
• Learn how to solve basic trigonometric equations
• Learn how to solve trigonometric equations with half-angles
## How to Solve Trigonometric Equations with Half-Angles
Over the course of the last few lessons, we learned how to solve trigonometric equations using linear methods, factoring, and squaring. Here, we will take the next step and learn how to work with trigonometric equations that have half-angles.
### Unit Circle
The unit circle will be given here for reference.
### Solving Trigonometric Equations with Half-Angles
In some cases, we will be asked to solve a trigonometric equation with a half-angle. Sometimes, we will need to use our half-angle identities. In other cases, this won't be necessary. Let's look at a few examples.
Example #1: Solve each equation over the interval $[0, 2π)$. $$\text{sin}\hspace{.1em}\frac{β}{2}=\sqrt{2}- \text{sin}\hspace{.1em}\frac{β}{2}$$ Let's first consider our interval written as an inequality: $$0 ≤ β < 2π$$ Since we have β/2, let's divide each part by 2: $$0 ≤ \hspace{.15em}\frac{β}{2}< π$$ Now, let's revisit our equation and find all values of β/2 over the interval $[0, π)$: $$\text{sin}\hspace{.1em}\frac{β}{2}=\sqrt{2}- \text{sin}\hspace{.1em}\frac{β}{2}$$ Add sin β/2 to both sides: $$2\text{sin}\hspace{.1em}\frac{β}{2}=\sqrt{2}$$ Divide both sides by 2: (Note: The angle is β/2) $$\text{sin}\hspace{.1em}\frac{β}{2}=\frac{\sqrt{2}}{2}$$ We want to find all values of β/2 over the interval $[0, π)$ that satisfy our equation. $$\text{sin}\hspace{.1em}\frac{π}{4}=\frac{\sqrt{2}}{2}$$ Sine is also positive in quadrant II.
What angle has a reference angle of $\frac{π}{4}$ or 45° in quadrant II? $$π - \frac{π}{4}=\frac{3π}{4}$$ $$\text{sin}\hspace{.1em}\frac{3π}{4}=\frac{\sqrt{2}}{2}$$ Our last step is to set β/2 equal to each and solve for β. $$\frac{β}{2}=\frac{π}{4}$$ $$\text{or}$$ $$\frac{β}{2}=\frac{3π}{4}$$ Let's start with the top equation. $$\frac{β}{2}=\frac{π}{4}$$ Multiply both sides by 2: $$β=\frac{π}{2}$$ Let's now work on the bottom equation. $$\frac{β}{2}=\frac{3π}{4}$$ Multiply both sides by 2: $$β=\frac{3π}{2}$$ Our solutions for β in our given interval: $$\left\{\frac{π}{2}, \frac{3π}{2}\right\}$$ Let's now look at an example that uses a half-angle identity.
Example #2: Solve each equation over the interval $[0, 2π)$. $$-\text{cos}\hspace{.1em}θ=-2 + 3\text{sin}\hspace{.1em}\frac{θ}{2}$$ Let's replace sin θ/2 using the half-angle identity: $$\text{sin}\hspace{.1em}\frac{A}{2}=\pm \sqrt{\frac{1 - \text{cos A}}{2}}$$ $$-\text{cos}\hspace{.1em}θ=-2 + 3\text{sin}\hspace{.1em}\frac{θ}{2}$$ $$-\text{cos}\hspace{.1em}θ=-2 \pm 3\sqrt{\frac{1 - \text{cos θ}}{2}}$$ Let's add 2 to both sides of the equation: $$2 -\text{cos}\hspace{.1em}θ=\pm 3\sqrt{\frac{1 - \text{cos θ}}{2}}$$ Square both sides: $$(2 -\text{cos}\hspace{.1em}θ)^2=\left(\pm 3\sqrt{\frac{1 - \text{cos}\hspace{.1em}θ}{2}}\right)^2$$ $$4 - 4\text{cos}\hspace{.1em}θ + \text{cos}^2 θ=9 \cdot \frac{1 - \text{cos}\hspace{.1em}θ}{2}$$ Multiply both sides by 2: $$8 - 8\text{cos}\hspace{.1em}θ + 2\text{cos}^2 θ=9(1 - \text{cos}\hspace{.1em}θ)$$ Distribute the 9 on the right-hand side: $$8 - 8\text{cos}\hspace{.1em}θ + 2\text{cos}^2 θ=9 - 9\text{cos}\hspace{.1em}θ$$ Subtract 9 away from each side and add 9 cos θ to both sides: $$-1 + \text{cos}\hspace{.1em}θ + 2\text{cos}^2 θ=0$$ Rearrange into $ax^2 + bx + c=0$: $$2\text{cos}^2 θ + \text{cos}\hspace{.1em}θ - 1=0$$ Factor the left-hand side: $$(2\text{cos}\hspace{.1em}θ - 1)(\text{cos}\hspace{.1em}θ + 1)=0$$ Use the zero-factor property: $$2\text{cos}\hspace{.1em}θ - 1=0$$ $$\text{or}$$ $$\text{cos}\hspace{.1em}θ + 1=0$$ Let's solve the top equation first: $$2\text{cos}\hspace{.1em}θ - 1=0$$ Add 1 to each side, then divide both sides by 2: $$\text{cos}\hspace{.1em}θ=\frac{1}{2}$$ $$θ=\frac{π}{3}, \frac{5π}{3}$$ Let's now solve the bottom equation: $$\text{cos}\hspace{.1em}θ + 1=0$$ $$\text{cos}\hspace{.1em}θ=-1$$ $$θ=π$$ Our solutions for θ in the given interval: $$\left\{\frac{π}{3}, π, \frac{5π}{3}\right\}$$ Note: when we square both sides of an equation, we need to check our solutions in the original equation. To keep this tutorial shorter, we will only show the check when we have an extraneous solution. Let's look at one more example using a half-angle identity.
Example #3: Solve each equation over the interval $[0, 2π)$. $$\sqrt{3}\text{cos}\hspace{.1em}\frac{θ}{2}=1 + \text{cos}\hspace{.1em}θ$$ Let's replace cos θ/2 using the half-angle identity: $$\text{cos}\hspace{.1em}\frac{θ}{2}=\pm \sqrt{\frac{1 + \text{cos}\hspace{.1em}θ}{2}}$$ $$\pm \sqrt{3}\sqrt{\frac{1 + \text{cos}\hspace{.1em}θ}{2}}=1 + \text{cos}\hspace{.1em}θ$$ Square both sides: $$\frac{3(1 + \text{cos}\hspace{.1em}θ)}{2}=1 + 2\text{cos}\hspace{.1em}θ + \text{cos}^2 θ$$ Multiply both sides by 2: $$3(1 + \text{cos}\hspace{.1em}θ)=2 + 4\text{cos}\hspace{.1em}θ + 2 \text{cos}^2 θ$$ Distribute the 3 on the left side: $$3 + 3\text{cos}\hspace{.1em}θ=2 + 4\text{cos}\hspace{.1em}θ + 2 \text{cos}^2 θ$$ Let's move all terms to the right and place in the form: $$0=ax^2 + bx + c$$ $$0=2\text{cos}^2 θ + \text{cos}\hspace{.1em}θ - 1$$ Flip Sides: $$2\text{cos}^2 θ + \text{cos}\hspace{.1em}θ - 1=0$$ Factor the left-hand side: $$(2\text{cos}\hspace{.1em}θ - 1)(\text{cos}\hspace{.1em}θ + 1)=0$$ Use the zero-product property: $$2\text{cos}\hspace{.1em}θ - 1=0$$ $$\text{or}$$ $$\text{cos}\hspace{.1em}θ + 1=0$$ Let's solve the top equation first: $$2\text{cos}\hspace{.1em}θ - 1=0$$ Add 1 to each side, then divide both sides by 2: $$\text{cos}\hspace{.1em}θ=\frac{1}{2}$$ $$θ=\frac{π}{3}, \frac{5π}{3}$$ Let's now solve the bottom equation: $$\text{cos}\hspace{.1em}θ + 1=0$$ Subtract 1 from each side of the equation: $$\text{cos}\hspace{.1em}θ=-1$$ $$θ=π$$ As we mentioned above and in previous tutorials, when we square both sides of an equation, it is possible to obtain extraneous solutions. Let's check our proposed solutions in the original equation to see if they work: $$\sqrt{3}\text{cos}\hspace{.1em}\frac{θ}{2}=1 + \text{cos}\hspace{.1em}θ$$ Replace θ with each proposed solution: Let's start with $\frac{π}{3}$: $$\sqrt{3}\text{cos}\hspace{.1em}\frac{\large{\frac{π}{3}}}{2}=1 + \text{cos}\hspace{.1em}\frac{π}{3}$$ $$\sqrt{3}\text{cos}\hspace{.1em}\frac{π}{6}=1 + \text{cos}\hspace{.1em}\frac{π}{3}$$ $$\sqrt{3}\cdot \frac{\sqrt{3}}{2}=1 + \frac{1}{2}$$ $$\frac{3}{2}=\frac{2}{2}+ \frac{1}{2}$$ $$\frac{3}{2}=\frac{3}{2}$$ $\frac{π}{3}$ is a valid solution.
Let's now check $\frac{5π}{3}$: $$\sqrt{3}\text{cos}\hspace{.1em}\frac{\large{\frac{5π}{3}}}{2}=1 + \text{cos}\hspace{.1em}\frac{5π}{3}$$ $$\sqrt{3}\text{cos}\hspace{.1em}\frac{5π}{6}=1 + \text{cos}\hspace{.1em}\frac{5π}{3}$$ $$\sqrt{3}\cdot -\frac{\sqrt{3}}{2}=1 + \frac{1}{2}$$ $$-\frac{3}{2}=\frac{2}{2}+ \frac{1}{2}$$ $$-\frac{3}{2}=\frac{3}{2}$$ $\frac{5π}{3}$ is not a valid solution. Lastly, let's check $π$: $$\sqrt{3}\text{cos}\hspace{.1em}\frac{π}{2}=1 + \text{cos}\hspace{.1em}π$$ $$\sqrt{3}\cdot 0=1 + (-1)$$ $$0=0$$ $π$ is a valid solution. $$\left\{\frac{π}{3}, π\right\}$$
#### Skills Check:
Example #1
Solve each equation for 0 ≤ θ < 2π $$-3=-4\text{cos}\hspace{.1em}\frac{θ}{2}+ \text{cos}\hspace{.1em}θ$$
A
$$\left\{\frac{5π}{6}\right\}$$
B
$$\left\{0\right\}$$
C
$$\left\{π\right\}$$
D
$$\left\{0, \frac{π}{2}\right\}$$
E
$$\left\{\frac{7π}{6}\right\}$$
Example #2
Solve each equation for 0 ≤ θ < 2π $$2\text{cos}\hspace{.1em}θ + 4\text{sin}\hspace{.1em}\frac{θ}{2}=3$$
A
$$\left\{\frac{π}{3}\right\}$$
B
$$\left\{\frac{2π}{3}, \frac{4π}{3}\right\}$$
C
$$\left\{\frac{π}{3}, \frac{5π}{3}\right\}$$
D
$$\left\{\frac{11π}{6}\right\}$$
E
$$\text{No Solution}$$
Example #3
Solve each equation for 0 ≤ θ < 2π $$\text{cos}\hspace{.1em}θ=3\text{sin}\hspace{.1em}\frac{θ}{2}+ 2$$
A
$$\left\{\frac{2π}{3}, \frac{4π}{3}\right\}$$
B
$$\left\{\frac{π}{3}, π, \frac{5π}{3}\right\}$$
C
$$\left\{π, \frac{3π}{2}\right\}$$
D
$$\text{No Solution}$$
E
$$\left\{\frac{π}{6}, \frac{11π}{6}\right\}$$ |
# Where will the hand of a clock stop if it starts at 2 and makes 3/4 of a revolution?
Contents
## Where will the hand of a clock stop if it starts at 3 and makes 3/4 of a revolution clockwise?
(d) starts at 5 and makes 3/4 of a revolution, clockwise? rotate by 180 degree and hence, it will stop at 6. rotate by 180 degree and hence, it will stop at 8. rotate by 90 degree and hence, it will stop at 8.
## Where will the hour hand of a clock stop if it starts at 12 and makes one by two of a revolution clockwise?
Understanding Elementary Shapes | Exercise 5.2
(d) starts at 5 and makes 3/4 of a revolution, clockwise? SOLUTION: (a) if a hand clock Starts at 12 and makes 1 2 frac{1}{2} 21of a revolution, clockwise, i.e two right angles, it reaches 6.
## Where will the hand of a clock stop if kavya starts from 5 and makes ¾ of a rotation anti clock wise 2?
where will the hand of a clock stop if kavya starts from 5 and make 3/4 of a rotation anti clock wise ? 1 rotation = 360° . so, → (3/4) of rotation = (3/4) * 360° = 270° .
## What is the measure of 3 by 4 of a revolution?
The measure of 3/4 of revolution is 270°.
## Where will the hour hand of a clock stop if it starts from 11 and turns through 3 right angles?
3. If the hour hand of a clock starts from 10 and turns through 3 right angles, then it will stop at 7. 4. If the hour hand of a clock starts from 7 and turns through 2 straight angles, then it will stop at 7.
Understanding Elementary Shapes.
Angle Measure Type
∠DOB
## Does a right angle have to be 90 degrees?
A right angle is 90 degrees. An acute angle is an angle that is less than 90 degrees. An obtuse angle is an angle that is more than 90 degrees. … Because a right triangle and an obtuse triangle will also contain at least one other angle that is acute.
## Where will the hand of a clock stop if kavya?
Answer: In one complete revolution the clock hand returns back to the same position. So, if it starts at 12 then it would again stop at 12, if it starts at 2 it would again stop at 2 and if it starts at 5 it would again stop at 5. Let’s understand the concept of turning clockwise and anti-clockwise.
## How many degrees are there in 2/3 of a right angle?
Therefore 2/3 of a right angle is 60° and it’s complement will be: (3/3[whole]-3/3)*90°.
## How many right angles will be turned by the hour hand of a clock when it goes from 2 to 11?
it starts from 2 and covers digits 3,4,5,6,7 and 8. so, the hour hand-turned by two right angles. here, the hour hand turns from 5 to 11.
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# Derivative of Parametric functions
### Parametric Equations:
• Parametric equations represent a relationship between two variables, often denoted as $x$ and $y$, where each variable is expressed in terms of a third variable, usually $t$, known as the parameter.
• Common parametric equations take the form: $x=f\left(t\right)$ $y=g\left(t\right)$
### Derivative of Parametric Functions:
• To find the derivative of a parametric curve given by $x=f\left(t\right)$ and $y=g\left(t\right)$, the derivative $\frac{dy}{dx}$ is calculated using the chain rule.
• The derivative $\frac{dy}{dx}$ represents the slope of the curve at any given point.
### Steps to Find $\frac{dy}{dx}$:
1. Derive both x and y with respect to $t$: $\frac{dx}{dt}={f}^{\mathrm{\prime }}\left(t\right)$ $\frac{dy}{dt}={g}^{\mathrm{\prime }}\left(t\right)$
2. Calculate $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{{g}^{\mathrm{\prime }}\left(t\right)}{{f}^{\mathrm{\prime }}\left(t\right)}$
### Example:
Given parametric equations $x=\mathrm{cos}\left(t\right)$ and $y=\mathrm{sin}\left(t\right)$:
1. Find $\frac{dy}{dx}$:
$\frac{dx}{dt}=-\mathrm{sin}\left(t\right)$
$\frac{dy}{dt}=\mathrm{cos}\left(t\right)$
2. Calculate $\frac{dy}{dx}$:
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\mathrm{cos}\left(t\right)}{-\mathrm{sin}\left(t\right)}=-\mathrm{tan}\left(t\right)$
### Special Cases:
• Vertical Tangents:
• When $\frac{dx}{dt}=0$ but $\frac{dy}{dt}\mathrm{\ne }0$, the curve has a vertical tangent.
• Horizontal Tangents:
• When $\frac{dy}{dt}=0$ but $\frac{dx}{dt}\mathrm{\ne }0$, the curve has a horizontal tangent.
### Tips:
• Parametric Differentiation Rule:
• Sometimes it's convenient to express $y$ explicitly as a function of $x$ and differentiate using implicit differentiation.
• Substitute for $t$ or Express in terms of $x$ or $y$:
• Substituting $t$ with expressions involving $x$ or $y$ can help simplify calculations or analyze the curve's behavior at specific points. |
# Ratio, Proportion, and Percent
## Presentation on theme: "Ratio, Proportion, and Percent"— Presentation transcript:
Ratio, Proportion, and Percent
Ratios A ratio is a comparison of numbers that can be expressed as a fraction. If there were 18 boys and 12 girls in a class, you could compare the number of boys to girls by saying there is a ratio of 18 boys to 12 girls. You could represent that comparison in three different ways: 18 to 12 18 : 12 18 12
Ratios The ratio of 18 to 12 is another way to represent the fraction
All three representations are equal. 18 to 12 = 18:12 = The first operation to perform on a ratio is to reduce it to lowest terms 18:12 = = 18:12 = = 3:2 18 12 18 12 ÷ 6 18 12 3 2 ÷ 6 3 2
Ratios A basketball team wins 16 games and loses 14 games. Find the reduced ratio of: Wins to losses – 16:14 = = Losses to wins – 14:16 = = Wins to total games played – 16:30 = = The order of the numbers is critical 16 14 8 7 14 16 7 8 16 30 8 15
Ratios A jar contains 12 white, 10 red and 18 blue balls. What is the reduced ratio of the following? White balls to blue balls? Red balls to the total number of balls? Blue balls to balls that are not blue?
Proportions A proportion is a statement that one ratio is equal to another ratio. Ex: a ratio of 4:8 = a ratio of 3:6 4:8 = = and 3:6 = = 4:8 = 3:6 = These ratios form a proportion since they are equal to other. 4 8 1 2 3 6 1 2 4 8 3 6
Proportions In a proportion, you will notice that if you cross multiply the terms of a proportion, those cross-products are equal. 4 8 3 6 = 4 x 6 = 8 x 3 (both equal 24) 3 2 = 18 12 3 x 12 = 2 x 18 (both equal 36)
Proportions Determine if ratios form a proportion 12 21 8 14 and 10 17
20 27 and 3 8 9 24 and
Proportions The fundamental principle of proportions enables you to solve problems in which one number of the proportion is not known. For example, if N represents the number that is unknown in a proportion, we can find its value.
Proportions N 12 3 4 = 4 x N = 12 x 3 4 x N = 36 4 x N 36 4 4
1 x N = 9 N = 9 Cross multiply the proportion Divide the terms on both sides of the equal sign by the number next to the unknown letter. (4) = That will leave the N on the left side and the answer (9) on the right side
Proportions Solve for N Solve for N 2 5 N 35 15 N 3 4 = =
5 x N = 2 x 35 5 x N = 70 5 x N 1 x N = 14 N = 14 6 7 102 N = 4 N 6 27 = =
Proportions At 2 p.m. on a sunny day, a 5 ft woman had a 2 ft shadow, while a church steeple had a 27 ft shadow. Use this information to find the height of the steeple. 2 x H = 5 x 27 2 x H = 135 H = 67.5 ft. 5 2 H 27 height shadow height shadow = = You must be careful to place the same quantities in corresponding positions in the proportion
Proportions If you drive 165 miles in 3 hours, how many miles can you expect to drive in 5 hours traveling at the same average speed? A brass alloy contains only copper and zinc in the ratio of 4 parts of copper to 3 parts zinc. If a total of 140 grams of brass is made, how much copper is used? If a man who is 6 feet tall has a shadow that is 5 feet long, how tall is a pine tree that has a shadow of 35 feet?
Percents Percent means out of a hundred
An 85% test score means that out of 100 points, you got 85 points. 25% means 25 out of 100 25% = = 0.25 137% means 137 out of 100 137% = = 1.37 6.5% means 6.5 out of 100 6.5% = = 25 100 137 100 6.5 100
Converting Percents to Fractions
To convert a percent to a fraction, drop the % sign, put the number over 100 and reduce if possible Express 30% as a fraction 30% = = (a reduced fraction) Express 125% as a fraction 125% = = = 1 (a reduced mixed number) 30 100 3 10 5 4 1 4 125 100
Converting Percents to Decimals
To convert a percent to a decimal, drop the % sign and move the decimal point two places to the left Express the percents as a decimal 30% = .30 125 % = 1.25
Converting Decimals to Fractions and Percents
Convert each percent to a reduced fraction or mixed number and a decimal 17% 5% 23% 236% 8%
Converting Decimals to Percents
To convert a decimal to a percent, move the decimal point two places to the right and attach a % sign. Ex: = 34% Ex: = 1%
Converting Fractions to Percents
To convert a fraction to a percent, divide the denominator of the fraction into the numerator to get a decimal number, then convert that decimal to a percent (move the decimal point two places to the right) .75 3 4 = = 75%
Converting Decimals and Fractions to Percents
Convert the Decimal to a percent .08 = ? 3.26 = ? .75 = ? Convert the Fraction to a percent 1 5 7 10
Percent of a Number Percents are often used to find a part of a number or quantity Ex: “60% of those surveyed” Ex: “35% discount” Ex: 8.25% sales tax” 60% of means 60% x 5690 35% of \$ means 35% x \$236 8.25% of \$180 means 8.25% x \$180 Change the percent into either a fraction or a decimal before you use it in multiplication
Percent of a Number Find 25% of 76 (as a decimal)
25% = .25 25% of 76 = .25 x 76 = 1 OR Find 25% of 76 (as a fraction) 25% = 25% of 76 = x 76 = 19 Find 60% of 3420 Find 30% of 50 Find 5% of 18.7 1 4 1 4
Percentage Problems On a test you got 63 out of 75 possible points. What percent did you get correct? Since “percent” means “out of a hundred”, 63 out of 75 is what number out of 100 63 75 P 100 (P is used to represent the percent or part out of 100) = Percent Proportion A P B 100 A is the amount B is the base (follows the word “of”) P is the percent (written with the word “percent” or the % sign) = 75 x P 75 6300 75 = P = 84
Percentage Problems 15 is what percent of 50?
16 is 22% of what number? 91 is what percent of 364? What is 9.5% of 75,000? Percent Proportion A P B 100 A is the amount B is the base (follows the word “of”) P is the percent (written with the word “percent” or the % sign) = |
# 009A Sample Final 1, Problem 1
In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity.
