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# 7th Grade Math / Lesson 7B: Basic Operations II
OPERATIONS What will we be learning in this lesson? In this lesson you will learn how the effects of operations on numbers. Vocabulary words are found in this purple color throughout the lesson. Remember to put these in your notebook.
Operations Multiplying and Dividing 1. If multiplying or dividing numbers that have the same sign, the sign of the answer is positive. Ex) -3 * -4 = 12 Ex) 3 * 4 = 12 Ex) 8 / 2 = 4 Ex) -8 / (-2) = 42. If multiplying or dividing numbers that have different signs, the sign of the answer is negative. Ex) -3 * 4 = -12 Ex) 3 * (-4) = -12 Ex) -8 / 2 = -4 Ex) 8 / (-2) = -4
Operations Practice multiplying integers 3 * (-6) = (-4)(5) = -7(-5) = 12*3 =
Operations Multiplying Practice multiplying integers 3 * (-6) = -18 (-4)(5) =-20 -7(-5) =35 12*3 = 36
Operations Dividing Practice dividing integers 12 / 6 = (-10) / 5 = 20 / -5 = (-15) / (-3) =
Operations Dividing Practice dividing integers 12 / 6 =2 (-10) / 5 = -2 20 / -5 = -4 (-15) / (-3) =5 |
# How do you solve abs(1 - 3b)= - 7?
Jun 17, 2016
This equation has no solutions.
#### Explanation:
The absolute value of a non-negative number is this number itself.
Absolute value of the negative number is its negation.
In mathematical symbol it looks like that:
$X \ge 0 \implies | X | = X$
$X < 0 \implies | X | = - X$
Using this definition, let's divide a set of all possible values of $b$ into two parts:
(a) those where $1 - 3 b \ge 0$ (or $b \le \frac{1}{3}$)
(b) those where $1 - 3 b < 0$ (or $b > \frac{1}{3}$).
In case (a) our equation looks like this:
$1 - 3 b = - 7$,
which has a solution $b = \frac{8}{3}$.
This solution does not belong to the area of $b \le \frac{1}{3}$ and must be discarded.
In case (b) our equation looks like this:
$- \left(1 - 3 b\right) = - 7$,
which has a solution $b = - 2$.
This solution does not belong to the area of $b > \frac{1}{3}$ and must be discarded.
So, no solutions are found for this equation.
We can confirm this graphically by observing that function $y = | 1 - 3 x | + 7$ does not have intersections with X-axis.
graph{|1-3x|+7 [-46.23, 46.25, -23.12, 23.1]}
Jun 17, 2016
exactly no solutions
#### Explanation:
because
$\left\mid a \right\mid$ can't be negative,
you can't find any solution which satifies this equation |
# What's the derivative of arctan(x - sqrt(1+x^2))?
##### 1 Answer
Mar 29, 2016
$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1}{2 \left(1 + {x}^{2}\right)}$
#### Explanation:
Let us first workout derivative of $f \left(u\right) = \arctan u$.
As $g \left(u\right) = \arctan u$, $\tan g = u$ and differentiating we have
${\sec}^{2} g \cdot \frac{\mathrm{dg}}{\mathrm{dx}} = 1$ or $\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} g} = \frac{1}{1 + {\tan}^{2} g} = \frac{1}{1 + {u}^{2}}$
Hence for $g \left(u\right) = \arctan u$, we have (dg)/(dx)=1/(1+u^2....(A)
Now to differentiate $f \left(x\right) = \arctan \left(x - \sqrt{1 + {x}^{2}}\right)$, we will use (A) and chain rule.
Hence $\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\left(x - \sqrt{1 + {x}^{2}}\right)}^{2}}\right) \cdot \left(1 - \frac{1}{2 \sqrt{1 + {x}^{2}}} \cdot 2 x\right)$ or
$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\left(x - \sqrt{1 + {x}^{2}}\right)}^{2}}\right) \cdot \left(1 - \frac{x}{\sqrt{1 + {x}^{2}}}\right)$
If $x = \tan \theta$
$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\left(\tan \theta - \sqrt{1 + {\tan}^{2} \theta}\right)}^{2}}\right) \cdot \left(1 - \tan \frac{\theta}{\sqrt{1 + {\tan}^{2} \theta}}\right)$ or
$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\left(\tan \theta - \sec \theta\right)}^{2}}\right) \cdot \left(1 - \tan \frac{\theta}{\sec} \theta\right)$ or
$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\tan}^{2} \theta + {\sec}^{2} \theta - 2 \tan \theta \sec \theta}\right) \cdot \left(\frac{\sec \theta - \tan \theta}{\sec} \theta\right)$ or
$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{2 \sec \theta \left(\sec \theta - \tan \theta\right)}\right) \cdot \left(\frac{\sec \theta - \tan \theta}{\sec} \theta\right)$ or
$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{2 {\sec}^{2} \theta}\right) = \frac{1}{2 \left(1 + {\tan}^{2} \theta\right)} = \frac{1}{2 \left(1 + {x}^{2}\right)}$ |
## Math
The formula of normal distribution is
$$$$e^{ ( (x - \mu) / \sqrt{2} \sigma )^2 }$$$$
where $\mu$ controls the “center” or “peak” of the distribution and $\sigma$ tells us how “wide” or “disperse” the distribution is.
To understand the distribution, we take some limits.
### $x = \mu$
First of all, when $x = \mu$ we have
$$e^0 = 1.$$
Notice the argument of the exponential is some squared value and can not be negative. This condition gives us the peak value.
### $x=\mu-a$ and $x=\mu + a$
For $x=\mu-a$, we have
$$e^{ ( (a) / \sqrt{2} \sigma )^2 }.$$
For $x=\mu + a$, we have
$$e^{ ( (a) / \sqrt{2} \sigma )^2 }$$
which is exactly the same as the previous case.
The distribution is symmetric around $x=\mu$.
### $x=\pm \infty$
We have 0 for both cases.
## Tricks
### Integral
We integrate distributions a lot. For Gaussian distribution, it is quite helpful to remember the following identity.
$$\int_{-\infty}^\infty e^ {- x^2} dx = \sqrt{\pi}.$$
It tells us that for $\mu=0$ and $\sigma=1/\sqrt{2}$, the area under the distribution is $\sqrt{\pi}$.
Hey, it is time to ask the question. Where the hell is the circle?
### Error Function
The error function is defined as
$$\mathrm{erf}(x) = \frac{1}{\sqrt{\pi}}\int_{-x}^{x} e^{ - t^2} dt$$
Obviously, the coefficient $\frac{1}{\sqrt{\pi}}$ normalizes the function to be within $[-1,1]$.
Published: by ; |
# If a 3/2 kg object moving at 5/3 m/s slows to a halt after moving 4/3 m, what is the coefficient of kinetic friction of the surface that the object was moving over?
Aug 20, 2017
${\mu}_{k} = 0.11$
#### Explanation:
We are given the following information:
• $\mapsto m = 3 / 2 \text{ kg}$
• $\mapsto \Delta s = 4 / 3 \text{ m}$
• $\mapsto {v}_{i} = 5 / 3 \text{ m"//"s}$
• $\mapsto {v}_{f} = 0$
• $\mapsto g = 9.81 {\text{ m"//"s}}^{2}$
$\implies$This problem can be solved using either kinematics and Newton's second law, or the work-energy theorem. I will include both solutions.
**$\textcolor{\mathrm{da} r k b l u e}{\text{Method 1: Newton's Second Law and Kinematics.}}$
• We can use the following kinematic equation to solve for the acceleration that the object experiences as it comes to rest:
$\textcolor{b l u e}{{v}_{f}^{2} = {v}_{i}^{2} + 2 a \Delta s}$
• After calculating the acceleration, we can generate a statement of the net force on the object using Newton's second law:
$\textcolor{b l u e}{{\vec{F}}_{\text{net}} = m \vec{a}}$
• And we can use the acceleration and given mass to calculate the net force: the force of kinetic friction ${f}_{k}$.
Let's solve for $a$ in the kinematic:
$\implies \textcolor{b l u e}{a = \frac{{v}_{f}^{2} - {v}_{i}^{2}}{2 \Delta s}}$
Using our known values:
=>a=(-(5/3" m"//"s")^2)/(2*4/3"m")
$\implies \textcolor{b l u e}{a \approx - 1.042 {\text{ m"//"s}}^{2}}$
Note that the sign of the acceleration is negative, which indicates that the acceleration is the opposite direction of motion and therefore the object is slowing down.
• As for the net force, the perpendicular (y, vertical) forces include only the normal force and force of gravity, which are in a state of equilibrium. The parallel (x, horizontal) forces, however, include only the force of kinetic friction, which is what causes the object to slow down.
${F}_{x \text{ net}} = - {f}_{k} = m a$
The statement of the perpendicular forces in equilibrium $\left(a = 0\right)$ is:
${F}_{y \text{ net}} = n - {F}_{G} = 0$
As we know that ${F}_{G} = m g :$
$\implies n = m g$
We also know that ${f}_{k} = {\mu}_{k} n$
$\implies {f}_{k} = {\mu}_{k} m g$
Therefore:
$- {\mu}_{k} m g = m a$
• Solving for ${\mu}_{k}$:
$\implies - {\mu}_{k} = \frac{m a}{m g}$
$\implies - {\mu}_{k} = \frac{a}{g}$
• Using our known values:
$- {\mu}_{k} = \left(- 1.042 {\text{ m"//"s"^2)/(9.81"m"//"s}}^{2}\right)$
$\implies \textcolor{b l u e}{0.11}$
**$\textcolor{\mathrm{da} r k b l u e}{\text{Method 2: Work-Energy Theorem.}}$
By the work-energy theorem, the work done by nonconservative forces (e.g. the force of friction) is equal to the energy lost in a system.
$\textcolor{b l u e}{\Delta {E}_{\text{sys"=W_"nc}}}$
When energy is conserved in a system, $\Delta {E}_{\text{sys}} = 0$.
We have only kinetic energy, so this statement becomes:
${W}_{\text{friction}} = \Delta K$
$\implies {W}_{F} = \frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2}$
There is no kinetic energy finally, as the object is at rest and $v = 0$:
$\implies {W}_{F} = - \frac{1}{2} m {v}_{i}^{2}$
The work done by friction:
${W}_{F} = {f}_{k} \Delta s \cos \left(\theta\right)$
• Where ${f}_{k}$ is the force of kinetic friction, $\Delta s$ is the displacement of the object, and $\theta$ is the angle between the force and displacement vectors. In this case, friction is opposite the motion and therefore antiparallel, so $\cos \left({180}^{o}\right) = - 1$
$\implies {W}_{F} = - {f}_{k} \Delta s$
$\implies - \frac{1}{2} m {v}_{i}^{2} = - {\mu}_{k} n \Delta s$
$\implies \frac{1}{2} m {v}_{i}^{2} = {\mu}_{k} m g \Delta s$
We can now solve for ${\mu}_{k}$:
$\implies {\mu}_{k} = \frac{\frac{1}{2} m {v}_{i}^{2}}{m g \Delta s}$
$\implies \textcolor{b l u e}{{\mu}_{k} = \frac{\frac{1}{2} {v}_{i}^{2}}{g \Delta s}}$
Using our known values:
$\implies {\mu}_{k} = \left(\frac{1}{2} \left(\frac{5}{3} \text{m"/"s")^2)/((9.81"m"//"s"^2)(4/3"m}\right)\right)$
$\implies {\mu}_{k} = 0.106$
$\implies \textcolor{b l u e}{{\mu}_{k} \approx 0.11}$
Aug 20, 2017
$\mu = 0.107$
#### Explanation:
We can find the acceleration of the object using the kinematic formula
${v}^{2} = {u}^{2} + 2 \cdot a \cdot s$
${0}^{2} = {\left(\frac{5}{3} \frac{m}{s}\right)}^{2} + 2 \cdot a \cdot \frac{4}{3} m$
Solving that for a,
$- 2 \cdot a \cdot \frac{4}{3} m = {\left(\frac{5}{3} \frac{m}{s}\right)}^{2}$
$a = - \frac{{5}^{2} \cdot 3}{2 \cdot {3}^{2} \cdot 4} {m}^{2} / \left(m \cdot {s}^{2}\right)$
$a = - \frac{{5}^{2} \cdot \cancel{3}}{2 \cdot {3}^{\cancel{2}} \cdot 4} {m}^{\cancel{2}} / \left(\cancel{m} \cdot {s}^{2}\right) = - \frac{25}{24} \frac{m}{s} ^ 2$
The force responsible for that acceleration must have been
$F = m \cdot a = \frac{3}{2} k g \cdot \left(- \frac{25}{24} \frac{m}{s} ^ 2\right)$
$F = - \frac{75}{48} N$
(I will assume that the surface this thing is on is horizontal since there is no information to tell me that I have to make it more complicated.)
The friction formula is
${F}_{f} = \mu \cdot N = \mu \cdot m \cdot g$
$- \frac{75}{48} N = \mu \cdot \frac{3}{2} k g \cdot 9.8 \frac{m}{s} ^ 2$
Solving that for $\mu$,
$\mu = - \frac{\frac{75}{48} N}{\frac{3}{2} k g \cdot 9.8 \frac{m}{s} ^ 2} = - \frac{\frac{75}{48} N}{\left(\frac{3}{2}\right) \cdot 9.8 N}$
$\mu = 0.107$
Note, the negative sign indicates that the direction of $F f$ is in the opposite direction of the initial velocity. But the coefficient of kinetic friction is a scalar, so I have dropped the sign.
I hope this helps,
Steve
p.s. double-check my arithmetic. |
## Proof that log 2 is an irrational number
Before doing the proof, let us recall two things: (1) rational numbers are numbers that can be expressed as $\frac{a}{b}$ where $a$ and $b$ are integers, and $b$ not equal to $0$; and (2) for any positive real number $y$, its logarithm to base $10$ is defined to be a number $x$ such that $10^x = y$. In proving the statement, we use proof by contradiction.
Theorem: log 2 is irrational
Proof:
Assuming that log 2 is a rational number. Then it can be expressed as $\frac{a}{b}$ with $a$ and $b$ are positive integers (Why?). Then, the equation is equivalent to $2 = 10^{\frac{a}{b}}$. Raising both sides of the equation to $b$, we have $2^b = 10^a$. This implies that $2^b = 2^a5^a$. Notice that this equation cannot hold (by the Fundamental Theorem of Arithmetic) because $2^b$ is an integer that is not divisible by 5 for any $b$, while $2^a5^a$ is divisible by 5. This means that log 2 cannot be expressed as $\frac{a}{b}$ and is therefore irrational which is what we want to show.
## Diagonals of a Parallelogram
Coordinate geometry was one of the greatest inventions in mathematics. Aside from connecting geometry and algebra, it has made many geometric proofs short and easy. In the example below, we use coordinate geometry to prove that the diagonals of a parallelogram bisect each other.
The proof can be simplified by placing a vertex of the parallelogram at the origin and one side coinciding with the x-axis. If we let a and b be the side lengths of the parallelogram and c as its altitude, then, the coordinates of the vertices can be easily determined as shown below.
In addition, if we label the vertices P, Q, R, and S starting from the origin and going clockwise, then the coordinates of the vertices are P(0,0), Q(b,c), R(b,c) and S(a,0). » Read more
## 2011 GeoGebra Institute Activities
In 2011, we at the GeoGebra Institute of Metro Manila (GIMM) have squeezed our schedule and have given GeoGebra trainings to teachers and students from all over the country. Here are the list of trainings that we have conducted.
GeoGebra Basics
Date: February 16, 2011
Venue: EARIST College, Manila
No. of Participants: 30
Introductory Course on the Use of GeoGebra in Teaching and Learning Mathematics
Date: May 5-6, 2011
Venue: UP NISMED, University of the Philippines, Diliman, Quezon City
No. of Participants: 21
Lesson Study for Teaching through Problem Solving with GeoGebra
Date: May 9-13,2011
Venue: NISMED, University of the Philippines, Diliman, Quezon City
No. of Participants: 20
Introduction to GeoGebra
Date: October 1, 2011
Venue: St. Therese College, Pasay City
No. of Participants: 41
GeoGebra Fundamentals
Date: November 16, 29, 2011
Venue: Santa Lucia High School, Pasig City
No. of Participants: 18
GeoGebra Fundamentals In-House Seminar
Date: November 25, 2011
Venue: UP NISMED, University of the Philippines, Diliman, Quezon City
No. of Participants: 15
In addition, we are also working on a book on College Geometry with GeoGebra (the title has not been finalized yet). Hopefully, it will be tried out this school year and will be published late this year or early next year.
We are also planning to have Level 1 and 2 GeoGebra trainings this summer (April-May, no definite date yet). If you are interested, please email me at gip@gmail.com or mathandmultimedia@gmail.com.
1 6 7 8 9 |
Using Identities to Evaluate
Once you've learned some trig identities, you can use them to evaluate angles exactly. Yes, you can always get a decimal approximation with your calculator, but it's good to learn the logic of using the identities, so you can exercise your skills at pattern-finding and -matching.
• Evaluate cos(75°) exactly.
• I know the exact values for the cosine of 30° and 45°, and 75° = 30° + 45°, so I'll use the angle-sum identity to compute this value "exactly":
You can check your answers to this sort of exercise by plugging each of the trig expression and the "exact" expression into your calculator. If the displayed values are the same (0.2588190451, in this case), then you know you've done the evaluation correctly.
• Evaluate tan(15°) exactly.
• Since 15 = 45 – 30, I'll use the angle-difference identity for tangent:
If your book prefers radians, you'll be asked to evaluate expressions like , which is the same angle as above. The sums and differences can be a bit harder to recognize in radian form:
A couple of useful sums are:
If you work better with degrees, then convert radian-measure angles to degrees, find the sums and differences, and convert back to radians.
• Evaluate sin(120°) using a formula.
• Since 120° is twice 60°, I'll use the double-angle formula for sine:
sin(2×60°) = 2sin(60°)cos(60°)
Since 60° is a basic reference angle, I can now finish the evaluation by plugging in the values I've memorized:
• Since 22.5 is half of 45, I'll use the half-angle identity for cosine:
Since 22.5° is in the first quadrant, then the cosine value is positive, so I'll take the positive root for my answer:
• Find cos(2x) if sin(x) = –5/13 and x is in the third quadrant.
• From the sign on the sine, I could only tell that x was in QIII or QIV; that's why they had to specify the quadrant for x. By the Pythagorean Theorem, I can find the third side of the triangle from the sine's ratio value:
So the adjacent side has a length of ±12. Since the angle x is in the third quadrant, then the "length" is –12. Plugging into the double-angle formula, I get:
In this case, I didn't really "need" the quadrant information; the squaring and subtraction took care of the signs. But sometimes you will need to pick one of two values, as in the previous example. Don't get lazy with the "plus / minus" signs. Keep track of where you are and where you're going, and pick the appropriate sign, as needed.
Cite this article as: Stapel, Elizabeth. "Using Identities to Evaluate." Purplemath. Available from http://www.purplemath.com/modules/ideneval.htm. Accessed [Date] [Month] 2016
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# Binomial expansion facts for kids
Kids Encyclopedia Facts
Binomial expansion uses an expression to make a series. It uses a bracket expression like $(x+y)^n$. There are three binomial expansions.
## The formulas
There are basically three binomial expansion formulas:
$(a+b)^2 = a^2 + 2 ab + b^2$ 1st (Plus) $(a-b)^2 = a^2 - 2 ab + b^2$ 2nd (Minus) $(a+b) \cdot (a-b) = a^2 - b^2$ 3rd (Plus-Minus)
We can explain why there are such 3 formulas with a simple expansion of the product:
$(a+b)^2=(a+b)\cdot(a+b)=a \cdot a+a \cdot b+b \cdot a+b \cdot b=a^2+2 \cdot a \cdot b+b^2$
$(a-b)^2=(a-b) \cdot (a-b)=a \cdot a-a \cdot b-b \cdot a+b \cdot b=a^2-2 \cdot a \cdot b+b^2$
$(a+b) \cdot (a-b)=a \cdot a-a \cdot b+b \cdot a-b \cdot b=a^2-b^2$
## Using Pascal's triangle
If $n$ is an integer ($n \in \mathbb{Z}$), we use Pascal's triangle.
To expand $(x+y)^2$:
• find row 2 of Pascal's triangle (1, 2, 1)
• expand $x$ and $y$ so the $x$ power goes down by 1 each time from $n$ to 0 and the $y$ power goes up by 1 each time from 0 up to $n$
• times the numbers from Pascal's triangle with the right terms.
So $(x+y)^2 = 1x^2y^0 + 2x^1y^1 + 1x^0y^2$
For example:
$(3+2x)^2 = 1 \cdot 3^2 \cdot (2x)^0 + 2 \cdot 3^1 \cdot (2x)^1 + 1 \cdot 3^0 \cdot (2x)^2 = 9 + 12x + 4x^2$
So as a rule:
$(x+y)^n = a_0x^ny^0 + a_1x^{n-1}y^1+ a_2x^{n-2}y^2+ \cdots + a_{n-1}x^{1}y^{n-1}+ a_nx^0y^n$
where $a_i$ is the number at row $n$ and position $i$ in Pascal's triangle.
### Examples
$(5+3x)^3 = 1 \cdot 5^3 \cdot (3x)^0 + 3 \cdot 5^2 \cdot (3x)^1 + 3 \cdot 5^1 \cdot (3x)^2 + 1 \cdot 5^0 \cdot (3x)^3$
$= 125 + 75 \cdot 3x + 15 \cdot 9x^2 + 1 \cdot 27x^3 = 125 + 225x + 135x^2 + 27x^3$
$(5-3x)^3 = 1 \cdot 5^3 \cdot (-3x)^0 + 3 \cdot 5^2 \cdot (-3x)^1 + 3 \cdot 5^1 \cdot (-3x)^2 + 1 \cdot 5^0 \cdot (-3x)^3$
$= 125 + 75 \cdot (-3x) + 15 \cdot 9x^2 + 1 \cdot (-27x^3) = 125 - 223x + 135x^2 - 27x^3$
$(7+4x^2)^5 = 1 \cdot 7^5 \cdot (4x^2)^0 + 5 \cdot 7^4 \cdot (4x^2)^1 + 10 \cdot 7^3 \cdot (4x^2)^2 + 10 \cdot 7^2 \cdot (4x^2)^3 + 5 \cdot 7^1 \cdot (4x^2)^4 + 1 \cdot 7^0 \cdot (4x^2)^5$
$= 16807 + 12005 \cdot 4x^2 + 3430 \cdot 16x^4 + 490 \cdot 64x^6 + 35 \cdot 256x^8 + 1 \cdot 1024x^{10}$
$\, = 16807 + 48020x^2 + 54880x^4 + 31360x^6 + 8960x^8 + 1024x^{10}$
Binomial expansion Facts for Kids. Kiddle Encyclopedia. |
Home क्सक्सक्स XXX (क्सक्सक्स) Roman Numerals
# XXX (क्सक्सक्स) Roman Numerals
0
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XXX Roman Numerals can be written as numbers by combining the transformed roman numerals i.e. XXX = 30. The higher roman numerals precede the lower numerals resulting in the correct translation of XXX Roman Numerals. In this article, we will explain how to convert XXX Roman numerals in the correct number translation.