(a) ${\displaystyle \lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}}$
(b) ${\displaystyle \lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}}$
(c) ${\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}}$
Foundations:
L'Hôpital's Rule
Suppose that ${\displaystyle \lim _{x\rightarrow \infty }f(x)}$ and ${\displaystyle \lim _{x\rightarrow \infty }g(x)}$ are both zero or both ${\displaystyle \pm \infty .}$
If ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}}$ is finite or ${\displaystyle \pm \infty ,}$
then ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.}$
Solution:
(a)
Step 1:
We begin by factoring the numerator. We have
${\displaystyle \lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,\lim _{x\rightarrow -3}{\frac {x(x-3)(x+3)}{2(x+3)}}.}$
So, we can cancel ${\displaystyle x+3}$ in the numerator and denominator. Thus, we have
${\displaystyle \lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,\lim _{x\rightarrow -3}{\frac {x(x-3)}{2}}.}$
Step 2:
Now, we can just plug in ${\displaystyle x=-3}$ to get
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}}&=&\displaystyle {\frac {(-3)(-3-3)}{2}}\\&&\\&=&\displaystyle {\frac {18}{2}}\\&&\\&=&\displaystyle {9.}\end{array}}}$
(b)
Step 1:
We proceed using L'Hôpital's Rule. So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {2\cos(2x)}{2x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\cos(2x)}{x}}.}\\\end{array}}}$
Step 2:
This limit is ${\displaystyle \infty .}$
(c)
Step 1:
We have
${\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}=\lim _{x\rightarrow -\infty }{\frac {3x}{{\sqrt {x^{2}(4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}})}}.}$
Since we are looking at the limit as ${\displaystyle x}$ goes to negative infinity, we have ${\displaystyle {\sqrt {x^{2}}}=-x.}$
So, we have
${\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}\,=\,\lim _{x\rightarrow -\infty }{\frac {3x}{-x{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}}.}$
Step 2:
We simplify to get
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {-3}{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}}\\&&\\&=&\displaystyle {-{\frac {3}{\sqrt {4}}}}\\&&\\&=&\displaystyle {-{\frac {3}{2}}.}\end{array}}}$
(a) ${\displaystyle 9}$
(b) ${\displaystyle \infty }$
(c) ${\displaystyle -{\frac {3}{2}}}$ |
# How do you use find the zeroes of f(x)=x^4-4x^2-45?
Aug 13, 2016
$f \left(x\right)$ has zeros: $3 , - 3 , \sqrt{5} i , - \sqrt{5} i$
#### Explanation:
Treat as a quadratic in ${x}^{2}$ then use the difference of squares identity:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
Note that $9 \cdot 5 = 45$ and $9 - 5 = 4$
Hence:
${x}^{4} - 4 {x}^{2} - 45$
$= \left({x}^{2} - 9\right) \left({x}^{2} + 5\right)$
$= \left({x}^{2} - {3}^{2}\right) \left({x}^{2} - {\left(\sqrt{5} i\right)}^{2}\right)$
$= \left(x - 3\right) \left(x + 3\right) \left(x - \sqrt{5} i\right) \left(x + \sqrt{5} i\right)$
Hence zeros:
$3 , - 3 , \sqrt{5} i , - \sqrt{5} i$ |
# GMAT at the Movies: What Austin Powers Can Teach You about Similar Triangles
In this series we return to classic movies to learn fundamental strategies for GMAT Success.
In the Austin Powers movies the character known as “Dr. Evil” creates an exact version of himself, only smaller, that he calls “Mini-me.” The two characters have identical proportions even though one evil villain is 8 times the size of the other. The hero, Austin Powers, quickly recognizes the similarity, despite the difference in size. This is something that you will need to be able to do on the GMAT!
If you are not familiar with “Dr. Evil” and “Mini-me, watch the following clip:
This is what similar triangles are all about! Not the evil villain stuff, but the “same proportions, different size.” When you have proven that you have similar triangles you know that any ratio of a side of one triangle to the corresponding side of the other triangle will hold true for each of the sides and even for the height of those triangles.
As you can see from the diagram below all three angles are equal. The ratio of the lengths of the triangle will remain constant. So if A:a = 2:1 then B:b and C:c and even H:h will stay at that same ratio of 2:1
Recognizing Similar Triangles
Often the biggest difficulty that people have with these similar triangle problems is simply recognizing that they are, in fact, “similar.”
Most similar triangles on the GMAT are not like the diagram above. They are actually overlapping triangles that have one angle in common. Be on the lookout for that “shared angle.” That is usually the first clue that you have similar triangles!
In addition to the shared angle look for one of these other two clues that similar triangles are present:
1) Parallel lines: If the triangle has a shared angle AND parallel lines then you have a similar triangle. For the diagram below you would be told that DE is parallel to AC. This creates similar triangles BDE and ABC.
2) Right angles: If the triangles each have a right angle AND a shared angle then you have a similar triangle. In the diagram below you see that angle “D” is shared and that angles DCE and ABC are right angles. This means that you have similar triangles ABD and CDE.
Don’t wait for the GMAT to make similar triangles as obvious as Dr. Evil and Mini-Me. Watch out for shared angles, parallel lines, and right angles. And remember that easily recognizing similar triangles is “groovy baby, yeah!”
If you plan on taking the GMAT soon, we have GMAT prep courses starting all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
David Newland has been teaching for Veritas Prep since 2006, and he won the Veritas Prep Instructor of the Year award in 2008. Students’ friends often call in asking when he will be teaching next because he really is a Veritas Prep and a GMAT rock star! Read more of his articles here. |
# NCERT Solutions For Class 9 Math Chapter – 8 Exercise – 8.1
## NCERT Solutions For Class 9 Math Chapter – 8 Exercise – 8.1
#### Experts at Sunstarup have created the NCERT Solutions after extensive search on each topic Students can refer to this study study material to boost their confidence and attempt the second term exam smartly. The concepts are explained with steps, shortcuts to remember formula , tips and tricks to solve the numerical problems wisely and quickly.
1.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Let the common ratio between the angles be = x.
We know that the sum of the interior angles of the quadrilateral = 360°
Now,
3x+5x+9x+13x = 360°
⇒ 30x = 360°
⇒ x = 12°
, Angles of the quadrilateral are:
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°
2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Given that,
AC = BD
To show that, ABCD is a rectangle if the diagonals of a parallelogram are equal
To show ABCD is a rectangle we have to prove that one of its interior angles is right angled.
Proof,
AB = BA (Common)
BC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD [SSS congruency]
∠A = ∠B [Corresponding parts of Congruent Triangles]
also,
∠A+∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Therefore, ABCD is a rectangle.
Hence Proved.
3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given that,
OA = OC
OB = OD
and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show that,
if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
i.e., we have to prove that ABCD is parallelogram and AB = BC = CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB [SAS congruency]
Thus, AB = BC [CPCT]
Similarly we can prove,
BC = CD
, AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.
Hence Proved.
4. Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show that,
AC = BD
AO = OC
and ∠AOB = 90°
Proof,
AB = BA (Common)
Thus,
AC = BD [CPCT]
diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
, ΔAOB ≅ ΔCOD [AAS congruency]
Thus,
AO = CO [CPCT].
, Diagonal bisect each other.
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
, ΔAOB ≅ ΔCOB [SSS congruency]
also, ∠AOB = ∠COB
∠AOB+∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90°
, Diagonals bisect each other at right angles
5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:
Given that,
Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O.
To prove that,
The Quadrilateral ABCD is a square.
Proof,
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite)
OB = OD (Diagonals bisect each other)
, ΔAOB ≅ ΔCOD [SAS congruency]
Thus,
AB = CD [CPCT] — (i)
also,
∠OAB = ∠OCD (Alternate interior angles)
⇒ AB || CD
Now,
In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Vertically opposite)
OD = OD (Common)
, ΔAOD ≅ ΔCOD [SAS congruency]
Thus,
AD = CD [CPCT] — (ii)
also,
⇒ AD = BC = CD = AB — (ii)
and ∠ADC+∠BCD = 180° (co-interior angles)
One of the interior angles is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.
Hence Proved.
6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Solution:
AD = CB (Opposite sides of a parallelogram)
DC = BA (Opposite sides of a parallelogram)
AC = CA (Common Side)
, ΔADC ≅ ΔCBA [SSS congruency]
Thus,
∠ACD = ∠CAB by CPCT
⇒ ∠ACD = ∠BCA
Thus,
AC bisects ∠C also.
(ii) ∠ACD = ∠CAD (Proved above)
⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)
Also, AB = BC = CD = DA (Opposite sides of a parallelogram)
Thus,
ABCD is a rhombus.
7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Solution:
Given that,
ABCD is a rhombus.
AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)
also, AB || CD
⇒∠DAC = ∠BCA (Alternate interior angles)
⇒∠DCA = ∠BCA
, AC bisects ∠C.
Similarly,
We can prove that diagonal AC bisects ∠A.
Following the same method,
We can prove that the diagonal BD bisects ∠B and ∠D.
8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) Diagonal BD bisects ∠B as well as ∠D.
Solution:
(i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)
⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)
also, CD = AB (Opposite sides of a rectangle)
,AB = BC = CD = AD
Thus, ABCD is a square.
(ii) In ΔBCD,
BC = CD
⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)
also, ∠CDB = ∠ABD (Alternate interior angles)
⇒ ∠CBD = ∠ABD
Thus, BD bisects ∠B
Now,
Thus, BD bisects ∠D
9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Solution:
(i) In ΔAPD and ΔCQB,
DP = BQ (Given)
∠ADP = ∠CBQ (Alternate interior angles)
AD = BC (Opposite sides of a parallelogram)
Thus, ΔAPD ≅ ΔCQB [SAS congruency]
(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.
(iii) In ΔAQB and ΔCPD,
BQ = DP (Given)
∠ABQ = ∠CDP (Alternate interior angles)
AB = CD (Opposite sides of a parallelogram)
Thus, ΔAQB ≅ ΔCPD [SAS congruency]
(iv) As ΔAQB ≅ ΔCPD
AQ = CP [CPCT]
(v) From the questions (ii) and (iv), it is clear that APCQ has equal opposite sides and also has equal and opposite angles. , APCQ is a parallelogram.
10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ
Solution:
(i) In ΔAPB and ΔCQD,
∠ABP = ∠CDQ (Alternate interior angles)
∠APB = ∠CQD (= 90o as AP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
, ΔAPB ≅ ΔCQD [AAS congruency]
(ii) As ΔAPB ≅ ΔCQD.
, AP = CQ [CPCT]
11. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF.
Solution:
(i) AB = DE and AB || DE (Given)
Two opposite sides of a quadrilateral are equal and parallel to each other.
Thus, quadrilateral ABED is a parallelogram
(ii) Again BC = EF and BC || EF.
Thus, quadrilateral BEFC is a parallelogram.
(iii) Since ABED and BEFC are parallelograms.
⇒ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)
Also, AD || BE and BE || CF (Opposite sides of a parallelogram are parallel)
(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.
(v) Since ACFD is a parallelogram
AC || DF and AC = DF
(vi) In ΔABC and ΔDEF,
AB = DE (Given)
BC = EF (Given)
AC = DF (Opposite sides of a parallelogram)
, ΔABC ≅ ΔDEF [SSS congruency]
12. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iv) diagonal AC = diagonal BD
[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Solution:
To Construct: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
, BC = CE
⇒∠CBE = ∠CEB
also,
∠A+∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)
∠B +∠CBE = 180° ( As Linear pair)
⇒∠A = ∠B
(ii) ∠A+∠D = ∠B+∠C = 180° (Angles on the same side of transversal)
⇒∠A+∠D = ∠A+∠C (∠A = ∠B)
⇒∠D = ∠C |
Algebra QuestionNovember 17, 2012 11:35 AM Subscribe
Can anyone help with an algebra question I am trying to help my neice with her homework.
(9/2m) - (m+4)/4m = 5/36
11th grade algebra- do you multiply entire equation by 2m? lowest common denominator? help. An explanation would be great and also any websites that help with algebra would be great too. Thanks.
posted by Summer Fall to Education (9 answers total)
I recently discovered this site: Kahn Academy. The way that equation is written out, I can't really help you.
posted by coolsara at 11:38 AM on November 17, 2012
Multiply the whole equation by 4m, that gets rid of the m's in the denominator. Then multiply the whole equation by 36 to get rid of the all remaining denominators. At this point you can simplify the equation to a point where solving it should be easy.
posted by Green With You at 11:42 AM on November 17, 2012
Here we go:
Multiply the top and bottom of the first term on the left hand side by 2 to get:
18/4m -(m+4)/4m = 5/36
So now both term on the left hand side have 4m as the denominator so we can write it as:
(18-(m+4))/4m = 5/36
Expanding the bracket in the numerator on the left hand side we get:
(18-m-4)/4m = 5/36
(14-m)/4m=5/36
Multiply both sides by 4 (to get rid of the 4 and make the denominator on the right hand side more tractable) and we get:
(14-m)/m = (5*4)/36
Now 36 = 9*4
Therefore we can cancel out the 4 in the numerator on the right hand side and change 36 to 9 giving us:
(14-m)/m = 5/9
Now cross-multiply the two sides:
9*(14-m) = 5m
Expanding:
126-9m = 5m
Take the 9m term to the other side:
126 = 14m
Now remember that 126 = 9*14
So m = 126/14 = 9.
posted by peacheater at 11:43 AM on November 17, 2012 [3 favorites]
FYI you can enter algebra into wolfram-alpha and get a step-by-step solution.
posted by idb at 12:05 PM on November 17, 2012 [3 favorites]
(9/2m) - (m+4)/4m = 5/36
It would sure be great if we didn't have that (1/4m) in there because it's hard to work with. So let's multiple both side of the equation by 4m.
4m((9/2m) - (m+4)/4m) = 4m(5/36)
Which reduces to
18 - m + 4 = (5/9)m
Add m to both sides to get
18 - 4 = (5/9)m + m
Or
14 = (14/9)m
Multiply both sides of the equation by (9/14)
9 = m
posted by deathpanels at 12:36 PM on November 17, 2012 [2 favorites]
When I teach this, I have them multiply by the LCD which is 36m.
You get:
162 - 9(m + 4) = 5m
162 - 36 = 9m + 5m
126 = 14m
so m = 9
posted by Obscure Reference at 1:50 PM on November 17, 2012
I actually went with multiplying the first term by 2 (so not changing the number, just how it's represented), which yields 18/4m - (m+4)/4m = 5/36... so 4m = 36... so m = 9... (and also 18-(m+4)=5... so m=9...) which is just a test to make sure the first value is correct.
It's been years since I took algebra so I started off with 1/2 - 1/4 = x (to refresh my memory of the rules of fraction subtraction)... 1/2 is the same as 2/4... so I rewrote as 2/4 - 1/4 = x... and so x is 1/4... realizing that the bottom remains consistent I was able to realize that once you change 9/2m to 18/4m you've got two equations that you can use to test the value...
Perhaps not the *best* way to solve, but just a demonstration that using what you know to figure out what you don't know works, sometimes.
posted by one4themoment at 6:22 PM on November 17, 2012
Also I loved this question, and I spent a few minutes destroying a piece of scrap paper with my scribbles :)
posted by one4themoment at 6:23 PM on November 17, 2012
Thank you all!
posted by Summer Fall at 6:44 PM on November 18, 2012
« Older once-off letterpress possible? | Can These Shoes Be Fixed? Newer » |
# 2.5 Addition of velocities (Page 3/12)
Page 3 / 12
Solution
Because ${\mathbf{\text{v}}}_{\text{tot}}$ is the vector sum of the ${\mathbf{\text{v}}}_{\text{w}}$ and ${\mathbf{\text{v}}}_{\text{p}}$ , its x - and y -components are the sums of the x - and y -components of the wind and plane velocities. Note that the plane only has vertical component of velocity so ${v}_{px}=0$ and ${v}_{py}={v}_{\text{p}}$ . That is,
${v}_{\text{tot}x}={v}_{\text{w}x}$
and
${v}_{\text{tot}y}={v}_{\text{w}y}+{v}_{\text{p}}\text{.}$
We can use the first of these two equations to find ${v}_{\text{w}x}$ :
${v}_{\text{w}x}={v}_{\text{tot}x}={v}_{\text{tot}}\text{cos 110º}\text{.}$
Because ${v}_{\text{tot}}=\text{38}\text{.}0 m/\text{s}$ and $\text{cos 110º}=–0.342$ we have
${v}_{\text{w}x}=\left(\text{38}\text{.}\text{0 m/s}\right)\left(\text{–0}\text{.}\text{342}\right)\text{=–13}\text{.}\text{0 m/s.}$
The minus sign indicates motion west which is consistent with the diagram.
Now, to find ${v}_{\text{w}\text{y}}$ we note that
${v}_{\text{tot}y}={v}_{\text{w}y}+{v}_{\text{p}}$
Here ; thus,
${v}_{\text{w}y}=\left(\text{38}\text{.}0 m/s\right)\left(0\text{.}\text{940}\right)-\text{45}\text{.}0 m/s=-9\text{.}\text{29 m/s.}$
This minus sign indicates motion south which is consistent with the diagram.
Now that the perpendicular components of the wind velocity ${v}_{\text{w}x}$ and ${v}_{\text{w}y}$ are known, we can find the magnitude and direction of ${\mathbf{\text{v}}}_{\text{w}}$ . First, the magnitude is
$\begin{array}{lll}{v}_{\text{w}}& =& \sqrt{{v}_{\text{w}x}^{2}+{v}_{\text{w}y}^{2}}\\ & =& \sqrt{\left(-\text{13}\text{.}0 m/s{\right)}^{2}+\left(-9\text{.}\text{29 m/s}{\right)}^{2}}\end{array}$
so that
${v}_{\text{w}}=\text{16}\text{.}0 m/s\text{.}$
The direction is:
$\theta ={\text{tan}}^{-1}\left({v}_{\text{w}y}/{v}_{\text{w}x}\right)={\text{tan}}^{-1}\left(-9\text{.}\text{29}/-\text{13}\text{.}0\right)$
giving
$\theta =\text{35}\text{.}6º\text{.}$
Discussion
The wind's speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in [link] . Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.
Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity.
## Relative velocities and classical relativity
When adding velocities, we have been careful to specify that the velocity is relative to some reference frame . These velocities are called relative velocities . For example, the velocity of an airplane relative to an air mass is different from its velocity relative to the ground. Both are quite different from the velocity of an airplane relative to its passengers (which should be close to zero). Relative velocities are one aspect of relativity , which is defined to be the study of how different observers moving relative to each other measure the same phenomenon.
Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879–1955), the greatest physicist of the 20th century. Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later chapters. The relative velocities in this section are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is, less than . Most things we encounter in daily life move slower than this speed.
Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving forward? The answer is that if air resistance is negligible, the binoculars will hit at the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See [link] .) To the observer on shore, the binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This observer sees the curved path shown in [link] . Although the paths look different to the different observers, each sees the same result—the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities relative to the observer.
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
how did you get the value of 2000N.What calculations are needed to arrive at it
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• Level: GCSE
• Subject: Maths
• Word count: 7894
# number grid
Extracts from this document...
Introduction
Algebraic Investigation 1: Square Boxes on a 10x10 Grid
In this first investigation, the difference in products of the alternate corners of a square, equal-sided box on a 10x10 gridsquare will be investigated. It is believed that the products and their differences should demonstrate a constant pattern no matter what dimensions are used; as long as they remain equal. In order to prove this, both a numeric and algebraic method will be used in order to calculate this difference. The numeric method will help establish a baseline set of numbers for testing, and to help in the establishment of a set of algebraic formulae for use on an n x n gridsquare.
In the example gridsquare below, the following method is used in order to calculate the difference between the products of opposite corners.
(a) (b) (c) (d)
Stage A: Top left number x Bottom right number = (a) multiplied by (d)
Stage B: Bottom left number x Top right number = (c) multiplied by (b)
Stage B – Stage A: (c)(b) - (a)(d) = The difference
The overall, 10 x 10 grid that is used for the first investigation will be a standard, cardinal gridsquare, which progresses in increments of 1. The formulae calculated will mainly be applicable to this grid, as other formats of gridsquares will require others formulae to provide valid results.
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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
This first investigation will focus only on gridsquares with equal widths and heights, which will at this stage be represented by the universal, constant term ‘w’.
The top left number in the grid (letter (a) in the above example) will be represented by the term ‘n’, which will be referred to in this manner in all proceeding investigations also.
This is only the first section of the investigation.
Middle
In this section of the investigation, additional variables (constant terms) will be used due to the change in the number of factors involved.
‘n’ will continue to represent the top left number in the grid. All other formulae in the grids will refer to n.
Due to the height and width of the boxes no longer being equal values, ‘w’ will represent the width of the number box, whereas ‘h’ will refer to the height,
Part B: Changing the Height, ‘h’
To provide a comparison for finding the height formula, summary boxes will be included, showing the algebraic results of the last section.
As in this part of the investigation, additional factors are concerned; the overall formula will contain numerous terms. The rules of brackets and of ‘bidmas’ will always apply to the formulae, helping to create a workable equation.
It should be possible to create an algebra grid detailing the numbers to be expected on any height x width grid on a 10 x 10 gridsquare. After multiplying the alternate corners and subtracting, an overall formula should be gained that can be used to calculate the difference in any h x w box; always on a 10 x 10 grid.
Part A:2 x Width Rectangles
2x2 Rectangles
Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.
15 16 25 26
n n+1 n+10 n+11
Stage A: Top left number x Bottom right number = n(n+11) = n2+11n
Stage B: Bottom left number x Top right number = (n+10)(n+1)= n2+1n+10n+10
= n2+11n+10
Stage B – Stage A: (n2+11n+10)-(n2+11n) = 10
When finding the general formula for any number (n), both answers begin with the equation n2+11n, which signifies that they can be manipulated easily.
Conclusion
Formula 1: Bottom Right (BR) = Top Right (TR) + Bottom Left (BL)
As also shown by the summary boxes and examples above, the formula for the top right number remains constant, and is linked with the width, w, of the box in the following way:
Formula 2: Top Right (TR) = n+ Increment size (s) x (Width -1)
It is also evident from the examples calculated that the bottom left number is also linked with the height, w, of the box using a formula that remains constant:
Formula 3: Bottom Left (BL) = n+ Increment size x Gridsize x (height -1)
Using these rules, it is possible to establish an algebraic box that could be used to calculate the difference for any hxw box on any gxg grid.
n ~ n+s(w-1) ~ ~ ~ n+gs(h-1) ~ n+s(w-1)+gs(h-1)
Which through simple algebraic process can simplify into:
n ~ n+sw-s ~ ~ ~ n+ghs-sg ~ n+sw+ghs-gs-s
Stage A: Top left number x Bottom right number
= n(n+sw+ghs-gs-s)
= n2+nsw+ghns-gns-ns
Stage B: Bottom left number x Top right number
= (n+ghs-gs)(n+sw-s)
= n2+nsw-ns+ghns+ghs2w-ghs2-gns+gs2w+gs2
= n2+nsw+ghns-gns-ns+ghs2w-ghs2+gs2w+gs2
Stage B – Stage A: (n2+nsw+ghns-gns-ns+ghs2w-ghs2+gs2w+gs2)-(n2+nsw+ghns-gns-ns)
= ghs2w-ghs2+gs2w+gs2
= s2ghw-s2gh+s2gw+s2g
When finding the general formula for any number (n) any height (h), any width (w), and with any gird size, both answers begin with the equation n2+nsw+ghns-gns-ns, which signifies that they can be manipulated easily. Because the second answer has +ghs2w-ghs2+gs2w+gs2 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of s2ghw-s2gh+s2gw+s2g will always be present.
Charankamal Singh Theora
This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.
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# Related GCSE Number Stairs, Grids and Sequences essays
1. ## Opposite Corners. In this coursework, to find a formula from a set of numbers ...
4 star(s)
� 3240�10 = (n - 1) � 19 = n Solution Check: y (n-1) � = difference 10 (19 - 1)� = difference 10 � 18� = difference Therefore difference = 3240 Below is a 13 by 13 grid. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
2. ## I am going to investigate the difference between the products of the numbers in ...
4 star(s)
A+L A+2L ? A+G(W-1) A+1+G(W-1) A+2+G(W-1) ? A+(L-1)+G(W-1) These new changes can now be put into the proof. Proof (A+(L-1)) x (A+G(W-1)) = A2+(GA(W-1))+ A(L-1) + G(L-1)(W-1) A x A+(L-1)+G(W-1) = A2+A(L-1)+(GA(W-1)) A2+(GA(W-1))+ A(L-1) + G(L-1)(W-1) ? (A2+A(L-1)+(GA(W-1)))= A2+(GA(W-1))+ A(L-1) + G(L-1)(W-1) - A2-(GA(W-1)) A2+(GA(W-1))+ A(L-1) + G(L-1)(W-1) - A2-A(L-1)-(GA(W-1))= G(L-1)(W-1) + A2-A2 +(GA(W-1))-(GA(W-1) + A(L-1)-A(L-1)
1. ## Number Grid Investigation.
25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74
2. ## How many squares in a chessboard n x n
3 The 45 degree diagonal line cutting through the corners of the squares in the chessboard give a mirror image (when folded diagonally ,lie on top of each other exactly), has a line of symmetry in the diagonal line as well as horizontally and vertically.
1. ## Number Grids Investigation Coursework
Grids of Different Widths and Rectangles I can see the similarities between this formula and my original formula for squares in 10 x 10 grids. My original formula was D = 10 (n - 1)2 and this later developed to D = w (n - 1)2, with w being the width of the grid.
2. ## &quot;Multiply the figures in opposite corners of the square and find the difference between ...
14 15 25 26 27 3 x 25 = 75 1 x 27 = 27 75 - 27 = 48 The difference is 48 This proves that my formula will work with any grid and any size square within it at any place on the grid.
1. ## Number Grid Investigation
So when: n = 5 m = 4 10 (n - 1) (m - 1) = 120 10 (5 - 1)(4 - 1) = 10(4 x 3) = 10 x 12 = 120 The formula worked. Grid Size By changing the size of the overall grid used, the numbers should
2. ## Number Grid Investigation.
- (x� + 33x) Difference = 90. What have we noticed so far? Within a 10 wide grid: 2 X 2 squares product difference is 10. 3 X 3 squares product difference is 40. 4 X 4 squares product difference is 90. Size of Square 2 X 2 3 X 3 4 X 4 Product Difference 10
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# Sequential Steps to Solve a Problem with Algorithm and Flowchart
In this article, I explain about how to use flowchart in our daily problems and present the List of Sequential Steps to Solve a Problem with Algorithm and Flowchart.
There are many tools for programmers for writing program such as algorithms, flowcharts, pseudocodes, data dictionary, decision table, data flow diagrams etc. They are effective tools for knowledge of programming tools
Any work that we do in our daily lives that work to solve the Problem. Problem can say and everyone has a certain order and under the certain order, we have a variety of Steps Use. Suppose for example that we have someone to Phone. It is also a problem because we have to do something. Phone us are now following tasks:
Suppose for example that we have someone to Phone. It is also a problem because we have to do something. Phone us are now following tasks:
Check for this, first we will Phone, the Phone is turned on or not. The Dial Tone coming or not.
If the Dial Tone is coming, Phone Number Dial the person is to us, so we want to talk.
After Phone Number Dial Phone us at Target person will have to wait to Bell.
Phone Attend the Bell's and Target person, then it will be.
Change the order of these Steps, we cannot, nor can we leave at any Step. Here is a summary of this example:
To obtain a definite and reasonable solution to any problem, we varied the problem Steps to define a group, which consists in a certain order. The Steps Follow the simple language of the group and its streamlined writing in order to be taken, which we receive written Description, as it is called Algorithm.
So, we have C = A + B. To accomplish this task, or to solve the Problem in order to follow us as follows:
Step 1: Start the process.
Step 2: Read the value of variable A.
Step 3: read the value of variable B.
Step 4: A and B, find the sum of values.
Step 5: The sum of the values A and B Variable C substitution.
Step 6: C to print the value.
Step 7: The process to end.
We can see that we have to add two numbers above are the various Steps to follow in the order in which depicts. If we do all of the above Steps to Follow When Exactly in the same order in which they were written above, we certainly receive the sum of two numbers. Since writing the above Description of the two numbers add to a Well Defined Working Algorithm.
An Algorithm to Solve the Problem of the way Description is written. Flowchart to Solve a Problem in exactly the same way is a featured Description. If we add the two numbers Flowchart Create a wish to Flowchart something we will be as follows
Flowchart of Addition of two Numbers.
From the Computer to Solve the Problem of any kind by us, you can use Algorithm and Flowcharts. Generally better Solve a Problem to both the problem Combined Algorithm and Flowchart is taken into use, so there is no chance of mistake. |
The Coordinate Plane The Coordinate Plane We define coordinate planes as a surface with the two dimensions where we are able to plot the points and locate the position for the particular point with respect to the given axes. We say that it is a plane with two scales called x – axis and y – axis, which are perpendicular to each other and thus form the right angles with each other. We must remember that axis word is the singular form and axes are the plural word for axis. When any point is marked on the plane, it is located with respect to x and y coordinates, which are measured starting from th point of intersection of the two lines called origin. The coordinates of origin are (0, 0 ) and it is the starting point of the measurement of both the axes. The labeling of the axes is done in the form of the scale, where each mark represents the number and are marked at equal distance. Know More About : Percentage Difference Calculator
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x- Axis is the scale which is placed horizontally on the plane Its marking starts with zero at the origin and as we move towards right direction the numbers increase by 1 unit and as we move towards left of origin , the number reduce by 1 unit. So we conclude that on the left of zero, we have negative numbers and on the right of the origin we have positive numbers. Now we look at the y- axis, as we move upward, it has positive numbers and as we move down from origin , it has negative numbers. All the measurements of the distances are done from the origin. Thus we say that the origin is the middle point on the Cartesian scale and it helps us to locate any point on the coordinate. Now if we look at the coordinate plane, we observe that the coordinate have two axes which are placed perpendicular to each other and intersect at the origin. These intersecting lines divide the coordinate plane in 4 quadrants. These quadrants are represented by the roman numerals I, II, III and IV. Any point on the plane can be located only if we know the location of the point with respect to x and y axis. We call this pair of points representing the location as ordered pairs. Thus the equations formed by the combination of x and y axis can be plotted on the coordinate plane. We call these types of equations as linear equations with two variables. We can work on the two variable equations and get the solution of the equations by working on the coordinate plane. Coordinate plane is also called as Cartesian plane. Read More About : Percentage Off Calculator
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Page No. : 2/4
If we have two different equations in the linear form of two variables, and we need to find the solution to the pair of equations, the simplest method to find the solution for such pair of equations is to plot the equations on the Cartesian plane and then locate the point of intersection of the two equations. This point of intersection will represent the solution for the pair of equations. In case we observe the lines do not intersect, we conclude that the pair of equations do not have any solution.
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Thank You For Watching
Presentation
The Coordinate Plane
We define coordinate planes as a surface with the two dimensions where we are able to plot the points and locate the position for the partic... |
### Hex
Explain how the thirteen pieces making up the regular hexagon shown in the diagram can be re-assembled to form three smaller regular hexagons congruent to each other.
### Triangular Tantaliser
Draw all the possible distinct triangles on a 4 x 4 dotty grid. Convince me that you have all possible triangles.
# Overlap
##### Stage: 3 Challenge Level:
Many thanks to those of you who sent in solutions to thie problem including Mai of The Chinese High School Singapore and Andrei of School 205 Bucharest. Thanks also to the pupils of Madras College who answered the first part of this problem as part of their solution to tilting triangles (November 2002).
$P$ is the centre of the square $ABCD$
## First
If the sides $PQ$ and $PS$ of the square $PQRS$ conincide with $PB$ and $PC$ then the overlapped area is $PBC$.
By symmetry (or congruence) $PBC$ has an area which is one quarter of the area of $ABCD$.
## Second
Rotating the square $PQRS$ about $P$ gives a diagram equivalent to the one opposite.
We know that $PBC$ is a quarter of $ABCD$ so if we can show that the area of $PXCY =$ area of $PBC$ then the overlap will always be a quarter of the square.
To prove this we need to show that triangle $PBX$ is congruent to triangle $PYC$.
$PB = PC$ ( half the diagonal of the square)
angle $PBX =$ angle $PCY$ (diagonals of square bisect the angles).
angle $CPY =$ angle $BPX =$ angle of rotation
Therefore triangles $BPX$ and $CPY$ are congruent (ASA)
Area $PXCY =$ Area $XPC +$ Area $CPY =$ Area $XPC +$ Area $BPX =$ Area $BPC$
Therefore $PXCY$ is a quarter of the square.
## Lastly
Let the length of the side of the large square is $x$ and the the length of the diagonal is $d$.
Using Pythagoras Theorem
$d^2 = x^2 + x^2 = 2x^2$
From this we know that the limit of the length of the side of the small square is:
$\frac {d}{2} = \frac {x\sqrt 2}{2}$ |
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# Transportation Problem | Set 4 (Vogel’s Approximation Method)
The North-West Corner method and the Least Cost Cell method has been discussed in the previous articles. In this article, the Vogel’s Approximation method will be discussed. Solution:
• For each row find the least value and then the second least value and take the absolute difference of these two least values and write it in the corresponding row difference as shown in the image below. In row O1, 1 is the least value and 3 is the second least value and their absolute difference is 2. Similarly, for row O2 and O3, the absolute differences are 3 and 1 respectively.
• For each column find the least value and then the second least value and take the absolute difference of these two least values then write it in the corresponding column difference as shown in the figure. In column D1, 2 is the least value and 3 is the second least value and their absolute difference is 1. Similarly, for column D2, D3 and D3, the absolute differences are 2, 2 and 2 respectively.
• These value of row difference and column difference are also called as penalty. Now select the maximum penalty. The maximum penalty is 3 i.e. row O2. Now find the cell with the least cost in row O2 and allocate the minimum among the supply of the respective row and the demand of the respective column. Demand is smaller than the supply so allocate the column’s demand i.e. 250 to the cell. Then cancel the column D1.
• From the remaining cells, find out the row difference and column difference.
• Again select the maximum penalty which is 3 corresponding to row O1. The least-cost cell in row O1 is (O1, D2) with cost 1. Allocate the minimum among supply and demand from the respective row and column to the cell. Cancel the row or column with zero value.
• Now find the row difference and column difference from the remaining cells.
• Now select the maximum penalty which is 7 corresponding to column D4. The least cost cell in column D4 is (O3, D4) with cost 2. The demand is smaller than the supply for cell (O3, D4). Allocate 200 to the cell and cancel the column.
• Find the row difference and the column difference from the remaining cells.
• Now the maximum penalty is 3 corresponding to the column D2. The cell with the least value in D2 is (O3, D2). Allocate the minimum of supply and demand and cancel the column.
• Now there is only one column so select the cell with the least cost and allocate the value.
• Now there is only one cell so allocate the remaining demand or supply to the cell
• No balance remains. So multiply the allocated value of the cells with their corresponding cell cost and add all to get the final cost i.e. (300 * 1) + (250 * 2) + (50 * 3) + (250 * 3) + (200 * 2) + (150 * 5) = 2850
Below is the implementation for the approach discussed above
## Python3
`grid ``=` `[[``3``, ``1``, ``7``, ``4``], [``2``, ``6``, ``5``, ``9``], [``8``, ``3``, ``3``, ``2``]] ``# table``supply ``=` `[``300``, ``400``, ``500``] ``# supply``demand ``=` `[``250``, ``350``, ``400``, ``200``] ``# demand``INF ``=` `10``*``*``3``n ``=` `len``(grid)``m ``=` `len``(grid[``0``])``ans ``=` `0` `# hepler function for finding the row difference and the column difference``def` `findDiff(grid):`` ``rowDiff ``=` `[]`` ``colDiff ``=` `[]`` ``for` `i ``in` `range``(``len``(grid)):`` ``arr ``=` `grid[i][:]`` ``arr.sort()`` ``rowDiff.append(arr[``1``]``-``arr[``0``])`` ``col ``=` `0`` ``while` `col < ``len``(grid[``0``]):`` ``arr ``=` `[]`` ``for` `i ``in` `range``(``len``(grid)):`` ``arr.append(grid[i][col])`` ``arr.sort()`` ``col ``+``=` `1`` ``colDiff.append(arr[``1``]``-``arr[``0``])`` ``return` `rowDiff, colDiff` `# loop runs until both the demand and the supply is exhausted``while` `max``(supply) !``=` `0` `or` `max``(demand) !``=` `0``:`` ``# finding the row and col difference`` ``row, col ``=` `findDiff(grid)`` ``# finding the maxiumum element in row difference array`` ``maxi1 ``=` `max``(row)`` ``# finding the maxiumum element in col difference array`` ``maxi2 ``=` `max``(col)` ` ``# if the row diff max element is greater than or equal to col diff max element`` ``if``(maxi1 >``=` `maxi2):`` ``for` `ind, val ``in` `enumerate``(row):`` ``if``(val ``=``=` `maxi1):`` ``# finding the minimum element in grid index where the maximum was found in the row difference`` ``mini1 ``=` `min``(grid[ind])`` ``for` `ind2, val2 ``in` `enumerate``(grid[ind]):`` ``if``(val2 ``=``=` `mini1):`` ``# calculating the min of supply and demand in that row and col`` ``mini2 ``=` `min``(supply[ind], demand[ind2])`` ``ans ``+``=` `mini2 ``*` `mini1`` ``# subtracting the min from the supply and demand`` ``supply[ind] ``-``=` `mini2`` ``demand[ind2] ``-``=` `mini2`` ``# if demand is smaller then the entire col is assigned max value so that the col is eliminated for the next iteration`` ``if``(demand[ind2] ``=``=` `0``):`` ``for` `r ``in` `range``(n):`` ``grid[r][ind2] ``=` `INF`` ``# if supply is smaller then the entire row is assigned max value so that the row is eliminated for the next iteration`` ``else``:`` ``grid[ind] ``=` `[INF ``for` `i ``in` `range``(m)]`` ``break`` ``break`` ``# if the row diff max element is greater than col diff max element`` ``else``:`` ``for` `ind, val ``in` `enumerate``(col):`` ``if``(val ``=``=` `maxi2):`` ``# finding the minimum element in grid index where the maximum was found in the col difference`` ``mini1 ``=` `INF`` ``for` `j ``in` `range``(n):`` ``mini1 ``=` `min``(mini1, grid[j][ind])` ` ``for` `ind2 ``in` `range``(n):`` ``val2 ``=` `grid[ind2][ind]`` ``if` `val2 ``=``=` `mini1:`` ``# calculating the min of supply and demand in that row and col`` ``mini2 ``=` `min``(supply[ind2], demand[ind])`` ``ans ``+``=` `mini2 ``*` `mini1`` ``# subtracting the min from the supply and demand`` ``supply[ind2] ``-``=` `mini2`` ``demand[ind] ``-``=` `mini2`` ``# if demand is smaller then the entire col is assigned max value so that the col is eliminated for the next iteration`` ``if``(demand[ind] ``=``=` `0``):`` ``for` `r ``in` `range``(n):`` ``grid[r][ind] ``=` `INF`` ``# if supply is smaller then the entire row is assigned max value so that the row is eliminated for the next iteration`` ``else``:`` ``grid[ind2] ``=` `[INF ``for` `i ``in` `range``(m)]`` ``break`` ``break` `print``(``"The basic feasible solution is "``, ans)`
My Personal Notes arrow_drop_up |
#### Exploring the binomial process
We start our exploration of the binomial process by looking at the probability of a certain number of successes s for a given number of trials n and probability of success p. Imagine we have two binomial trials, there are four possible outcomes, as shown below, namely SS, SF, FS, and FF, where SF means success followed by failure, etc.
These outcomes have probabilities p2, p(1-p), (1-p)p and (1-p)2 respectively. The binomial process considers each success to be identical and therefore does not differentiate between the two events SF and FS: they are both just one success in two trials. The probability of one success in two trials is then just 2p(1-p): the 2 in this equation is the number of different paths that result in 1 success in 2 trials.
Now imagine that we have three trials:
The eight outcomes are: SSS, SSF, SFS, SFF, FSS, FSF, FFS, and FFF. There is thus 1 event producing 3 'successes', 3 events producing 2 successes, 3 events producing 1 success and 1 event producing no successes for three trials.
#### The binomial distribution equation
In general, the number of ways that we can get s successes from n trials can be calculated directly using the binomial coefficient nCs, which is given by:
We can check this is right, by choosing n =3, (remembering that 0! = 1) then:
which match the number of combinations we have already calculated. Each of the ways of getting s successes in n trials has the same probability, namely ps(1-p)n-s, so the probability of observing x successes in n trial is given by:
which is the probability mass function of the Binomial(p,n) distribution. In other words, the number of successes s one will observe in n trials, where each trial has the same probability of success is given by:
s = Binomial(p,n)
#### Examples of the Binomial
The four figures below show the distribution for a number of different p's (probabilities) and n's (trials).
The first figure above shows the number of tails (here called success (s)) if we toss a coin five times could be 0, 1, 2, 3, 4 or 5 with the probabilities displayed on the y-axes. The figure to the right shows the same situation, but now with 40 tosses. The most likely outcome is 20 tails , but it could be as low as 10 or as high as 30. Note that this figure looks somewhat like a bell-shaped Normal distribution. |
Worksheet Solutions: Ratio & Proportion
# Ratio and Proportion Class 6 Worksheet Maths Chapter 12
Q1: Fill in the box :
Solution: 2, 9
Explanation: To find the missing values, we simplify the given ratio 14/21 to 2/3.
So, the first missing value is 2.
For the second missing value in the numerator of 6/___, we note that 2/3 is equivalent to 6/9, indicating the second missing value is 9.
Hence, the missing values are 2 and 9 respectively.
Q2: Find the ratio of the following :
(i) 21 hours to 49 hours __________
(ii) 75 cm to 3 m __________
(iii) A dozen to a score __________
(iv) 1 hour to 20 minutes __________
(v) A dozen to a gross __________
Sol:
(i) 21 : 49 = 3 : 7
(ii) 75 : 300 = 1 : 4
(iii) 12 : 20 = 3 : 5
(iv) 60 : 20 = 3 : 1
(v) 12 : 144 = 1 : 12
Explanation:
(i) To find the ratio of 21 hours to 49 hours, both numbers are divided by their greatest common divisor, which is 7. So, 21 ÷ 7 = 3 and 49 ÷ 7 = 7, resulting in the ratio 3 : 7.
(ii) To express 75 cm to 3 m in the same unit, we convert 3 m to cm (1 m = 100 cm), giving us 300 cm. Then, we simplify the ratio by dividing both numbers by their greatest common divisor, which is 75. So, 75 ÷ 75 = 1 and 300 ÷ 75 = 4, resulting in the ratio 1 : 4.
(iii) A dozen equals 12 and a score equals 20. We simplify the ratio by dividing both numbers by their greatest common divisor, which is 4. So, 12 ÷ 4 = 3 and 20 ÷ 4 = 5, giving us the ratio 3 : 5.
(iv) Converting 1 hour to minutes (1 hour = 60 minutes), we get 60 minutes. Then, we simplify the ratio by dividing both numbers by their greatest common divisor, which is 20. So, 60 ÷ 20 = 3 and 20 ÷ 20 = 1, resulting in the ratio 3 : 1.
(v) A dozen is 12 and a gross is 144. Simplifying the ratio by dividing both numbers by their greatest common divisor, which is 12, we get 12 ÷ 12 = 1 and 144 ÷ 12 = 12, giving us the ratio 1 : 12.
Q3: Write True or False :
(i) 2 : 8 :: 4 : 16 ( )
(ii) 500 : 200 :: 150 : 60 ( )
(iii) 50 : 45 :: 30 : 20 ( )
Sol:
(i) True
(ii) True
(iii) False
Explanation:
(i) This statement is true because both ratios simplify to 1:4.
(ii) This statement is true because both ratios simplify to 5:2.
(iii) This statement is false because the ratio 50:45 doesn't simplify to 30:20. They are not in proportion.
Q4: Fill in the blanks so that the numbers are in proportion :
(i) 20, 18, 40, ..............
(ii) ............., 35, 3, 15
(iii) 25, 100, ..........., 160
(iv) 32, ..........., 6, 12
Sol:
(i) 36
(ii) 7
(iii) 40
(iv) 64
Explanation:
(i) To find the missing number, we notice that 20 is 90% of 40 and 18 is 80% of 36. Applying this pattern, we find that 36 is 90% of 40, making it the missing number.
(ii) The missing number can be found by observing that 35 is 7 times greater than 5 and 3 is 7 times less than 21. So, 7 is the missing number.
(iii) By observing the pattern, we see that 25 is 25% of 100 and 160 is 25% more than 100. Thus, the missing number is 40.
(iv) The missing number can be found by realizing that 32 is 8 times greater than 4 and 6 is 8 times less than 48. Hence, 64 is the missing number.
Q5: Find x, if the numbers are in proportion :
(i) 3, 9, 9, x __________
(ii) 25, x, 1 , 4 __________
Sol:
(i) 27
(ii) 100
Explanation:
(i) The ratio between 3 and 9 is 1:3. Since the ratio between 9 and x is also 1:3, x must be 27 to maintain proportionality.
(ii) The ratio between 25 and 1 is 25:1, which simplifies to 25:1. To maintain proportionality, the ratio between x and 4 must also be 25:1. Hence, x equals 100.
Q6: Divide 60 in the ratio 1:2 between Bulbul and Kanika :
(i) Bulbul's share = __________
(ii) Kanika's share = __________
Sol:
(i) 20
(ii) 40
Explanation:
(i) Since the ratio between Bulbul and Kanika's shares is 1:2, Bulbul gets one-third (1/3) of 60, which is 20.
(ii) Kanika's share is two-thirds (2/3) of 60, which equals 40.