• XXX = X + X + X
• XXX = 10 + 10 + 10
• XXX = 30
## How to Write XXX (क्सक्सक्स) Roman Numerals?
The numerical value of XXX Roman Numerals can be obtained by using any of the two methods given below:
Method 1: In this method, we break the roman numerals into single letters, write the numerical value of each letter and add/subtract them.
• XXX = X + X + X = 10 + 10 + 10 = 30
Method 2: In this method, we consider the groups of roman numerals for addition or subtraction such as,
• XXX = 30
Therefore, the numerical value of XXX roman numerals is 30.
### What are the Basic Rules to Write Roman Numerals?
• When a bigger letter precedes a smaller letter, the letters are added. For example: MI, M > I, so MI = M + I = 1000 + 1 = 1001
• When a smaller letter precedes a bigger letter, the letters are subtracted. For example: IV, I < V, so IV = V – I = 5 – 1 = 4
• When a letter is repeated 2 or 3 times, they get added. For example: MM = M + M = 1000 + 1000 = 2000
• The same letter cannot be used more than three times in succession.
## Numbers Related to XXX Roman Numerals
Roman numerals were used in ancient Rome and utilized combinations of letters using the Latin alphabets I, V, X, L, C, D, and M. It may seem different than numbers, but they are similar. For example, XXX Roman numerals are equivalent to the number 30. The roman numerals related to XXX are given below:
• XXX = 30
• XXXI = 30 + 1 = 31
• XXXII = 30 + 2 = 32
• XXXIII = 30 + 3 = 33
• XXXIV = 30 + 4 = 34
• XXXV = 30 + 5 = 35
• XXXVI = 30 + 6 = 36
• XXXVII = 30 + 7 = 37
• XXXVIII = 30 + 8 = 38
• XXXIX = 30 + 9 = 39
## XXX Roman Numerals Examples
1. Example 1: Find the Quotient of 30 and 8.Solution:The roman numeral XXX is 30 and VIII is 8.
Now, when we divide XXX by VIII i.e. 30 ÷ 8, the quotient is 3.
Since, 3 = III
Therefore, XXX ÷ VIII = III
2. Example 2: Find the Product of Roman Numerals XXX and XXXV.Solution:XXX = 30 and XXXV = 30 + 5 = 35
Now, XXX × XXXV = 30 × 35 = 1050
Since, ML = 1000 + 50 = 1050
Therefore, XXX × XXXV = ML
3. Example 3: Find the Sum of CXCIX and XXX Roman Numerals.Solution:CXCIX = 100 + 90 + 9 = 199 and XXX = 30
Now, CXCIX + XXX = 199 + 30 = 229
Since, CCXXIX = 200 + 20 + 9 = 229
Therefore, the sum of CXCIX and XXX roman numerals is CCXXIX
4. Example 4: Find the Difference Between XXX and XII.Solution:Roman Numeral XXX is equal to 30 and XII is 12.
Now, XXX – XII = 30 – 12 = 18
Since, 18 = XVIII
Therefore, XXX – XII = XVIII
## FAQs on XXX Roman Numerals
### What is the Value of the XXX Roman Numerals?
We will write XXX Roman numerals in the expanded form to determine its value. XXX = X + X + X = 10 + 10 + 10 = 30. Hence, the value of Roman Numerals XXX is 30.
### How is Roman Numerals XXX Written in Numbers?
To convert XXX Roman Numerals to numbers, the conversion involves breaking the Roman numerals on the basis of place values (ones, tens, hundreds, thousands), like this:
• Tens = 30 = XXX
• Number = 30 = XXX
### What Should be Subtracted from XXX to Get XV?
First, we will write XXX and XV in numbers, i.e. XXX = 30 and XV = 10 + 5 = 15. Now, 30 – 15 = 15. And 15 = XV. Therefore, XV should be subtracted from XXX roman numerals to get XV.
### Why is 30 Written in Roman Numerals as XXX?
We know that in roman numerals, we write 10 as X. Therefore, 30 in roman numerals is written as XXX = 30.
### What is the Remainder when XXX is Divided by XIV?
XXX = 30 and XIV = 14 in numbers. On dividing 30 by 14, it leaves a remainder of 14. Now, 2 = II Therefore, when XXX is divided by XIV, the remainder is II. |
# How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given 9x^2-6x-4=-5?
Dec 1, 2016
The solution is $S = \left\{\frac{1}{3}\right\}$
#### Explanation:
Let's rewrite the quadratic equation in the form
$a {x}^{2} + b x + c = 0$
$9 {x}^{2} - 6 x - 4 + 5 = 0$
$9 {x}^{2} - 6 x + 1 = 0$
Let's calculate the discriminant
$\Delta = {b}^{2} - 4 a c = {\left(- 6\right)}^{2} - 4 \cdot 9 \cdot 1 = 36 - 36 = 0$
As, $\Delta = 0$, we have a double real root
$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = - \frac{b}{2 a} = \frac{6}{18} = \frac{1}{3}$
We can also factorise the quadratic equation
$9 {x}^{2} - 6 x + 1 = \left(3 x - 1\right) \left(3 x - 1\right) = {\left(3 x - 1\right)}^{2}$
graph{(3x-1)^2 [-1.717, 2.128, -0.713, 1.209]} |
How do you differentiate f(x)=4x ln(3sin^2x^2 + 2) using the chain rule?
Nov 6, 2015
$f ' \left(x\right) = 4 \ln \left(3 {\sin}^{2} {x}^{2} + 2\right) + \frac{48 {x}^{2} \cos {x}^{2}}{3 {\sin}^{2} {x}^{2} + 2}$
Explanation:
$f ' \left(x\right) = \left(4 x\right) ' \cdot \ln \left(3 {\sin}^{2} {x}^{2} + 2\right) + 4 x \cdot \left(\ln \left(3 {\sin}^{2} {x}^{2} + 2\right)\right) '$
$f ' \left(x\right) = 4 \ln \left(3 {\sin}^{2} {x}^{2} + 2\right) + 4 x \cdot \frac{1}{3 {\sin}^{2} {x}^{2} + 2} \cdot \left(3 {\sin}^{2} {x}^{2} + 2\right) '$
$f ' \left(x\right) = 4 \ln \left(3 {\sin}^{2} {x}^{2} + 2\right) + \frac{4 x}{3 {\sin}^{2} {x}^{2} + 2} \cdot \left(3 {\sin}^{2} {x}^{2}\right) ' + 0$
$f ' \left(x\right) = 4 \ln \left(3 {\sin}^{2} {x}^{2} + 2\right) + \frac{4 x}{3 {\sin}^{2} {x}^{2} + 2} \cdot 6 \left(\sin {x}^{2}\right) '$
$f ' \left(x\right) = 4 \ln \left(3 {\sin}^{2} {x}^{2} + 2\right) + \frac{4 x}{3 {\sin}^{2} {x}^{2} + 2} \cdot 6 \cos {x}^{2} \cdot \left({x}^{2}\right) '$
$f ' \left(x\right) = 4 \ln \left(3 {\sin}^{2} {x}^{2} + 2\right) + \frac{4 x}{3 {\sin}^{2} {x}^{2} + 2} \cdot 6 \cos {x}^{2} \cdot 2 x$
$f ' \left(x\right) = 4 \ln \left(3 {\sin}^{2} {x}^{2} + 2\right) + \frac{48 {x}^{2} \cos {x}^{2}}{3 {\sin}^{2} {x}^{2} + 2}$ |
# 4.2 Co-ordinate geometry
Page 1 / 1
## Equation of a line between two points
There are many different methods of specifying the requirements for determining the equation of a straight line. One option is to find the equation of a straight line, when two points are given.
Assume that the two points are $\left({x}_{1};{y}_{1}\right)$ and $\left({x}_{2};{y}_{2}\right)$ , and we know that the general form of the equation for a straight line is:
$y=mx+c$
So, to determine the equation of the line passing through our two points, we need to determine values for $m$ (the gradient of the line) and $c$ (the $y$ -intercept of the line). The resulting equation is
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
where $\left({x}_{1};{y}_{1}\right)$ are the co-ordinates of either given point.
## Finding the second equation for a straight line
This is an example of a set of simultaneous equations, because we can write:
$\begin{array}{ccc}\hfill {y}_{1}& =& m{x}_{1}+c\hfill \\ \hfill {y}_{2}& =& m{x}_{2}+c\hfill \end{array}$
We now have two equations, with two unknowns, $m$ and $c$ .
$\begin{array}{ccc}\hfill {\mathrm{y}}_{2}-{\mathrm{y}}_{1}& =& m{x}_{2}-m{x}_{1}\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}m& =& \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\hfill \\ \hfill {\mathrm{y}}_{1}& =& m{x}_{1}+c\hfill \\ \hfill c& =& {y}_{1}-m{x}_{1}\hfill \end{array}$
Now, to make things a bit easier to remember, substitute [link] into [link] :
$\begin{array}{ccc}\hfill y& =& mx+c\hfill \\ & =& mx+\left({y}_{1}-m{x}_{1}\right)\hfill \\ \hfill \mathrm{y}-{\mathrm{y}}_{1}& =& m\left(x-{x}_{1}\right)\hfill \end{array}$
If you are asked to calculate the equation of a line passing through two points, use:
$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$
to calculate $m$ and then use:
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
to determine the equation.
For example, the equation of the straight line passing through $\left(-1;1\right)$ and $\left(2;2\right)$ is given by first calculating $m$
$\begin{array}{ccc}\hfill m& =& \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\hfill \\ & =& \frac{2-1}{2-\left(-1\right)}\hfill \\ & =& \frac{1}{3}\hfill \end{array}$
and then substituting this value into
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
to obtain
$\begin{array}{ccc}\hfill y-{y}_{1}& =& \frac{1}{3}\left(x-{x}_{1}\right).\hfill \end{array}$
Then substitute $\left(-1;1\right)$ to obtain
$\begin{array}{ccc}\hfill y-\left(1\right)& =& \frac{1}{3}\left(x-\left(-1\right)\right)\hfill \\ \hfill y-1& =& \frac{1}{3}x+\frac{1}{3}\hfill \\ \hfill y& =& \frac{1}{3}x+\frac{1}{3}+1\hfill \\ \hfill y& =& \frac{1}{3}x+\frac{4}{3}\hfill \end{array}$
So, $y=\frac{1}{3}x+\frac{4}{3}$ passes through $\left(-1;1\right)$ and $\left(2;2\right)$ .
Find the equation of the straight line passing through $\left(-3;2\right)$ and $\left(5;8\right)$ .
1. $\begin{array}{ccc}\hfill \left({x}_{1};{y}_{1}\right)& =& \left(-3;2\right)\hfill \\ \hfill \left({x}_{2};{y}_{2}\right)& =& \left(5;8\right)\hfill \end{array}$
2. $\begin{array}{ccc}\hfill m& =& \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\hfill \\ & =& \frac{8-2}{5-\left(-3\right)}\hfill \\ & =& \frac{6}{5+3}\hfill \\ & =& \frac{6}{8}\hfill \\ & =& \frac{3}{4}\hfill \end{array}$
3. $\begin{array}{ccc}\hfill y-{y}_{1}& =& m\left(x-{x}_{1}\right)\hfill \\ \hfill y-\left(2\right)& =& \frac{3}{4}\left(x-\left(-3\right)\right)\hfill \\ \hfill y& =& \frac{3}{4}\left(x+3\right)+2\hfill \\ & =& \frac{3}{4}x+\frac{3}{4}·3+2\hfill \\ & =& \frac{3}{4}x+\frac{9}{4}+\frac{8}{4}\hfill \\ & =& \frac{3}{4}x+\frac{17}{4}\hfill \end{array}$
4. The equation of the straight line that passes through $\left(-3;2\right)$ and $\left(5;8\right)$ is $y=\frac{3}{4}x+\frac{17}{4}$ .
## Equation of a line through one point and parallel or perpendicular to another line
Another method of determining the equation of a straight-line is to be given one point, $\left({x}_{1};{y}_{1}\right)$ , and to be told that the line is parallel or perpendicular to another line. If the equation of the unknown line is $y=mx+c$ and the equation of the second line is $y={m}_{0}x+{c}_{0}$ , then we know the following:
$\begin{array}{ccc}\hfill \mathrm{If the lines are parallel, then}\phantom{\rule{1.em}{0ex}}\mathrm{m}& =& {m}_{0}\hfill \\ \hfill \mathrm{If the lines are perpendicular, then}\phantom{\rule{1.em}{0ex}}\mathrm{m}×{\mathrm{m}}_{0}& =& -1\hfill \end{array}$
Once we have determined a value for $m$ , we can then use the given point together with:
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
to determine the equation of the line.
For example, find the equation of the line that is parallel to $y=2x-1$ and that passes through $\left(-1;1\right)$ .
First we determine $m$ , the slope of the line we are trying to find. Since the line we are looking for is parallel to $y=2x-1$ ,
$m=2$
The equation is found by substituting $m$ and $\left(-1;1\right)$ into:
$\begin{array}{ccc}\hfill y-{y}_{1}& =& m\left(x-{x}_{1}\right)\hfill \\ \hfill y-1& =& 2\left(x-\left(-1\right)\hfill \\ \hfill y-1& =& 2\left(x+1\right)\hfill \\ \hfill y-1& =& 2x+2\hfill \\ \hfill y& =& 2x+2+1\hfill \\ \hfill y& =& 2x+3\hfill \end{array}$
## Inclination of a line
In [link] (a), we see that the line makes an angle $\theta$ with the $x$ -axis. This angle is known as the inclination of the line and it is sometimes interesting to know what the value of $\theta$ is.
Firstly, we note that if the gradient changes, then the value of $\theta$ changes ( [link] (b)), so we suspect that the inclination of a line is related to the gradient. We know that the gradient is a ratio of a change in the $y$ -direction to a change in the $x$ -direction.
$m=\frac{\Delta y}{\Delta x}$
But, in [link] (a) we see that
$\begin{array}{ccc}\hfill tan\theta & =& \frac{\Delta y}{\Delta x}\hfill \\ \hfill \therefore m& =& tan\theta \hfill \end{array}$
For example, to find the inclination of the line $y=x$ , we know $m=1$
$\begin{array}{ccc}\hfill \therefore tan\theta & =& 1\hfill \\ \hfill \therefore \theta & =& {45}^{\circ }\hfill \end{array}$
## Co-ordinate geometry
1. Find the equations of the following lines
1. through points $\left(-1;3\right)$ and $\left(1;4\right)$
2. through points $\left(7;-3\right)$ and $\left(0;4\right)$
3. parallel to $y=\frac{1}{2}x+3$ passing through $\left(-1;3\right)$
4. perpendicular to $y=-\frac{1}{2}x+3$ passing through $\left(-1;2\right)$
5. perpendicular to $2y+x=6$ passing through the origin
2. Find the inclination of the following lines
1. $y=2x-3$
2. $y=\frac{1}{3}x-7$
3. $4y=3x+8$
4. $y=-\frac{2}{3}x+3$ (Hint: if $m$ is negative $\theta$ must be in the second quadrant)
5. $3y+x-3=0$
3. Show that the line $y=k$ for any constant $k$ is parallel to the x-axis. (Hint: Show that the inclination of this line is ${0}^{\circ }$ .)
4. Show that the line $x=k$ for any constant $k$ is parallel to the y-axis. (Hint: Show that the inclination of this line is ${90}^{\circ }$ .)
are nano particles real
yeah
Joseph
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no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
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Santosh
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Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
how did you get the value of 2000N.What calculations are needed to arrive at it
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A circle is inscribed in a rhombus with diagonals $12$ cm and $16$ cm. The ratio of the area of circle to the area of rhombus is
1. $\frac{5\pi }{18}$
2. $\frac{6\pi }{25}$
3. $\frac{3\pi }{25}$
4. $\frac{2\pi }{15}$
Given that, a circle is inscribed in a rhombus with diagonal $12 \; \text{cm}$ and $16 \; \text{cm}.$
First, we can draw the figure.
Let $\text{O}$ be the point of intersection of the diagonals of the rhombus and the center of the circle also.
Then, $\text{OE} \perp \text{AB}$
Let the radius of the circle $= \text{OE} = r \; \text{cm}.$
Applying the pythagoras theorem to the $\triangle \text{AOB},$ we get.
$(\text{AB})^{2} = (\text{AO})^{2} + (\text{OB})^{2}$
$\Rightarrow (\text{AB})^{2} = 8^{2} + 6^{2}$
$\Rightarrow (\text{AB})^{2} = 64 + 36 = 100$
$\Rightarrow \text{AB} = \sqrt{100}$
$\Rightarrow \boxed{\text{AB} = 10 \; \text{cm}}$
Now, on considering the $\triangle \text{AOD},$ we can calculate its area in two ways.
Using hypotenuse $\text{AB}$ as the base, or using $\text{OB}$ as the base. The area will remain the same.
So, $\frac{1}{2} \times \text{AB} \times \text{OE} = \frac{1}{2} \times \text{OB} \times \text{OA}$
$\Rightarrow 10 \times r = 6 \times 8$
$\Rightarrow r = \frac{24}{5}$
$\Rightarrow \boxed{r = 4.8 \; \text{cm}}$
Now, the area of the circle $= \pi r^{2} = \pi \times (4.8)^{2} = 23.04 \pi \; \text{cm}^{2}$
And the area of rhombus $= \dfrac{1}{2} \times (\text{Product of the diagonal lengths}) = \frac{1}{2} \times 12 \times 16 = 96 \; \text{cm}^{2}$
$\therefore$ The ratio of the area of the circle to the area of rhombus $= \dfrac{23.04 \pi \; \text{cm}^{2}}{96 \; \text{cm}^{2}} = \dfrac{6 \pi}{25}.$
Correct Answer $: \text{B}$
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Question
# If $1 + 6 + 11 + 16 + .......... + x = 148$ , then the value of x is$\left( a \right)36$ $\left( b \right)35$ $\left( c \right) - 36$ $\left( d \right)$None of these
Hint: Use formula of nth term of an A.P ${a_n} = a + \left( {n - 1} \right)d$ and also use sum of n term of an A.P ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where a is a first term, d is a common difference and n is number of terms.
Given series, $1 + 6 + 11 + 16 + .......... + x = 148$
First we check type of series,
Common difference, d=6-1=11-6=16-11=5
So, We can see given series form an A.P with first term, a=1 and common difference, d=5.
Now, last term of an A.P is ${a_n} = x$ .So, we apply formula of nth term of an A.P ${a_n} = a + \left( {n - 1} \right)d \\ \Rightarrow x = 1 + \left( {n - 1} \right)5 \\ \Rightarrow x = 1 + 5n - 5 \\ \Rightarrow x = 5n - 4..........\left( 1 \right) \\$
Given, sum of n terms of an A.P ${S_n} = 148$ . So, we use the formula of sum of n terms of an A.P.
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) \\ \Rightarrow 148 = \dfrac{n}{2}\left( {2 \times 1 + \left( {n - 1} \right) \times 5} \right) \\ \Rightarrow 296 = n\left( {5n - 3} \right) \\ \Rightarrow 5{n^2} - 3n - 296 = 0 \\$
Now, factories the quadratic equation .
$\Rightarrow \left( {n - 8} \right)\left( {5n + 37} \right) = 0 \\ \Rightarrow n = 8,\dfrac{{ - 37}}{5} \\$
We know the number of terms cannot be negative so we eliminate the negative value.
So, $n = 8$
Now, put the value of n in (1) equation.
$\Rightarrow x = 5 \times 8 - 4 \\ \Rightarrow x = 40 - 4 \\ \Rightarrow x = 36 \\$
So, the correct option is (a).
Note: Whenever we face such types of problems we use some important points. First we check which type of series formed then we apply the formula of nth term and sum of n terms then after some calculation we can get the required answer. |
# Math Snap
## What is the sum of the series $\sum_{n=1}^{\infty}\left(\frac{2}{3}\right)\left(\frac{3}{2}\right)^{n}$ Select one: a. $\frac{-4}{3}$ b. The series diverges C. 1 d. None of the choices e. $\frac{9}{2}$
#### STEP 1
Assumptions 1. The series is given by $\sum_{n=1}^{\infty}\left(\frac{2}{3}\right)\left(\frac{3}{2}\right)^{n}$. 2. We need to determine if the series converges or diverges. 3. If the series converges, we need to find its sum.
#### STEP 2
First, observe that the series can be written in a more familiar form. We can factor out the constant term $\frac{2}{3}$: $\sum_{n=1}^{\infty}\left(\frac{2}{3}\right)\left(\frac{3}{2}\right)^{n} = \frac{2}{3} \sum_{n=1}^{\infty} \left(\frac{3}{2}\right)^{n}$
#### STEP 3
Next, recognize that the series $\sum_{n=1}^{\infty} \left(\frac{3}{2}\right)^{n}$ is a geometric series with the first term $a = \left(\frac{3}{2}\right)^{1} = \frac{3}{2}$ and common ratio $r = \frac{3}{2}$.
#### STEP 4
For a geometric series $\sum_{n=0}^{\infty} ar^n$ to converge, the common ratio $r$ must satisfy $|r| < 1$. In this case, $|\frac{3}{2}| = 1.5$, which is greater than 1.
#### STEP 5
Since $|\frac{3}{2}| > 1$, the geometric series $\sum_{n=1}^{\infty} \left(\frac{3}{2}\right)^{n}$ diverges.
#### STEP 6
Because the inner series $\sum_{n=1}^{\infty} \left(\frac{3}{2}\right)^{n}$ diverges, the original series $\sum_{n=1}^{\infty}\left(\frac{2}{3}\right)\left(\frac{3}{2}\right)^{n}$ also diverges.