Q7: Give two equivalent ratios of 6 : 4
Sol: 3 : 2, 12 : 8 (or any other two equivalent ratios)
Explanation: Equivalent ratios are obtained by multiplying or dividing both parts of a ratio by the same number. For example, to find another equivalent ratio for 6:4, we can divide both parts by 2, resulting in 3:2. Similarly, we can multiply both parts by 2 to get 12:8.
Q8: Weight of 80 books in 160 kg. What is the weight of 25 books?
Sol: 50 kg
Explanation: To find the weight of 25 books, we divide the total weight of 80 books (160 kg) by 80 to get the weight of one book (2 kg). Then, we multiply the weight of one book by 25 to find the weight of 25 books, which equals 50 kg.
Q9: Divide 1200 among A, B and C in the ratios 4 : 5 : 6.
(i) A's share = __________
(ii) B's share = __________
(iii) C's share = __________
In a class of 40 students, 15 like cricket, 20 like football and 5 like both cricket and football. (Now answer question number 10 and 11) :
Sol:
(i) 320
(ii) 400
(iii) 480
Explanation: To divide 1200 among A, B, and C in the ratios 4:5:6, we first find the total number of parts in the ratio, which is 4 + 5 + 6 = 15.
(i) To find A's share, we calculate (4/15) * 1200 = 320. This means A's share is 320.
(ii) To find B's share, we calculate (5/15) * 1200 = 400. This means B's share is 400.
(iii) To find C's share, we calculate (6/15) * 1200 = 480. This means C's share is 480.
So, A's share is 320, B's share is 400, and C's share is 480. Each person's share is in proportion to the ratio given.
Q10: Find the ratio of number of students who like cricket to those who like football.
Sol: 20 : 25 = 4 : 5
Explanation: Out of 40 students, 15 like cricket and 20 like football. To find the ratio between the number of students who like cricket to those who like football, we simplify the ratio between 15 and 20, which is 3:4. Since both numbers are divisible by 5, we divide each by 5 to get 4:5.
Q11: Find the ratio of number of students who like football to those who like both.
Sol: 25 : 5 = 5 : 1
Explanation: Out of 40 students, 20 like football, and 5 like both cricket and football. To find the ratio between the number of students who like football to those who like both, we simplify the ratio between 20 and 5, which is 4:1. Since both numbers are divisible by 5, we divide each by 5 to get 5:1.
Q12: If 2, 5 and x are in proportion, find x __________ [Hind 2 : 5 :: 5: x]
Sol:
Explanation: If 2, 5, and x are in proportion, it means 2 is to 5 as 5 is to x. Mathematically, we can express this as 2/5 = 5/x. Solving for x, we get x = (5 * 5) / 2 = 25 / 2 = 10.
Q13: Find the ratio of 5 days to 2 weeks.
Sol: 5 : 14
Explanation: To find the ratio between 5 days and 2 weeks, we first convert weeks to days. Since 1 week = 7 days, 2 weeks = 2 * 7 = 14 days. Then, we simplify the ratio between 5 days and 14 days to get 5:14.
Q14: The cost of one dozen bananas is 60. Find the cost of 8 bananas.
Sol: 40
Explanation: Since one dozen equals 12 bananas and costs 60, the cost of one banana is 60 / 12 = 5. To find the cost of 8 bananas, we multiply the cost of one banana by 8, giving us 5 * 8 = 40.
Q15: In a class of 48 students, there are 12 girls and rest are boys. Find the ratio of :
(i) Girls to Boys. __________
(ii) Girls to total number of students. __________
Sol:
(i) 12 : 36 = 1 : 3
(ii) 12 : 48 = 1 : 4
Explanation:
(i) Out of 48 students, 12 are girls. The rest, which is 48 - 12 = 36, are boys. So, the ratio of girls to boys is 12:36, which simplifies to 1:3.
(ii) The number of girls (12) compared to the total number of students (48) gives the ratio of girls to the total number of students, which is 12:48, simplifying to 1:4.
The document Ratio and Proportion Class 6 Worksheet Maths Chapter 12 is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6
## Mathematics (Maths) Class 6
134 videos|323 docs|42 tests
## FAQs on Ratio and Proportion Class 6 Worksheet Maths Chapter 12
1. What is ratio and proportion in mathematics?
Ans. Ratio is a comparison between two or more quantities, while proportion is an equation that states two ratios are equal. In mathematics, ratio and proportion are used to compare and relate quantities in various situations.
2. How do you find the ratio between two numbers?
Ans. To find the ratio between two numbers, divide the first number by the second number. The ratio can be expressed as a fraction, decimal, or simplified whole numbers.
3. What are the applications of ratio and proportion in real life?
Ans. Ratio and proportion are widely used in everyday life. They are used in cooking recipes to determine ingredient quantities, in construction to calculate dimensions and measurements, and in finance to analyze financial statements and make comparisons.
4. How can we solve problems involving ratio and proportion?
Ans. To solve problems involving ratio and proportion, it is important to set up a proportion equation and solve it using cross-multiplication. The given ratios can be compared and related to find the unknown quantity or to solve for a missing value in the proportion.
5. Can ratio and proportion be used in geometry?
Ans. Yes, ratio and proportion are used in geometry as well. They are used to find similar triangles, solve problems involving similar shapes, and determine the scale factor between two figures. Ratio and proportion help in understanding the relationships between lengths, areas, and volumes in geometric figures.
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# Calculate the Area of a Shape
Extracts from this document...
Introduction
## Calculating the area of a shape
In this project I will investigate the area of shapes; I will investigate shapes with 3,4,5,6 and 7 dots joined and I will look at building up the dots inside the shapes.
My prediction is that shapes with the same amount of dots joined and number of dots inside, will have the same area.
After drawing the shapes I have notice that, shapes with the equal number of dots have the same area, this can be seen on the previous diagram pages. So from now on I will only draw 1 of each type of diagram.
I will now put all of my results for no dots inside into a table.
Table 1
Number of dotsN AreaA 3 ½ cm2 4 1 5 1 ½ cm2 6 2 7 2 ½ cm2
• After creating a table 1, I have noticed a pattern in the area column.
• The pattern is going up in ½ cm2 each time a dot is added to the edge.
• By looking at the pattern I have created formula in both words and algebra:
Area = the number of dots - 1 A=N – 1
A= Area 2 2
N= Number of dots
To prove I’m right
When= N = 6 A= 6 - 1 = 3-1
2 = 2cm2
If you look at my table you will see this correct.
Now I know I am
Middle
Table 2
A=N+0
2
Table 3
A=N +1
2
Final formula
A=N+D -1
2
A=N-1 0 dots inside
2
A=N+0 1 dot inside (I have added the 0 just to show the pattern works)
2
A=N+1 2 dots inside
2
As you can see the formulas are going up in 1, every time the dots inside the shape are increased.
By doing this I have created 1 formula which is the three formulas combined as 1
This is:
Area= number of dots + dots inside -1
2
A=N + D -1
2
Now I’m correct with my formula I can find area of a shape of any size with any number of dots inside and joint.
EG:
When= N= 12 D= 5 A=12+5-1 = 6+5-1=10cm2
2
## Calculating the area of a shape
In this project I have investigated the area of shapes; I have investigated shapes with 3,4,5,6 and 7 dots joined and I will look at building up the dots inside the shapes later on.
After drawing the shapes I have noticed that, shapes with the equal number of dots have the same area, this can be seen on the previous diagram pages. So from now on I will only draw 1 of each type of diagram.
I will now put all of my results for no dots inside into a table.
Table 1
Number of dotsN AreaA 3 ½ cm2 4 1 cm2 5 1 ½ cm2 6 2 cm2 7 2 ½ cm2
• After creating a table 1, I have noticed a pattern in the area column.
• The pattern is going up in ½ cm2 each time a dot is added to the edge.
• By looking at the pattern I have created formula in both words and algebra:
Area = the number of dots - 1 A=N – 1
A= Area 2 2
N= Number of dots
To prove I’m right
When N = 3 A=3 - 1 = 1 ½ -1 When N=4 A=4 -1 =2-1
2 = ½ cm2 2 =1 cm2
When N=5 A=5 -1 = 2 ½ -1 When N=6 A=6 -1 =3-1 When N=7 A=7 – 1 =3 ½ -1
2 = 1 ½ cm2 2 =2cm2 2 =2 ½ cm2
If you look at my table you will see this correct.
Now I know I am definitely correct with my formula I can find area of a shape of any size as long as there are no dots inside and this is shown on back of the diagram sheet.
E.G
N=10 A=10 – 1 = 5 - 1
2 = 4cm2
Now I have investigated shapes with 1 dot inside I will put my answers into a table.
Table 2
Number of dotsN AreaA 3 1 ½cm2 4 2cm2 5 2 ½cm2 6 3cm2 7 3 ½cm2
Conclusion
E.G
When =N=10 A=10 +1 =5+1
1. =6cm2
After completing all the diagrams, tables and formulas I have noticed a pattern between the three formulas:
I have put all the formulas into a table:
Table 4
Table Formula Table 1 A=N -1 2 Table 2 A=N+0 2 Table 3 A=N +12 Final formula A=N+D -1 2
A=N-1 0 dots inside
2
A=N+0 1 dot inside (I have added the 0 just to show the pattern works)
2
A=N+1 2 dots inside
2
As you can see the formulas are going up in 1, every time the dots inside the shape are increased.
By doing this I have created 1 formula which is the three formulas combined as 1
This is:
A=Area D=Dots inside N=number of dots joined
Area= number of dots + dots inside -1
2
A=N + D -1
2
Now I’m correct with my formula I can find area of a shape of any size with any number of dots inside.
EG:
When= N= 10 D=2 A=10+2-1 = 5+2-1=6cm2
2
Now I will show some examples for the final formula for 0 dots, 1 dot and 2 dots in side on some of the past diagrams I have drew , but with the final formula not the individual formulas.
EG 1:
When N=5 D=0 A=5 +0 - 1 =1 ½ cm2
2
EG 2:
When N=7 D=1 A=7 +1-1 = 3 ½ cm2
2
EG 3:
When N=3 D=2 A=3 +2 -1 =2 ½ cm2
2
Mathematics
Coursework
Area of a shape
Leigh Bevan
This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.
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# A software sales woman is paid a base salary and she makes a commission of 10% on everything she sales. How much sales does she need to have an annual salary of $120,000 if she has a base salary of$85,000.?
Dec 6, 2016
This sales person must have $350,000 in sales to have a total annual salary of$120,000.
#### Explanation:
we can describe the salary this salesperson makes as:
t = b + (10%*s) where $t$ is the total sales, $b$ is the base salary and $s$ is the total sales. Using the information we are provided in the problem we can substitute and solve for the "total sales, $s$. We can also substitute $\frac{10}{100}$ or $0.1$ for 10# as % means per 100.:
$120000 = 85000 + \left(0.1 \cdot s\right)$
$120000 - 85000 = 85000 - 85000 + \left(0.1 \cdot s\right)$
$35000 = 0 + \left(0.1 \cdot s\right)$
$35000 = 0.1 \cdot s$
$\frac{35000}{0.1} = \frac{0.1 \cdot s}{0.1}$
$350000 = \frac{\cancel{0.1} \cdot s}{\cancel{0.1}}$
$s = 350000$ |
# How do you find the derivative of sin(x^3)?
May 29, 2015
We use the chain rule.
http://socratic.org/calculus/basic-differentiation-rules/chain-rule
Using the notation provided there, if we define $y \left(x\right) = \sin \left({x}^{3}\right)$ and $u = {x}^{3}$, we may rewrite $y \left(x\right)$ as $y \left(u\right) = \sin \left(u\right)$
From the chain rule we know that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \frac{\mathrm{dy}}{\mathrm{du}}$. Recall that $u \left(x\right) = {x}^{3}$ and $y \left(u\right) = \sin \left(u\right)$. Therefore, by the power rule, $\frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2}$, and by the definitions of trigonometric derivatives, $\frac{\mathrm{dy}}{\mathrm{du}} = \cos \left(u\right)$. Thus:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(3 {x}^{2}\right) \left(\cos \left(u\right)\right)$
Substituting ${x}^{3}$ back for u yields:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} \cos \left({x}^{3}\right)$ |
# Fractions. Cavendish Community Primary School
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1 Fractions Children in the Foundation Stage should be introduced to the concept of halves and quarters through play and practical activities in preparation for calculation at Key Stage One. Y Understand that a fraction can describe part of a whole. Understand that a unit fraction represents one EQUAL part of a whole. Recognise, find and name a half as one of two EQUAL parts of an object, shape or quantity (including measure). Recognise, find and name a quarter as one of four EQUAL parts on a object, shape or quantity (including measure).
2 This can be formally recorded like this: The above image is an example of how the children may record. Any images presented to the children should always have EQUAL parts exactly.
3 Y2 Understand and use the terms numerator and denominator. Understand that a fraction can describe part of a set. Understand that the larger the denominator is, the more pieces it is split into and therefore the smaller each part will be. Recognise, find, name and write fractions:,, 2 and 3 of a length, shape, set of objects or quantity. Count on and back in steps of 2 and 4 Write simple fractions for example, 2 of 2 4 and 2. of 6 = 3 and recognise the equivalence This can be formally recorded with the symbols but it must be pictorially represented.
4 Challenge: Once children are secure with finding one quarter and one half of numbers and shapes children can find three quarters. This must be represented pictorially. The above image on the right is an example of how the children may record. Any images presented to the children (image on the left) should always have EQUAL parts exactly.
5 Y3 Show practically and pictorially that a fraction is one whole number divided by another e.g. 3 4 is 3 4 Understand that finding a fraction relates to division Recognise, find and write fractions of a discrete set of objects with small denominators Recognise that tenths arise from dividing an object into 0 equal parts and in dividing one-digit numbers or quantities by 0 Continue to count in steps of 2 and 4 and also count in steps of 3 Compare and order unit fractions and fractions with the same denominator Recognise and show, using diagrams, equivalent fractions with small denominators Add and subtract fractions with the same denominator within one whole using diagrams
7 Y4 Show practically and pictorially that a fraction is one whole number divided by another e.g. 3 4 is 3 4 Recognise, find and write fractions of a discrete set of objects with small denominators Recognise that hundredths arise from dividing an object into 00 equal parts and in dividing tenths by 0 Continue to count in steps of unit fractions Recognise and write decimal equivalents to quarter, half and three-quarters Add and subtract fractions with the same denominator within one whole using diagrams Continue to reinforce steps from Y3 throughout Y4. Y5 Add and subtract fractions with the same denominator and denominators that are multiples of the same number (using diagrams). Children should have had lots of practical experience of equivalence prior to this stage. Adding Unlike Fractions These have different denominators. Work out the sum below. Shade in one fifth of this shape. Now shade in another seven tenths. What have you got altogether?
8 Subtracting Step. The bottom numbers are different. See how the slices are different sizes? We need to make them the same before we can continue, because we can t subtract them like this: Step 2. Using our knowledge of equivalence, convert one half into sixths. So now we Have 3 sixths minus sixth = 6 Step 3. Simplify the fraction. 2 = 6 3
9 Write mathematical statements > as a mixed number. A mixed number is made up of a whole number and a fraction. Eg We say 3 and two fifths. This means As a picture this is 3 whole ones and 2 5 of one = 7 5 The mixed number is the same as the improper fraction 7 5 What is eight fifths as a mixed number?
10 Multiply proper fractions and mixed numbers by whole numbers, supported by materials and diagrams. Multiplying a Fraction by a Whole Number We know that multiply means `lots of` so 4 x 2 3 means 4 lots of 2 3 There are 8 thirds. 8 3 Now write this as a mixed number. 8 3 = 2 2 3
11 Y6 Adding and subtract fractions with different denominators and mixed numbers, using the concept of equivalent fractions building on from Y Write down the number statement. 2. Add together the whole numbers, leaving the fractions at the end. 3. Write fractions with the same denominators so we can add them (using knowledge of equivalence). 4. Add the fractions. 5. If your fraction is improper, write it as a mixed number.. Write down the number statement. 2. Split the second fraction into a whole number and a fraction. Subtract them in turn. 3. First subtract the whole number. 4. Write the first fraction as an improper fraction so you can do it. 5. Write a common denominator. 6. Subtract. 7. Now finish off by writing as a mixed number.
12 Multiplying two fractions This is 4 of a circle. Remember that 3 x 4 means: 3 lots of 4 or 3 of 4 This is 3 of our 4 of a circle. This is 2 of our whole. 3 x 4 is the same as 3 of 4 3 x 4 = x 3 x 4 = 2
13 Dividing a proper fraction by a whole number using diagrams Dividing a fraction by a whole number This is 3 What does 3 2 mean? This is 3 2. What fraction is this part? It is 6 of the whole. 3 2 is the same as 2 of 3 3 of 2 = 2 of 3 = 2 x 3 = x 2 x 3 = 6
### Number: Fractions (including Decimals and Percentages) COUNTING IN FRACTIONAL STEPS Year 1 Year 2 Year 3 Year 4 Year 5 Year 6
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### count up and down in tenths count up and down in hundredths
Number: Fractions (including Decimals and Percentages COUNTING IN FRACTIONAL STEPS Pupils should count in fractions up to 10, starting from any number and using the1/2 and 2/4 equivalence on the number
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Differential calculus - derivatives
Derivatives of basic or elementary functions
The derivative of the cosine function
The derivative of the exponential function
We use the limit definition
to find the derivative of a function.
The derivative of the sine function
We use the limit of the difference quotient to find the derivative of the function f (x) = sin x.
Let rewrite the difference quotient applying the sum to product formula,
Since, the derivative is the limit of the difference quotient as h tends to zero then,
Therefore, if f(x) = sin x then
The derivative of the cosine function
We use the limit of the difference quotient to find the derivative of the function f (x) = cos x.
Let rewrite the difference quotient applying the sum to product formula,
Since the derivative is the limit of the difference quotient as h tends to zero then,
Therefore, if f (x) = cos x then
The derivative of the exponential function
Let use the limit of the difference quotient to find the derivative of the function f (x) = ax.
Since the difference quotient is
then, the derivative as the limit of the difference quotient as h tends to zero
That is, by plugging t = ah - 1, then t ® 0 as h ® 0, and
Therefore, if f (x) = ax then
or when the base a is substituted by the natural base e obtained is the exponential function ex, thus
if f (x) = ex then
The derivative of the logarithmic function
Let's use the limit of the difference quotient to find the derivative of the function f (x) = loga x.
The difference quotient applied to the given function
As the derivative is the limit of the difference quotient as h tends to zero, then
Then, applying the base change identity and substituting a = e
if f (x) = loga x then
if f (x) = ln x then
Calculus contents C |
Collection of recommendations and tips
# Where is a constant in math?
## Where is a constant in math?
A constant in math is a value that doesn’t change. All numbers are constants. Some letters, like e, or symbols, such as π, are also constants. Additionally, a variable can be a constant if the problem assigns a specific value to it.
## How do you find a constant?
Since k is constant (the same for every point), we can find k when given any point by dividing the y-coordinate by the x-coordinate. For example, if y varies directly as x, and y = 6 when x = 2, the constant of variation is k = = 3. Thus, the equation describing this direct variation is y = 3x.
What are constants in math?
A constant, sometimes also called a “mathematical constant,” is any well-defined real number which is significantly interesting in some way. A function, equation, etc., is said to “be constant” (or be a constant function) if it always assumes the same value independent of how its parameters are varied.
What are the constant numbers?
In mathematics, a constant is a specific number or a symbol that is assigned a fixed value. In other words, a constant is a value or number that never changes in expression. Its value is constantly the same. Examples of constant are 2, 5, 0, -3, -7, 2/7, 7/9 etc.
### What is the constant term in algebra?
From Wikipedia, the free encyclopedia. In mathematics, a constant term is a term in an algebraic expression that does not contain any variables and therefore is constant. For example, in the quadratic polynomial. the 3 is a constant term.
### What do you mean by constant?
: something invariable or unchanging: such as. a : a number that has a fixed value in a given situation or universally or that is characteristic of some substance or instrument.
What is a coefficient in math?
A coefficient is a number multiplied by a variable. Examples of coefficients: In the term 14 c 14c 14c , the coefficient is 14. In the term g, the coefficient is 1.
What are constants?
In math and science, a constant is a number that is fixed and known, unlike a variable which changes with the context. That idea crosses over to real life. If a friend is a constant in your life, that means they have always been with you and there for you.
## What are the constants in an experiment?
A constant is a quantity that does not change. Although you can measure a constant, you either cannot alter it during an experiment or else you choose not to change it. Contrast this with an experimental variable, which is the part of an experiment that is affected by the experiment.
## Where are constants stored in C?
As per the memory layout of C program ,constant variables are stored in the Initialized data segment of the RAM.
How do you find a constant in an experiment?
Where do you find constants in a math problem?
Other constants in math include e, or Euler’s number, which is approximately 2.71828. Often times, constants can be found next to variables. It’s important to distinguish constants from variables when solving math problems, so you’ll know what to solve for, as shown in this example:
### How to use math constants in Microsoft Docs?
To use them, you must first define _USE_MATH_DEFINES and then include cmath or math.h. The file ATLComTime.h includes math.h when your project is built in Release mode. If you use one or more of the math constants in a project that also includes ATLComTime.h, you must define _USE_MATH_DEFINES before you include ATLComTime.h.
### Which is the best definition of a mathematical constant?
From Wikipedia, the free encyclopedia A mathematical constant is a key number whose value is fixed by an unambiguous definition, often referred to by a symbol (e.g., an alphabet letter), or by mathematicians’ names to facilitate using it across multiple mathematical problems.
How to define math constants in C + + project?
Math Constants are not defined in Standard C/C++. To use them, you must first define _USE_MATH_DEFINES and then include cmath or math.h. The file ATLComTime.h includes math.h when your project is built in Release mode. |
# Problems on Percentage
Problems on percentage:
1. A team lost 25 % of the matches it played. If it won 15 matches, find the number of matches it played.
Solution:
Percentage of matches lost = 25 %
Therefore Percentage of matches won (100 - 25) % = 75 %
Let the number of matches played be m.
Then 75 % of m = 15
75/100 × m = 15
m = (15 × 100)/75 %
m = (1500)/75 %
m = 20 %
Therefore, the total number of matches played is 20.
2. In a plot of 6000 sq. m., only 4500 sq. m. is allowed for construction. What percent of the plot is to be left without construction?
Solution:
Percentage of plot allowed for construction = (4500/6000 × 100) % = 75 %.
Thus, the percentage of plot to be left without construction = 100 % - 75 % = 25 %.
3. A number is reduced by 100 %. Its present value is 270. What was its original value?
Solution:
Original value is percentage = 100 %.
Reduce amount in percentage = 10 %
Therefore, Percent value in percentage = 100 % - 10 % = 90 %.
According to the problem,
90 % of original value = 270.
Therefore, 100 % of original value = 270/90 × 100 = 300.
Thus, the original value was 300.
4. A girl is scored 60 out of 75 in English, 60 out of 90 in mathematics and 80 out of 100 in Science. Find girls score as percentage:
(i) in Mathematics
(ii) in all the three subjects (on the whole).
Solution:
(i) Percentage scored in Mathematics = 60/90 × 100 %
= 6000/90 %
= 200/3 %
= 662/3 %
(ii) Total maximum of all the three subjects = 75 + 90 + 100 = 265 and
Total score in the three subjects = 60 + 60 + 80 = 200
Therefore, percentage on the whole = (200/265 × 100) %
= (20000/265) %
= 4000/65 %
= 7525/53 %
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CBSE Class 10 Maths Question Paper | 2018
Class 10 Board Paper Solution | 3 Mark Questions
Section C contains 10 questions of 3 marks each. Scroll down for explanatory answer and video solution to all ten 3-mark questions that appeared in Class 10 Maths CBSE board exam in 2018.
1. Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.
Hint to solve this Real Numbers problem
Step 1: Prime factorize 404 and 96.
Step 2: Compute LCM of 404 and 96. LCM is the product of the highest power of all prime factors.
Step 3: Compute HCF of 404 and 96. HCF is the product of the lowest power of common prime factors of 404 and 96.
Step 4: Find product of 404 and 96. Verify that the value is the same as the product of the LCM and HCF of 404 and 96.
2. Find all zeroes of the polynomial (2x4 – 9x3 + 5x2 + 3x – 1) if two of its zeroes are (2 + $$sqrt3$) and$2 - $$sqrt3$). Hint to solve this Polynomials problem Step 1: Two of its zeroes are given. If the zeroes are p and q, then$x - p)(x - q) will divide the polynomial.
Step 2: Compute the quotient of dividing the given polynomial by (x - p)(x - q)
Step 3: The quotient of the division will be a quadratic expression. Factorize the quadratic expression to find the remaining two zeroes.
3. A(-2, 1), B(a, 0), C(4, b) D(1, 2) are the vertices of a parallelogram ABCD. Find the values of a and b. Hence, find the lengths of its sides.
Hint to solve this Coordinate Geometry Parallelogram problem
Concept: In a parallelogram, the diagonals bisect each other.
Step 1: The diagonals are AC and BD. Using coordinates of A and C, compute coordinates of midpoint of AC.
Step 2: Using coordinates of B and C, compute the coordinates of midpoint of BD.
Step 3: The midpoint of AC and midpoint of Bd are the same point. So equate the cooridantes obtained in step 1 and step 2 to find the values of 'a' and 'b'
Step 4: Now that we have the coordinates of all 4 vertices, compute length of AB and BC to find the lengths of the sides of the parallelogram.
OR
If A(- 5, 7), B(- 4, - 5), C(- 1, - 6) and D(4, 5) are the vertices of a quadrilateral. Find the area of the quadrilateral ABCD.
Hint to solve this Coordinate Geometry problem
Step 1: Join AC. The quadrilateral gets divided into two triangles ABC and ADC.
Step 2: Using coordinates of vertices A, B, and C compute area of triangle ABC.
Step 3: Using coordinates of vertices A, D, and C compute area of triangle ADC.
Step 4: Add areas computed in steps 2 and 3 to compute area of quadrilateral.
4. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed, Find its usual speed.
Hint to solve this Quadratic Equations problem
Step 1: Assign variables for usual speed and usual time taken.
Step 2: Frame equation for distance travelled at usual speed.
Step 3: Frame equation for distance travelled at higher speed when the plane reached its destination in 30 minutes lesser time.
Step 4: Equate expressions in steps 2 and 3 and solve for the usual speed of the plane.
5. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonals.
Hint to solve this Similar Triangles problem
Step 1: Compute ratio of the lengths of the side and the diagonal of a square.
Step 2: Ratio of the sides of the two equilateral triangles (both will be similar by AAA rule of similarity) will be in the ratio computed in step 1.
Step 3: Use theorem "Ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides" to compute ratio of areas of the two triangles.
OR
If the area of two similar triangles are equal, prove that they are congruent.
Hint to solve this Similar & Congruent Triangles problem
Step 1: Use theorem "Ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides".
Step 2: Because their areas are equal, the ratio of their sides, which is the square root of the ratio of the areas will also be equal.
6. Prove that the lengths of tangents drawn from an external point to a circle are equal.
Hint to solve this Circles & Trangents problem
Step 1: The two tangents, corresponding radii drawn from the tangents where they meet the circle and the line joining the centre of the circle and the point where the tangents meet will form two congruent right triangles by RHS test of congruence.
Step 2: By CPCT, we can prove that the tangents are equal in length.
7. If 4 tanθ = 3, evaluate $$frac{\text{4sinθ - cosθ + 1}}{\text{4sinθ + cosθ - 1}}$. Hint to solve this Trigonometry problem Step 1: Compute value of tan θ. Step 2: Compute hypotenuse of the right triangle. Step 3: Compute values of sin θ and cos θ. Step 4: Compute value of the expression given in the question. OR If tan 2A = cot$A - 18°), where 2A is an acute angle, find the value of A.
Hint to solve this Trigonometry - Complementary Angles problem
Step 1: By trigonometric ratios of complementary angles, tan 2A can be written as cot (90° - 2A).
Step 2: Establish that (90° - 2A) and (A - 18°) are acute angles.
Step 3: Because (90° - 2A) and (A - 18°) are acute angles, both will be positive as the angles are in the first quadrant.
Step 4: Equate (90° - 2A) and (A - 18°) to find the value of A.
8. Find the area of the shaded region in figure given below, where areas drawn with centres A, B, C, and D intersect in pairs at mid – points P, Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm.
Hint to solve this Areas Related to Circles & Squares problem
Step 1: Area of shaded region is the difference between the area of the square and the sum of the areas of the 4 sectors.
Step 2: Compute area of the given square.
Step 3: The sectors subtend an angle of 90°. The radii of all 4 sectors are equal. So, they are 4 equal quarter circles. Sum of their area equals area of 1 circle. Compute the area of the resultant circle.
Step 4: Difference between value obtained in step 2 and step 3 is the answer to the question.
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the given figure. If the height of the cylinder is 10 cm and its base is of radius 3. 5 cm. Find the total surface area of the article.
Hint to solve this Surface Areas & Volumes problem
Step 1: Total Surface Area of wooden article is the sum of the CSA of the cylinder and the CSA of the two hemispheres scooped from the top and bottom of the cylinder.
Step 2: Compute Curved Surface Area (CSA) of the cylinder.
Step 3: Compute CSA of the two equal hemispheres.
Step 4: Compute Total Surface Area of the wooden article by adding values obtained in step 2 and 2 times the value obtained in step 3.
OR
A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap ?
Hint to solve this Surface Areas & Volumes problem
Step 1: Using base radius and height compute volume of cone.
Step 2: Compute slant height using base radius and height of the cone.
Step 3: Compute curved surface area (CSA) of cone to find area of canvas required to cover the heap.
10. The table below shows the salaries of 280 persons. Calculate the median salary of the data.
Salary (In thousand) No. of Persons
5 - 10 49
10 - 15 133
15 - 20 63
20 - 25 15
25 - 30 6
30 - 35 7
35 - 40 4
40 - 45 2
45 - 50 1
Hint to solve this Statistics problem
Step 1: Compute cumulative frequency for the given data.
Step 2: Identify the class interval in which the median lies.
Step 3: Use the formula to compute median for the given data.
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# AP Statistics Curriculum 2007 Hypothesis Basics
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## General Advance-Placement (AP) Statistics Curriculum - Fundamentals of Hypothesis Testing
### Fundamentals of Hypothesis Testing
A (statistical) Hypothesis Test is a method of making statistical decisions about populations or processes based on experimental data. Hypothesis testing just answers the question of how well the findings fit the possibility that the chance alone might be responsible for the observed discrepancy between the theoretical model and the empirical observations. This is accomplished by asking and answering a hypothetical question. What is the likelihood of the observed summary statistics of interest, if the data did come from the distribution specified by the null-hypothesis? One use of hypothesis-testing is deciding whether experimental results contain enough information to cast doubt on conventional wisdom.
• Example: Consider determining whether a suitcase contains some radioactive material. Placed under a Geiger counter, the suitcase produces 10 clicks (counts) per minute. The null hypothesis is that there is no radioactive material in the suitcase and that all measured counts are due to ambient radioactivity typical of the surrounding air and harmless objects in a suitcase. We can then calculate how likely it is that the null hypothesis produces 10 counts per minute. If it is likely, for example if the null hypothesis predicts on average 9 counts per minute, we say that the suitcase is compatible with the null hypothesis (which does not imply that there is no radioactive material, we just can't determine from the 1-minute sample we took using this specific method!); On the other hand, if the null hypothesis predicts for example 1 count per minute, then the suitcase is not compatible with the null hypothesis and there must be other factors responsible to produce the increased radioactive counts.
The Hypothesis Testing is also known as Statistical Significance Testing. The null hypothesis is a conjecture that exists solely to be disproved, rejected or falsified by the sample-statistics used to estimate the unknown population parameters. Statistical significance is a possible finding of the test, that the sample is unlikely to have occurred in this process by chance given the truth of the null hypothesis. The name of the test describes its formulation and its possible outcome. One characteristic of hypothesis testing is its crisp decision about the null-hypothesis: reject or do not reject (which is not the same as accept).
### Null and Alternative (Research) Hypotheses
A Null Hypothesis is a thesis set up to be nullified or refuted in order to support an Alternate (research) Hypothesis. The null hypothesis is presumed true until statistical evidence, in the form of a hypothesis test, indicates otherwise. In science, the null hypothesis is used to test differences between treatment and control groups, and the assumption at the outset of the experiment is that no difference exists between the two groups for the variable of interest (e.g., population means). The null hypothesis proposes something initially presumed true, and it is rejected only when it becomes evidently false. That is, when a researcher has a certain degree of confidence, usually 95% to 99%, that the data do not support the null hypothesis.
### Example 1: Gender effects
If we want to compare the test scores of two random samples of men and women, a null hypothesis would be that the mean score of the male population was the same as the mean score of the female population:
H0 : μmen = μwomen
where:
H0 = the null hypothesis
μmen = the mean of the males (population 1), and
μwomen = the mean of the females (population 2).
Alternatively, the null hypothesis can postulate that the two samples are drawn from the same population, so that the center, variance and shape of the distributions are equal.
Formulation of the null hypothesis is a vital step in testing statistical significance. Having formulated such a hypothesis, one can establish the probability of observing the obtained data from the prediction of the null hypothesis, if the null hypothesis is true. That probability is what commonly called the significance level of the results.
In many scientific experimental designs we predict that a particular factor will produce an effect on our dependent variable — this is our alternative hypothesis. We then consider how often we would expect to observe our experimental results, or results even more extreme, if we were to take many samples from a population where there was no effect (i.e. we test against our null hypothesis). If we find that this happens rarely (up to, say, 5% of the time), we can conclude that our results support our experimental prediction — we reject our null hypothesis.
### Type I Error, Type II Error and Power
Directly related to hypothesis testing are the following 3 concepts:
• Type I Error: The false positive (Type I) Error of rejecting the null hypothesis given that it is actually true; e.g., A court finding a person guilty of a crime that they did not actually commit.
• Type II Error: The Type II Error (false negative) is the error of failing to reject the null hypothesis given that the alternative hypothesis is actually true; e.g., A court finding a person not guilty of a crime that they did actually commit.
• Statistical Power: The Power of a Statistical Test is the probability that the test will reject a false null hypothesis (that it will not make a Type II Error). As power increases, the chances of a Type II error decrease. The probability of a Type II error is referred to as the false negative rate (β). Therefore power is equal to 1 − β. You can also see this SOCR Power Activity.
Actual condition
Absent (Ho is true) Present (H1 is true)
Test
Result
Negative
(fail to reject Ho
Condition absent + Negative result = True (accurate) Negative (TN, 0.98505) Condition present + Negative result = False (invalid) Negative (FN, 0.00025)
Type II error $$(\beta)$$
Positive
(reject Ho)
Condition absent + Positive result = False Positive (FP, 0.00995)
Type I error $$(\alpha)$$
Condition Present + Positive result = True Positive (TP, 0.00475)
Test
Interpretation
Power = 1-FN=
1-0.00025 = 0.99975
Specificity: TN/(TN+FP) =
0.98505/(0.98505+ 0.00995) = 0.99
Sensitivity: TP/(TP+FN) =
0.00475/(0.00475+ 0.00025)= 0.95
• Remarks:
• A Specificity of 100% means that the test recognizes all healthy individuals as (normal) healthy. The maximum is trivially achieved by a test that claims everybody is healthy regardless of the true condition. Therefore, the specificity alone does not tell us how well the test recognizes positive cases.
• False positive rate (α)= $$\frac{FP}{FP+TN} = \frac{0.00995}{0.00995 + 0.98505}=0.01$$= 1 - Specificity.
• Sensitivity is a measure of how well a test correctly identifies a condition, whether this is medical screening tests picking up on a disease, or quality control in factories deciding if a new product is good enough to be sold. $Sensitivity=\frac{TP}{TP+FN} =\frac{0.00475}{0.00475+ 0.00025}= 0.95.$
• False Negative Rate (β)= $$\frac{FN}{FN+TP} = \frac{0.00025}{0.00025+0.00475}=0.05$$= 1 - Sensitivity.
• Power = 1 − β= 0.95. Power $= P(\mbox{reject null hypothesis} | \mbox{null hypothesis is false})$. For example, see Power_Analysis_for_Normal_Distribution.
• Both (Type I ($\alpha$) and Type II ($\beta$)) errors are proportions in the range [0,1], so they represent error-rates. The reason they are listed in the corresponding cells in the table is that they are directly proportionate to the numerical values of the FP and FN, respectively.
• The two alternative definitions of power are equivalent:
power$=1-\beta$, and
power=sensitivity
This is because power=$1-\beta=1-\frac{FN}{FN+TP}=\frac{FN+TP}{FN+TP} - \frac{FN}{FN+TP}=-\frac{TP}{FN+TP}=$ sensitivity.
### Example 2: Sodium content in hot-dogs
Use the Hot-dog dataset to see if there are statistically significant differences in the sodium content of the poultry vs. meat hotdogs.
• Formulate Hypotheses: $$H_o: \mu_p = \mu_m$$ vs. $$H_1: \mu_p \not= \mu_m$$, where $$\mu_p, \mu_m$$ represent the mean sodium content in poultry and mean hotdogs.
• Plug in the data in SOCR Analyses under the Two Independent Sample T-Test (Unpooled) will generate results as shown in the figure below (Two-Sided P-Value (Unpooled) = 0.196, which does not provide strong evidence to reject the null hypothesis that the two types of hot-dogs have the same mean sodium content).
### Example 3: Rapid testing in strep-throat
This study investigated the accuracy of rapid diagnosis of group A $$\beta$$-streptococcal pharyngitis by commercial immunochemical antigen test kits in the setting of recent streptococcal pharyngitis. Specifically, it explored whether the false-positive rate of the rapid test was increased because of presumed antigen persistence.
Study used 443 patients who had clinical pharyngitis diagnosed as group A $$\beta$$-hemolytic streptococcus infection in the past 28 days and compared them with 232 control patients who had symptoms of pharyngitis but no recent diagnosis of streptococcal pharyngitis. The aim was narrowly focused to compare the rapid strep test with the culture method used in clinical practice.
The study found that the rapid strep test in this setting showed no difference in specificity (0.96 vs. 0.98). Hence, the assertion that rapid antigen testing had higher false-positive rates in those with recent infection was not confirmed. It also found that in patients who had recent streptococcal pharyngitis, the rapid strep test appears to be more reliable (sensitivity 0.91 vs 0.70, P < .001) than in those patients who had not had recent streptococcal pharyngitis. These findings indicated that the rapid strep test is both sensitive and specific in the setting of recent group A $$\beta$$-hemolytic streptococcal pharyngitis, and its use might allow earlier treatment in this subgroup of patients.
Table 1. Sensitivity and Specificity of Laboratory Culture and Rapid Strep Test in Patients With Recently Treated Cases of Streptococcal Pharyngitis (N=443).
Results Culture Negative Culture Positive
Rapid strep test negative 93 10
Rapid strep test positive 4 104
Estimate 95% CI
Sensitivity 104/(104+10) = 0.91 0.84, 0.96
Specificity 93/(93+4) = 0.96 0.90, 0.99
Positive predictive value 0.96 0.91, 0.99
Negative predictive value 0.90 0.83, 0.95
False-positive rate 0.04 0.01, 0.10
False-negative rate 0.09 0.04, 0.15
Table 2. Sensitivity and Specificity of Laboratory Culture and Rapid Strep Test in Patients With No Recently Treated Cases of Streptococcal Pharyngitis (N=232).
Results Culture Negative Culture Positive
Rapid strep test negative 165 19
Rapid strep test positive 4 44
Estimate 95% CI
Sensitivity 44/(44+19) = 0.70 0.57, 0.81
Specificity 168/(165+4) = 0.98 0.94, 0.99
Positive predictive value 0.92 0.80, 0.99
Negative predictive value 0.90 0.84, 0.94
False-positive rate 0.02 0.01, 0.06
False-negative rate 0.30 0.19, 0.43
### References
Robert D. Sheeler, MD, Margaret S. Houston, MD, Sharon Radke, RN, Jane C. Dale, MD, and Steven C. Adamson, MD. (2002) Accuracy of Rapid Strep Testing in Patients Who Have Had Recent Streptococcal Pharyngitis. JABFP, 2002, 15(4), 261-265.
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# Intervals and neighborhoods, extended real line
In this section we introduce basic tools for working on the real line and the extended real line.
In analysis we often talk about things being close. The key notion when talking about closeness is the distance. On a real line we use the usual distance, when the distance between points a and b is given by |a − b|. Note that the inequality |a − b| > 0 is equivalent to saying that a and b are not equal. We will use this soon.
Definition.
Consider real numbers a < b. We define
open interval (a,b) = {x∈ℝ; a < x < b},
closed interval [a,b] = {x∈ℝ; a ≤ x ≤ b},
half-closed interval (or half-open interval) [a,b) = {x∈ℝ; a ≤ x < b},
half-closed interval (or half-open interval) (a,b] = {x∈ℝ; a < x ≤ b}.
We also define
open interval (a,∞) = {x∈ℝ; a < x},
open interval (−∞,b) = {x∈ℝ; x < b},
closed interval [a,∞) = {x∈ℝ; a ≤ x},
closed interval (−∞,b] = {x∈ℝ; x ≤ b},
The points a, b are called endpoints of these intervals, we talk of intervals from a to b. In the last four cases also the infinity or negative infinity is considered an endpoint.
To have a better idea of properties of points and sets we usually draw them on a real line. For intervals we use filled dots for closed endpoints and empty circles for open endpoints. In the same order as in the definition:
Intervals are very popular, since they have many useful properties. Probably the most important property is that they are connected sets, that is, if we have two points from an interval I, then we can pass from one to another without ever getting out of I. In fact, every connected subset of a real line is an interval. We will see in the next part where the open/closed terminology came from. We will also see that those intervals with infinity are in fact nothing special, but for that we will have to extend real numbers. By the way, talking about infinity, the set of all real numbers is also an interval, namely (−∞,∞).
There are two kinds of rather funny intervals that are not exactly welcome in many applications, namely for any a we have (a,a) = {} and [a,a] = {a}. Analogous half-closed intervals are also empty. These empty and one-point intervals are called degenerate, all the other intervals are called non-degenerate. In many applications we only want to work with non-degenerate intervals, since these have non-empty interior (see the next section).
Intervals are important because many methods we will use only work on intervals, so they are, in a sense, our workplace. Now we will introduce another kind of a set which will be our working tool in analysis.
Definition.
Consider a real number a and some ε > 0.
We define the ε-neighborhood of a as
Uε(a) = {x∈ℝ; |x-a| < ε}.
We define the reduced ε-neighborhood of a as
Pε(a) = {x∈ℝ; 0 < |x-a| < ε}.
By a neighborhood of a we mean an ε-neighborhood of a for some ε > 0. By a reduced neighborhood of a we mean a reduced ε-neighborhood of a for some ε > 0.
The difference is that a neigborhood contains its reference point a, while a reduced neighborhood does not. These neighborhoods can be also expressed using intervals, for instance
Uε(a) = (a − ε,a + ε); Pε(a) = (a − ε,a) ∪ (a,a + ε).
We use neighborhoods to express some important ideas. In many situations we need to know what is happening just next to a given point, and this is best expressed using neighborhoods, as we will see below. Sometimes we do not care what happens on both sides of a, but we only worry about one side. For that we use one-sided neighborhoods.
Definition.
Consider a real number a and some ε > 0.
We define the right ε-neighborhood of a as
Uε+(a) = [a,a + ε).
We define the left ε-neighborhood of a as
Uε-(a) = (a − ε,a].
We define the right reduced ε-neighborhood of a as
Pε+(a) = (a,a + ε).
We define the left reduced ε-neighborhood of a as
Pε-(a) = (a − ε,a).
For instance, the left neighborhood looks like this:
### Extended real line.
In some situations it pays off to treat infinity as just another object. For instance, when we ask about a limit at some particular number a and about a limit at infinity, it turns out that the procedures used in answering such questions are basically the same. Thus it saves time if we can talk of some a, which can be a real number but also infinity, or perhaps negative infinity as the two "ends" of the real line.
We therefore introduce the extended real line (denoted ℝ*) as the real line plus two abstract elements ∞ and −∞. Points of this set now fall into two categories, real numbers are proper points and the two elements ±∞ are improper points.
In order to be able to work with extended reals we need to extend also the operations and ordering. We define that for every real number a we have −∞ < a < ∞. We also put |−∞| = ∞. Concerning operations the situation is a bit delicate. We have to define them in such a way that they make sense and fit fell with operations we already have. We naturally arrive at the following:
• Addition: ∞ + ∞ = ∞, ∞ + a = ∞ for any real a, ∞ − a = ∞ for any real a, −∞ + a = −∞ for any real a, −∞ − a = −∞ for any real a.
• Multiplication: ∞ ⋅ ∞ = ∞, ∞ ⋅ a = ∞ for any real a > 0 and ∞ ⋅ a = −∞ for any real a < 0, −∞ ⋅ a = −∞ for any real a > 0 and −∞ ⋅ a = ∞ for any real a < 0, ∞ / a = ∞ for any real a > 0 and ∞ / a = −∞ for any real a < 0, −∞ / a = −∞ for any real a > 0 and −∞ / a = ∞ for any real a < 0. We also set −∞ = (−1) ⋅ ∞.
• a = ∞ for any a > 0, a = 0 for any a < 0, a = ∞ for any real a > 1, a = 0 for any a satisfying |a| < 1.
Note that some operations are not listed, for instance ∞ − ∞, 0 ⋅ ∞, ∞ / ∞, 1 etc. These operations cannot be defined to make sense, this will be seen when working with limits (see for instance the note on indeterminate expressions).