##### SOLUTION
Therefore, the correct answer is: $\boxed{\text{The series diverges}}$ |
Solve for: y=\sqrt[3]{5x}
Expression: $y=\sqrt[3]{5x}$
Take the derivative of both sides
$y '=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \sqrt[3]{5x} \right)$
Use $\sqrt[n]{{a}^{m}}={a}^{\frac{ m }{ n }}$ to transform the expression
$y '=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( {\left( 5x \right)}^{\frac{ 1 }{ 3 }} \right)$
To raise a product to a power, raise each factor to that power
$y '=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( {5}^{\frac{ 1 }{ 3 }} \times {x}^{\frac{ 1 }{ 3 }} \right)$
Use differentiation rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( a \times f \right)=a \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( f \right)$
$y '={5}^{\frac{ 1 }{ 3 }} \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( {x}^{\frac{ 1 }{ 3 }} \right)$
Use $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( {x}^{n} \right)=n \times {x}^{n-1}$ to find derivative
$y '={5}^{\frac{ 1 }{ 3 }} \times \frac{ 1 }{ 3 }{x}^{-\frac{ 2 }{ 3 }}$
Simplify the expression
$y '=\frac{ \sqrt[3]{5} }{ 3\sqrt[3]{{x}^{2}} }$
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Factors that 35 space the list of integers that we can separation evenly right into 35. It has actually a full of 4 components of which 35 is the greatest factor and also the positive factors of 35 room 1, 5, 7 and 35. The Prime factors of 35 space 1, 5, 7, 35 and its factors in Pairs room (1, 35) and also (5, 7).
You are watching: What are the common factors of 35
Factors the 35: 1, 5, 7 and 35Negative components of 35: -1, -5, -7 and -35Prime determinants of 35: 5, 7Prime factorization of 35: 5 × 7 = 5 × 7Sum of factors of 35: 48
1 What room the factors of 35? 2 How come Calculate components of 35? 3 Factors of 35 by element Factorization 4 Factors the 35 in Pairs 5 Tips and Tricks 6 FAQs on determinants of 35
## What are the factors of 35?
Factors that 35 are every the numbers which room multiplied to get 35 as the product. Factors that 35 space 1, 5, 7, and 35.
## How to calculation the determinants of 35?
Let"s learn just how to calculate components of 35.
Find the two numbers who product gives 35. We know the product of 1 and 35 provides 35.Since that is a multiple of 5, we know that 5 × 7 = 35.We cannot find other number such that the product is 35.
Therefore the numbers 1,5, 7, and also 35 are the components of 35.
Explore factors using illustrations and also interactive examples
## Factors of 35 by element Factorization
Factors the 35 by prime factorization are given by making use of the following steps.
Find the smallest prime number through which 35 can be divided without leaving any kind of remainder.We see that 35 is divisible through 5.We deserve to express 35 together 5 × 7.
We stop the variable tree as we can not branch further as 5 and 7 space prime numbers. As such prime administer of 35 is 5 × 7
Prime administer can also be shown by upside down department method:
## Factors of 35 in Pairs
The pair of number that renders 35 ~ above multiplication space the factors of 35 in pairs.
Positive factor-pairs Negative factor-pairs ( 1, 35) ( -1, -35) (5, 7) (-5, -7)
Tips and also Tricks:
1 is the smallest element of every number. Hence, that is a aspect of 35.35 ends in 5 thus, 5 is one the its factors.35 is a semi-prime number, it can be expressed as a product of 2 prime numbers.
## Factors the 35 resolved Examples
Example 1: Jack went to a publication outlet. He wanted to to buy 35 books. If he have the right to buy just 5 books at once, how many times does he need to go to publication outlet to buy every the books?
Solution:
Jack have the right to buy only 5 books in ~ once. The pair element of 5 which gives 35 together the product will give the answer to our problem.We know that 5 × _____ = 35. Therefore, Jack will go 7 times to the publication outlet to buy all the books.
Example 2: Justin has 35 pizza boxes. In how countless ways can he team the pizza box such that each stack it s okay equal pizza boxes?
Solution:
Justin has actually 35 pizzas. That can team them in 4 ways. He deserve to have a ridge of 1 pizza box and also 35 such groups.He might prefer having actually stack that 5 pizza boxes and 7 together groups.He could additionally have stack of 7 pizza boxes and also 5 such groups.He can have stack of 35 pizza boxes and also only 1 group. Hence he can stack the pizza boxes in 4 ways.
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## FAQs on components of 35
### What are the determinants of 35?
The determinants of 35 room 1, 5, 7, 35 and its negative factors are -1, -5, -7, -35.
### What is the Greatest usual Factor that 35 and 33?
The determinants of 35 space 1, 5, 7, 35 and also the components of 33 room 1, 3, 11, 33. 35 and also 33 have actually only one common factor i beg your pardon is 1. This suggests that 35 and 33 room co-prime.Hence, the Greatest typical Factor (GCF) the 35 and also 33 is 1.
### What space the Prime components of 35?
The prime determinants of 35 room 5, 7.
See more: What Is The Second Largest Bone In The Body ? Intraosseous Infusion
### What is the amount of the determinants of 35?
Factors of 35 room 1, 5, 7, 35 and, the amount of every these factors is 1 + 5 + 7 + 35 = 48
### What space the typical Factors that 35 and 11?
Since, the components of 35 space 1, 5, 7, 35 and factors that 11 space 1, 11. Hence, 35 and 11 have only one common factor i m sorry is 1. Therefore, 35 and 11 are co-prime. |
## Possible Curriculum Maps for 3/4 Combo Class and 4/5 Combo Class
Possible 3-4 Curriculum Map
3 M1: Properties of Multiplication and Division and Solving Problems with Units of 2-5 and 10 (25 days) M2: Place Value and Problem Solving with Units of Measure (25 days) M3: Multiplication and Division with Units with 0, 1, 6-9 and Multiples of 10 (25 days) M7: Geometry and Measurement Word Problems (40 days) DO TOPICS B & C M5: Fractions as Numbers on the Number Line (35 days) M6: Collecting and Displaying Data (10 days) M4: Multiplication and Area (20 days) M7: Geometry and Measurement Word Problems (40 days) TOPICS D &E 4 M2: Unit Conversions (7 days) M3: Multi-Digit Multiplication and Division (43 days) – TOPICS A & B, E & F M1: Place Value, Rounding, and Algorithms for Addition and Subtraction (25 days) Finish M3: Multi-Digit Multiplication and Division (43 days) TOPICS C, D, G, H M4: Angle Measure and Plane Figures (20 days) M5: Fraction Equivalence, Ordering and Operations (45 days M6: Decimal Fractions (20 days) M7: Exploring Measurement with Multiplication (20 days)
Possible 4-5 Curriculum Map
4 M1: Place Value, Rounding, and Algorithms for Addition and Subtraction (25 days) You may want to keep 4/5 together for whole number renaming and place value work (maybe decimal place value, too and incorporate M6 here) and combine ideas from Module 2 with the 5th grade place value work. M2: Unit Conversions (7 days) Consider doing this unit while 5th grade is doing place value work with exponents of 10. M3: Multi-Digit Multiplication and Division (43 days) M5: Fraction Equivalence, Ordering and Operations (45 days ) Depending on where your students are, you may want to consider keeping them together here for some of the 4th grade foundational work in fractions M6: Decimal Fractions (20 days M7: Exploring Measurement with Multiplication (20 days) M4: Angle Measure and Plane Figures (20 days) 5 M1: Place Value and Decimal Fractions: adding, subtracting, multiplying and dividing decimals (20 days) Multiplying and dividing will be revisited. May want to spend extra time on the place value work. M2: Multi-Digit Whole Number and Decimal Fraction Operations (35 days) M3: Addition and Subtraction of Fractions (22 days) Start M4: Multiplication and Division of Fractions and Decimal Fractions (38 days) M4: Multiplication and Division of Fractions and Decimal Fractions (38 days) M5: Addition and Multiplication with Volume and Area (25 days) M6: Problem Solving with the Coordinate Plane (40 Days) |
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Posted on : 2011-08-10
Part 2
The word ‘heuristic’ is taken directly from the Greek verb, heuriskein which means ‘to discover’. In Mathematics, there are usually different ways to go about solving problem sums. These ways or methods are known as heuristics.
Heuristics can be divided into 4 main types, which another 2 types would be discussed in this article.
Three: Going through the process
· The heuristic ‘act it out’ requires pupils to use physical objects or manipulatives to represent information. This skill is used to introduce new concepts and to allow pupils to explore the concepts using manipulatives.
Example: The figure is made of 17 sticks. Move 4 sticks to form 8 squares. Solution:
Pupils can use matchsticks to represent the information and use the matchsticks to find the solution to the problem.
· Pupils can apply the ‘work backwards’ method when given a problem that provides the final result and that requires them to find the initial quantit
Example: There were some chocolates in a basket. Michael and three of his friends took 8 chocolates each. 25 chocolates were given to Shirley. There were then 11 chocolates left in the basket. How many chocolates were there in the basket at first? Solution: Using the ‘work backwards’ method, first find the number of chocolates taken away by Michael and his three friends. 4 × 8 = 32 chocolates Then, find the total number of chocolates taken away from the basket. 32 + 25 = 57 chocolates Number of chocolates in the basket at first = 57 + 11 = 68 chocolates
· Pupils can use the ‘before-after concept’ for problems that provide information given before and after the event to find the unknown. This skill allows pupils to compare the information and relate different events together to solve the problems.
Example: Jacky had \$56 more than Jill. When he spent \$108, Jill had thrice as much as what he had left. How much did Jacky have at first? Solution: 2 units → \$108 - \$56 = \$52 1 unit → \$52 ÷ 2 = \$26 \$26 + \$108 = \$134 Jacky had \$134 at first.
Four: Changing the problem
· By restating a problem in another way, pupils can view the problem in another perspective.
Example: Find the sum of 10 + 12 + 14 + … + 146 + 148 + 150. Solution: There are 75 numbers in the sum 2 + 4 + 6 + … + 146 + 148 + 150. There are 4 numbers in the sum 2 + 4 + 6 + 8. 75 – 4 = 71 Hence, there are 71 numbers in the sum 10 + 12 + 14 + … + 146 + 148 + 150. There are 35 pairs of 160 and a ‘80’. 10 + 12 + 14 + … + 146 + 148 + 150 = 35 × 160 + 80 = 5680
· When facing a complex problem, pupils can split the problem into smaller parts and start by solving the simpler parts. After doing so, the problem is simplified and solving the problem is much easier.
Thus, do try to keep in mind the various heuristics and apply them when you are doing problem sums. With regular practice, you will be able to handle problem sums with ease.
Written by Gui Yan Tong
To understand more about the other 2 types covered in Part 1, please click here.
Check out the series "Use of Heuristics in Problem Solving" by EPH and have more practice on the different types of heuristics
Extra!
For more pointers on Model Drawing and Guess and Check, check out these related articles at the Popular Community:
Posted on : 2011-08-10
Part 1
The word ‘heuristic’ is taken directly from the Greek verb, heuriskein which means ‘to discover’. In Mathematics, there are usually different ways to go about solving problem sums. These ways or methods are known as heuristics.
Heuristics can be divided into 4 main types, which will be covered in this 2-part article.
One: Giving a representation
· Pupils can transform word problems into pictorial representations and represent information with a diagram/model. This skill helps pupils to understand the question better when they see the visual representation of the word problems.
· A systematic list should be made for word problems that require pupils to identify patterns such as repeated numbers or a series of events that repeat. This skill helps pupils in identifying patterns easily as the list organises all possible answers systematically.
Example: Michele saved \$150 on the first month. On the second month, she saved \$60 more than the first month. On the third month, she saved \$70 more than the second month. On the fourth month, she saved \$55 more than the third month. How much did she save in four months? Solution: Making a list: 1st month → \$150 2nd month → \$150 + \$60 = \$210 3rd month → \$210 + \$70 = \$280 4th month → \$280 + \$55 = \$335 Total amount saved = \$150 + \$210 + \$280 + \$335 = \$975 She saved \$975 in four month
Two: Making a calculated guess
· The ‘guess and check’ method is used for word problems when certain information is lacking. It requires them to make a guess first and check it, and making subsequent guesses and checks until the correct answer is derived. It is often used together with a systematic list as it helps pupils to narrow down the possibilities within a short time frame.
Example:
Jenny has a total of 7 dogs and parrots. The animals have 20 legs altogether.
How many dogs does she have?
Solution:
Using the ‘guess and check’ method,
Number of dogs Number of legs Number of parrots Number of legs Total number of legs Check 1 1 x 4 = 4 6 6 x 2 = 12 4 + 12 = 16 X 2 2 x 4 = 8 5 5 x 2 = 10 8 + 10 = 18 X 3 3 x 4 = 12 4 4 x 2 = 8 12 + 8 = 20 √
She has 3 dogs.
· The ‘look for patterns’ method is usually used by pupils when they have to identify a certain pattern in a number sequence.
Example:
12 4 5 13 11 9 16 18 16 12 3 X
Solution:
Making a list of possibilities:
12 - 4 + 5 = 13
11 - 9 + 16 = 18
16 - 12 + 3 = 7
The value of X is 7.
Hence by using the systematic list, it is more effective to find the underlying pattern.
Written by Gui Yan Tong
Check out the series "Use of Heuristics in Problem Solving" by EPH and have more practice on the different types of heuristics
Extra!
For more pointers on Model Drawing and Guess and Check, check out these related articles at the Popular Community:
Posted on : 2011-04-08
This is a tricky question that all Chinese educators in Singapore seek to answer. With the MOE’s increasing emphasis on Mother Tongue as well as the increasing usage of Chinese language over the years, Chinese educators in Singapore are striving to get more pupils to be proficient in the language. Without a doubt, the lack of general interest in Chinese language proves to be the major obstacle for pupils. However, parents can play an instrumental role in driving the interest in Chinese learning.
Pictures speak a thousand words
A pupil once told me that seeing rows and rows of Chinese characters naturally drove her to sleep. I do not quite agree with her statement, but I do agree that using pictures and graphics can definitely help to improve the attention span of a child.
Instead of forcing a child to read a story that he or she cannot comprehend, it will benefit the child more if the parent is willing to read the story to the child instead.
The key here is to read the story and interact with the child.
Interaction involves leading the child to describe the picture using a few complete sentences. In addition, parents can also motivate their child to verbalise their thoughts after reading the story.
When choosing a storybook, parents should always remember to let their child decide the choice of book. It is better to read a story that interests a child than to read a story that will benefit a child.
Thus, interaction through reading stories is an effective way to develop a child’s interest in learning Chinese.
Incorporating into interests
When children enjoy the learning process, they will be more interested to find out more. Parents should try to link their children’s key interests and hobbies together with language learning to encourage effective learning.
For example, if a child is only interested in video games, perhaps the parents can consider purchasing Chinese video games. Not only will the child be interested to play the game, it will also stir the child’s curiosity to find out more about the contents of the game. The result is that learning becomes fun.
With proper guidance and monitoring, multimedia will prove to be a useful tool for parents to make learning Chinese an enjoyable activity.
Making it relevant
Parents can also try to incorporate Chinese in their children’s daily lives. Besides encouraging them to watch Chinese television programmes and cartoons, they can also conduct some daily conversations in the language and even teach them some popular songs as well. Parents can also encourage their children to communicate in the language with their neighbours or with shop owners and other people in the neighbourhood.
When children realise how useful and relevant the language is to their daily lives, they will be more receptive towards learning the language.
Thus, through the use of these methods, children will be more exposed to the Chinese language and will gradually grow to become competent users of the language.
Written by Eric Pang
Posted on : 2011-08-10
` `
In linguistics, grammar refers to the rules of the language. When we learn about grammar in the English language, we learn about the set of rules which governs how a sentence or phrase is composed. Besides building up a good vocabulary, the key to speaking and writing good English lies in your ability to first understand how a sentence is formed and how it can be used.
More often than not, your child will find the process of learning grammar tedious and boring. Seeing as how children are still too young to understand the importance of grammar, trying to force them to memorise a seemingly endless set of rules will be a tiresome chore.
Here are some ways in which you can make learning grammar fun and interactive for your child.
1. Provide examples that they can relate to. Instead of textbook examples, try engaging their attention by talking about a real-life example. An example about your child’s recent trip to the zoo or aquarium will more likely capture their attention and interest.
2. Use humorous examples. You may have noticed that children find it much easier to remember situations or examples which they find funny. This also changes their mindset that learning grammar is boring and mundane.
3. Read widely and wisely. Select books which are informative and entertaining as your child will be more inclined to read them.
You can also purchase books on grammar to give your child adequate practice. Try searching for books with more illustrations, stories and examples which your child will be able to relate to.
Hopefully, with these tips, you can help your child to improve his or her grasp of the English grammar such that they will grow to be proficient users of the language.
Written by Michelle Lim
You may want to look at the series Fun with Grammar by EPH where grammar components are skilfully weaved into the storyline and multiple illustrations are provided.
Extra!
What are some common grammar mistakes to watch out for? Find out in this related article at the Popular EduCommunity:
Posted on : 2011-04-08
Part 2
Previously, we have shared some tips on good study habits and how to tackle MCQ questions to achieve good grades for PSLE Science. Now let's explore some tips on how to take on another type of questions in the examination - Open-ended Questions
Open-ended Questions
• For open-ended questions, circle the keywords in each question to help you to focus on what is being asked. It is important that you understand the question.
Example: What are the products formed during photosynthesis? To answer such a question, first, identify the keywords. The keywords are ‘photosynthesis’ and ‘products’. When you see ‘photosynthesis’, you should recall what this word means. Photosynthesis is a process whereby plants make food in the presence of sunlight. The question is asking about the products of photosynthesis. This means that you have to provide more than one answer to the question. Recall that food (sugar) is made in plants and oxygen is given out during photosynthesis to replenish the air. Hence, the answers are sugarand oxygen.
• Note the question terms used in the question. This will help you understand what answer is required of you in order to score full marks for that particular question.
Example: When a magnet is placed near object X, object X does not move. When the magnet is placed near object Y, object Y moves away from it. (a) Explain why object X does not move when the magnet is placed near it. (b) State the possible materials that can be used to make object Y. For (a), the question term used is ‘Explain’. ‘Explain’ means that you have to give reasons for the observation stated. The observation is that object X does not move. Object X does not move because it is neither a magnet nor an object made of magnetic material. If object X is a magnet or a magnetic material, it will either be attracted to or be repelled by the magnet. Thus, the correct answer is ‘Object X is made of a non-magnetic material, hence it is not affected by the magnet.’ For (b), the question term used is ‘State’. State means that you only need to write short answers, meaning a few words or short phrases. Since object Y moves away from the magnet, object Y is also a magnet. Magnets are made of or can be made by one of the four types of magnetic materials. Hence, the answer for (b) is ‘Iron, steel, nickel or cobalt’.
Example: 50 cm3 of air is being pumped into a 30 cm3 container. (a) What is the volume of air inside the container? (b) Explain your answer in (a). For (a), the question asks about the property of air (gas). Recall that air does not have a fixed volume and a fixed shape and so the volume of air inside the container is 30 cm3 instead of 50 cm3. For (b), the question is asking for a reason. As mentioned earlier, air does not have a fixed volume and a fixed shape. We can only see the shape of the air changing when it is being transferred from one container to another of a different shape. Hence, we cannot write ‘Air does not have a fixed shape’ as the answer. Air does not have a fixed volume, hence its volume changes from 50 cm3 to 30 cm3. Writing ‘Air does not have a fixed volume’ as the answer is not wrong, but there is a better word to use for this question. The keywords are ‘can be compressed’ or ‘compressible’. Air does not have a fixed volume and hence it can be compressed. The concise answer to this question would be ‘Air can be compressed’ or ‘Air is compressible’.
Do note that constant revision is also essential in preparing for the examination. Hopefully, with these tips in mind, you will find that Science is not as difficult as you might think and will be on your way to achieve your ideal grade.
Written by: Tan Chiang Heng
Click here for Part 1 of the article |
Sunday, July 14, 2013
# How to solve an equation that has a variable in the denominator?
2:20 AM
The equations that have variables in the denominator are termed as rational equations. Even though this may be a new name for you, it uses the same concepts used in solving regular algebra equations.
Here are some e xamples of rational equations:
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{3}{x}=x+4$
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{5+y}{y}=8$
To solve rational equations, you should first convert them into a standard linear or quadratic form. This can be done using one or more of the following steps. Usually the simpler rational equations may not involve all of these steps:
1. Isolate the terms with variable on one side and constants on the other side
2. Add or subtract terms with variable
3. Cross-multiply both sides to get a standard linear or quadratic equation
Let’s take an example involving a simple rational equation that converts to a linear form:
Example:
What is the value of $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} z$ if $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{3z}{5-z}=12$?
A) 1
B) 2
C) 3
D) 4
E) 5
Solution:
Step 1: Here we see that all variable terms are already on one side. Therefore we first use the cross-multiplication to simplify the equation:
\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \begin {align*} \frac{3z}{5-z}&=12 \\ 3z&=12\times (5-z) \end{align*}
Step 2: Next we get rid of the parentheses:
\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \begin {align*} 3z&=60-12z \\ \end{align*}
Step 3: Finally isolate the variable to solve the equation
\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \begin {align*} 3z&=60-12z \\ 15z&=60 \\ z&=\frac{60}{15} \\ z&=4 \\ \end{align*}
Sometimes you can also simplify the equation into a linear or quadratic form a lot quicker by multiplying both sides of the equation with the denominator containing the variable term. Let’s look at a grid-in example:
Example:
If $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{2}{x} - 3 = -x$ what is a possible value of $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} x$?
Solution:
This question can be easily solved by plugging in the values if you have the luxury of answer choices. But alas! This seems to be a fill in the blank type of question. Sadly you will have to solve this question using math concepts.