To take the advantage of extended real numbers requires that we also generalize the notions that we had above. First, note that in the definition of intervals we listed four separate cases with infinite endpoints, but now that we have extended real numbers they are not needed. We can simply use appropriate definitions from among the first four and apply them also to cases when one or both endpoints are improper. Note however that the infinities themselves are never elements of intervals (intervals only include real numbers), in other words, we never close intervals at their improper endpoints.
The next important notion is neighborhood and there we will have to do some extra work. What do we expect of neighborhoods? They are sets that are connected and extend to the left and to the right of the given point a. With infinities this is not possible, since we can go to only one side, so we will have to make a modification here. In fact, for neighborhoods of improper points the both-sided and one-sided notions of neighborhood coincide, we do not have much choice there. Also, the distinction between neighborhoods and reduced neighborhoods becomes irrelevant, since a neighborhood of infinity (or minus infinity) will contain only real numbers and therefore they are automatically reduced.
Finally, a neighborhood of a proper point is given by a parameter ε that says how far one can go from the given point. This does not help us when a is improper, since every real number is infinitely far from infinity. Therefore we use another point of view at neighborhoods of proper points. If we take all possible neighborhoods of a proper point a and let ε go to zero (from the right, since ε can only be positive), then the ends of neighborhoods go to the center a. And this is something that we can emulate for neighborhoods of infinity and negative infinity. Now we are ready to understand the following definition.
Definition.
Consider ε > 0.
We define the ε-neighborhood of ∞ as
Uε(∞) = (1/ε,∞).
We define the ε-neighborhood of −∞ as
Uε(−∞) = (−∞,−1/ε).
Now we can work with neighborhoods and we do not really care whether the reference point a is proper or improper.
Once we can use infinity as a point, many definitions that would have to be done twice can be done just once. For a simple example see supremum at the end of the next section, a big simplification appears in the definition of the limit, see Definition of a limit in Functions - Theory - Limit of a function. |
# Distributive property and binomial multiplication
### Algebra 1
The distributive property can be used even when there are two sets of parentheses with two terms each. It’s called binomial multiplication (remember that a bicycle has two wheels and a binomial has two terms).
Binomial Multiplication:
Notice that ???a??? is multiplied by both terms in the second set of parentheses and then ???b??? is multiplied by both terms in the second set of parentheses.
We can also make a chart in which the terms ???a??? and ???b??? from the first set of parentheses go across the top, and the terms ???c??? and ???d??? from the second set of parentheses go along the left side. Then we multiply each row by each column to get a result. The four results all get added together to make the expanded polynomial.
When we add all the results in the chart together, we get
When we have negative signs in the binomials, we keep the negative sign with the term that follows it, and our chart looks like
When we add all the results in the chart together, we get
These charts are another way to keep track of the different multiplications that happen during binomial multiplication.
Example
Use the distributive property to expand the expression.
???5(x-2)(x+3)???
Start by distributing the ???5??? across ???x-2???.
???[5(x)-5(2)](x+3)???
???(5x-10)(x+3)???
Now distribute both of the terms in the first set of parentheses across both of the terms in the second set of parentheses. You may use a chart to help organize your work.
When we add all the results in the chart together, we get
???5x^2+15x-10x-30???
Combine like terms ???15x-10x???.
???5x^2+5x-30???
The distributive property can be used even when there are two sets of parentheses with two terms each. It’s called binomial multiplication
Let’s try another example of binomial multiplication.
Example
Use the distributive property to expand the expression.
???3x(x+4)(x+1)(x-2)???
Start by distributing the ???3x??? across ???x+4???.
???(3x^2+12x)(x+1)(x-2)???
Now distribute ???3x^2+12x??? across ???x+1???. You may use a chart to help organize your work.
When we add all the results in the chart together, we get
???3x^3+3x^2+12x^2+12x???
Combine like terms ???3x^2+12x^2???.
???3x^3+15x^2+12x???
Then distribute ???3x^3+15x^2+12x??? across ???x-2???. You may use a chart to help organize your work.
When we add all the results in the chart together, we get
???3x^4+15x^3+12x^2-6x^3-30x^2-24x???
Combine like terms ???15x^3-6x^3??? and ???12x^2-30x^2???.
???3x^4+9x^3-18x^2-24x??? |
# SSAT Upper Level Math : How to find the equation of a perpendicular line
## Example Questions
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
What line is perpendicular to x + 3y = 6 and travels through point (1,5)?
y = 3x + 2
y = 6x – 3
y = 2x + 1
y = 2/3x + 6
y = –1/3x – 4
y = 3x + 2
Explanation:
Convert the equation to slope intercept form to get y = –1/3x + 2. The old slope is –1/3 and the new slope is 3. Perpendicular slopes must be opposite reciprocals of each other: m1 * m2 = –1
With the new slope, use the slope intercept form and the point to calculate the intercept: y = mx + b or 5 = 3(1) + b, so b = 2
So y = 3x + 2
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
What line is perpendicular to and passes through ?
Explanation:
Convert the given equation to slope-intercept form.
The slope of this line is . The slope of the line perpendicular to this one will have a slope equal to the negative reciprocal.
The perpendicular slope is .
Plug the new slope and the given point into the slope-intercept form to find the y-intercept.
So the equation of the perpendicular line is .
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
What is the equation of a line that runs perpendicular to the line 2x + = 5 and passes through the point (2,7)?
2x – y = 6
2x + y = 7
x/2 – y = 6
x/2 + y = 6
x/2 + y = 5
x/2 + y = 6
Explanation:
First, put the equation of the line given into slope-intercept form by solving for y. You get y = -2x +5, so the slope is –2. Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6. Thus, the equation of the line is y = ½x + 6. Rearranged, it is –x/2 + y = 6.
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
Line m passes through the points (1, 4) and (5, 2). If line p is perpendicular to m, then which of the following could represent the equation for p?
x y = 3
2x + y = 3
3x + 2y = 4
2x y = 3
4x 3y = 4
2x y = 3
Explanation:
The slope of m is equal to y2-y1/x2-x1 = 2-4/5-1 -1/2
Since line p is perpendicular to line m, this means that the products of the slopes of p and m must be 1:
(slope of p) * (-1/2) = -1
Slope of p = 2
So we must choose the equation that has a slope of 2. If we rewrite the equations in point-slope form (y = mx + b), we see that the equation 2x y = 3 could be written as y = 2x – 3. This means that the slope of the line 2x – y =3 would be 2, so it could be the equation of line p. The answer is 2x – y = 3.
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
What is the equation for the line that is perpendicular to through point ?
Explanation:
Perpendicular slopes are opposite reciprocals.
The given slope is found by converting the equation to the slope-intercept form.
The slope of the given line is and the perpendicular slope is .
We can use the given point and the new slope to find the perpendicular equation. Plug in the slope and the given coordinates to solve for the y-intercept.
Using this y-intercept in slope-intercept form, we get out final equation: .
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
Which line below is perpendicular to ?
Explanation:
The definition of a perpendicular line is one that has a negative, reciprocal slope to another.
For this particular problem, we must first manipulate our initial equation into a more easily recognizable and useful form: slope-intercept form or .
According to our formula, our slope for the original line is . We are looking for an answer that has a perpendicular slope, or an opposite reciprocal. The opposite reciprocal of is . Flip the original and multiply it by
Our answer will have a slope of . Search the answer choices for in the position of the equation.
(As an aside, the negative reciprocal of 4 is . Place the whole number over one and then flip/negate. This does not apply to the above problem, but should be understood to tackle certain permutations of this problem type where the original slope is an integer.)
### Example Question #7 : How To Find The Equation Of A Perpendicular Line
If a line has an equation of , what is the slope of a line that is perpendicular to the line?
Explanation:
Putting the first equation in slope-intercept form yields .
A perpendicular line has a slope that is the negative inverse. In this case, .
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
Given a line defined by the equation , which of the following lines is perpendicular to ?
Explanation:
For a given line defined by the equation , any line perpendicular to must have a slope that is the negative reciprocal of 's slope .
In this instance, the slope of line is , so . The only line provided with an equation that has this slope is
### Example Question #2 : How To Find The Equation Of A Perpendicular Line
A given line is defined by the equation . What is the slope of any line that is perpendicular to ?
Not enough information provided
Explanation:
For a given line defined by the equation , any line perpendicular to must have a slope that is the negative reciprocal of 's slope .
Since in this case
.
### Example Question #11 : How To Find The Equation Of A Perpendicular Line
Which of the following equations represents a line that goes through the point and is perpendicular to the line ?
Explanation:
In order to solve this problem, we need first to transform the equation from standard form to slope-intercept form:
Transform the original equation to find its slope.
First, subtract from both sides of the equation.
Simplify and rearrange.
Next, divide both sides of the equation by 6.
The slope of our first line is equal to . Perpendicular lines have slopes that are opposite reciprocals of each other; therefore, if the slope of one is x, then the slope of the other is equal to the following:
Let's calculate the opposite reciprocal of our slope:
The slope of our line is equal to 2. We now have the following partial equation:
We are missing the y-intercept, . Substitute the x- and y-values in the given point to solve for the missing y-intercept.
Add 4 to both sides of the equation.
Substitute this value into our partial equation to construct the equation of our line: |
# Test: Ratio And Proportion- 3
## 15 Questions MCQ Test Practice Questions for GMAT | Test: Ratio And Proportion- 3
Description
Attempt Test: Ratio And Proportion- 3 | 15 questions in 15 minutes | Mock test for GMAT preparation | Free important questions MCQ to study Practice Questions for GMAT for GMAT Exam | Download free PDF with solutions
QUESTION: 1
### Which of the following fractions is at least twice as great as 11/50?
Solution:
First, let us rephrase the question. Since we need to find the fraction that is at least twice greater than 11/50, we are looking for a fraction that is equal to or greater than 22/50. Further, to facilitate our analysis, note that we can come up with an easy benchmark value for this fraction by doubling both the numerator and the denominator and thus expressing it as a percent: 22/50 = 44/100 = 44%. Thus, we can rephrase the question: “Which of the following is greater than or equal to 44%?”
Now, let’s analyze each of the fractions in the answer choices using benchmark values:
2/5: This fraction can be represented as 40%, which is less than 44%.
11/34: This value is slightly less than 11/33 or 1/3. Therefore, it is smaller than 44%.
43/99: Note that the fraction 43/99 is smaller than 44/100, since fractions get smaller if the same number (in this case integer 1) is subtracted from both the numerator and the denominator.
8/21: We know that 8/21 is a little less than 8/20 or 2/5. Thus, 8/21 is less than 44%.
9/20: Finally, note that by multiplying the numerator and the denominator by 5, we can represent this fraction as 45/100, thus concluding that this fraction is greater than 44%.
QUESTION: 2
### At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4. During the year, 10 juniors and twice as many seniors transferred to another high school, while no new students joined high school X. If, at the end of the year, the ratio of juniors to seniors was 4 to 5, how many seniors were there in high school X at the beginning of the year?
Solution:
Let’s denote the number of juniors and seniors at the beginning of the year as j and s, respectively. At the beginning of the year, the ratio of juniors to seniors was 3 to 4:
j/s = 3/4. Therefore, j = 0.75s
At the end of the year, there were (j - 10) juniors and (s - 20) seniors. Additionally, we know that the ratio of juniors to seniors at the end of the year was 4 to 5. Therefore, we can create the following equation:
Let’s solve this equation by substituting j = 0.75s:
(j - 10) = 0.8(s - 20)
(0.75s - 10) = 0.8s - 16
0.8s - 0.75s = 16 - 10
0.05s = 6
s = 120
Thus, there were 120 seniors at the beginning of the year.
QUESTION: 3
### 3/5 of a certain class left on a field trip. 1/3 of the students who stayed behind did not want to go on the field trip (all the others did want to go). When another vehicle was located, 1/2 of the students who did want to go on the field trip but had been left behind were able to join. What fraction of the class ended up going on the field trip?
Solution:
For a fraction question that makes no reference to specific values, it is best to assign a smart number as the "whole value" in the problem. In this case we'll use 30 since that is the least common denominator of all the fractions mentioned in the problem.
If there are 30 students in the class, 3/5 or 18, left for the field trip. This means that 12 students were left behind.
1/3 of the 12 students who stayed behind, or 4 students, didn't want to go on the field trip.
This means that 8 of the 12 who stayed behind did want to go on the field trip.
When the second vehicle was located, half of these 8 students or 4, were able to join the other 18 who had left already.
That means that 22 of the 30 students ended up going on the trip. 22/30 reduces to 11/15 so the correct answer is C.
QUESTION: 4
The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If the number of boys in Class B is one less than the number of boys in Class A, and if the number of girls in Class B is two less than the number of girls in Class A, how many girls are in Class A?
Solution:
The ratio of boys to girls in Class A is 3 to 4. We can represent this as an equation: b/g = 3/4. We can isolate the boys:
4b = 3g
b = (3/4)g
Let's call the number of boys in Class B x, and the number of girls in Class B y
We know that the number of boys in Class B is one less than the number of boys in Class A. Therefore, x = b – 1. We also know that the number of girls in Class B is two less than the number of girls in Class A. Therefore, y = g – 2. We can substitute these in the combined class equation:
The combined class has a boy/girl ratio of 17 to 22: (b + x)/(g + y) = 17/22.
(b + b – 1)/(g + g – 2) = 17/22
(2b – 1)/(2g – 2) = 17/22
Cross-multiplying yields:
44b – 22 = 34g – 34
Since we know that b = (3/4)g, we can replace the b:
44(3/4)g – 22 = 34g – 34
33g – 22 = 34g – 34
12 = g
Alternatively, because the numbers in the ratios and the answer choices are so low, we can try some real numbers. The ratio of boys to girls in Class A is 3:4, so here are some possible numbers of boys and girls in Class A:
B:G
3:4
6:8
9:12
The ratio of boys to girls in Class B is 4:5, so here are some possible numbers of boys and girls in Class A:
B:G
4:5
8:10
12:15
We were told that there is one more boy in Class A than Class B, and two more girls in Class A than Class B. If we look at our possibilities above, we see that this information matches the case when we have 9 boys and 12 girls in Class A and 8 boys and 10 girls in Class B. Further, we see we would have 9 + 8 = 17 boys and 12 + 10 = 22 girls in a combined class, so we have the correct 17:22 ratio for a combined class. We know now there are 12 girls in Class A.
QUESTION: 5
John's front lawn is 1/3 the size of his back lawn. If John mows 1/2 of his front lawn and 2/3 of his back
lawn, what fraction of his lawn is left unmowed?
Solution:
We can solve this problem by choosing a smart number to represent the size of the back lawn. In this case, we want to choose a number that is a multiple of 2 and 3 (the denominators of the fractions given in the problem). This way, it will be easy to split the lawn into halves and thirds. Let's assume the size of the back lawn is 6.
(size back lawn) = 6 units
(size front lawn) = (1/3)(size back lawn) = 2 units
(size total lawn) = (size back lawn) + (size front lawn) = 8 units
Now we can use these numbers to calculate how much of each lawn has been mowed:
Front lawn: (1/2)(2) = 1 unit
Back lawn: (2/3)(6) = 4 unit
So, in total, 5 units of lawn have been mowed. This represents 5/8 of the total, meaning 3/8 of the lawn is left unmowed.
Alternatively, this problem can be solved using an algebraic approach. Let's assume the size of the front lawn is x and size of the back lawn is y. So, John has mowed (1/2)x and (2/3)y, for a total of (1/2)x + (2/3)y. We also know that x = (1/3)y. Substituting for x gives:
(1/2)x + (2/3)y
(1/2)(1/3)y + (2/3)y
(1/6)y + (2/3)y
(5/6)y = lawn mowed
The total lawn is the sum of the front and back, x + y. Again, substituting for x gives (1/3)y + y = (4/3)y. So, the fraction of the total lawn mowed is:
= (5/6) × (3/4) = 15/24 = 5/8.
This leaves 3/8 unmowed.
QUESTION: 6
At Jefferson Elementary School, the number of teachers and students (kindergarten through sixth grade) totals 510. The ratio of students to teachers is 16 to 1. Kindergarten students make up 1/5 of the student population and fifth and sixth graders account for 1/3 of the remainder. Students in first and second grades account for 1/4 of all the students. If there are an equal number of students in the third and fourth grades, then the number of students in third grade is how many greater or fewer than the number of students in kindergarten?
Solution:
We know that the student to teacher ratio at the school is 16 to 1, and the total number of people is 510. Therefore:
Number of students = (16/17)(510) = 480
Number of teachers = (1/17)(510) = 30
Kindergarten students make up 1/5 of the student population, so:
Number of kindergarten students = (1/5)(480) = 96
Fifth and sixth graders account for 1/3 of the remainder (after kindergarten students are subtracted from the total), therefore:
Number of 5th and 6th grade students = (1/3)(480 – 96) = (1/3)(384) = 128
Students in first and second grades account for 1/4 of all the students, so:
Number of 1st and 2nd grade students = (1/4)(480) = 120
So far, we have accounted for every grade but the 3rd and 4th grades, so they must consist of the students left over:
Number of 3rd and 4th grade students = Total students – students in other grades
Number of 3rd and 4th grade students = 480 – 96 – 128 – 120 = 136
If there are an equal number of students in the third and fourth grades, then:
Number of 3rd grade students = 136/2 = 68
The number of students in third grade is 68, which is fewer than 96, the number of students in kindergarten. The number of students in 3rd grade is thus 96 – 68 = 28 fewer than the number of kindergarten students.
QUESTION: 7
A certain galaxy is known to comprise approximately 4 x 1011 stars. Of every 50 million of these stars, one is larger in mass than our sun. Approximately how many stars in this galaxy are larger than the sun?
Solution:
50 million can be represented in scientific notation as 5 x 107. Restating this figure in scientific notation will enable us to simplify the division required to solve the problem. If one out of every 5 x 107 stars is larger than the sun, we must divide the total number of stars by this figure to find the solution:
= 4/5 x 10(11-7)
= 0.8 x 104
The final step is to move the decimal point of 0.8 four places to the right, with a result of 8,000.
QUESTION: 8
A lemonade stand sold only small and large cups of lemonade on Tuesday. 3/5 of the cups sold were small and the rest were large. If the large cups were sold for 7/6 as much as the small cups, what fraction of Tuesday's total revenue was from the sale of large cups?
Solution:
For a fraction word problem with no actual values for the total, it is best to plug numbers to solve.
Since 3/5 of the total cups sold were small and 2/5 were large, we can arbitrarily assign 5 as the number of cups sold.
Total cups sold = 5
Small cups sold = 3
Large cups sold = 2
Since the large cups were sold at 7/6 as much per cup as the small cups, we know:
Pricelarge = (7/6)Pricesmall
Let's assign a price of 6 cents per cup to the small cup.
Price of small cup = 6 cents
Price of large cup = 7 cents
Now we can calculate revenue per cup type:
Large cup sales = quantity × cost = 2 × 7 = 14 cents
Small cup sales = quantity × cost = 3 × 6 = 18 cents
Total sales = 32 cents
The fraction of total revenue from large cup sales = 14/32 = 7/16.
QUESTION: 9
Miguel is mixing up a salad dressing. Regardless of the number of servings, the recipe requires that 5/8 of the finished dressing mix be olive oil, 1/4 vinegar, and the remainder an even mixture of salt, pepper and sugar. If Miguel accidentally doubles the vinegar and forgets the sugar altogether, what proportion of the botched dressing will be olive oil?
Solution:
This problem can be solved most easily by picking smart numbers and assigning values to the portion of each ingredient in the dressing. A smart number in this case would be one that enables you to add and subtract ingredients without having to deal with fractions or decimals. In a fraction problem, the ‘smart number’ is typically based on the least common denominator among the given fractions.
The two fractions given, 5/8 and 1/4, have a least common denominator of 8. However, we must also consider the equal parts salt, pepper and sugar. Because 1/4 = 2/8, the total proportion of oil and vinegar combined is 5/8 + 2/8 = 7/8. The remaining 1/8 of the recipe is split three ways: 1/24 each of salt, pepper, and sugar. 24 is therefore our least common denominator, suggesting that we should regard the salad dressing as consisting of 24 units. Let’s call them cups for simplicity, but any unit of measure would do. If properly mixed, the dressing would consist of
5/8 × 24 = 15 cups of olive oil
1/4 × 24 = 6 cups of vinegar
1/24 × 24 = 1 cup of salt
1/24 × 24 = 1 cup of sugar
1/24 × 24 = 1 cup of pepper
Miguel accidentally doubled the vinegar and omitted the sugar. The composition of his bad salad dressing would therefore be
15 cups of olive oil
12 cups of vinegar
1 cup of salt
1 cup of pepper
The total number of cups in the bad dressing equals 29. Olive oil comprises 15/29 of the final mix.
QUESTION: 10
Harold and Millicent are getting married and need to combine their already-full libraries. If Harold, who has 1/2 as many books as Millicent, brings 1/3 of his books to their new home, then Millicent will have enough room to bring 1/2 of her books to their new home. What fraction of Millicent's old library capacity is the new home's library capacity?
Solution:
This problem never tells us how many books there are in any of the libraries. We can, therefore, pick numbers to represent the quantities in this problem. It is a good idea to pick Smart Numbers, i.e. numbers that are multiples of the common denominator of the fractions given in the problem.
In this problem, Harold brings 1/3 of his books while Millicent brings 1/2. The denominators, 2 and 3, multiply to 6, so let's set Harold's library capacity to 6 books. The problem tells us Millicent has twice as many books, so her library capacity is 12 books. We use these numbers to calculate the size of the new home's library capacity. 1/3 of Harold's 6-book library equals 2 books. 1/2 of Millicent's 12-book library equals 6 books. Together, they bring a combined 8 books to fill their new library.
The fraction we are asked for, (new home's library capacity) / (Millicent's old library capacity), therefore, is 8/12, which simplifies to 2/3.
QUESTION: 11
In a certain pet shop, the ratio of dogs to cats to bunnies in stock is 3 : 5 : 7. If the shop carries 48 cats and bunnies total in stock, how many dogs are there?
Solution:
The ratio of dogs to cats to bunnies (Dogs: Cats: Bunnies) can be expressed as 3x: 5x : 7x. Here, x represents an "unknown multiplier." In order to solve the problem, we must determine the value of the unknown multiplier.
Cats + Bunnies = 48
5x + 7x = 48
12x = 48
x = 4
Now that we know that the value of x (the unknown multiplier) is 4, we can determine the number of dogs.
Dogs = 3x = 3(4) = 12
QUESTION: 12
A foreign language club at Washington Middle School consists of n students, 2/5 of whom are boys. All of the students in the club study exactly one foreign language. 1/3 of the girls in the club study Spanish and 5/6 of the remaining girls study French. If the rest of the girls in the club study German, how many girls in the club, in terms of n, study German?
Solution:
Boys = 2n/5, girls = 3n/5
3n
Girls studying Spanish =
3n
Girls not studying Spanish =
2n
Girls studying French =
Girls studying German = (all girls) – (girls studying Spanish) – (girls studying French)
3n
Girls studying German =
2n
Girls studying German =
QUESTION: 13
A certain ball team has an equal number of right- and left-handed players. On a certain day, two-thirds of the players were absent from practice. Of the players at practice that day, one-third were left handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of lefthanded players who were not at practice?
Solution:
Since the problem deals with fractions, it would be best to pick a smart number to represent the number of ball players. The question involves thirds, so the number we pick should be divisible by 3. Let's say that we have 9 right-handed players and 9 left-handed players (remember, the question states that there are equal numbers of righties and lefties).
Two-thirds of the players are absent from practice, so that is (2/3)(18) = 12. This leaves 6 players at practice. Of these 6 players, one-third were left-handed. This yields (1/3)(6) = 2 left-handed players at practice and 9 – 2 = 7 left-handed players NOT at practice. Since 2 of the 6 players at practice are lefties, 6 – 2 = 4 players at practice must be righties, leaving 9 – 4 = 5 righties NOT at practice.
The question asks us for the ratio of the number of righties not at practice to the number of lefties not at practice. This must be 5 : 7 or 5/7.
QUESTION: 14
Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A?
Solution:
We are told that bag B contains red and white marbles in the ration 1:4. This implies that WB, the number of white marbles in bag B, must be a multiple of 4.
What can we say about WA, the number of white marbles in bag A? We are given two ratios involving the white marbles in bag A. The fact that the ratio of red to white marbles in bag A is 1:3 implies that WA must be a multiple of 3. The fact that the ratio of white to blue marbles in bag A is 2:3 implies that WA must be a multiple of 2. Since WA is both a multiple of 2 and a multiple of 3, it must be a multiple of 6.
We are told that WA + WB = 30. We have already figured out that WA must be a multiple of 6 and that WB must be a multiple of 4. So all we need to do now is to test each candidate value of WA (i.e. 6, 12, 18, and 24) to see whether, when plugged into WA + WB = 30, it yields a value for WB that is a multiple of 4. It turns out that WA = 6 and WA = 18 are the only values that meet this criterion.
Recall that the ratio of red to white marbles in bag A is 1:3. If there are 6 white marbles in bag A, there are 2 red marbles. If there are 18 white marbles in bag A, there are 6 red marbles. Thus, the number of red marbles in bag A is either 2 or 6. Only one answer choice matches either of these numbers.
QUESTION: 15
The ratio by weight, measured in pounds, of books to clothes to electronics in Jorge's suitcase initially stands at 8 to 5 to 3. Jorge then removes 4 pounds of clothing from his suitcase, thereby doubling the ratio of books to clothes. Approximately how much do the electronics in the suitcase weigh, to the nearest pound?
Solution:
Initially the ratio of B: C: E can be written as 8x: 5x: 3x. (Recall that ratios always employ a common multiplier to calculate the actual numbers.)
After removing 4 pounds of clothing, the ratio of books to clothes is doubled. To double a ratio, we double just the first number; in this case, doubling 8 to 5 yields a new ratio of 16 to 5. This can be expressed as follows:
Cross multiply to solve for x:
40x = 80x – 64
40x = 64
x = 8/5
The question asks for the approximate weight of the electronics in the suitcase. Since there are 3x pounds of electronics there are 3 × (8/5) = 24/5 or approximately 5 pounds of electronics in the suitcase. |
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# Which of the following quadratic polynomial having zeros $1$ and $-2$: A) ${{x}^{2}}-x+2$ B) ${{x}^{2}}-x-2$C) ${{x}^{2}}+x-2$D) None of these
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Hint: The given question is related to quadratic equations. Try to recall the formulae related to the relation between the coefficients and sum and product of the roots of a quadratic equation.
Before proceeding with the solution, we must know about the relation between the coefficients and sum and product of the roots of the quadratic equation given by $a{{x}^{2}}+bx+c=0$ .
We know, the roots of the equation $a{{x}^{2}}+bx+c=0$ are given by the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Let $\alpha$ and $\beta$ be the roots of the equation. So, $\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$ and $\beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$. The sum of the roots is given as $\alpha +\beta =\left( \dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \right)+\left( \dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \right)=\dfrac{-2b}{2a}=\dfrac{-b}{a}$ .
So, the sum of the roots is related to the coefficients as $\alpha +\beta =\dfrac{-b}{a}$ .
The product of the roots is given as $\alpha \beta =\left( \dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \right)\left( \dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \right)=\dfrac{{{b}^{2}}-\left( {{b}^{2}}-4ac \right)}{4{{a}^{2}}}=\dfrac{c}{a}$ .
So, the product of the roots is related to the coefficients as $\alpha \beta =\dfrac{c}{a}$ .
Now, we have $a{{x}^{2}}+bx+c=0$. On dividing the equation by $a$ , we get ${{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=0.....(i)$.
We have $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ . So, we can rewrite equation $(i)$ with coefficients in the form sum and product of roots as ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0$.
Now, coming to the question , we are given the zeros of a quadratic polynomial as $1$ and $-2$. So, the sum of zeroes is equal to $-2+1=-1$ and the product of zeroes is equal to $-2\times 1=-2$ .
Hence, the quadratic polynomial having zeros $1$ and $-2$ is given as ${{x}^{2}}+x-2$ .
Note: The quadratic equation with coefficients in the form sum and product of roots is given as ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0$ and not ${{x}^{2}}+\left( \alpha +\beta \right)x+\alpha \beta =0$. Students often get confused and make a mistake. Such mistakes should be avoided as they can lead to wrong answers. |
# What is the limit of (1 / (4x - 4x^2) ) - (1 / (4x - 5x^2) ) as x approaches 0?
Apr 14, 2016
$- \frac{1}{16}$
#### Explanation:
You wish to find
${\lim}_{x \rightarrow 0} \left(\frac{1}{4 x - 4 {x}^{2}} - \frac{1}{4 x - 5 {x}^{2}}\right)$
Note first that plugging in $0$ for $x$ will cause fractions with denominators of $0$, so we can't determine the limit without doing some manipulation.
First, factor the fractions' denominators:
$= {\lim}_{x \rightarrow 0} \left(\frac{1}{x \left(4 - 4 x\right)} - \frac{1}{x \left(4 - 5 x\right)}\right)$
Now, find a common denominator of $x \left(4 - 4 x\right) \left(4 - 5 x\right)$.
$= {\lim}_{x \rightarrow 0} \left(\frac{4 - 5 x}{x \left(4 - 4 x\right) \left(4 - 5 x\right)} - \frac{4 - 4 x}{x \left(4 - 4 x\right) \left(4 - 5 x\right)}\right)$
Combine the fraction.
$= {\lim}_{x \rightarrow 0} \frac{\left(4 - 5 x\right) - \left(4 - 4 x\right)}{x \left(4 - 4 x\right) \left(4 - 5 x\right)}$
Simplify the numerator.
$= {\lim}_{x \rightarrow 0} \frac{4 - 5 x - 4 + 4 x}{x \left(4 - 4 x\right) \left(4 - 5 x\right)}$
$= {\lim}_{x \rightarrow 0} \frac{- x}{x \left(4 - 4 x\right) \left(4 - 5 x\right)}$
Cancel the $x$ term in the numerator and denominator.
$= {\lim}_{x \rightarrow 0} \frac{- 1}{\left(4 - 4 x\right) \left(4 - 5 x\right)}$
We can now evaluate the limit by plugging in $0$ for $x$.
$= \frac{- 1}{\left(4 - 4 \left(0\right)\right) \left(4 - 5 \left(0\right)\right)} = \frac{- 1}{\left(4 - 0\right) \left(4 - 0\right)} = - \frac{1}{16}$
If we graph the original function, we should see that the graph approaches $- \frac{1}{16}$ at $x = 0$, even though the function is undefined at $x = 0$:
graph{1/(4x-4x^2)-1/(4x-5x^2) [-1.662, 2.664, -1.718, 0.445]} |
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In the given figure, the points A, B, C, D are concyclic, when $x = {80^ \circ }, z = y - {20^ \circ }$ then $z = ?$A. ${100^ \circ }$B. ${80^ \circ }$C. ${40^ \circ }$D. ${140^ \circ }$
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Hint: We use the property of concyclic points that they all lie on the circle and joining these points forms a quadrilateral. Using the property of the sum of opposite interior angles of a quadrilateral we calculate the value of y and substitute in to find the value of z.
We know points $A,B,C,D$ are concyclic, which means they all lie on the circle.
Now joining the points on the circle we can form a quadrilateral $ABCD$.
From the property of opposite angles in a quadrilateral we know that the sum of opposite pairs of angles in a quadrilateral is equal to ${180^ \circ }$.
Now from the diagram we see that pairs of opposite angles are $\angle ADC,\angle ABC$ and $\angle BCD,\angle BAD$.
Now we will consider the set of opposite angles $\angle ADC,\angle ABC$.
So, using the property we can write
$\Rightarrow \angle ADC + \angle ABC = {180^ \circ }$
We know the value of $\angle ADC = y,\angle ABC = {80^ \circ }$
$\Rightarrow y + {80^ \circ } = {180^ \circ }$
Shift all constant terms in degree to the right hand side of the equation.
$\Rightarrow y = {180^ \circ } - {80^ \circ } \\ \Rightarrow y = {100^ \circ } \\$
Therefore the value of $\angle ADC = {100^ \circ }$
Now we have to find the value of $z$ from the equation $z = y - {20^ \circ }$
We will substitute the value of $y = {100^ \circ }$ in the equation.
$\Rightarrow z = {100^ \circ } - {20^ \circ } = {80^ \circ }$
Therefore, value of $z = {80^ \circ }$
So, the correct answer is “Option B”.
Note: Students many times make the mistake of writing the final answer as the value of y as in figure there is no mention of z, so they look at the variable and write its value, but keep in mind the question is asking for the value of z. Many students write their final answer without a degree sign, always writing the symbol along with the value. |
# Applications of Linear Systems
## Word problems using two equations and two unknowns
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Point of Intersection
Learning Goal
By the end of this lesson you will be able to determine the point of intersection in context and state when to choose each scenario.
What if you were playing a game in which you collected houses and hotels. Three houses and one hotel are worth $1750. One house and three hotels are worth$3250. How could you find the value of each house and each hotel? After completing this Concept, you'll be able to solve real-world applications like this one that involve linear systems.
### Guidance
In this section we will use Real Life Applications of Linear Relations to determine under what conditions we should choose each scenario.
#### Example A
The movie rental store CineStar offers customers two choices. Customers can pay a yearly membership of $45 and then rent each movie for$2 or they can choose not to pay the membership fee and rent each movie for 3.50. Determine under which conditions each option would be a better choice. Solution Let’s translate this problem into algebra. Since there are two different options to consider, we can write two different equations and form a system. The choices are “membership” and “no membership.” We’ll call the number of movies you rent x\begin{align*}x\end{align*} and the total cost of renting movies for a year y\begin{align*}y\end{align*}. flat fee rental fee total membership45 2x\begin{align*}2x\end{align*} y=45+2x\begin{align*}y = 45 + 2x\end{align*}
no membership 0 3.50x\begin{align*}3.50x\end{align*} y=3.5x\begin{align*}y = 3.5x\end{align*} The flat fee is the dollar amount you pay per year and the rental fee is the dollar amount you pay when you rent a movie. For the membership option the rental fee is 2x\begin{align*}2x\end{align*}, since you would pay2 for each movie you rented; for the no membership option the rental fee is 3.50x\begin{align*}3.50x\end{align*}, since you would pay 3.50 for each movie you rented. Our two equations are: y=45+2xy=3.50x\begin{align*}y = 45 + 2x\!\\ y = 3.50x\end{align*} Here’s a graph of the system: Now we need to find the exact intersection point. We can use the graph to find the Point of Intersection. The Point of Intersection occurs where the two lines cross each other. Looking at the graph you can see at 30 movies per year both options would cost105.
We need to report both numbers.
Now we need to decide when each option would be a better choice.
BEFORE the point of intersection:
If you are going to rent less than 30 movies than being a Non-Member would cost less and therefore be the better choice.
AFTER the point of intersection:
If you are going to rent more than 30 movies than being a Member would cost less and therefore be the better choice.
AT the point of intersection:
### Notes/Highlights Having trouble? Report an issue.
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Sum of the Interior Angles
Curriculum Goal
Junior: Geometry and Spatial Sense
• Identify geometric properties of triangles and construct different types of triangles when given side or angle measurements.
Junior: Measurement
• Use the properties of supplementary angles, complementary angles, opposite angles, and interior and exterior angles to solve for unknown angle measures.
Context
• Group-based exploratory activity to explore the relationship between the sums of interior angles in different polygons.
Materials
• Chart paper with a triangle in the middle
• Various hexagon cut-outs (Appendix A)
• Printout with all hexagons on one page (Appendix B)
• Heptagon and decagon (Appendix C)
• Organization table (Appendix D)
• Protractors (for students to check their work)
• Markers
Lesson
• Divide the class into six groups and provide each group with a piece of chart paper.
• Have students discuss their pre-existing knowledge of angles and write down their thoughts. Encourage students to think about angles in the context of polygons.
• Some students will remember that a triangle has three angles that add up to 180°. This fact is necessary to solve the next challenge.
• Help groups that do not have that understanding by giving them several construction paper triangles. Demonstrate that by ripping off the corners of a triangle and placing them side-by-side with the vertices meeting at one point, the edge created is a straight line.
• Introduce polygons (a closed figure with at least three straight sides and three angles) with visuals to support the instruction. Show examples, such as the triangle on their paper, and include relevant terminology (e.g., hexagon – polygon with six sides, interior angles – the angles that are inside the closed figure).
• Distribute a hexagon cut-out (Appendix A) to each group. Tell students they will work together to find a strategy to determine the sum of its interior angles without a protractor.
• Potential strategy: split the hexagon into triangles. Because a triangle’s interior angles add up to 180°, the sum of the hexagon’s interior angles is the number of triangles × 180°. This is also the case for the sum of the interior angles of a quadrilateral.
• Ask groups that finish early to test out their method on the other groups’ hexagons.
• Assist groups that are having difficulty. Strategies:
• If students are having difficulty using the polygons they have drawn inside their hexagon to find a solution, ask, “Do all of your drawn polygons’ internal angles also make up the hexagon’s interior angles? Is there a way you could make that so?”
• Suggest they line their interior angles with a colour to help visualize the problem.
• Invite groups to share the sum they determined, their strategy, and any problems they encountered.
• Provide each group with a copy of Appendix B to help students visualize the angles while listening to their peers.
• Students should recognize that every group had a hexagon with its interior angles totalling 720°.
• Repeat the exercise and discussion with heptagons and decagons (Appendix C). Remind students of their previous strategies to determine whether there is a pattern that allows them to find the solution for any polygon.
• Ask struggling groups, “Do you notice a pattern between the number of sides in your polygon and the sum of its interior angles? How could that pattern help you find the sum of any polygon’s interior angles quickly?
• Not all groups will arrive at the same understanding – listening to other groups and allow students to consolidate the concepts that have been explored.
• Help students recognize patterns by inviting them to complete the organization table (Appendix D). Students can be provided the other polygons if needed.
Look Fors
• Are students able to explain why the sum of a triangle’s corners is 180°? Do students use this knowledge to discover the sum of the angles in a hexagon?
• Are students using any other strategies to understand the sum of the angles in a hexagon? How about heptagons and decagons?
• Can students recognize the pattern between the number of sides in a polygon and the sum of its interior angles?
Extension
• Have students use their pattern to write an algorithm that can be applied to any polygon to determine the sum of its interior angles. |
## Intermediate Algebra (12th Edition)
$x=\left\{ -6,9 \right\}$
Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the given equation, $\left|\dfrac{2}{3}x-1\right|=5 ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{2}{3}x-1=5 \text{ OR } \dfrac{2}{3}x-1=-5 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{2}{3}x-1=5 \\\\ 3\left(\dfrac{2}{3}x-1\right)=(5)3 \\\\ 2x-3=15 \\\\ 2x=15+3 \\\\ 2x=18 \\\\ x=\dfrac{18}{2} \\\\ x=9 \\\\\text{ OR }\\\\ 3\left(\dfrac{2}{3}x-1\right)=(-5)3 \\\\ 2x-3=-15 \\\\ 2x=-15+3 \\\\ 2x=-12 \\\\ x=-\dfrac{12}{2} \\\\ x=-6 .\end{array} Hence, the solutions are $x=\left\{ -6,9 \right\} .$ |
Computers and Technology
# What is Standard Form?
If you’re questioning what standard form in math is, you’re proper place.
Standard shape is used to lessen the issue in analyzing very massive or very small numbers. Standard shape of a number of is any variety among 1.zero and 10.zero improved via way of means of electricity 10,
For example, 1.2× 102
For an indication of widespread shape, test beneath examples:
Number: 85500000000000
Standard shape: 8.55 × 1013
Number: zero.000458912
Standard shape: 4.58 × 10-4
The fashionable shape calculator is used to transform the numbers into fashionable shape with the aid of using putting the decimal fee with inside the range. It converts a protracted range into an effortlessly readable fashionable shape.
It is a write in fashionable shape calculator which takes the range from the person and converts to traditional shape.
In this content, we can give an explanation for what fashionable shape are, the way to use our fashionable shape calculator, and the way to calculate fashionable shape as well.
## How to use standard form calculator?
To use this fashionable notation calculator, comply with the under steps:
• Enter the range with inside the given enter box.
• Press the Calculate button to peer the result.
• You can reset the values with the aid of using the use of the Reset
This fashionable shape equation calculator will right away display you the transformed fashionable shape of the given range. You also can use our medical notation calculator and medical notation converter to calculate the medical notations. And you can use this calculator by clicking the below link.
https://www.allmath.com/standard-form.php
### How to write in Standard Form?
You can use our convert to traditional shape calculator to calculate the usual shape of any variety. However, we are able to give an explanation for how you could convert various to traditional shape manually. To convert various to traditional shape, observe the under steps:
1. Write down the variety.
2. Identify the decimal factor with inside the variety. If there’s no decimal factor with inside the given variety, it’s far taken into consideration as on the proper facet of the variety after the closing digit.
3. After figuring out the decimal factor, flow the decimal to the first non-0 digit with inside the variety.
4. Count the overall range of digits you’ve got moved the decimal factor. Multiply the range with 10 and lift the electricity of 10 with the overall range of digits decimal have moved. If the decimal is moved from proper to left, electricity can be positive, and if the decimal is moved from left to proper, electricity can be negative.
Example:
Convert 0.0009 to the same old shape.
Solution:
Follow the stairs to discover the same old shape of the given range.
Step 1: Write down the range.
0.0009
Step 2: Identify the decimal factor with inside the range. You can see the decimal factor is mendacity after four digits from the left side.
Step 3: After figuring out the decimal factor, flow the decimal to the first non-0 digit with inside the range.
It will become 9. Because there’s no non-0 digit after 9, we don’t want to write the decimal factor after 9.
Step 4: Count the whole variety of digits you’ve got moved the decimal factor. We have moved the decimal four locations further. Multiply the variety with 10 and lift the strength of 10 with the whole variety of digits decimal have moved. As we’ve got moved the decimal factor from left to right, the strength could be negative.
9 × 10-four
So, the usual shape of the variety 0.0009 is 9 × 10-four. |
## Solve Exercise 3. Combination (C5 – Math 10 Kite)
### Solution of Exercises Lesson 1, page 17, Math Textbook 10, Kite episode 2
Given 8 points such that no 3 points are collinear. How many triangles with 3 vertices are 3 out of 8 given points?
Solution method
+) A triangle is made up of 3 non-collinear points, so to have a triangle we will choose 3 out of 8 non-collinear points.
=> triangle number is a convolution of 3 of 8 elements.
Solution guide
The number of triangles with 3 vertices is 3 of the given 8 points which is a convolution of 3 of 8 elements, so the number of triangles is: $$C_8^3$$ (triangle)
### Solve the exercise Exercise 2 page 17 Math textbook 10 Kite episode 2
There are 10 teams participating in a Soccer Exercise Tournament. How many ways are there to arrange a round-robin match so that the two teams meet exactly once?
Solution method
+) To have a match, there must be 2 teams participating. Therefore, to have a match, we will choose 2 teams out of 10 teams.
+) So the number of matches is a convolution 2 of 10 elements.
Solution guide
The number of ways to arrange the score round match so that the two teams only meet exactly once is a convolution 2 of 10 elements, so the number of ways to arrange the match is: $$C_{10}^2 = 45$$ ( arrangement)
### Solution of Exercises Lesson 3, page 17, Math Textbook 10, Kite episode 2
Grade 10 has 16 girls and 18 boys participating in the Green Summer volunteering. The school delegation plans to set up a tree planting team consisting of 3 students, both male and female. How many ways are there to set up such a tree nest?
Solution method
*) Analysis: (Number of ways to choose 3hs including both men and women in 34hs) + (Number of ways to choose 3hs for men in 34hs) + (Number of ways to choose 3hs for women in 34hs) = Number of ways to choose any 3hs in 34hs.
+) Step 1: Calculate the number of ways to choose any 3hs in 34hs.
+) Step 2: Calculate the number of ways to choose 3 male students in 34 hours.
+) Step 3: Calculate the number of ways to choose 3 female students in 34 hours.
+) Step 4: Consider the difference to calculate the number of ways to choose 3hs including both men and women in 34hs.
Solution guide
+) The number of ways to choose any 3hs in 34hs is: $$C_{34}^3$$ (how to choose)
+) Number of ways to choose 3 male students in 34 hours is: $$C_{18}^3$$ (choose method)
+) The number of ways to choose 3 female students in 34 hours is: $$C_{16}^3$$ (how to choose)
+) The number of ways to choose 3hs including both men and women in 34hs is: $$C_{34}^3 – C_{18}^3 – C_{16}^3 = 4608$$ (how to choose)
### Solving Exercises Lesson 4, page 17, Math Textbook 10 Kite, episode 2
A small shop selling flowers has 50 roses and 60 daisies. Uncle Ngoc wants to buy 5 flowers including both of the above flowers. How many ways does Uncle Ngoc have to choose flowers?
Solution method
*) Analysis: ( Number of ways to choose 5 flowers including 2 types out of 110 flowers) + (Number of ways to choose 5 roses out of 50 roses) + (Number of ways to choose 5 daisies out of 60 daisies) = Number How to choose any 5 flowers out of 110 flowers.
+) Step 1: Calculate the number of ways to choose any 5 flowers out of 110 flowers.
+) Step 2: Calculate the number of ways to choose 5 chrysanthemums out of 60 chrysanthemums.
+) Step 3: Calculate the number of ways to choose 5 roses out of 50 roses.