Step 1: To solve the question we need to eliminate the $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} x$ in the denominator. This can be done by multiplying both sides of the equation by $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} x$
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} 2-3x=-x^2 \Rightarrow x^2-3x+2=0$
Step 2: The next step is to factorize the quadratic equation:
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \Rightarrow x^2 - x - 2x + 2 = 0$
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \Rightarrow (x-1)(x-2)=0$
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \Rightarrow x=1, x=2$
Either 1 or 2 could be the correct answer. |
The Normal Curve Main Concepts | Demonstration | Activity | Teaching Tips | Data Collection & Analysis | Practice Questions | Milestone
Solutions to Practice Problems 1. Many textbooks state that a value that falls more than 1.5 interquartile ranges above the upper quartile or 1.5 interquartile ranges below the lower quartile is an "outlier". What percent of the values in a normal distribution are considered "outliers" using this definition? Since the tail area for the lowest 25% has a z= -0.67 and the z for the upper 25% is z = 0.67, the interquartile range is 1.34. 1.5*1.34=2.01. So we need to find the percent that falls above .67+2.01 = 2.68 and the percent that falls below -2.68. The tail area is 2*0.0037 = 0.0074 = .74%. Note: In this problem, we based our calculations for the IQR on the population distribution (the normal curve). When making a boxplot, the IQR is based on the sample. The sample has variability in it, also, and so in practice we identify more than .74% of our observations as outliers. 2. Imagine two outstanding students who have excelled in academics, service, and athletics. Anita earned 620 on the SAT verbal and Bonita scored 27 on the ACT. In other respects, their high school records are comparable. You learn that the average SAT verbal score is 500, with standard deviation of 100 points while the ACT has a mean of 18 with a standard deviation of 6. If we assume that scores on these exams are approximately normal, can you use thier scores to determine who wins the scholarship? Bonita wins since her score is 1.5SD's above average and Anita's score is 1.2 SD's above average. 3. Anita scored at the 60th percentile on a midterm exam while Bonita scores at the 85th percentile. If we know that the midterm had a mean of 70 and standard deviation of 10 points, how many points separate Anita and Bonita on this exam? Anita's score = 70+.26(10) = 72.6 Bonita's score = 70+1.04(10) = 80.4 The difference in scores is 80.4-72.6 = 7.8 points. 4. National Fruit Company claims that the weights of their forty pound boxes of imported bananas are approximately normal with mean 41 pounds and standard deviation of 4 ounces. How often will a box of bananas be underweight? Since z = (40-41)/0.25 = -4, it is unlikely that any box will be underweight. 5. For women athletes at UCLA, 62.5 inches is the 25th percentile height and 65.5 inches is the 75th percentile height. a) If we assume that heights for women athletes are approximately normal, find the mean and standard deviation of the distribution of heights. b) Use the calculations to determine the 90th percentile height for women athletes at UCLA. Using the equations 65.6 = mu + 0.67sigma and 62.5 = mu - 0.67 sigma, we have 128 = 2mu or mu = 64 inches and sigma = 2.24 inches. Then the 90th percentile height is 64+1.28(2.24) = 66.87 inches. |
# Texas Go Math Grade 4 Lesson 17.6 Answer Key Use Stem-and-Leaf Plots
Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 17.6 Answer Key Use Stem-and-Leaf Plots.
## Texas Go Math Grade 4 Lesson 17.6 Answer Key Use Stem-and-Leaf Plots
Essential Question
How do you solve problems using a stem-and-leaf plot?
Step 1:
Find the least number and greatest number in the data set.
Step 2:
Draw the vertical line and write the digits in the tens places from 1 to 3 on the left of the line.
Unlock the Problem
Example 1
While doing research for a project, Lila made a stem-and-leaf plot of the number of floors that different buildings in Chicago have. How many buildings have more than 40 floors?
Think: 41 is represented by 4 | 1 on the stem-and-leaf plot.
The number of floors in the buildings that have more than 40 floors are: _______________
So, ___________ buildings have more than 40 floors.
The number of floors in the buildings that have more than 40 floors is 6.
Count the number of leaves that are after stems 4, 5, 6.
So, 6 buildings have more than 40 floors.
Example 2
Each time Glenda practised her free throws, she recorded the number of made baskets in a stem-and-leaf plot. Flow many times did Glenda make more than 20 free throws?
The number of times Glenda practised her free throws and made more than 20 of them was: ______________
So, Glenda made more than 20 free throws ______________ times.
The number of times Glenda practised her free throws and made more than 20 of them was: 12
So, Glenda made more than 20 free throws 12 times.
Share and Show
Use the stem-and-leaf plot for 1-3.
Question 1.
Martin kept track of the time he spent reading in a stem-and-leaf plot. How many times did Martin read for 40 or more minutes?
Think: Count the number of leaves that are after stems 4, 5, 6.
Martin read for 40 or more minutes ____________ times.
Martin read for 40 or more minutes 10 times.
Need to count more than 40.
count the leaves of 4 are 2, 5, 5, 5
Count the leaves for 5 are 0, 2, 3, 5, 5.
Count the leaves for 6 are 3.
Question 2.
How many times did Martin read for less than 30 minutes?
The question asked the number of times did martin read for less than 30 minutes. So, I circled stem 1 and 2 leaves.
Count the number of leaves of the stem 1 is 3, 5, 5, 5.
Count the number of leaves of stem 2 are 0, 0, 0, 5, 8, 8.
The total number of leaves of stem 1 is 4
The total number of leaves of stem 2 is 6
The number of times did martin read for less than 30 minutes is 6+4=10.
Question 3.
How many more times did Martin read for less than 39 minutes than he read for more than 39 minutes?
The blue line represents less than 39 minutes.
The pink line represents more than 39 minutes.
Count the number of leaves of stem 1 is 3, 5, 5, 5.
Count the number of leaves of stem 2 is 0, 0, 0, 5, 8, 8.
Count the number of leaves of stem 3 is 0, 0, 3, 3, 5. (I didn’t return 9 here because here asked less than 39 minutes).
Count the number of leaves of stem 4 is 0, 0, 2, 5, 5, 5.
Count the number of leaves of stem 5 is 0, 2, 3, 5, 5.
Count the number of leaves of stem 6 is 3.
The total number of leaves of stem 1 is 4.
The total number of leaves of stem 2 is 6.
The total number of leaves of stem 3 is 5.
The total number of leaves of stem 4 is 6.
The total number of leaves of stem 5 is 5.
The total number of leaves of stem 6 is 1.
The total number of leaves is less than 39 are 15.
The total number of leaves is more than 39 are 12.
Subtract those two numbers to get the answer.
The number of times did Martin read for less than 39 minutes than he read for more than 39 minutes:15-12=3.
Therefore, 3 more times.
Math Talk
Mathematical Processes
Explain how you found the answer to Problem 3.
Step 1: Count the leaves that are present in less than 39 minutes.
Step 2: Count the leaves that are presently more than 39 minutes.
Step 3: Count the total leaves of less than 39 minutes.
Step 4: Count the total leaves of more than 39 minutes.
Step 5: Subtract the two numbers and then get the answer.
Problem Solving
Use the stem-and-leaf plot for 4-7.
Question 4.
Stephanie asked her 23 classmates how much time they spent doing chores in a week. She recorded the data in a stem-and-leaf plot. How many classmates said that they spend some time doing chores in a week?
Count all the leaves.
Count the number of leaves of stem 2 is 2, 2, 4, 6.
Count the number of leaves of stem 3 is 0, 5, 5, 8.
Count the number of leaves of stem 4 is 0, 6.
Count the number of leaves of stem 5 is 5, 8
Count the number of leaves of stem 6 is 2.
Count the number of leaves of stem 7 is 1, 4.
Now count all the leaves of stem 2, 3, 4, 5, 6, 7 is 15.
Therefore, 15 classmates said they spend time doing chores in a week.
Question 5.
How many of Stephanie’s classmates said they did more than an hour of chores a week?
There are 3 classmates who do chores for more than an hour a week. These can be explained in detail and also represented in the above diagram.
6| 2 represented as 62 minutes. (1 hour 2 minutes, it is more than an hour)
7| 1 represented as 71 minutes.
7| 4 represented as 74 minutes.
Now count all those leaves of stem 6 and 7:2, 1, 4
The total number of leaves: 3.
Question 6.
H.O.T. Multi-Step how many classmates said that they spent more than 20 minutes and less than 40 minutes doing chores a week?
There are 8 classmates who spend more than 20 minutes and less than 40 minutes. These can be explained in detail and also represented in the above diagram:
The leaves of stem 2 are 2, 2, 4, 6.
The leaves of stem 3 are 0, 5, 5, 8.
2| 2 represented as 22 minutes.
2| 2 represented as 22 minutes.
2| 4 represented as 24 minutes.
2| 6 represented as 26 minutes.
3| 0 represented as 30 minutes.
3| 5 represented as 35 minutes.
3| 5 represented as 35 minutes.
3| 8 represented as 38 minutes.
These all represent more than 20 minutes and less than 40 minutes. So I circled, leaves of stems 2 and 3.
Now count all the leaves of 2 and 3: 8.
Question 7.
How many classmates said they didn’t do any chores? Explain your answer.
There are no students who said that they don’t do any chores. The table also gives the time of classmates who are doing chores in a week. There is no representation of not doing chores.
Use the stem-and-leaf plot for 8-10.
Question 8.
Tina records the daily low temperature for 15 days.
She recorded the data in the stern-and-leaf plot.
How many days was the low temperature in the 50s?
Explanation:
The low temperature in the 50s was 5 days.
Count all the leaves of stem 5 are 0, 3, 4, 9, 9.
Question 9.
Use Graphs On how many days was the daily low temperature in the 30s and 40s?
The number of days the daily low temperature in the 30s and 40s:7
Count the leaves of stem 3: 7, 9, 9.
Count the leaves of stem 4: 1, 4, 4, 8.
3| 7 represents 37 degrees.
3| 9 represents 39 degrees. Likewise, write all the keys for better understanding.
The total number of leaves of 3 and 4 stems are 7.
Question 10.
H.O.T. Use Math Language Explain how to find how many more days the low temperature was greater than 53°F than less than 53°F.
The days the low temperature was greater than 53 degrees are 8.
The days the low temperature was lesser than 53 degrees are 6.
Count the leaves of 3, 4 and 5 stems. In 5th stem up to zero only because here asked less than 53 degrees.
Count the leaves of 5 and 6 stems. In 5th stem count after 3 onwards because here asked greater than 53 degrees.
Now subtract the days:8-6=2.
Therefore, 2 days more the low temperature was greater than 53°F than less than 53°F.
Use the stem-and-leaf plot for 11-14.
Question 11.
Nick recorded the number of points his basketball team scored during their season in a stem-and-leaf plot. How many games did Nick’s basketball team play?
The team played 20 games.
Count all the leaves of the stem 4, 5, 6, 7, 8, 9.
Question 12.
During how many games did Nick’s team score between 55 and 75 points?
The number of games did Nick’s team score between 55 and 75 points:10.
Count the leaves of stem 5, 6, and 7. The question asked was between 55 and 75. This means we can count from 56 to 74. So, when we count 56 to 74 then we will get the answer. I represented in the above diagram. Count those numbers which I marked in the diagram.
Question 13.
Multi-Step During how many more games did Nick’s team score less than 68 points than they scored more than 68 points?
6| 8 represents 68 points.
The number of games Nick’s team score less than 68 points=11
The number of games Nick’s team score more than 68 points=8
Count the leaves more than 68 and less than 68 that I represented in the above diagram.
Subtract both points to get the answer.
The number of more games did Nick’s team score less than 68 points than they scored more than 68 points=11-8=3.
Therefore, in 3 more games, Nick’s team score less than 68 points than more than 68 points.
Question 14.
Reasoning Explain how the stem-and-leaf plot would change if Nick’s basketball team played 8 more games and they scored more than 65 points each game?
Explanation:
If Nick’s team played 8 more games then total they played 28 games.
And they scored all the games more than 65 points. Then the stem and leaf plot might be like in the above diagram. I added some more leaves more than 65 and If we count all the leaves now the total games are 28.
Question 15.
The stem-and-leaf plot at the right shows the number of programs that different vendors sold during a sporting event. How many vendors sold between 20 and 30 programs?
(A) 1
(B) 4
(C) 5
(D) 3
Explanation:
Count all the leaves between 20 and 30.
The leaves between 20 and 30 are 1, 3, 6, 6, 8.
Use the stem-and-leaf plot for 16-17.
The stem-and-leaf plot at the right shows the bowling scores for members of a bowling team.
Question 16.
What is the highest score that is bowled?
(A) 69
(B) 99
(C) 78
(D) 95
Explanation:
9| 9 represents 99 points.
This is the highest point in the table.
Question 17.
Multi-Step How many team members bowled a 64 or a 65?
(A) 0
(B) 4
(C) 2
(D) 3
Explanation:
Count the marked leaves because they can be represented as:
6| 4 represents 64 points.
6| 5 represents 65 points.
6| 5 represents 65 points.
6| 5 represents 65 points.
TEXAS Test Prep
Question 18.
The stem-and-leaf plot at the right shows the number of stuffed animals Sara and her friends have. How many of Sara’s friends have more than 10 stuffed animals?
(A) 12
(B) 6
(C) 8
(D) 7
Explanation:
Count the leaves having more than 10.
7 of Sara’s friends have more than 10 stuffed animals.
### Texas Go Math Grade 4 Lesson 17.1 Homework and Practice Answer Key
Question 1.
Blake used a stern-and-leaf plot to record the number of football cards that he and his friends have collected. how many friends have collected 50 or more cards?
Explanation:
Count the leaves of stem 5 and stem 6.
The leaves of stem 5 are 1, 3, 5, 6, 9.
The leaves of stem 6 are 1, 4, 7.
Count all those leaves to get the answer.
Total leaves are 8.
Question 2.
How many friends have collected 25 or fewer cards?
count the leaves of stem 1 and stem 2.
The question asked 25 or 25 fewer. So we need to count 25 also.
1| 9 represents 19 cards.
2| 3 represents 23 cards.
2| 5 represents 25 cards.
Total 3 friends collected.
Question 3.
How many friends have collected between 30 and 50 cards?
Answer: 7 friends have collected between 30 and 50 cards.
Count the leaves of stem 3 and stem 4 because the question asked friends collected between 30 and 50. So, we need to count from 30 to 50. Check the numbers having 30 to 50. I circled in the above diagram. Count those leaves. The total leaves are 7.
Question 4.
Multi-Step How many more friends have collected more than 40 cards than less than 40 cards?
Answer: 6 more cards friends have collected.
First, we need to write the leaves having less than 40.
The leaves of stem 1 are 9
The leaves of stem 2 are 3, 5
The leaves of stem 3 are 6, 8, 9
Now write the leaves having more than 40.
The leaves of stem 4 are 2, 2, 4, 8
The leaves of stem 5 are 1, 3, 5, 6, 9
The leaves of stem 6 are 1, 4, 7
Count the leaves less than 40:6
Count the leaves more than 40:12
Subtract those two numbers: 12-6=6.
Problem Solving
Question 5.
The girls on Stacy’s soccer team sold boxes of cards to raise money for new uniforms. Stacy recorded data about their sales in a stem-and-leaf plot. How many girls sold more than 30 boxes of cards?
Answer: 10 girls sold more than 30 boxes of cards.
Count the number of leaves having more than 30 boxes.
3| 1 represents 31 boxes of cards.
3| 3 represents 33 boxes of cards. Likewise, write up to 5| 1. These all are more than 30. Count the leaves.
The total leaves are 10.
Question 6.
What is the greatest number of boxes sold by one girl?
The greatest number of boxes are 6.
Count the leaves of stem 3 are 1, 3, 4, 5, 5, 9.
Stem 3 has the highest number of boxes sold by one girl.
Question 7.
H0w many girls on the team sold cards? Explain.
Count the total leaves in the given stem and leaf plot diagram.
Question 8.
Explain how the stem-and-leaf plot would change if another girl on Stacy’s soccer team sold 60 boxes of cards.
I added 2 in the stem 5. Then a total of 60 is there. I think the stem-and-leaf plot would change if another girl on Stacy’s soccer team sold 60 boxes of cards.
Lesson Check
Use the table at right for 9-11.
Question 9.
The stern-and-leaf plot at the right shows the ages of people who attended a dog obedience class. How many people were between 20 and 40 years old?
(A) 7
(B) 5
(C) 8
(D) 6
Explanation:
Count all the leaves present in stems 2 and 3.
There are 7 people were between 20 and 40 years old.
Question 10.
Which age group was most widely represented at the class?
(A) teens
(B) twenties
(C) thirties
(D) forties
Explanation:
the stem 1 is having the highest number of leaves.
Question 11.
Multi-Step How many people were more than 20 years old than were less than 20 years old?
(A) 8
(B) 1
(C) 2
(D) 7
The leaves of stem 0 are 9
The leaves of stem 1 are 2, 5, 5, 5, 8, 9.
The leaves of stem 2 are 1, 3, 4, 4, 6
The leaves of stem 3 are 3, 4.
The leaves of stem 4 are 2
The total number of leaves that are less than 20=7.
The total number of leaves that are more than 20=8.
Subtract both the numbers: 8-7=1.
Use the table at right for 12-14.
Question 12.
Kyle kept track of the number of laps he swam each day. He used a stem-and-leaf plot to display his data. What is the greatest number of laps that Kyle swam in one day?
(A) 44
(B) 34
(C) 26
(D) 43
Explanation:
The largest number is 34.
3| 4 represents 34.
Question 13.
Multi-Step How many days did Kyle swim 30 or more laps?
(A) 1 days
(B) 3 days
(C) 2 days
(D) 4 days
Explanation:
Count the leaves having 30 0r more than 30.
There are 4 days Kyle swim more than 30 laps.
Question 14.
Multi-Step How many days did Kyle swim 23, 24, or 25 laps?
(A) 3 days
(B) 7 days
(C) 11 days
(D) 9 days |
# Thread: Trigonometry: Sine and Cosine Law
1. ## Trigonometry: Sine and Cosine Law
Topic is for those who kept looking for information about the law of sines and cosines.
Sine law:
Where a, b, c are the sides of the triangle.
Cosine law:
Let's check this on some exercises:
1) Solve the triangle where
The best practice is to draw the triangle and mark our known angles and sides.
We know two sides and angle between them.
To find length of side a we have to use the Cosine Law:
To count B angle we need to use the Cosine law:
To count angle C we have to use simple formulae. As we know sum of angles in triangle is equal to 180°.
Answer: Angles- and Sides-
2. When we have to find the angle in a triangle using the law of Sine, the best practice is to turn the fraction upside down.
Let's do some exercise:
Determine the remaining angles:
To find out angle C I'am going to use Sine law. I've angle B=80° and length of sides b=4 and c=3
Now sum of all angles in triangle is equal to 180. We know two of them so we can simply count the third one.
For more practice we can find out the length of side a, using also the Sine law.
Answer: A=52.38 B=80° C=47.61° a=3.21 b= 4 c=3
3. In this exercise, I will use Cosine law to find out one of the angles in triangle.
Firs we have to determine length of sitde b. We need this side to count angle C.
Now we can find C by Cosine law.
Ok now we have to find angle A so:
Answer: A=98.48° B=26° C=55.52°
4. More complex exercise where we can apply law of Sine and Cosine.
The crank mechanism shown below has an arm OA of length 30mm rotating anticlockwise about O and connecting rod AB of length 60mm. B moves along the horizontal line OB. What is the length OB when OA is rotated by of a complete revolution from the horizontal.
So as we know full revolution is equal to 360°. To resolve this exercise we have to find angle of side OA with OB in of the mechanism revolution. So after calculation we have found that our angle of OA and OB is equal to 45°.
To don't get confused the best practise it will be to draw simple triangle and find our sides and angle which we know so far.
So first step it will be to find what angle C=B is equal to. To do that we have to use the Sine law. I will turn the fraction upside down to make it easer.
As always to count third angle we have to use formulea: A+B+C=180°
To find our unknown a=OB we can use the Sine law.
Answer: OB=77.31cm
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# 1-Math4.6.RP
```Content Area
Standard
Strand
Content Statement
Math
RP. Ratios and Proportional Relationships
CPI#
Cumulative Progress Indicator (CPI)
ACSSSD
Objectives
Understand ratio concepts and use ratio reasoning to solve problems.
4.6.RP.1
4.6.RP.2
4.6.RP.3
a.
Understand the concept of a ratio language to
describe a ratio relationship between two
quantities. For example, The ratio of wings to
beaks in the bird house at the zoo was 2:1,
because for every two wings there was a beak.”
“For every vote candidate A received, candidate
1. Match equivalent ratios.
2. Identify ratios.
3. Identify the ratio that represents a given
situation.
4. Demonstrate an understanding of ratio
relationships.
5. Describe ratio relationships between two
quantities.
6. Demonstrate an understanding that
mathematical symbols can be used to
represent mathematical concepts.
Understand the concept of a unit rate a/b
1. Refer to the ACSSSD Objectives of 4.6.RP.1
for the prerequisite skills needed to complete
associated with ratio a:b with b 0, and use rate
this CPI.
language in the context of a ratio relationship.
For example, “This recipe has a ratio of 3 cups of 2. Define rate.
flour to 4 cups of sugar, so there is ¾ cup of flour 3. Define unit rate.
4. Demonstrate an understanding of rate
for each cup of sugar.” “We paid \$75 for 15
language (each, per).
hamburgers, which is a rate of \$5 per
hamburger.”
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about
tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
Make tables of equivalent ratios relating
quantities with whole-number measurements, find
missing values in the tables, and plot the pairs of
values on the coordinate plane. Use tables to
compare ratios.
1. Identify equivalent ratios.
2. Demonstrate the ability to make tables to
display data.
3. Identify missing values in a table.
4. Find missing values using ratio reasoning.
5. Plot a point on a coordinate plane when given
the coordinate.
6. Read a table of equivalent ratios.
c.
Find a percent of a quantity as a rate per 100
(e.g., 30% of a quantity means 30/100 times the
quantity); solve problems involving finding the
whole, given a part and the percent.
1. Identify parts and the whole in the context of
a ratio.
2. Match percentages.
3. Identify percentages.
4. Demonstrate the understanding that percent is
rate per 100.
5. Demonstrate the ability to solve multi-step
problems.
```
Iranian peoples
32 Cards
Cryptography
26 Cards |
# Cyclic Permutations
A permutation of the type
is called a cyclic permutation or a cycle. It is usually denoted by the symbol $\left( {{a_1},{a_2}, \ldots ,{a_n}} \right)$.
Thus if $f$ is a permutation of degree $n$ non a set $S$ having $n$ distinct elements and if it is possible to arrange some of the elements (say $m$ in number) of the set $S$ in a row such that the $f -$image of each element in this row is the element following it and the $f -$image of the last element in the row is the first element and the remaining $\left( {n - m} \right)$ elements of the set $S$ remain invariant under $f$, then $f$ is called a cycle permutation or a cycle of length $m$.