+) Step 4: Consider the difference to calculate the number of ways to choose 5 flowers including 2 types out of 110 flowers.
Solution guide
+) The number of ways to choose any 5 flowers out of 110 flowers is: $$C_{110}^5$$ (how to choose)
+) Number of ways to choose 5 chrysanthemums out of 60 chrysanthemums is: $$C_{60}^5$$ (how to choose)
+) The number of ways to choose 5 roses out of 50 roses is: $$C_{50}^5$$ (how to choose)
+) The number of ways to choose 5 flowers including 2 types out of 110 flowers is: $$C_{110}^5 – C_{60}^5 – C_{50}^5$$ (how to choose)
### Solve the exercise Exercise 5 page 17 Math textbook 10 Kite episode 2
Sum $$C_{15}^{12} + C_{15}^{13} + C_{16}^{14}$$
Solution method
$$C_{n – 1}^{k – 1} + C_{n – 1}^k = C_n^k\left( {1 \le k < n} \right)$$
Solution guide
We have: $$(C_{15}^{12} + C_{15}^{13} )+ C_{16}^{14} = C_{16}^{13} + C_{16}^{14 } = C_{17}^{14} = 680$$ |
# Demystifying 12.8 As A Fraction: Converting And Simplifying The Decimal Value
Demystifying 12.8 As A Fraction: Converting And Simplifying The Decimal Value. Numbers can be expressed in different forms, including whole numbers, decimals, and fractions. Fractions offer a precise way to represent parts of a whole or a ratio. In this blog post, we will explore the process of converting and simplifying 12.8 as a fraction. By understanding the steps involved, we can unravel the fraction’s true representation and its significance in mathematical calculations. Let’s dive into the world of fractions and explore 12.8 as a fraction!
## Understanding The Basics Of Fractions
Before we delve into converting 12.8 into a fraction, let’s refresh our understanding of fractions. Fractions are mathematical expressions that represent a part of a whole or a ratio. They consist of a numerator and a denominator, with the numerator indicating the number of parts and the denominator indicating the total number of equal parts in the whole.
## Converting 12.8 Into A Fraction
To convert 12.8 into a fraction, we need to consider the digits after the decimal point. The digit 1 is in the tens place, and the digit 2 is in the tenths place. To express 12.8 as a fraction, we can write it as the sum of its whole number part and the fractional part. The whole number part is 12, and the fractional part is 8/10.
## Simplifying The Fraction
Once we have expressed 12.8 as a fraction, we can simplify it if possible. To simplify a fraction, we divide both the numerator and the denominator by their greatest common divisor (GCD). In this case, the GCD of 8 and 10 is 2. Dividing both the numerator and denominator by 2, we get 4/5. Therefore, 12.8, when expressed as a fraction, simplifies to 4/5.
## Understanding The Value Of 12.8 As 4/5
Representing 12.8 as the fraction 4/5 provides us with a deeper understanding of its value. The fraction 4/5 signifies that we have 4 parts out of a total of 5 equal parts. This fraction can be interpreted in various contexts. For example, if we consider a rectangular shape divided into 5 equal sections, having 4/5 of the shape means we have occupied four sections.
## Applications And Significance
Understanding fractions, including the conversion of decimals like 12.8, is essential in various fields. Fractions play a crucial role in mathematics, science, engineering, and finance. They allow for precise representation, comparison, and calculations of values. Fractional form is particularly useful in ratio analysis, proportions, and problem-solving scenarios that involve part-to-whole relationships.
You can search for more about similar topics like these on Tipsfeed.
## FAQ
### What Is 12.8 As A Decimal?
Shortcut: When you divide anything by 100, just move the decimal point two places to the left. Here is the next percent in our How-to Catalog that we converted to a decimal. Remember This: 12.8 as a decimal is 0.128 and you can multiply 0.128 by a number to get 12.8 percent of that number.
### What Is 12.92 As A Fraction?
12.92 as a fraction is 323/25.
### Is 12.5 In Fraction 1 8?
Therefore, the fraction 18 is equivalent to 12.5% .
### What Is 12.5 As A Fraction Out Of 100?
The whole number to produce 12.5 and 100 can be different. In this case the highest common factor is 12.5 as 12.5 times 1 is 12.5 and 12.5 times 8 is 100. Now you divide the top and bottom of the fraction by 12.5 to produce 1/8.
## Conclusion:
In conclusion, expressing 12.8 as a fraction, specifically as 4/5, provides us with a more accurate representation and a deeper understanding of its value. Fractions serve as powerful tools in a wide range of applications. By understanding fractions and their conversion from decimals, we enhance our mathematical skills and broaden our problem-solving abilities. So, the next time you come across a decimal like 12.8, remember that it can be represented as the fraction 4/5.
I Have Covered All The Following Queries And Topics In The Above Article
What Is 12.8 As A Fraction
-12.8 As A Fraction
12.8 As A Mixed Fraction
12.8 As A Decimal
12.8 As A Percent
12.8 Inches
64/5 Simplified
64/5 As A Mixed Number
12.8 Inches On A Ruler
64/5 As A Fraction
12.8 As A Fraction
What is 12.8 converted into a fraction in simplest form
How to convert 12.8 as a fraction? |
Tough Maths problem
This is a very tough question.
Try this and see if you can solve it without using algebra.
Andy, Betty, Candy and Danny each have some money.
The amount of money Andy has is 1/3 of the total amount of money Betty, Candy and Danny have. The amount of money Betty has is 1/4 of the total amount of money Andy, Candy and Danny have. The amount of money Candy has is 1/5 of the total amount of money Andy, Betty and Danny have. If Danny has \$1725, how much do they have altogether?
Here’s the solution.
Step 1: Observe that if Andy has 1/3 of the amount of money as the rest, then the total must be divisible by 4 parts (so that Andy gets 1/4 and the rest gets 3/4.)
Step 2: Similarly, if Betty gets 1/4 of the rest, then the total must be divisible by 5 parts (so that she gets 1/5 and the rest gets 4/5). Similarly, the total is also divisible by 6 parts so that Candy gets 1/6 of the total.
Step 3: We need a number of units to represent the total. This number must be divisible by 4, 5 and 6. The smallest number for that is 60.
Step 4: Andy has 1/4 of the total, so that is 15 units (1/4 * 60 = 15). Betty has 1/5 of the total, so that is 12 units (1/5 * 60 = 12). Candy has 1/6 of the total, and that is 10 units (1/6 * 60 = 10). This means that Danny has 23 units (60-15-12-10 = 23).
Step 5: 23 units is \$1725. 1 unit is \$75. They have altogether 60 units = 60 * \$75 = \$4500. |
# PROBABILITY! Let’s learn about probability and chance!
## Presentation on theme: "PROBABILITY! Let’s learn about probability and chance!"— Presentation transcript:
PROBABILITY! Let’s learn about probability and chance!
What is probability? Probability is the measure of how likely an event or outcome is. Different events have different probabilities!
How do we describe probability? You can describe the probability of an event with the following terms: –certain (the event is definitely going to happen) –likely (the event will probably happen, but not definitely) –unlikely (the event will probably not happen, but it might) –impossible (the event is definitely not going to happen) Can you think of examples of each type of event?
How do we express probabilities? Usually, we express probabilities as fractions. –The numerator is the number of ways the event can occur. –The denominator is the number of possible events that could occur. Let’s look at an example!
What is the probability the spinner will land on the number 3? 1 4 3 2
Ask yourself the following questions: 1. How many 3’s are on the spinner? 2. How many possible numbers could the spinner land on? 1 4 3 2 1 4
What is the probability the die will land on an even number? Remember, a die has six sides. Numbers 1, 2, 3, 4, 5, and 6 are each depicted once on the die.
Ask yourself the following questions: 1. How many even numbers are on the die? 2. How many possible numbers could the die land on? 3 6
What is the probability that I will choose a red marble? In this bag of marbles, there are: –3 red marbles –2 white marbles –1 purple marble –4 green marbles
Ask yourself the following questions: 1. How many red marbles are in the bag? 2. How many marbles are in the bag IN ALL? 3 10
Probability Is Fun! Next time you’re playing a board game or a carnival game, think about the probability of the situation! In the next few days, you will be conducting probability experiments and designing probability games! Get ready to use your probability power! |
Math Central - mathcentral.uregina.ca
An Arc Midpoint Computation Lesson
Gregory V. Akulov, teacher,
Luther College High School, Regina
and
Oleksandr (Alex) G. Akulov, student,
University of Waterloo
February 25, 2011
Resource Room
The well-known midpoint formula tells that, if the straight line’s segment has the ends at $x = a, b,$ and midpoint at $x = \mu$, then $2 \mu = a + b \mbox{ or } \mu = \large \frac{a + b}{2}.$ It also tells that the same relationship holds for y-values. Circular analogue of midpoint formula is a new topic.
# Arc Midpoint Computation
Let origin-centered arc of radius $r$ (see Figure 2) have the ends at $x = a, b,$ and midpoint at $x = \mu.$ Then
$2 \small \mu = \pm \sqrt{(r + a)(r + b)} \pm \sqrt{(r - a)(r - b)}\; ,$
where the first radical gets “−” iff1 the arc has a negative x-intercept, and
the second radical gets “+” iff the arc has a positive x-intercept.
### The same relationship holds for y-values.
#### Examples
A . Origin-centered arc of radius 50, located as shown at Figure 2, has the ends at $x = 14$ and $x = 25$. Find x-coordinate of its midpoint.
B . An arc of radius 40 is centered at origin. It starts and ends a $x = \;‒24$ and $x = 9$. What is x-coordinate of arc’s midpoint, if the arc is located in quadrants I and II?
C . An arc has center at $(0, 0)$ and radius 82. It starts at $y = 18$ in quadrant II, passes through quadrants III and ends in quadrant IV at $y = \;‒1$. What is y-value of arc’s midpoint?
#### Solutions
A . $r = 50, a = 14, b = 25,$ both radicals go with “+”, and $20 \sqrt{3} + 15$ is the answer;
B
. $r = 40, a = ‒24, b = 9,$ first radical goes with “+”, second radical goes with “−”, and $14 - 4 \sqrt{31}$ is the answer;
C .
$r = 82, a = 18, b = \;‒1,$ both radicals go with “−”, and $-45 - 4 \sqrt{83}$ is the answer.
Arc midpoint computation was suggested by first attempt shown in [1]. Details of proof involved identities received from [2].
References
1. Oleksandr (Alex) G. Akulov (2009). Dot product finds arc midpoint. Math News, U of W. Volume 111, Issue 6, p. 5.
2. Gregory V. Akulov (2010). The slope of the angle bisector relationship in applied and theoretical problems. Vinculum, SMTS. Volume 2, Number 1, p. 49.
______________________
1The word iff first appeared in 1950s as a shortening of conjunction if and only if.
Copyright © January, 2011 by Gregory V. Akulov, Oleksandr (Alex) G. Akulov
# Arc Midpoint Calculator 1
Oleksandr G. Akulov, MM in OR, BMath, Vancouver, British Columbia
Gregory V. Akulov, teacher, Luther College High School, Regina, Saskatchewan
Cartesian x-values of the endpoints, a, b, and the centre, c,
a = b = c =
If the arc has a point with x-value cr, enter −1, otherwise 1
If the arc has a point with x-value c + r, enter 1, otherwise −1
μ =
Irrelevant output indicates inaccurate input.
To calculate the y-value assume y instead of x in the above.
Copyright © May 19, 2016 by Oleksandr G. Akulov, Gregory V. Akulov
Math Central is supported by the University of Regina and the Pacific Institute for the Mathematical Sciences.
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# Functions and Sets: Get Help Understanding Them
• cam875
In summary, understanding functions using sets can be helpful in fully grasping their meaning. Sets can be used as a foundation for defining other mathematical objects, such as functions, and understanding this relationship can provide a deeper understanding of mathematics. Sets are the building blocks of mathematics, and by using them, we can define and understand various mathematical concepts.
cam875
Ive never been taught functions using sets and I have been told that to really understand the real meaning of them it is helpful to work them out and understand them with sets and stuff. I am not sure if I am on the right track or confused but I am sure someone here can help me out. I have a basic understanding of sets and elements and all that so i should be able to follow along.
I am not sure how often the set definition of functions is helpful. There is also a notion of a function inducing a function from the power set of the domain into the power set of the range. It is helpful to first give the definition of a relation.
Definition: Catesian Product
Let A and B be sets. AxB is called the Cartesian product of A and B.
AxB contains all ordered pairs of the form (a,b) were a and be are respectively elements of A and B.
Definition: relation from A to B (set version)
Let A and B be sets.
R is a relation from A to B if R is a subset of AxB
Definition: Domain (of a relation)
Let R be a relation from A to B
The domain of R, Dom(R) is the set of all elements a in A such that there exists a b in B so that
(a,b) is in R.
Definition: range (of a relation)
Let R be a relation from A to B
The range of R, Rng(R) is the set of all elements b in B such that there exists a in A so that
(a,b) is in R.
Definition: function from A to B (set version)
let A and B be sets. A function f from A to B is a relation from A to B such that for each a in A there exist exactly one b in B such that (a,b) is in f.
Often is is helpful to break this condition in two parts.
Dom(f)=A (for each a in A there exist at least one b in B such that (a,b) is in f)
if (a,b) and (a,b') are both in f, then b=b' (for each a in A there exist at most one b in B such that (a,b) is in f)
would you be able to give an example of a cartesian product between two example sets called A and B so that I could see the resulting set from that?
cam875 said:
would you be able to give an example of a cartesian product between two example sets called A and B so that I could see the resulting set from that?
{1, 2, 3} x {a, b, c} = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c), (3, a), (3, b), (3, c)}
The set definitions of pairs, tuples, relations, functions, and integers only serves to show that set theory can be used as a formal basis for pretty much all of mathematics. The point of set theory was to reduce as many ideas as possible to as few as possible. It makes everything it can into a set.
If you have done any computer programming before, it's essentially the same concept. Computer languages only understand a limited set of objects: integers, floating point numbers, unicode characters, functions, and lists of other objects (or some similar-looking list). But then, how do you program a 3-D game? There are no "vectors" in the computer's eyes! The answer is you have to "build" vectors yourself out of what's given. A vector might be a list of floating point numbers, coupled with functions that can act on them (addition, scaling, dot products). What about a matrix? Well, now that we've defined vectors, a matrix is simply a list of vectors and functions for how to multiply them, transpose them, take their determinants, etc.
We're doing the same thing here with sets. You start off with only sets. Nothing but sets. Well, how do we make ordered pairs? We define them in terms of sets: (a, b) = {{a}, {a, b}} or whatnot. What about relations? We define a~b to be a set of pairs. What about functions? We define a function as a relationship with a condition that no two pairs (a, b) and (a', b') in the relation can have a = a'.
In a sense, we are merely "programming" the rest of mathematics with set theory. After we have defined a function in terms of sets, we can treat it as if functions are their own "real" entites. Once we have proven basic theorems about functions by their definition (composition is associative, and basic principles behind "onto" and "one-to-one" functions), we can more or less completely forget about the definition, knowing that as long as we defined it properly and made no mistakes in our core theorems, evereything else we prove with them will be correct, regardless of the details of the definition.
The take home message is, if sets are the only "primitive" object you have at your disposal, you can define many other mathematical objects in terms of them. Functions are not *really* sets. But if you went to a planet where functions were not yet discovered, but where the natives understood sets, you could give a precise description of what Earth mathematicians mean by "function."
make sense and yes I am an avid programmer so it does help me understand with the analogy u gave me, thanks
## What are functions and sets?
Functions and sets are mathematical concepts used to describe relationships between elements in a collection. A function maps each input element to a unique output element, while a set is a collection of distinct elements.
## What is the difference between a function and a set?
The main difference between a function and a set is that a function describes a relationship between elements, while a set is simply a collection of elements. A function can be thought of as a special type of set where each input has only one corresponding output.
## How do I represent a function or a set?
Functions and sets can be represented in various ways, such as using mathematical notation, diagrams, or tables. For functions, the most common notation is f(x) = y, where x is the input and y is the output. Sets can be represented using braces { } and listing the elements inside, or using set-builder notation {x | x satisfies a certain condition}.
## What is the domain and range of a function?
The domain of a function is the set of all possible input values, while the range is the set of all possible output values. In other words, the domain is the set of x-values and the range is the set of y-values in a function.
## How are functions and sets used in real life?
Functions and sets are used in various fields such as mathematics, physics, economics, and computer science. They can be used to model relationships between quantities, analyze data, and solve problems. For example, in economics, functions can be used to describe the relationship between supply and demand, while sets can be used to represent different categories of goods or services.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Calculus Go to the latest version.
# 6.6: L’Hôpital’s Rule
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
A student will be able to:
• Learn how to find the limit of indeterminate form (0/0)\begin{align*}(0/0)\end{align*} by L’Hospital’s rule.
If the two functions f(x)\begin{align*}f(x)\end{align*} and g(x)\begin{align*}g(x)\end{align*} are both equal to zero at x=a,\begin{align*}x = a,\end{align*} then the limit
limx af(x)g(x)
cannot be found by directly substituting x=a.\begin{align*}x = a.\end{align*} The reason is because when we substitute x=a,\begin{align*}x = a,\end{align*} the substitution will produce (0/0),\begin{align*}(0/0),\end{align*} known as an indeterminate form, which is a meaningless expression. To work around this problem, we use L’Hospital’s rule, which enables us to evaluate limits of indeterminate forms.
L’Hospital’s Rule
If limxaf(x)=limxag(x)=0\begin{align*} \lim_{x\to a}f(x) = \lim_{x\to a}g(x) = 0\end{align*}, and f(a)\begin{align*} f'(a)\end{align*} and g(a)\begin{align*} g'(a)\end{align*} exist, where g(a)0\begin{align*}g'(a) \neq 0\end{align*}, then
limx af(x)g(x)=limx af(x)g(x).
The essence of L’Hospital’s rule is to be able to replace one limit problem with a simpler one. In each of the examples below, we will employ the following three-step process:
1. Check that limxaf(x)g(x)\begin{align*} \lim_{x \rightarrow a} \frac{f(x)} {g(x)}\end{align*} is an indeterminate form 0/0.\begin{align*}0/0.\end{align*} To do so, directly substitute x=a\begin{align*}x = a\end{align*} into f(x)\begin{align*}f(x)\end{align*} and g(x).\begin{align*}g(x).\end{align*} If you get f(a)=g(a)=0,\begin{align*}f(a) = g(a) = 0,\end{align*} then you can use L’Hospital’s rule. Otherwise, it cannot be used.
2. Differentiate f(x)\begin{align*}f(x)\end{align*} and g(x)\begin{align*}g(x)\end{align*} separately.
3. Find limxaf(x)g(x).\begin{align*} \lim_{x \rightarrow a} \frac{f'(x)} {g'(x)}.\end{align*} If the limit is finite, then it is equal to the original limit limxaf(x)g(x)\begin{align*}\lim_{x \rightarrow a} \frac{f(x)} {g(x)}\end{align*} .
Example 1:
Find limx02+x2x.\begin{align*}\lim_{x \rightarrow 0} \frac{\sqrt{2 + x} - \sqrt{2}} {x}.\end{align*}
Solution:
When x=0\begin{align*}x = 0\end{align*} is substituted, you will get 0/0.\begin{align*}0/0.\end{align*}
Therefore L’Hospital’s rule applies:
limx02+x2x=limx0ddx(2+x2)ddx(x)=[1/(22+x)1]x=0=122=24.
Example 2:
Find limx01cos2xx2+2x\begin{align*} \lim_{x \rightarrow 0} \frac{1 - \cos 2x} {x^2 + 2x}\end{align*}
Solution:
We can see that the limit is 0/0\begin{align*}0/0\end{align*} when x=0\begin{align*}x = 0\end{align*} is substituted.
Using L’Hospital’s rule,
Example 3:
Use L’Hospital’s rule to evaluate \begin{align*} \lim_{x \rightarrow 3} \frac{x^2 - 9} {x - 3}\end{align*} .
Solution:
Example 4:
Evaluate \begin{align*} \lim_{x \rightarrow 0} \frac{\sin 3x} {x}.\end{align*}
Solution:
Example 5:
Evaluate \begin{align*} \lim_{x \rightarrow \pi/2} \frac{5 - 5 \sin x} {\cos x}\end{align*} .
Solution:
We can use L’Hospital’s rule since the limit produces the \begin{align*}0/0\end{align*} once \begin{align*}x = \pi/2\end{align*} is substituted. Hence
A broader application of L’Hospital’s rule is when \begin{align*}x = a\end{align*} is substituted into the derivatives of the numerator and the denominator but both still equal zero. In this case, a second differentiation is necessary.
Example 6:
Evaluate \begin{align*} \lim_{x \rightarrow 0} \frac{e^x -\cos x} {x^2}.\end{align*}
Solution:
As you can see, if we apply the limit at this stage the limit is still indeterminate. So we apply L’Hospital’s rule again:
## Review Questions
Find the limits.
1. \begin{align*} \lim_{\theta \rightarrow 0} \frac{\tan \theta}{\theta}\end{align*}
2. \begin{align*} \lim_{x \rightarrow 1} \frac{\ln x} {\tan \pi x}\end{align*}
3. \begin{align*} \lim_{x \rightarrow 0} \frac{e^{10x} - e^{6x}} {x}\end{align*}
4. \begin{align*} \lim_{x \rightarrow \pi} \frac{\sin x} {x - \pi}\end{align*}
5. \begin{align*} \lim_{x \rightarrow 0} \frac{xe^x} {1 - e^x}\end{align*}
6. If \begin{align*}k\end{align*} is a nonzero constant and \begin{align*}x > 0.\end{align*}
1. Show that \begin{align*} \int_1^x \frac{1} {t^{1 - k}} dt = \frac{x^k - 1} {k}.\end{align*}
2. Use L’Hospital’s rule to find \begin{align*} \lim_{k \rightarrow 0} \frac{x^k - 1} {k}.\end{align*}
7. Cauchy’s Mean Value Theorem states that if the functions \begin{align*}f\end{align*} and \begin{align*}g\end{align*} are continuous on the interval \begin{align*}(a, b)\end{align*} and \begin{align*}g' \neq 0,\end{align*} then there exists a number \begin{align*}c\end{align*} such that
Find all possible values of \begin{align*}c\end{align*} in the interval \begin{align*}(a, b)\end{align*} that satisfy this property for
on the interval
1. \begin{align*}1\end{align*}
2. \begin{align*}1/\pi\end{align*}
3. \begin{align*}4\end{align*}
4. \begin{align*}-1\end{align*}
5. \begin{align*}-1\end{align*}
6. \begin{align*}\ln x\end{align*}
7. \begin{align*} \frac{\pi} {4}\end{align*}
## Texas Instruments Resources
In the CK-12 Texas Instruments Calculus FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9731.
Feb 23, 2012
Aug 19, 2015 |
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