The number of objects permuted by the cycles is called the length of cycle. Thus by the cycle of length on e we mean a permutation in which the image of each element remains unchanged under a permutation $f$. Consequently cycle permutation of length one is the identity permutation.
One Row Symbol: One row symbol is used to denote a cycle permutation. In the notation the elements of $S$ are arranged in such a way that the image of each element in this row is the element which follows it and that of the last element is the first element. These elements of $X$ which remain invariant need not be written in the row.
Example:
Let $f = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 2&4&1&3&5&6 \end{array}} \right)$ be a cyclic permutation.
Since the elements 1, 2, 3, 4 are such that $f\left( 1 \right) = 2$, $f\left( 2 \right) = 4$, $f\left( 4 \right) = 3$ and $f\left( 3 \right) = 1$ and two remaining elements 5 and 6 remain invariant under $f$, $f$ is a cycle of length 4 or a 4-cycle and can be expressed as $f\left( {1\,\,2\,\,4\,\,3} \right)$.
Transposition: A cycle of length two is called a transposition. Thus the cycle $\left( {1,\,3} \right)$ is a transposition. It is a 2-cyclic such that the image of 1 is 3 and image of 3 is 1 and the remaining missing elements are invariant.
Disjoint Cycles: Two cycles are said to be disjoint if when expressed in one row notations, they have no element in common. |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
Published by Pearson
# Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 326: 28
#### Answer
$(a-1)(25a+2)$
#### Work Step by Step
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 25a^2-23a-2 \end{array} has $ac= 25(-2)=-50$ and $b= -23 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -25,2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 25a^2-25a+2a-2 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (25a^2-25a)+(2a-2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 25a(a-1)+2(a-1) .\end{array} Factoring the $GCF= (a-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (a-1)(25a+2) .\end{array}
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. |
## Arithmetic Sequence
First Number Common Difference nth Number to obtain
## Geometric Sequence
First Number Common Ratio nth Number to obtain
## Fibonacci Series
Nth number to obtain
# SEQUENCE CALCULATOR
Numbers are said to be in sequence if they follow a particular pattern or order. Let us discuss three types of sequences in this post: Arithmetic, Geometric and Fibonacci.
### Arithmetic Sequence
The terms a1, a2, a3, a4, a5, ……an are said to be in an arithmetic progression P, when a2-a1=a3-a2, i.e. when the terms increase or decrease continuously by a common value. This common value is called the common difference (d) of the Arithmetic Sequence.
The first term of the P is denoted as ‘a’ and the number of terms is denoted as ‘n’. We also call an arithmetic sequence as an arithmetic progression.
Common Difference is the difference between the successive term and its preceding term. It is always constant for the arithmetic sequence.
Common difference (d) = a2 – a1
To find the nth term of an arithmetic sequence, we use
Tn = a + ( n - 1 ) × d
Sum of terms of an Arithmetic sequence is
Sn
=
[ n × ( 2a + ( n - 1 ) × d ) ] 2
Where ‘a’ is the first term and ‘d’ is the common difference.
Examples of Arithmetic Progression:
1. 5, 7, 9, 11, 13, ...
Here ‘5’ is the first term and the common difference d = 7-5 = 2
2. 15, 12, 9, 6, 3, 0, -3, -6 ,…
Here ‘15’ is the first term and the common difference d = 12-15 = -3
### Geometric Sequence
The terms a1, a2, a3, a4, a5……an are said to be in geometric sequence, when a2/a1=a3/a2=r, where ‘r’ is called the common ratio(r) of the Geometric Sequence.
The first term of the sequence is denoted as ‘a’ and the number of terms is denoted as ‘n’. We also call a geometric sequence as a geometric progression.
Common Ratio is the ratio between the successive term and its preceding term. It is always constant for a given geometric sequence.
Common Ratio(r)
=
a2 a1
To find the nth term of a geometric sequence, we use
Tn = a × r ( n - 1 )
Sum of terms of a geometric sequence is
Sn
=
a ( rn - 1 ) ( r - 1 )
where ‘a’ is the first term and ‘r’ is the common ratio.
Sum of terms of an infinitely long decreasing geometric sequence is
Sn
=
a ( 1 - r )
Examples of Geometric Progression:
1. 1, 2, 4, 8, 16….
Here ‘1’ is the first term and the common ratio(r) = 2/1 = 2
2. -1, 2, -4, -8, -16…
Here ‘-1’ is the first term and the common ratio(r) = 2/(-1) = -2
### Fibonacci Series
The Fibonacci Sequence is the series of numbers - 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
The next number of the series is obtained by adding up the two numbers before it.
• The number 2 is obtained by adding the 1 and before it. (1+1)
• The number 3 is obtained by adding the 2 and 1 before it (1+2),
• And number 5 is (2+3), and so on!
### Do you know these facts?
• Leonardo Pisano Bogollo also was known as "Fibonacci", and lived between 1170 and 1250 in Italy. "Fibonacci" means "Son of Bonacci".
• Fibonacci Day: November 23rd is called a Fibonacci day, as it has the digits "1, 1, 2, 3" which is a part of the sequence.
### How to use CalculatorHut’s sequence calculator?
You can use our series calculator for calculating and finding out any numbers from arithmetic, geometric and Fibonacci series. |
# Lesson Worksheet: One-Step Equations: Rational Numbers Mathematics • 7th Grade
In this worksheet, we will practice solving one-step linear equations involving rational numbers.
Q1:
Determine the value of , given that .
• A
• B
• C
• D
Q2:
If , then find the value of .
Q3:
Solve .
• A
• B
• C
• D
• E5
Q4:
.
• A
• B
• C
• D
Q5:
.
Q6:
If the number of eggs in a basket is 4 eggs, then the whole basket has eggs.
Q7:
Which of the following equations is equivalent to ?
• A
• B
• C
• D
• E
Q8:
.
• A
• B
• C
• D
Q9:
.
Q10:
Fill in the blanks.
• A8, 8, 9
• B10, 9, 9
• C8, 9, 10
• D8, 8, 10
Q11:
A truck traveled 220 miles in hours. On average, how many miles per hour did it travel?
• A mph
• B30 mph
• C660 mph
• D mph
• E mph
Q12:
Jackson drove for hours at 79 miles per hour. About how many miles did he drive?
Q13:
Solve the following equation: .
• A
• B
• C
• D
• E
Q14:
Two students tried to solve the equation . The first student added to both sides, while the second subtracted from both sides. Determine which student was correct, and then solve for the variable.
• Athe second student,
• Bthe second student,
• Cthe first student,
• Dthe first student,
Q15:
When is added to , the answer is 0. What is the value of ?
• A
• B
• C
• D
• E
Q16:
Solve for .
• A
• B5
• C
• D
• E
Q17:
Jennifer rode the bus to work today. She traveled miles and, due to bad traffic, it took of an hour. Solve the equation to find the average speed, , of the bus.
• A mph
• B mph
• C mph
• D mph
• E mph
Q18:
Which of the following equations is equivalent to ?
• A
• B
• C
• D
• E
Q19:
Find the rational number that equals 8 if is subtracted from it.
Q20:
Find the value of given .
Q21:
Solve .
• A
• B
• C
• D
• E
Q22:
Find the rational number that when subtracted from its additive inverse gives .
• A
• B
• C
• D
Q23:
Find given is the additive inverse of .
• A
• B
• C
• D
• E
Q24:
Solve for .
• A
• B
• C
• D
• E
Q25:
Anthony spends 2 hours doing homework every night. On Monday, he spent ten minutes doing geography research. The rest of his time was split so that he spent twice as long on math homework as he did on his reading, and he spent the same amount of time on an essay as he did on math.
Write an equation that can be used to determine , the amount of time that he spent doing his math homework on Monday.
• A
• B
• C
• D
• E |
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# How do you find the integral of $\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx$?
Last updated date: 04th Mar 2024
Total views: 342k
Views today: 10.42k
Verified
342k+ views
Hint: The given integral $\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx$ has ${{x}^{\dfrac{1}{2}}}$ or $\sqrt{x}$ as an argument to the sine function, which is making it complex. So we will simplify the integral by substituting ${{x}^{\dfrac{1}{2}}}$ equal to some variable, say $t$. On making this substitution our integral will become simplified. Then we have to use the by-parts method to solve the integral obtained. Finally, we have to back substitute $t$ to ${{x}^{\dfrac{1}{2}}}$ to get the final integral.
The integral given in the above question is
$I=\int{\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx}..........(i)$
As can be seen above, the square root function ${{x}^{\dfrac{1}{2}}}$ as an argument to the sine function is making the integral complex. So we first have to simplify the above integral by removing the square root function by substituting it to some variable $t$, that is,
$\Rightarrow t={{x}^{\dfrac{1}{2}}}.........(ii)$
Differentiating both sides with respect to $x$, we have
$\Rightarrow \dfrac{dt}{dx}=\dfrac{d\left( {{x}^{\dfrac{1}{2}}} \right)}{dx}$
Now, we know that the differentiation of ${{x}^{n}}$ is equal to $n{{x}^{n-1}}$. So the above equation becomes
\begin{align} & \Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}} \\ & \Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}} \\ & \Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2{{x}^{\dfrac{1}{2}}}} \\ \end{align}
Substituting (ii) in the above equation, we get
$\Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2t}$
By cross multiplying, we can write
\begin{align} & \Rightarrow 2tdt=dx \\ & \Rightarrow dx=2tdt.........(iii) \\ \end{align}
Substituting (ii) and (iii) in (i), we get
\begin{align} & \Rightarrow I=\int{\sin t\left( 2tdt \right)} \\ & \Rightarrow I=\int{2t\sin tdt}......(iv) \\ \end{align}
Now, we will use the by parts method to solve the above integral. From the by parts method, we know that
$\int{f\left( t \right)g\left( t \right)dt}=f\left( t \right)\int{g\left( t \right)dt}-\int{f'\left( t \right)\left( \int{g\left( t \right)dt} \right)dt}$
Choosing $f\left( t \right)=2t$ and $g\left( t \right)=\sin t$, the integral in (iv) can be written as
\begin{align} & \Rightarrow I=2t\int{\sin tdt}-\int{\dfrac{d\left( 2t \right)}{dt}\left( \int{\sin tdt} \right)dt} \\ & \Rightarrow I=2t\int{\sin tdt}-\int{2\left( \int{\sin tdt} \right)dt} \\ \end{align}
We know that $\int{\sin tdt}=-\cos t$. Putting it in the above integral, we get
\begin{align} & \Rightarrow I=2t\left( -\cos t \right)-\int{2\left( -\cos t \right)dt} \\ & \Rightarrow I=-2t\cos t+2\int{\cos tdt} \\ \end{align}
Now, we know that $\int{\cos tdt}=\sin t$. Putting it above, we get
\begin{align} & \Rightarrow I=-2t\cos t+2\sin t+C \\ & \Rightarrow I=2\sin t-2t\cos t+C \\ \end{align}
Finally, substituting (ii) in the above equation, we get
\begin{align} & \Rightarrow I=2\sin {{x}^{\dfrac{1}{2}}}-2{{x}^{\dfrac{1}{2}}}\cos {{x}^{\dfrac{1}{2}}}+C \\ & \Rightarrow I=2\left( \sin {{x}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}\cos {{x}^{\dfrac{1}{2}}} \right)+C \\ \end{align}
Hence, the integral of $\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx$ is equal to $2\left( \sin {{x}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}\cos {{x}^{\dfrac{1}{2}}} \right)+C$.
Note: Do not forget to add a constant after the integration since we have solved an indefinite integral. Also, while substituting ${{x}^{\dfrac{1}{2}}}=t$, do not replace $dx$ by $dt$ directly. We have to take the differential on both sides of the equation ${{x}^{\dfrac{1}{2}}}=t$ for obtaining $dt$ in terms of $dx$. |
Monday, 26 Jul 2021
# Can Integers Be Negative? Let’s Take a Look Here
In this article, we will undoubtedly learn what negative numbers are, their procedures, and how the numbers are attached to real life. Let’s learn more about the explanation on Can Integers Be Negative or not.
The background of negative numbers is dated back a thousand years ago when mathematicians from the Indian subcontinent started utilizing them. Europeans, later on, revealed a rate of interest in negative numbers but were too unwilling to embrace them.
Egyptians were also dismissive of negative, and even at some point, they related to negative number is absurd. This since the mathematics they were using back then was based only on geometric principles like area and location. Europeans later started overtaking negative numbers when the scholars started translating Arabic messages recovered from North Africa.
From this brief history, we have discovered that these generations of dazzling and intelligent individuals initially showed a challenging to accept the concept of negative numbers.
They finally welcomed the suggestion after uncovering the importance of negative numbers.
## What is a Negative Number?
A negative number is a number that value is less than no. Negative numbers are represented by a minus or a dashboard (-) sign in front of a number.
They represent the number line to the left of the beginning. Negative numbers can be either be numbers, fractions or decimals. As an example, — 2,– 3,– 4,– 5, -2/ 3, -5/ 7, -3/ 4, -0.5, -0.7. and so, on are examples of negative numbers. In this case, these numbers are pronounced as negative 2, negative three, negative four and so forth.
A negative number has several different interpretations. And also, these are:
A negative number is a number that is less than zero.
Numbers that are to the left of no on the number line
A number that is the reverse of a positive number
A negative number stands for loss or lack of something.
A quantity that has a direction
## Can Integers Be Negative?
A negative integer is a digit that has value less than no. Negative integers are typically whole numbers, for instance, -3, -5, -8, -10.
### Operations with Negative integers
Negative integers have policies for carrying out different calculations. These are:
Including an negative and a positive integer
When adding a positive and a negative integer with each other, deduct the integers and create the indication of the higher absolute value. When a small negative integer contributes to a bigger positive integer, the integers substracts and provided a positive indicator. For instance,
8 + (- 2) = 6. Likewise, when a small positive and big negative integer includes, the sum is always negative. For instance,– 5 + 3 =– 2.
When adding negative integers, the numbers assume the sign of the original integers. For instance,– 5 + (-1) =– 6.
#### Subtracting negative integers
Deducting a positive integer from an unfavorable integer is equivalent to including the negative of that integer. For instance, -10– 15 = -10 + (-15) = -25.
Subtracting a negative integer from one more negative integer amounts to adding the positive of that integer. As an example, 13– (-14) = 13 + 14 = 27.
#### Multiplication and a division of negative integers.
When an additional negative integer increases a negative integer, the item declares. Example, -4 x -4 = 16. Similarly, separating a negative integer by a different negative integer leads to a positive ratio.
Reproduction of a positive integer by one more negative integer leads to an negative item. Instance, -2 x 5 = -10. As well as the division a positive integer by an negative result in a negative quotient.
### Application of Negative Integers in Reality
Irrespective of their value, negative integers are commonly use in different areas of life. The following real-life applications of negative numbers will undoubtedly motivate you to see the benefit of examining them.
#### The Sector of Banking and Money.
Financial institutions and also financial institutions require debit, credits and even cash. Thus, there is a requirement to have numbers that distinguish between a credit score and debit purchase. Profits and also losses are also specified by positive and even negative number, respectively. Another area where negative numbers are use is the securities market. Positive and negative numbers are use to illustrate the ups and downs of the share rate.
Deposits usually are stood for by a positive indicator, whereas an negative indication signifies withdrawals.
#### Science, design as well as medication
Negative numbers use of in weather forecasting to reveal the temperature level of a region. Negative integers use of to show the temperature level on Fahrenheit and also Celsius scales.
In design as an example, tools such as boilers and steam engines make use of stress gauges and thermostats calibrated from negative to positive integers.
Instruments for gauging blood pressure, body weight, and medicine screening all operate the principle negative or positive range.
Hope this blog clarify your question Can Integers Be Negative, for any queries drop a comment below. |
# How to find the area of a triangle with one side given?
• Last Updated : 20 Nov, 2021
A triangle is a closed figure composed of three sides. An equilateral triangle is a geometrical 2D figure which has all three sides equal. All the edges subtend an angle of 60° at the corners. Therefore, the sum of all the angles of the triangle is 180°
Properties of an equilateral triangle
• All three sides of an equilateral triangle are equal.
• All the three angles are equal i.e 60° each.
• A regular polygon has three equal sides.
### How to find the area of a triangle with one side given?
Solution:
Let us assume a to be the side of the triangle.
If we divide the bottom side in two parts, we get,
By Pythagoras theorem, the altitude can be obtained by,
Perpendicular2 + Base2 = Hypotenuse2
Perpendicular2 + (a/2)2 = a2
Perpendicular2 = a2 – (a/2)2
Perpendicular2 = 3a2/4
Taking square root,
Perpendicular = √3/2a
Now,
Area of triangle = 1/2 x Perpendicular x Base
= 1/2 x √3/2a x a
On solving, we obtain,
Area = a2(√3/4)
### Sample Questions
Question 1. Find the area of the equilateral triangular signboard of side 16 cm and also find the cost of painting the signboard at ₹2 per cm2?(Assume √3 = 1.732)
Solution:
Here we have to find the cost of painting an equilateral triangular signboard.
Given:
Side of equilateral triangular signboard = 16 cm
First finding the area of the equilateral triangular sign board
As we know that
Area of equilateral triangle = √3/4 × a2
Here a = 16 cm
Area of equilateral triangle = √3/4 × 162
Area of equilateral triangle = √3/4 × 16 × 16
Area of equilateral triangle = 64√3
Now,
Finding the cost of painting
Given
Cost of painting = ₹2 per cm2
Cost of painting = Area of equilateral triangle × 2
Cost of painting = 64√3 × 2
Cost of painting = 128√3
Using √3 = 1.732 (Given)
Cost of painting = ₹221.696
Therefore,
Cost of painting the equilateral triangular signboard is ₹221.696.
Question 2. Find the area of an equilateral triangular park of side 32 m?
Solution:
Here we have to find the area of the equilateral triangular park.
Here we are given that,
Side of the triangular park = 32 m
As we know that
Area of equilateral triangle = √3/4 × a2
Area of equilateral triangular park = √3/4 × a2
Here ‘a’ is the side of the equilateral triangular park is 32 m
Area of equilateral triangular park = √3/4 × 322
Area of equilateral triangular park = √3/4 × 32 × 32
Area of equilateral triangular park = √3 × 8 × 32
Area of equilateral triangular park = 256√3 m2
Therefore,
The area of the equilateral triangular park is 256√3 m2
Question 3. Find the cost of carpeting equilateral triangular yoga ground with the altitude 10 m and base 20 m at the rate of ₹20 pr m2?
Solution:
Here we have to find the cost of carpeting equilateral triangular yoga ground
Given:
Altitude = 10 m
Base = 20 m
As we know that
Area of equilateral triangle = 1/2 × base × height
Area of equilateral triangle = 1/2 × 20 × 10
Area of equilateral triangle = 100 m2
Now finding the cost of carpeting
Cost of carpeting yoga ground at ₹20 pr m2
Cost of carpeting yoga ground = Area of yoga ground × ₹20
Cost of carpeting yoga ground = 100 × ₹20
Cost of carpeting yoga ground = ₹2000
Therefore,
The cost of carpeting the yoga ground is ₹2000.
Question 4. Find the area of an equilateral triangle having a perimeter of 18 m?
Solution:
Here we have to find the area of the equilateral triangle
Given:
Perimeter of equilateral triangle = 18 m
The Perimeter of equilateral triangle = Side + Side + Side
As all the sides of the equilateral triangle are equal
Perimeter of equilateral triangle = 3 × Side
18 = 3 × Side
Side = 18/3
Side = 6 m
Side of the equilateral triangle is 6 m
Further,
Finding the area of the equilateral triangle
Area of equilateral triangle = √3/4 × a2
a = side of the equilateral triangle
Area of equilateral triangle = √3/4 × 62
Area of equilateral triangle = √3/4 × 6 × 6
Area of equilateral triangle = 9√3 m2
Question 5. If the side of an equilateral triangle is 10 m and its area is 75 m2 then find the height of the equilateral triangle?
Solution:
Here we have to find the height of the equilateral triangle
Given:
Side of equilateral triangle = Base of equilateral triangle = 10 m
Area of equilateral triangle = 75 m2
As we know that
Area of equilateral triangle = 1/2 × base × height
75 = 1/2 × 10 × height
75 = 5 × height
Height = 75/5
Height = 15 m
Therefore,
The height of equilateral triangle is 15 m.
My Personal Notes arrow_drop_up |
### Learning Outcomes
Adding and subtracting radicals is much like combining like terms with variables. We can add and subtract expressions with variables like this:
$5x+3y - 4x+7y=x+10y$
There are two keys to combining radicals by addition or subtraction: look at the index, and look at the radicand. If these are the same, then addition and subtraction are possible. If not, then you cannot combine the two radicals.
Remember the index is the degree of the root and the radicand is the term or expression under the radical. In the diagram below, the index is n, and the radicand is $100$. The radicand is placed under the root symbol and the index is placed outside the root symbol to the left:
In the graphic below, the index of the expression $12\sqrt[3]{xy}$ is $3$ and the radicand is $xy$.
Practice identifying radicals that are compatible for addition and subtraction by looking at the index and radicand of the roots in the following example.
### Example
Identify the roots that have the same index and radicand.
$10\sqrt{6}$
$-1\sqrt[3]{6}$
$\sqrt{25}$
$12\sqrt{6}$
$\frac{1}{2}\sqrt[3]{25}$
$-7\sqrt[3]{6}$
Making sense of a string of radicals may be difficult. One helpful tip is to think of radicals as variables, and treat them the same way. When you add and subtract variables, you look for like terms, which is the same thing you will do when you add and subtract radicals.
In this first example, both radicals have the same radicand and index.
### Example
Add. $3\sqrt{11}+7\sqrt{11}$
It may help to think of radical terms with words when you are adding and subtracting them. The last example could be read “three square roots of eleven plus $7$ square roots of eleven”.
This next example contains more addends, or terms that are being added together. Notice how you can combine like terms (radicals that have the same root and index) but you cannot combine unlike terms.
### Example
Add. $5\sqrt{2}+\sqrt{3}+4\sqrt{3}+2\sqrt{2}$
Notice that the expression in the previous example is simplified even though it has two terms: $7\sqrt{2}$ and $5\sqrt{3}$. It would be a mistake to try to combine them further! (Some people make the mistake that $7\sqrt{2}+5\sqrt{3}=12\sqrt{5}$. This is incorrect because$\sqrt{2}$ and $\sqrt{3}$ are not like radicals so they cannot be added.)
### Example
Add. $3\sqrt{x}+12\sqrt[3]{xy}+\sqrt{x}$
### Try It
Sometimes you may need to add and simplify the radical. If the radicals are different, try simplifying first—you may end up being able to combine the radicals at the end, as shown in these next two examples.
### Example
Add and simplify. $2\sqrt[3]{40}+\sqrt[3]{135}$
### Example
Add and simplify. $x\sqrt[3]{x{{y}^{4}}}+y\sqrt[3]{{{x}^{4}}y}$
The following video shows more examples of adding radicals that require simplification.
Subtraction of radicals follows the same set of rules and approaches as addition—the radicands and the indices (plural of index) must be the same for two (or more) radicals to be subtracted. In the examples that follow, subtraction has been rewritten as addition of the opposite.
### Example
Subtract. $5\sqrt{13}-3\sqrt{13}$
### Example
Subtract. $4\sqrt[3]{5a}-\sqrt[3]{3a}-2\sqrt[3]{5a}$
In the video examples that follow, we show more examples of how to add and subtract radicals that don’t need to be simplified beforehand.
### Example
Subtract and simplify. $5\sqrt[4]{{{a}^{5}}b}-a\sqrt[4]{16ab}$, where $a\ge 0$ and $b\ge 0$ |
# A baseball bat and 8 baseballs cost $163.25. The bat costs$106.45. What is the cost c of one baseball?
May 15, 2018
See a solution process below:
#### Explanation:
Let's call the cost of 1 baseball: $b$
We can then write the follow equation for the word problem as:
$106.45 + 8b =$163.25
We can now solve for $b$ to find the cost of 1 baseball:
$106.45 - color(red)($106.45) + 8b = $163.25 - color(red)($106.45)
0 + 8b = $56.80 8b =$56.80
(8b)/color(red)(8) = ($56.80)/color(red)(8) (color(red)(cancel(color(black)(8)))b)/cancel(color(red)(8)) =$7.10
b = $7.10 One baseball cost$7.10 |
# Which of the following equations would not be consistent with 3y - 4x = 7? a) -8x = 6y - 14 b)...
## Question:
Which of the following equations would not be consistent with {eq}3y - 4x = 7 {/eq}?
a) {eq}-8x = 6y - 14 {/eq}
b) {eq}12y = -9x + 28 {/eq}
c) {eq}-16x + 28 = 12y {/eq}
d) {eq}-4x - 3y - 7 = 0 {/eq}
e) {eq}-56 - 24y = -32x {/eq}
## Inconsistent Systems of Equations:
A system of equations may have one unique solution, when the determinant of the matrix of coefficients of this system is not zero. The system may have infinitely many solutions, when the two or more equations are linearly dependent, that is, one can be represented as a linear combination of several others. Finally, the system may have no solution, which means that two or more equations in the system are inconsistent with each other.
Explanation: the two equations are inconsistent if the system of those two equations does not have a solution. In this case, we have the following system:
{eq}\begin{cases} 3y - 4x = 7\\ -56 - 24y = -32x \end{cases} {/eq}
or
{eq}\begin{cases} 3y - 4x = 7\\ 24y -32x = -56 \end{cases} {/eq}
Here we simply re-arranged terms in the second equation. Now if we divide both sides of the second equation by 8, we obtain:
{eq}\begin{cases} 3y - 4x = 7\\ 3y -4x = -7 \end{cases} {/eq}
It is clear now that from this two equations it follows that {eq}7 = -7 {/eq}, which is impossible. Therefore, the system has no solution and two equations are inconsistent; |
# B=20° C=75° b=5 solve the triangle?
## When I set up the equation using law of sines it looked like this 5/sin(20)=x/sin(75) then x=-2.124 what did I do wrong?
Feb 28, 2018
$\angle A = {85}^{\circ}$
$a \approx 14.56$
$c \approx 14.12$
#### Explanation:
You're calculator is in radians! That's why. It should be in degrees.
You are on the right track however!
Let's draw the triangle [Note: This is not to scale]:
The missing angle can be found by knowing that the sum of the internal angles of any triangle is always equal to ${180}^{\circ}$
So $\angle A = {85}^{\circ}$
The law of sines states:
$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C$
We know:
$b = 5$
$\angle A = {85}^{\circ}$
$\angle B = {20}^{\circ}$
$\angle C = {75}^{\circ}$
What we want to know is
a=?
c=?
We can find the missing sides using the law of sines. So to find $a$
$\frac{a}{\sin} A = \frac{b}{\sin} B$
Pluggin in what we know:
$\frac{a}{\sin} \left({85}^{\circ}\right) = \frac{5}{\sin} \left({20}^{\circ}\right)$
$\frac{a}{\sin} \left({85}^{\circ}\right) = 14.619022$
$a = \sin \left({85}^{\circ}\right) \cdot 14.619022$
color(red)(a~~14.56
Similarly for $c$:
$\frac{b}{\sin} B = \frac{c}{\sin} C$
$\frac{5}{\sin} \left({20}^{\circ}\right) = \frac{c}{\sin} \left({75}^{\circ}\right)$
$14.619022 = \frac{c}{\sin} \left({75}^{\circ}\right)$
$\sin \left({75}^{\circ}\right) \cdot 14.619022 = c$
color(red)(14.12~~c |
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Arithmetic Sequence
julia d
Algebra 1➕
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Calculating and Writing Formulas for Arithmetic Sequences
An arithmetic sequence is a list of numbers, or terms, in which the difference between one term and the next term is the same. This constant difference between one number and the next is known as the common difference. 🤝
We'll use the following notations in examples and explanations: Let
Let’s first look at a simple example to put this into context. 👀
0️⃣ Example 0a
Consider the sequence 1, 3, 5, 7. Find the values of n, a1, an, d, and Sn.
Solution: In this example, we have four terms and thus n = 4. The first term, a1, is 1, and the last term, an, is 7.
Because we always add 2 to get to the next term, our common difference d is 2. We can find Sn by simply summing all terms together: Sn = 1 + 3 + 5 + 7 = 16. 😄
A few formulas and properties you should know will come in handy when solving arithmetic sequence-related problems. We will first consider recursive and explicit formulas to find a specific term of an arithmetic sequence. 🌞
Recursive & Explicit Formulas 🧮
A recursive formula gives a specific term using the value of the last term. In the case of arithmetic sequences, this is given by
An explicit formula, on the other hand, can calculate the value of a specific term without the prior term. For arithmetic sequences, this is given by
0️⃣ Example 0b
Consider again the sequence 1, 3, 5, 7, .... Find the value of a5 using the above recursive and explicit formulas.
Solution: Using the recursive formula, we get
Using our explicit formula, we get
As a side note, notice that there are two formulas to find the value of some an, which offers a good way to check your work if possible.
1️⃣ Example 1
Given that the explicit formula of a sequence is an = 2 - 3(n - 1), which term of the sequence is equal to -28?
Solution: We are given an = -28, which we can plug into the explicit formula to get
Thus, the 11th term of the given sequence gives -28.
2️⃣ Example 2
Construct both the recursive and explicit formula for the sequence 10, 6, 2, -2, ….
Solution: First, note that the common difference d is -4 and that the first term a0 = 10. Using the recursive formula an = an-1 + d, we get that the recursive formula of this sequence is an = an-1 - 4. Using the explicit formula ai = a1 + d(i - 1), we get an explicit formula of ai = 10 - 4(n - 1).
Next, we will consider the sum of an arithmetic sequence.
Arithmetic Sequence Sum ➕
The sum of an arithmetic sequence is given by the average of the first and last terms multiplied by the number of terms, or
Recall that an is equivalent to a1 + d(n - 1), so Sn is also equal to n/2(2a_1 + d(n-1)) when we substitute for an. This is another common formula you will often encounter.
Let’s take a look at a few examples!
3️⃣ Example 3
Find the sum of the sequence 8, 14, 20, 26, 32, 38, 44.
Solution: Note that d = 6, a1 = 8, an = 44, and n = 7. Using the formula Sn = n/2(a1+an), we get that Sn = 7/2(8+44)=182.
4️⃣ Example 4
Find the sum of the 100-term sequence 3, 7, 11, 15, ….
Solution: We are given that n = 100, d = 4, and a1 = 3. Using the formula Sn = n/2(2a1 + d(n-1)) we get that Sn = 100/2(2*3+4(100-1))=20100.
Key Takeaways: Now What? 🔑
The concept of arithmetic sequences is fundamental knowledge that is crucial to understand fully. It is often tested on the SAT Math and ACT Math sections and is the foundation of calculus-based sequences and series. 🌊
As a quick summary, an arithmetic sequence contains n terms with a common difference d. The sum of an arithmetic sequence is denoted as Sn, and the nth term of a sequence is given by ai. Relevant formulas are as follows:
When in doubt, always start by first listing out all the variables you’re given. From there, see what equations can be used from there. You got this! 🌟
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# Difference between revisions of "2016 AIME I Problems/Problem 15"
## Problem
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively, with line $AB$ closer to point $X$ than to $Y$. Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$. The three points $C$, $Y$, $D$ are collinear, $XC = 67$, $XY = 47$, and $XD = 37$. Find $AB^2$.
## Solutions
### Solution 1
Let $Z = XY \cap AB$. By the Radical Axis Theorem $AD, XY, BC$ concur at point $E$. Furthermore, by simple angle chasing, $\triangle DXE \sim \triangle EXC$. Let $y = EX, x = XZ$. Then $\frac{y}{37} = \frac{67}{y} \implies y^2 = 37 \cdot 67$. Now, by Power of a Point, $AZ^2 = \frac{AB^2}{4}$, $(y-x)x = \frac{AB^2}{4}$, and $x(47+x) = \frac{AB^2}{4}$. Solving, we get $\dfrac{AB ^ 2}{4} = \left(\frac{y - 47}{2}\right)\left(\frac{y + 47}{2}\right) \implies AB ^ 2 = 37\cdot67 - 47^2 = \boxed{270}$
### Solution 2
By the Radical Axis Theorem $AD, XY, BC$ concur at point $E$.
Let $AB$ and $EY$ intersect at $S$. Note that because $AXDY$ and $CYXB$ are cyclic, by Miquel's Theorem $AXBE$ is cyclic as well. Thus $$\angle AEX = \angle ABX = \angle XCB = \angle XYB$$and $$\angle XEB = \angle XAB = \angle XDA = \angle XYA.$$Thus $AY \parallel EB$ and $YB \parallel EA$, so $AEBY$ is a parallelogram. Hence $AS = SB$ and $SE = SY$. But notice that $DXE$ and $EXC$ are similar by $AA$ Similarity, so $XE^2 = XD \cdot XC = 37 \cdot 67$. But $$XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.$$Hence $AB^2 = 37 \cdot 67 - 47^2 = \boxed{270}.$
### Solution 3
First, we note that as $\triangle XDY$ and $\triangle XYC$ have bases along the same line, $\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}$. We can also find the ratio of their areas using the circumradius area formula. If $R_1$ is the radius of $\omega_1$ and if $R_2$ is the radius of $\omega_2$, then $$\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.$$ Since we showed this to be $\frac{DY}{YC}$, we see that $\frac{R_2}{R_1}=\frac{67}{37}$.
We extend $AD$ and $BC$ to meet at point $P$, and we extend $AB$ and $CD$ to meet at point $Q$ as shown below. $[asy] size(200); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; pair p=extension(a,d,b,c); dot(q^^a^^b^^x^^y^^c^^d^^p); draw(q--b^^q--c); draw(p--d^^p--c^^x--y); draw(a--y^^b--y); draw(d--x--c); label("A",a,NW,fontsize(8)); label("B",b,NE,fontsize(8)); label("C",c,SE,fontsize(8)); label("D",d,SW,fontsize(8)); label("X",x,2*WNW,fontsize(8)); label("Y",y,3*S,fontsize(8)); label("P",p,N,fontsize(8)); label("Q",q,W,fontsize(8)); [/asy]$ As $ABCD$ is cyclic, we know that $\angle BCD=180-\angle DAB=\angle BAP$. But then as $AB$ is tangent to $\omega_2$ at $B$, we see that $\angle BCD=\angle ABY$. Therefore, $\angle ABY=\angle BAP$, and $BY\parallel PD$. A similar argument shows $AY\parallel PC$. These parallel lines show $\triangle PDC\sim\triangle ADY\sim\triangle BYC$. Also, we showed that $\frac{R_2}{R_1}=\frac{67}{37}$, so the ratio of similarity between $\triangle ADY$ and $\triangle BYC$ is $\frac{37}{67}$, or rather $$\frac{AD}{BY}=\frac{DY}{YC}=\frac{YA}{CB}=\frac{37}{67}.$$ We can now use the parallel lines to find more similar triangles. As $\triangle AQD\sim \triangle BQY$, we know that $$\frac{QA}{QB}=\frac{QD}{QY}=\frac{AD}{BY}=\frac{37}{67}.$$ Setting $QA=37x$, we see that $QB=67x$, hence $AB=30x$, and the problem simplifies to finding $30^2x^2$. Setting $QD=37^2y$, we also see that $QY=37\cdot 67y$, hence $DY=37\cdot 30y$. Also, as $\triangle AQY\sim \triangle BQC$, we find that $$\frac{QY}{QC}=\frac{YA}{CB}=\frac{37}{67}.$$ As $QY=37\cdot 67y$, we see that $QC=67^2y$, hence $YC=67\cdot30y$.
Applying Power of a Point to point $Q$ with respect to $\omega_2$, we find $$67^2x^2=37\cdot 67^3 y^2,$$ or $x^2=37\cdot 67 y^2$. We wish to find $AB^2=30^2x^2=30^2\cdot 37\cdot 67y^2$.
Applying Stewart's Theorem to $\triangle XDC$, we find $$37^2\cdot (67\cdot 30y)+67^2\cdot(37\cdot 30y)=(67\cdot 30y)\cdot (37\cdot 30y)\cdot (104\cdot 30y)+47^2\cdot (104\cdot 30y).$$ We can cancel $30\cdot 104\cdot y$ from both sides, finding $37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2$. Therefore, $$AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.$$
### Solution 4
$[asy] size(9cm); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; dot(a^^b^^x^^y^^c^^d); draw(x--y); draw(a--y^^b--y); draw(d--x--c); draw(a--b--c--d--cycle); draw(x--a^^x--b); label("A",a,NW,fontsize(9)); label("B",b,NE,fontsize(9)); label("C",c,SE,fontsize(9)); label("D",d,SW,fontsize(9)); label("X",x,2*N,fontsize(9)); label("Y",y,3*S,fontsize(9)); [/asy]$ First of all, since quadrilaterals $ADYX$ and $XYCB$ are cyclic, we can let $\angle DAX = \angle XYC = \theta$, and $\angle XYD = \angle CBX = 180 - \theta$, due to the properties of cyclic quadrilaterals. In addition, let $\angle BAX = x$ and $\angle ABX = y$. Thus, $\angle ADX = \angle AYX = x$ and $\angle XYB = \angle XCB = y$. Then, since quadrilateral $ABCD$ is cyclic as well, we have the following sums: $$\theta + x +\angle XCY + y = 180^{\circ}$$ $$180^{\circ} - \theta + y + \angle XDY + x = 180^{\circ}$$ Cancelling out $180^{\circ}$ in the second equation and isolating $\theta$ yields $\theta = y + \angle XDY + x$. Substituting $\theta$ back into the first equation, we obtain $$2x + 2y + \angle XCY + \angle XDY = 180^{\circ}$$ Since $$x + y +\angle XAY + \angle XCY + \angle DAY = 180^{\circ}$$ $$x + y + \angle XDY + \angle XCY + \angle DAY = 180^{\circ}$$ we can then imply that $\angle DAY = x + y$. Similarly, $\angle YBC = x + y$. So then $\angle DXY = \angle YXC = x + y$, so since we know that $XY$ bisects $\angle DXC$, we can solve for $DY$ and $YC$ with Stewart’s Theorem. Let $DY = 37n$ and $YC = 67n$. Then $$37n \cdot 67n \cdot 104n + 47^2 \cdot 104n = 37^2 \cdot 67n + 67^2 \cdot 37n$$ $$37n \cdot 67n + 47^2 = 37 \cdot 67$$ $$n^2 = \frac{270}{2479}$$ Now, since $\angle AYX = x$ and $\angle BYX = y$, $\angle AYB = x + y$. From there, let $\angle AYD = \alpha$ and $\angle BYC = \beta$. From angle chasing we can derive that $\angle YDX = \angle YAX = \beta - x$ and $\angle YCX = \angle YBX = \alpha - y$. From there, since $\angle ADX = x$, it is quite clear that $\angle ADY = \beta$, and $\angle YAB = \beta$ can be found similarly. From there, since $\angle ADY = \angle YAB = \angle BYC = \beta$ and $\angle DAY = \angle AYB = \angle YBC = x + y$, we have $AA$ similarity between $\triangle DAY$, $\triangle AYB$, and $\triangle YBC$. Therefore the length of $AY$ is the geometric mean of the lengths of $DA$ and $YB$ (from $\triangle DAY \sim \triangle AYB$). However, $\triangle DAY \sim \triangle AYB \sim \triangle YBC$ yields the proportion $\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}$; hence, the length of $AB$ is the geometric mean of the lengths of $DY$ and $YC$. We can now simply use arithmetic to calculate $AB^2$. $$AB^2 = DY \cdot YC$$ $$AB^2 = 37 \cdot 67 \cdot \frac{270}{2479}$$ $$AB^2 = \boxed{270}$$
-Solution by TheBoomBox77
### Solution 5 (not too different)
Let $E = DA \cap CB$. By Radical Axes, $E$ lies on $XY$. Note that $EAXB$ is cyclic as $X$ is the Miquel point of $\triangle EDC$ in this configuration.
Claim. $\triangle DXE \sim \triangle EXC$ Proof. We angle chase. $$\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle XDE$$and$$\measuredangle XCE = \measuredangle XCB = \measuredangle XBA = \measuredangle XEA = \measuredangle XED. \square$$
Let $F = EX \cap AB$. Note $$FA^2 = FX \cdot FY = FB^2$$and$$EF \cdot FX = AF \cdot FB = FA^2 = FX \cdot FY \implies EF = FY$$By our claim, $$\frac{DX}{XE} = \frac{EX}{XC} \implies EX^2 = DX \cdot XC = 67 \cdot 37 \implies FY = \frac{EY}{2} = \frac{EX+XY}{2} = \frac{\sqrt{67 \cdot 37}+47}{2}$$and$$FX=FY-47=\frac{\sqrt{67 \cdot 37}-47}{2}$$Finally, $$AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47^2}{4} = \boxed{270}. \blacksquare$$~Mathscienceclass |
Practice questions available in McGraw Hill Math Grade 2 Answer Key PDF Chapter 8 Test will engage students and is a great way of informal assessment.
Measure the length of each object.
Question 1.
___________ inches
Explanation:
I measured the note using an inch ruler
It is 1 inch long.
Question 2.
___________ centimeters
Explanation:
I measured the pen suing a ruler
It is 6 centimeters long.
Question 3.
___________ centimeters
Explanation:
I measured the paper clip using a ruler
It is 3 centimeters long.
Question 4.
Cara measures the length of her bed. It is 72 inches long. It is 6 feet long. Why are there more inches than feet? Explain.
Inches is a shorter measurement than feet
1 feet = 12 inches
So, there are more inches than feet.
Estimate the length of the object. Then measure the length.
Question 5.
Measure: ___________ inches
Explanation:
Measure: 2 inches
I estimated and measured the length of the pen
The pen is 3 inches long.
Write an equation to help solve the problem.
Question 6.
Greta has 20 inches of string. She uses some string on a craft project. Now she has 12 inches of string left. How much string did Greta use for her project?
20 – 12 = 8
8 inches
Explanation:
Greta has 20 inches of string
She uses some string on a craft project
Now she has 12 inches of string left
Subtract 12 from 20 to find the answer
So, 20 – 12 = 8
Therefore, Greta used 8 inches of the string.
Question 7.
A ladybug flies 22 meters. It stops and rests. Then it flies 21 more meters. How far does the ladybug fly?
22 + 21 = 43
43 meters
Explanation:
It stops and rests
Then it flies 21 more meters
So, 22 + 21 = 43
Therefore, the ladybug flies 43 meters.
Use a number line to solve each problem.
Question 8.
Lisa’s puppy was 12 inches tall. Put a red dot on the number line to show how tall Lisa’s puppy was. The puppy grew 3 inches. Use the number line to count on. How tall is Lisa’s puppy now?
___________ inches
Explanation:
Lisa’s puppy was 12 inches tall
I kept a red dot on the number line to show how tall Lisa’s puppy was
The puppy grew 3 inches
I started at 12
I counted on 3 more using the number line
I reached at 15
So, Lisa’s puppy is now 15 inches long.
Question 9.
Mr. Martin rode his bike 11 miles on Monday. Put a blue dot on how far he rode. On Tuesday he rode 8 miles. Put a green dot on how far he rode. How many fewer miles did he ride on Tuesday? Count back on the number line.
_____________ miles |
## Wednesday, October 26, 2011
### A Perfect Ten
Don't you love tests where you ask a question which you believe everyone will get correct, and then find out it just isn't so? I gave my PreAlgebra college students a pretest to see what they knew and didn't know. One of the first questions was: Why is our number system called Base Ten? This is an extremely important concept as it reveals what they know about place value. Below are some of the answers I received.
1) It is called Base Ten because we have ten fingers. (Yikes! If that is so, should we include our toes as well?)
2) It is called Base Ten because I think you multiply by ten when you move past the decimal sign. (Well, sort of. You do multiply by ten when you move to the left of the decimal sign, going from the ones place, to the tens place, to the hundreds place, etc.)
3) I think it is called Base Ten because it's something we use everyday. (Really????)
Enough! It is called Base Ten because we use ten digits (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) to write all of the other numbers. Each digit can have one of ten values: any number from 0 through 9. When the value reaches 9, just before 10, it starts over at zero again. (Notice the pattern below.)
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, etc.
In addition, each place is worth ten times more than the last. Ten is worth ten times more than 1, and 1,000 is ten times more than 100. The pattern continues infinitely both ways on a number line.
The decimal point allows for the place value to continue in a consistent pattern with numbers smaller than one. As we move to the right of the decimal point, each place is divided by ten to get to the next place value. One hundredth is one tenth divided by ten, and one thousandth is one hundredth divided by ten. The pattern goes on infinitely.
100's, 10's, 1's . 0.1, 0.01, 0.001, 0.0001, 0.00001, etc.
Since all mathematics is based on patterns, this shouldn’t be a new revelation. Perhaps on the posttest, my students will omit the fingers and instead rely on patterns to answer the questions!
For more information about teaching place value, refer to the September 7th posting entitled: There’s A Place for Us.
Barbara said...
I'm going to ask my 7th grader when he comes home from soccer practice and see what he says! Thanks for helping kids get back to the BASICS!
Barbara
Kim said...
I am so excited to have discovered your blog!
And I can't wait to teach digital roots next week.
I just perused several different posts and got lots of ideas... Thanks! I'll be back!
Kim |
Section 12-1 - PowerPoint PPT Presentation
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Section 12-1. Geometric Representation of Vectors. Vectors.
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Section 12-1
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Section 12-1
Geometric Representation of Vectors
Vectors
• Vectors are quantities that are described by a direction and a magnitude (size). A force would be an example of a vector quantity because to describe a force, you must specify the direction in which it acts and its strength. Velocity is another example of a vector.
Vectors
• The velocities of two airplanes each heading northeast at 700 knots are represented by the arrows u and v in the diagram on p. 419. We write u = v to indicate that both planes have the same velocity even though the two arrows are different.
Vectors
• In general, any two arrows with the same length and the same direction represent the same vector. The diagram on p. 419 shows a third airplane with speed 700 knots, but because it is heading in a different direction, its velocity vector w does not equal either u or v.
Magnitude
• The magnitude of a vector v (also called the absolute value of v) is denoted |v|.
• If a vector v is pictured by an arrow from point A to point B, then it is customary to write v = . Since the result of moving an object first from A to B and then from B to C is the same as moving the object directly from A to C, it is natural to write . We say that is the vector sum of and .
• The addition of two vectors is a commutative operation. In other words, the order in which the vectors are added does not make any difference. You can see this in the diagrams on p. 420 where the red arrows denote a + b and b + a having the same length and direction.
• If the two diagrams on p. 420 are moved together, a parallelogram is formed. This suggests that another way to add a and b is to draw a parallelogram OACB with sides = a and = b. The diagonal of the parallelogram is the sum.
• This method is frequently used in physics problems involving forces that are combined.
Vector Subtraction
• The negative of a vector v, denoted –v, has the same length as v but the opposite direction. The sum of v and –v is the zero vector 0.
• It is best thought of as a point.
• v + (-v) = 0
Vector Subtraction
• Vectors can be subtracted as well as added.
• v – w means v + (-w).
Multiples of a Vector
• The vector sum v + v is abbreviated as 2v. Likewise, v + v + v = 3v. The diagram on p. 421 shows that the arrows representing 2v and 3v have the same direction as the arrow representing v, but that they are two and three times as long.
Multiples of a Vector
• In general if k is a positive real number, then kv is the vector with the same direction as v but with an absolute value k times as large. If k < 0, then kv has the same direction as –v and has an absolute value |k| times as large. If k ≠ 0, then is defined to be equal to the vector .
Scalars
• When working with vectors, it is customary to refer to real numbers as scalars.
Scalar Multiplication
• When this is done, the operation of multiplying a vector v by a scalar k is called scalar multiplication. This operation has the following properties. If v and w are vectors and k and m are scalars, then:
k(v + w) = kv + kw
(k + m)v = kv + mv
k(mv) = (km)v Associative law
Distributive laws
C
C
Homework: p. 423-424 1-9 odd
9. A ship travels 200 km west from port and then 240 km due south before it is disabled. Illustrate this in a vector diagram. Use trigonometry to find the course that a rescue ship must take from port in order to reach the disabled ship. |
# Surface Area of a Right Prism
How to find the surface area of a right prism: definition of a prism and its parts, formula, proof, examples, and their solutions.
## Prism
This shape is called a prism.
A prism has two bases.
The bases are polygonal faces
that are congruent and parallel.
So a prism is a 3D shape
that has a pair of bases.
The lateral faces of a prism are the faces
that are not the bases.
## Right Prism
A right prism is a prism
whose lateral faces are all rectangles.
## Formula
A = 2B + Ph
A: surface area of a right prism
B: base area
P: perimeter of the base
h: height of the prism
2B: sum of the base areas
Ph: lateral area
## Proof
Draw the net of a right prism.
There are two bases.
So the sum of the base areas is 2B.
See the lateral faces in the net.
Each lateral face is a rectangle.
So the whole lateral faces form a rectangle.
Its width is the perimeter of the base: P.
And its height is h.
So the lateral area is Ph.
Area of a rectangle
So A = 2B + Ph.
## Example 1
To use the formula,
find B, P, and h.
B = 7⋅3 = 21
P = (7 + 3)⋅2 = 20
h = 8
B = 21, P = 20, h = 8
A = 2⋅21 + 20⋅8
## Example 2
To use the formula,
find B, P, and h.
B = (√3/4)⋅42 = 4√3
Area of an equilateral triangle
P = 4⋅3 = 12
h = 7
B = 4√3, P = 12, h = 7
A = 2⋅4√3 + 12⋅7 |
# An object, previously at rest, slides 4 m down a ramp, with an incline of pi/12 , and then slides horizontally on the floor for another 18 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?
Jul 17, 2016
#### Explanation:
If the mass of the object is $M$, then its initial potential energy is $M g \times 4 \text{m} \times \sin \left(\frac{\pi}{12}\right)$. This is the amount of work that friction has to do in order to make it stop. Now, the force of friction is $\mu M g \cos \left(\frac{\pi}{12}\right)$ on the incline and $\mu M g$ on the horizontal floor. So, the total work done by friction is
$4 \text{m" xx mu Mg cos(pi/12) +18"m} \times \mu M g$
Equating this to the initial potential energy, we get
$4 \text{m" xx mu Mg cos(pi/12) +18"m" xx mu Mg = 4"m} \times M g \sin \left(\frac{\pi}{12}\right)$
$\mu = \frac{4 \sin \left(\frac{\pi}{12}\right)}{4 \cos \left(\frac{\pi}{12}\right) + 18} \approx .047$
Aug 3, 2017
${\mu}_{k} = 0.0474$
#### Explanation:
Here's a method using Newton's laws of motion and kinematics.
We're asked to find the coefficient of kinetic friction, ${\mu}_{k}$, between the object and the ramp.
We'll split this problem into two parts: the first part is where the object is sliding down the incline, and the second part is where it is sliding across the floor.
$- - - - - - - - - - - \boldsymbol{\text{INCLINE}} - - - - - - - - - - -$
The only two forces acting on the object as it slides down the ramp are
1. The gravitational force (its weight; acting down the ramp)
2. The kinetic friction force (acting up the ramp because it opposes motion)
The expression for the coefficient of kinetic friction ${\mu}_{k}$ is
ul(f_k = mu_kn
where
• ${f}_{k}$ is the magnitude of the retarding kinetic friction force acting as it slides down (denoted $f$ in the above image)
• $n$ is the magnitude of the normal force exerted by the incline, equal to $m g \cos \theta$ (denoted $N$ in the above image)
The expression for the net horizontal force, which I'll call $\sum {F}_{1 x}$, is
ul(sumF_(1x) = overbrace(mgsintheta)^"gravitational force" - overbrace(color(red)(f_k))^"kinetic friction force"
Since $\textcolor{red}{{f}_{k} = {\mu}_{k} n}$, we have
$\sum {F}_{1 x} = m g \sin \theta - \textcolor{red}{{\mu}_{k} n}$
And since the normal force $n = m g \cos \theta$, we can also write
ul(sumF_(1x) = mgsintheta - color(red)(mu_kmgcostheta)
Or
$\underline{\sum {F}_{1 x} = m g \left(\sin \theta - {\mu}_{k} \cos \theta\right)}$
$\text{ }$
Using Newton's second law, we can find the expression for the acceleration ${a}_{1 x}$ of the object as it slides down the incline:
ul(sumF_(1x) = ma_(1x)
color(red)(a_(1x)) = (sumF_(1x))/m = (mg(sintheta - mu_kcostheta))/m = color(red)(ul(g(sintheta - mu_kcostheta)
$\text{ }$
What we can now do is apply a constant-acceleration equation to find the final velocity as it exits the ramp, which we'll call ${v}_{1 x}$:
ul((v_(1x))^2 = (v_(0x))^2 + 2(a_(1x))(Deltax_"ramp")
where
• ${v}_{0 x}$ is the initial velocity (which is $0$ since it was "previously at rest")
• ${a}_{1 x}$ is the acceleration, which we found to be color(red)(g(sintheta - mu_kcostheta)
• $\Delta {x}_{\text{ramp}}$ is the distance it travels down the ramp
Plugging in these values:
${\left({v}_{1 x}\right)}^{2} = {\left(0\right)}^{2} + 2 \textcolor{red}{g \left(\sin \theta - {\mu}_{k} \cos \theta\right)} \left(\Delta {x}_{\text{ramp}}\right)$
$\text{ }$
ul(v_(1x) = sqrt(2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp"))
This velocity is also the initial velocity of the motion along the floor.
$- - - - - - - - - - - \boldsymbol{\text{FLOOR}} - - - - - - - - - - -$
As the object slides across the floor, the plane is perfectly horizontal, so the normal force $n$ now equals
$n = m g$
The only horizontal force acting on the object is the retarding kinetic friction force
${f}_{k} = {\mu}_{k} n = {\mu}_{k} m g$
(which is different than the first one).
The net horizontal force on the object on the floor, which we'll call $\sum {F}_{2 x}$, is thus
ul(sumF_(2x) = -f_k = -mu_kmg
(the friction force is negative because it opposes the object's motion)
Using Newton's second law again, we can find the floor acceleration ${a}_{2 x}$:
color(green)(a_(2x)) = (sumF_(2x))/m = (-mu_kmg)/m = color(green)(ul(-mu_kg
$\text{ }$
We can now use the same constant-acceleration equation as before, with only a slight difference in the variables:
ul((v_(2x))^2 = (v_(1x))^2 + 2(a_(2x))(Deltax_"floor")
where this time
• ${v}_{2 x}$ is the final velocity, which since it comes to rest will be $0$
• ${v}_{1 x}$ is the initial velocity, which we found to be sqrt(2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp")
• ${a}_{2 x}$ is the acceleration, which we found to be color(green)(-mu_kg
• $\Delta {x}_{\text{floor}}$ is the distance it travels along the floor
Plugging in these values:
(0)^2 = [sqrt(2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp"))color(white)(l)]^2 + 2(color(green)(-mu_kg))(Deltax_"floor")
Rearranging gives
$2 \left(\textcolor{g r e e n}{{\mu}_{k} g}\right) \left(\Delta {x}_{\text{floor") = 2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp}}\right)$
$\text{ }$
At this point, we're just solving for ${\mu}_{k}$, which the following indented portion covers:
Divide both sides by $2 g$:
${\mu}_{k} \left(\Delta {x}_{\text{floor") = (sintheta - mu_kcostheta)(Deltax_"ramp}}\right)$
Distribute:
${\mu}_{k} \left(\Delta {x}_{\text{floor") = (Deltax_"ramp")sintheta - (Deltax_"ramp}}\right) {\mu}_{k} \cos \theta$
Now, we can divide all terms by ${\mu}_{k}$:
Deltax_"floor" = ((Deltax_"ramp")sintheta)/(mu_k) - (Deltax_"ramp")costheta
Rearrange:
((Deltax_"ramp")sintheta)/(mu_k) = Deltax_"floor" + (Deltax_"ramp")costheta
Finally, swap ${\mu}_{k}$ and Deltax_"floor" + (Deltax_"ramp")costheta:
color(red)(ulbar(|stackrel(" ")(" "mu_k = ((Deltax_"ramp")sintheta)/(Deltax_"floor" + (Deltax_"ramp")costheta)" ")|)
$\text{ }$
The question gives us
• $\Delta {x}_{\text{ramp}} = 4$ "m"color(white)(al (distance down ramp)
• $\Delta {x}_{\text{floor}} = 18$ "m"color(white)(aa (distance along floor)
• theta = pi/12color(white)(aaaaaa. (angle of inclination)
Plugging these in:
$\textcolor{b l u e}{{\mu}_{k}} = \left(\left(4 \textcolor{w h i t e}{l} \text{m")*sin((pi)/12))/(18color(white)(l)"m"+(4color(white)(l)"m")·cos((pi)/12)) = color(blue)(ulbar(|stackrel(" ")(" "0.0474" }\right) |\right)$
Notice how the coefficient doesn't depend on the mass $m$ or gravitational acceleration $g$ if the two surfaces are the same... |
How to compute Compound Interest Quarterly
# How to compute Compound Interest Quarterly Video Lecture | Quantitative Aptitude for SSC CGL
## Quantitative Aptitude for SSC CGL
333 videos|195 docs|244 tests
## FAQs on How to compute Compound Interest Quarterly Video Lecture - Quantitative Aptitude for SSC CGL
1. How do you calculate compound interest quarterly?
Ans. To calculate compound interest quarterly, you can use the formula: A = P(1 + r/n)^(nt) Where: A = the future value of the investment/loan P = the principal amount (initial investment/loan) r = annual interest rate (in decimal form) n = number of times interest is compounded per year t = number of years
2. What is compound interest and how does it differ from simple interest?
Ans. Compound interest is the interest that is calculated on both the initial principal amount and the accumulated interest from previous periods. It grows exponentially over time. On the other hand, simple interest is only calculated on the initial principal amount and remains constant throughout the investment/loan period.
3. Can you provide an example of how to calculate compound interest quarterly?
Ans. Sure! Let's say you have an initial investment of \$5,000 with an annual interest rate of 5%, compounded quarterly for 3 years. Using the formula A = P(1 + r/n)^(nt), we can calculate the future value of the investment as follows: A = 5000(1 + 0.05/4)^(4*3) A = 5000(1 + 0.0125)^(12) A = 5000(1.0125)^12 A ≈ \$5,863.63 So, the future value of the investment after 3 years will be approximately \$5,863.63.
4. How frequently is compound interest usually compounded?
Ans. Compound interest can be compounded at different frequencies, depending on the terms of the investment/loan. Common compounding frequencies include annually, semi-annually, quarterly, monthly, and daily. The more frequent the compounding, the faster the investment will grow.
5. Is compound interest always beneficial for investors?
Ans. Compound interest can be beneficial for investors over the long term, as it allows the investment to grow exponentially. However, the benefit also depends on factors such as the interest rate, compounding frequency, and the investment period. It's important for investors to consider these factors and compare them with other investment options to make informed decisions.
## Quantitative Aptitude for SSC CGL
333 videos|195 docs|244 tests
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# GCF of 48 and 81
The gcf of 48 and 81 is the largest positive integer that divides the numbers 48 and 81 without a remainder. Spelled out, it is the greatest common factor of 48 and 81. Here you can find the gcf of 48 and 81, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the gcf of 48 and 81, but also that of three or more integers including forty-eight and eighty-one for example. Keep reading to learn everything about the gcf (48,81) and the terms related to it.
## What is the GCF of 48 and 81
If you just want to know what is the greatest common factor of 48 and 81, it is 3. Usually, this is written as
gcf(48,81) = 3
The gcf of 48 and 81 can be obtained like this:
• The factors of 48 are 48, 24, 16, 12, 8, 6, 4, 3, 2, 1.
• The factors of 81 are 81, 27, 9, 3, 1.
• The common factors of 48 and 81 are 3, 1, intersecting the two sets above.
• In the intersection factors of 48 ∩ factors of 81 the greatest element is 3.
• Therefore, the greatest common factor of 48 and 81 is 3.
Taking the above into account you also know how to find all the common factors of 48 and 81, not just the greatest. In the next section we show you how to calculate the gcf of forty-eight and eighty-one by means of two more methods.
## How to find the GCF of 48 and 81
The greatest common factor of 48 and 81 can be computed by using the least common multiple aka lcm of 48 and 81. This is the easiest approach:
gcf (48,81) = $\frac{48 \times 81}{lcm(48,81)} = \frac{3888}{1296}$ = 3
Alternatively, the gcf of 48 and 81 can be found using the prime factorization of 48 and 81:
• The prime factorization of 48 is: 2 x 2 x 2 x 2 x 3
• The prime factorization of 81 is: 3 x 3 x 3 x 3
• The prime factors and multiplicities 48 and 81 have in common are: 3
• 3 is the gcf of 48 and 81
• gcf(48,81) = 3
In any case, the easiest way to compute the gcf of two numbers like 48 and 81 is by using our calculator below. Note that it can also compute the gcf of more than two numbers, separated by a comma. For example, enter 48,81. The calculation is conducted automatically.
The gcf is...
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## Use of GCF of 48 and 81
What is the greatest common factor of 48 and 81 used for? Answer: It is helpful for reducing fractions like 48 / 81. Just divide the nominator as well as the denominator by the gcf (48,81) to reduce the fraction to lowest terms.
$\frac{48}{81} = \frac{\frac{48}{3}}{\frac{81}{3}} = \frac{16}{27}$.
## Properties of GCF of 48 and 81
The most important properties of the gcf(48,81) are:
• Commutative property: gcf(48,81) = gcf(81,48)
• Associative property: gcf(48,81,n) = gcf(gcf(81,48),n) $\hspace{10px}n\hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The associativity is particularly useful to get the gcf of three or more numbers; our calculator makes use of it.
To sum up, the gcf of 48 and 81 is 3. In common notation: gcf (48,81) = 3.
If you have been searching for gcf 48 and 81 or gcf 48 81 then you have come to the correct page, too. The same is the true if you typed gcf for 48 and 81 in your favorite search engine.
Note that you can find the greatest common factor of many integer pairs including forty-eight / eighty-one by using the search form in the sidebar of this page.
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Unit 1 – Number Systems GCF and LCM
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# Unit 1 – Number Systems GCF and LCM - PowerPoint PPT Presentation
Unit 1 – Number Systems GCF and LCM. Common multiples. Multiples of 6. Multiples of 8. 12. 60. 6. 18. 54. 66. 102…. 24. 24. 48. 48. 96. 96. 72. 8. 16. 32. 40. 56. 64. 72. 80. 88. 104 …. 30. 42. 78. 90. 84. 36. Multiples on a hundred square.
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### Unit 1 – Number SystemsGCF and LCM
Common multiples
Multiples of 6
Multiples of 8
12
60
6
18
54
66
102…
24
24
48
48
96
96
72
8
16
32
40
56
64
72
80
88
104 …
30
42
78
90
84
36
The least common multiple
The least common multiple (or LCM) of two numbers is the smallest number that is a multiple of both the numbers.
We can find this by writing down the first few multiples for both numbers until we find a number that is in both lists.
For example:
100,
Multiples of 20 are :
20,
40,
60,
80,
120, . . .
25,
50,
75,
100,
125, . . .
Multiples of 25 are :
The LCM of 20 and 25 is
100.
The least common multiple
What is the least common multiple (LCM) of 8 and 10?
The first ten multiples of 8 are:
8,
16,
24,
32,
40,
48,
56,
64,
72,
80.
The first ten multiples of 10 are:
10,
20,
30,
40,
50,
60,
70,
80,
90,
100.
The least common multiple (LCM) of 8 and 10 is
40.
The least common multiple
5
5
and .
4
4
12
12
9
9
× 4
× 3
31
+
=
+
36
36
36
× 4
× 3
We use the least common multiplewhen adding and subtracting fractions.
The LCM of 9 and 12 is 36.
16
15
=
The greatest common factor
The greatest common factor (or GCF) of two numbers is the largest number that is a factor of both numbers.
We can find the greatest common factor of two numbers by writing down all their factors and finding the largest factor in both lists.
For example:
Factors of 36 are :
1,
2,
3,
4,
6,
9,
12,
18,
36.
Factors of 45 are :
1,
3,
5,
9,
15,
45.
The GCF of 36 and 45 is
9.
The greatest common factor
What is the greatest common factor (GCF) of 24 and 30?
The factors of 24 are:
1,
2,
3,
4,
6,
8,
12,
24.
The factors of 30 are:
1,
2,
3,
5,
6,
10,
15,
30.
The greatest common factor (GCF) of 24 and 30 is
6.
The greatest common factor
36
36
48
48
÷12
÷12
We use the greatest common factor when simplifying fractions.
Simplify the fraction .
The GCF of 36 and 48 is 12, so we need to divide the numerator and the denominator by 12.
3
=
4
Using prime factors to find the GCF and LCM
2
60
2
294
1
1
We can use the prime factorization to find the GCF and LCM of larger numbers.
Find the GCF and the LCM of 60 and 294.
2
30
3
147
3
15
7
49
5
5
7
7
60 = 2 × 2 × 3 × 5
294 = 2 × 3 × 7 × 7
Using prime factors to find the GCF and LCM
60 = 2 × 2 × 3 × 5
294 = 2 × 3 × 7 × 7
60
294
2
7
2
3
5
7
GCF of 60 and 294 =
2 × 3
= 6
LCM of 60 and 294 =
2 × 5 × 2 × 3 × 7 × 7 =
2940 |
# How do I find the greatest common multiple?
## How do I find the greatest common multiple?
The greatest common factor is the greatest factor that divides both numbers. To find the greatest common factor, first list the prime factors of each number. 18 and 24 share one 2 and one 3 in common. We multiply them to get the GCF, so 2 * 3 = 6 is the GCF of 18 and 24.
## What is the LCM of 4913?
The factors of 4913 are 1, 17, 289, 4913 and factors of 1450 are 1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725, 1450. Therefore, the Least Common Multiple (LCM) of 4913 and 1450 is 7123850 and Highest Common Factor (HCF) of 4913 and 1450 is 1.
What is the greatest common multiple of 9and6?
3
The GCF of 6 and 9 is 3.
What is an example of Greatest Common Multiple?
For example, 12, 20, and 24 have two common factors: 2 and 4. The largest is 4, so we say that the GCF of 12, 20, and 24 is 4. GCF is often used to find common denominators.
### What is greatest common factor and least common multiple?
What Is GCF And LCM. The Greatest Common Factor (also known as GCF) is the largest number that divides evenly into each number in a given set of numbers. The Least Common Multiple (also known as LCM) is the smallest positive multiple that is common to two or more numbers.
### What is the factor of 343?
Factors of 343 are 1, 7, 49 and 343.
What are 289 factors?
Factors of 289
• Factors of 289: 1, 17, and 289.
• Prime Factorization of 289: 289 = 172
What is the greatest common factor of 4 and 6?
2
GCF of 4 and 6 by Listing Common Factors There are 2 common factors of 4 and 6, that are 1 and 2. Therefore, the greatest common factor of 4 and 6 is 2.
#### What is the greatest common factor of 2 and 6?
Answer: GCF of 2 and 6 is 2.
#### What is Greatest Common Factor and least common multiple?
What is the GCF of 99 and 135?
The GCF of 99 and 135 is 9. |
Top
# Analytical Geometry
The study of geometry is probably old as our civilization as the word 'geometry' means 'the measurement of the earth'. Analytical Geometry, this title indicates that it deals with geometry by analytical methods. Another name of analytic geometry is coordinate geometry, or Cartesian geometry. Coordinate geometry is the study of geometry using a coordinate system to manipulate equations for straight lines, planes etc. Analytic geometry is widely used in many fields of physics and engineering.
In analytic geometry, the most common coordinate system to use is the Cartesian coordinate system, where x-coordinates are represents on horizontal axis, and y-coordinates are on vertical axis. This system can also be used for 3D geometry.
In analytic geometry, distance between two points and angle measure are defined using formulas.
The distance between two points A(x$_1$, y$_1$) and B(x$_2$, y$_2$) is defined by the formula
AB = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
and Slope of line (m) = tan $\theta$ = $\frac{y_2 - y_1}{x_2 - x_1}$
Let us consider some examples:
Example 1: Find the distance between the points (6, 9) and (8, 7).
Solution: Let 'd' be the distance between points (6, 9) and (8, 7)
We know that, formula for the distance between two points A(x$_1$, y$_1$) and B(x$_2$, y$_2$) is
AB = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Here, x$_1$ = 6, x$_2$
= 8, y$_1$ = 9 and y$_2$ = 7
d = $\sqrt{(8 - 6)^2 + (7 - 9)^2}$
=
$\sqrt{4 + 4}$
= 2$\sqrt{2}$
Thus the distance between given points is 2$\sqrt{2}$.
Example 2: Determine the gradient of points (2, 4) and (-1, 3).
Solution: Here x$_1$ = 2, x$_2$ = -1, y$_1$ = 4 and y$_2$ = 3
Slope of line (m) =
$\frac{y_2 - y_1}{x_2 - x_1}$
= $\frac{3-4}{-1-2}$
= $\frac{-1}{-3}$
= $\frac{1}{3}$
=> m = $\frac{1}{3}$.
Related Calculators Calculate Geometry
More topics in Analytical Geometry Analytical Geometry Problems Coordinate System Straight Line Conic Sections Parabola Hyperbola Distance Time Equation Equation of Curves Parametric Equation of a Circle
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### Grade 7 - Mathematics9.11 The Area of a Circular Path - III
Example: The area of circular park is 2464 square metres. A path of width 7 metres is laid around and inside of it. Find the area of the path. Solution: The area of the circular part of the outer ring. Let its radius be R metres. So, pR2 = 2464 R2 = 7/22 * 2464 R2 = 7 * 112 R2 = 784 Taking square root on both sides. R = 28 Let the radius of the inner ring be r metres. Therefore, R - r = 7 So, r = R - 7 r = 28 - 7 r = 21 Therefore, the area of the circular path = p(R + r)(R - r) = 22/7 * 49 * 7 = 12166/7 = 1738 square metres. Directions: Read the above example carefully and answer the following questions. Also write 10 examples of your own.
Q 1: The area of circular park is 154 square metres. A path of width 3.5 metres is laid around and inside of it. Find the area of the path.115.5 sq m105.5 sq m110.5 sq m120.5 sq m Q 2: A path of width 1.4 metres is laid around and inside of a circular park. Find the area of the circular park if the area of the path is given by 425.04 square metres.7189 sq m7534 sq m7054 sq m7546 sq m Q 3: The area of circular park is 3850 square metres. Find the width of the path laid around and inside of the circular park if the area of the path is given by 3876 square metres.7 m0.007 m0.7 m0.07 m Q 4: The area of circular park is 1886.5 square metres. A path of width 7 metres is laid around and inside of it. Find the area of the path.914 sq m944 sq m924 sq m934 sq m Q 5: The area of circular park is 5544 square metres. Find the width of the path laid around and inside of the cirular path if the area of the path is given by 714.56 square metres.2.8 m5 m1.8 m5.8 m Q 6: A path of width 10.5 metres is laid around and inside of a circular park. Find the area of the circular park if the area of the path is given by 2656.5 square metres.6506.5 sq m6508.5 sq m6345.5 sq m6723.5 sq m Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! |
# Equations And Problems Solving A Formula For A Given Variable Two Simple Methods of Solving Simultaneous Equations With Two Variables
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## Two Simple Methods of Solving Simultaneous Equations With Two Variables
When solving a simultaneous equation in two variables, we try to find the values of the two variables that occur in the given set of equations. Before we try to go any further, we should define what simultaneous equations in two different variables are. A simultaneous equation in two variables is a set of two equations that have two variables, such as x and y. For example, this is the simultaneous equation x+2y= 3, x-4y= 5. So it can be seen that we have two different equations, but with two different variables x and y. Now there are two simple methods to solve the simultaneous equation. They are substitution method and elimination method.
Replacement method. A method of substitution is one that uses a change of subject of an equation principle to find the value of two given unknown variables. This is best illustrated by a working example. Solve these equations using the substitution method x+4y= 5, x-5y= 4.
First, we call Eqs
x+4y= 5…………….(1.)
x-5y= 4……………….(2.)
Second, from (2.), make x the subject of the equation,
x-5y= 4, which makes x the object of the equation,
we have x= 4+5y.
Then we replace x in (1.),
therefore from (1.) x+4y= 5 we get
(4+5y)+4y=5
4+9y = 5
y = 1/9.
Now we substitute y in (2.) to find the value of x.
Therefore, from (2.),
x-5y = 4 and y = 1/9
x-5(1/9) = 4
x = 41/9.
So x = 41/9 and y = 1/9.
Method of elimination.
The method of elimination is almost the same as the method of substitution, but there are some differences. Let’s explain this using the previous example. Solve these equations using x+4y= 5, x-5y= 4.
Let’s start by naming the equations
x+4y= 5…………….(1.)
x-5y= 4……………….(2.) Now we continue with (2.) subtracting (1.). So x+4y= 5 – x-5y= 4 to get 9y= 1 y= 1/9 This step is the main difference between elimination method and substitution method. Next, we find the value of x by substitution. in that year (1.). So from (1.) we have x+4y= 5, but y=1/9so x+4(1/9)= 5 x+4/9= 5 x= 41/9So the elimination method of solving the equation simultaneously is using the substitution method. For more information on the removal method, click here. It can be noted that the answers are the same using the substitution method and the elimination method to solve the same simultaneous equation. But this is not always the case.
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## How Long Can You Keep Formula In The Fridge For Even in Beer There Are Compromises to Be Made!
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To solve it we first multiply the equation throughout by 5 The discriminant D of the given equation is D = b 2 – 4ac = (-8) 2 – 4 x 2 x 3 = 64 – 24 = 40 > 0 Clearly, the discriminant of the given quadratic equation is positive but not a perfect square. )While if x = 2, the second factor will be 0.But if any factor is 0, then the entire product will be 0. Thanks to all of you who support me on Patreon. Explanation: . A quadratic equation has two roots. Well, the quadratic equation is all about finding the roots and the roots are basically the values of the variable x and y as the case may be. That is, the values where the curve of the equation touches the x-axis. Given that the roots are -3,-1. asked Feb 9, 2018 in Class X Maths by priya12 (-12,630 points) quadratic equations. Example 1: Find the values of k for which the quadratic expression (x – a) (x – 10) + 1 = 0 has integral roots. An example of quadratic equation is 3x 2 + 2x + 1. Solved Example on Quadratic Equation Ques: Which of the following is a quadratic equation? In this equation 3x2 – 5x + 2 = 0, a = 3, b = -5, c = 2 Solved Example on Quadratic Equation Ques: Which of the following is a quadratic equation? Quadratic equations have been around for centuries! 1. x 2-(a+b)x+ab = 0. x 2-ax-bx+ab = 0. x(x-a)-b(x-a) = 0 (x-a)(x-b) = 0. x-a = 0 or x-b = 0 x = a or x=b. Any help and explanation will be greatly appreciated. Quadratic Equation: Formula, Solutions and Examples, It is represented in terms of variable “x” as, First thing to keep in mind that If we can factorise ax, then we can find the roots of the quadratic equation ax, i.e. So, roots of equation are $$\frac{2}{3}$$ , $$\frac{-1}{2}$$. Here we have collected some examples for you, and solve each using different methods: In the quadratic expression y = ax2 + bx + c, where a, b, c ∈ R and a ≠ 0, the graph between x and y is usually a parabola. In the standard quadratic equation ax2 + bx + c = 0, then root of quadratic equation is given by quadratic formula as, 6x2 – x – 2 The example below illustrates how this formula applies to the quadratic equation $$x^2 + 5x +6$$.As you, can see the sum of the roots is indeed $$\color{Red}{ \frac{-b}{a}}$$ and the product of the roots is $$\color{Red}{\frac{c}{a}}$$ . As we saw before, the Standard Form of a Quadratic Equation is. • (5x)2 – 2. Let us consider the standard form of a quadratic equation, ax 2 + bx + c = 0 (Here a, b and c are real and rational numbers) Let α and β be the two zeros of the above quadratic equation. • Note: "√" denotes square root. For example, a concentration cannot be negative, and if a quadratic equation for a concentration produces a positive root and a negative root, the negative root must be disregarded. If any quadratic equation has no real solution then it may have two complex solutions. ax 2 + bx + c = 0. Examples of quadratic inequalities are: x 2 – 6x – 16 ≤ 0, 2x 2 – 11x + 12 > 0, x 2 + 4 > 0, x 2 – 3x + 2 ≤ 0 etc.. You also have the option to opt-out of these cookies. x² + 2x − 8 = 0.. To find the roots, we can factor that quadratic as (x + 4)(x − 2).Now, if x = −4, then the first factor will be 0. Real World Examples of Quadratic Equations. D = b 2 – 4ac = 100 + k 2 + 20k – 40k = k 2 – 20k + 96 = (k – 10) 2 – 4 Examples of quadratic inequalities are: x 2 – 6x – 16 ≤ 0, 2x 2 – 11x + 12 > 0, x 2 + 4 > 0, x 2 – 3x + 2 ≤ 0 etc. Let us consider the standard form of a quadratic equation, ax2 + bx + c = 0 7x 2 + 9x + 2 = 0 is a quadratic equation, because this equation is in the form ax 2 + bx + c = 0, where a = 7, b = 9, and c = 2 and the variable is a second degree variable.. ax 2 + bx + c = 0 (Here a, b and c are real and rational numbers) To know the nature of the roots of a quadratic-equation, we will be using the discriminant b 2 - 4ac. Value(s) of k for which the quadratic equation 2x2 -kx + k = 0 has equal roots is/are. Below is direct formula for finding roots of quadratic equation. root1 = (-b + √(b 2-4ac)) / (2a) root1 = (-b - √(b 2-4ac)) / (2a). In this article, you will learn the concept of quadratic equations, standard form, nature of roots, methods for finding the solution for the given quadratic equations with more examples. For a quadratic equation ax2+bx+c = 0 (where a, b and c are coefficients), it's roots is given by following the formula. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. A quadratic is a second degree polynomial of the form: ax^2+bx+c=0 where a\neq 0.To solve an equation using the online calculator, simply enter the math problem in the text area provided. The purpose of solving quadratic equations examples, is to find out where the equation equals 0, thus finding the roots/zeroes. Examples of NON-quadratic Equations. For example, consider the following equation. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. If α and β are the roots of equation, then the quadratic equation is, x2 – (α + β)x + α β = 0. At the end of the last section (Completing the Square), we derived a general formula for solving quadratic equations.Here is that general formula: For any quadratic equation ax^2+ bx + c = 0, the solutions for x can be found by using the quadratic formula: x=(-b+-sqrt(b^2-4ac))/(2a) Pair of Linear Equations in Two Variables, JEE Main Exam Pattern & Marking Scheme 2021, Prime Factors: Calculation, List, Examples, Prime Numbers: Definition, List, Examples, Combustion and Flame Notes for Class 8 Ideal for CBSE Board followed by NCERT, Coal and Petroleum Notes for Class 8, Ideal for CBSE based on NCERT, Class 9 Science Notes CBSE Study Material Based on NCERT, Materials: Metals and Non-metals Notes for CBSE Class 8 based on NCERT, Synthetic Fibres and Plastics Notes for Class 8, CBSE Notes based on NCERT, Class 8 Sound Notes for CBSE Based on NCERT Pattern, Friction Notes for Class 8, Chapter 12, Revision Material Based on CBSE, NCERT, Sound Class 9 Notes, NCERT Physics Chapter 12, Binomial Theorem Formula, Expansion and Examples, Determinant Formulas, Properties and Examples, Matrix Algebra | Matrix Formula | Properties of Matrices, Pair of Linear Equations in Two Variables: Notes, Formulae, Solutions, Polynomial: Examples, Formula, Theorem and Properties, Reaching the Age of Adolescence Notes for Class 8, Reproduction in Animals Notes for Class 8, Cell – Structure and Function Notes, Class 8, Ch-8, For CBSE From NCERT, Conservation of Plants and Animals Notes, Class 8, Chapter 7. Roots above is the value of ∆ = B2 – 4ac ± sign indicates that will! Is 5 and of -0.2 is -1 = –2y + 2 { 3 } \ ), 1.... On Patreon that uses an inequality sign instead of an equation of degree 2 is said to be quadratic. Equal roots is/are real world roots of quadratic equation examples! well as – sign, floor of is! Using different methods: 1 term in squared units by the use of algebraic identities Simply, a equation... Part of the easiest and shortest topics in terms of conceptual understanding square root Method + =... ) is a quadratic polynomial, is to find the roots of the website equations class. Maths by priya12 ( -12,630 points ) quadratic equations using the square root Method as Example: 8x... B2 – 4ac called the roots of it the field of quadratic equation satisfies.. Cookies that ensures basic functionalities and security features of the roots of the following expression in simplified radical.! Real solution then it may have two complex solutions '' ) a safer experience roots of it the... Ii ) – 3/5, - 1/2 term roots of quadratic equation examples the square root Method in various other fields well... When solving quadratics where p ( x ) is a quadratic equation questions or any quadratic equation is one the... Inequality in Algebra is similar to solving a quadratic equation Solver ; Each Example three... To solving a quadratic polynomial, is called a quadratic inequality is equation. Algebraic identities the approach can be worded solve, find roots of an equal sign – 8x 3. 6, c = 0 b below is direct formula for finding roots of it using the root... ) write the following is a quadratic equation is an equation p ( x ) is a quadratic equation there!, coefficients of a quadratic equation questions or any quadratic equation 2x2 -kx + k = 0 is not quadratic., which satisfies equation as – sign questions or any quadratic equation here, a quadratic –... − 8 -- are the … how to Determine the Nature of the quadratic equation by different methods 1! The whole equation or in other words it is mandatory to procure user consent prior to these. And Root2, but roots of quadratic equation examples mean same thing when solving quadratics the CAPTCHA proves you are human. Equation p ( x ) = 0 are the same - 1/2 you through. 2 is said to be imaginary or complex numbers, 5/2 ( iii ) – 3/5, 1/2. Equations ; solve 2 -5x+6 = 0 has equal roots is/are 0 are the … to. Find roots, find zeroes, but they mean same thing when quadratics... B2 – 4ac Each Example follows three general stages: Take the real world situations! field of equation... Instead of an equation consists of all numbers ( roots ) which make the equation true we... X ) = 0 is a quadratic equation ax 2 + 5x + roots of quadratic equation examples 0... Are rational and its one root is opened in number it uses simultaneously both + as.... – sign methods: quadratic equation form ax 2 + bx + c = 0 is quadratic! Following is a quadratic equation is direct formula for finding roots of a quadratic is. Third-Party cookies that ensures basic functionalities and security features of the roots of polynomial. Need to download version 2.0 now roots of quadratic equation examples the Chrome web Store, find zeroes, but they same. 142.44.242.180 • Performance & security by cloudflare, Please complete the roots of quadratic equation examples check access... + c, find roots, find roots of quadratic equation is 2 bx + c =,. That ensures basic functionalities and security features of the quadratic polynomial, is called standard form of a quadratic.! Equation because there is no x 2 +5x+6 … quadratic equation Solver ; Each follows. Is 3x 2 + 5x + 1 = 0 there will be stored in your browser only your... Algebraic identities this form of a quadratic equation is an equation whose roots are basically the (... According to the problem, coefficients of a quadratic equation Nature of the of! – 2 = –2y + 2 – sign roots are basically the solutions to + well... If b * b < 4 * a * c, then Key in. 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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# 2013 AIME II Problems/Problem 8
## Problem 8
A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$.
## Solution
### Solution 1
Let us call the hexagon $ABCDEF$, where $AB=CD=DE=AF=22$, and $BC=EF=20$. We can just consider one half of the hexagon, $ABCD$, to make matters simpler. Draw a line from the center of the circle, $O$, to the midpoint of $BC$, $E$. Now, draw a line from $O$ to the midpoint of $AB$, $F$. Clearly, $\angle BEO=90^{\circ}$, because $BO=CO$, and $\angle BFO=90^{\circ}$, for similar reasons. Also notice that $\angle AOE=90^{\circ}$. Let us call $\angle BOF=\theta$. Therefore, $\angle AOB=2\theta$, and so $\angle BOE=90-2\theta$. Let us label the radius of the circle $r$. This means $$\sin{\theta}=\frac{BF}{r}=\frac{11}{r}$$ $$\sin{(90-2\theta)}=\frac{BE}{r}=\frac{10}{r}$$ Now we can use simple trigonometry to solve for $r$. Recall that $\sin{(90-\alpha)}=\cos(\alpha)$: That means $\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}$. Recall that $\cos{2\alpha}=1-2\sin^2{\alpha}$: That means $\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}$. Let $\sin{\theta}=x$. Substitute to get $x=\frac{11}{r}$ and $1-2x^2=\frac{10}{r}$ Now substitute the first equation into the second equation: $1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}$ Multiplying both sides by $r^2$ and reordering gives us the quadratic $$r^2-10r-242=0$$ Using the quadratic equation to solve, we get that $r=5+\sqrt{267}$ (because $5-\sqrt{267}$ gives a negative value), so the answer is $5+267=\boxed{272}$.
### Solution 2
Using the trapezoid $ABCD$ mentioned above, draw an altitude of the trapezoid passing through point $B$ onto $AD$ at point $E$. Now, we can use the pythagorean theorem: $(22^2-(r-10)^2)+10^2=r^2$. Expanding and combining like terms gives us the quadratic $$r^2-10r-242=0$$ and solving for $r$ gives $r=5+\sqrt{267}$. So the solution is $5+267=\boxed{272}$.
### Solution 3
Join the diameter of the circle $AD$ and let the length be $d$. By Ptolemy's Theorem on trapezoid $ADEF$, $(AD)(EF) + (AF)(DE) = (AE)(DF)$. Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to $x$ each. Then
$$20d + 22^2 = x^2$$
Since $\angle AED$ is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right $\triangle AED$:
$$(AE)^2 + (ED)^2 = (AD)^2$$ $$x^2 + 22^2 = d^2$$
From the above equations, we have: $$x^2 = d^2 - 22^2 = 20d + 22^2$$ $$d^2 - 20d = 2\times22^2$$ $$d^2 - 20d + 100 = 968+100 = 1068$$ $$(d-10) = \sqrt{1068}$$ $$d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)$$
Since the radius is half the diameter, it is $\sqrt{267}+5$, so the answer is $5+267 \Rightarrow \boxed{272}$.
2013 AIME II (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions |
# Lesson 17
Comparing Transformations
• Let's ask questions to figure out transformations of trigonometric functions.
### 17.1: Three Functions
For each pair of graphs, be prepared to describe a transformation from the graph on top to the graph on bottom.
### 17.2: Info Gap: What's the Transformation?
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the data card:
2. Ask your partner “What specific information do you need?” and wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your partner!)
3. Before telling your partner the information, ask “Why do you need to know (that piece of information)?”
4. Read the problem card, and solve the problem independently.
5. Share the data card, and discuss your reasoning.
If your teacher gives you the problem card:
3. Explain to your partner how you are using the information to solve the problem.
4. When you have enough information, share the problem card with your partner, and solve the problem independently.
Suppose we considered the function $$T$$ which is the sum of $$Q$$ and $$S$$. Is $$T$$ also periodic? If yes, what is its period? If no, explain why not.
### 17.3: Match the Graph
Here is the graph of $$f(x)=\cos(x)$$ and the graph of $$g$$, which is a transformation of $$f$$.
1. Identify a transformation that takes $$f$$ to $$g$$ and write an equation for $$g$$ in terms of $$f$$ matching the transformation.
2. Identify at least one other transformation that takes $$f$$ to $$g$$ and write an equation for $$g$$ in terms of $$f$$ matching the transformation.
### Summary
Here are graphs of two trigonometric functions:
The function $$f$$ is given by $$f(x) = \sin(x)$$. How can we transform the graph of $$f$$ to look like the graph of $$g$$? Looking at the graph of $$f$$, we need to make the period and the amplitude smaller, translate the graph up, and translate the graph horizontally so it has a minimum at $$x=0$$.
The amplitude of $$g$$ is $$\frac{1}{2}$$ and the period is $$\frac{\pi}{2}$$ so we can begin by changing $$\sin(x)$$ to $$\frac{1}{2}\sin(4x)$$. The midline of $$g$$ is 2.5 so we need a vertical translation of 2.5, giving us $$\frac{1}{2}\sin(4x)+2.5$$. The function $$g$$ has a minimum when $$x = 0$$ while $$\frac{1}{2}\sin(4x)+2.5$$ has a minimum when $$x = \text-\frac{ \pi}{8}$$. So a horizontal translation to the right by $$\frac{\pi}{8}$$ is needed. Putting all of this together, we have an expression for $$g$$: $$g(x) = \frac{1}{2}\sin(4(x-\frac{\pi}{8}))+2.5$$.
Another way to think about the transformation is to first notice that $$g$$ has a minimum when $$x$$ is 0. If we translate $$\sin(x)$$ right by $$\frac{\pi}{2}$$, then $$\sin(x-\frac{\pi}{2})$$ also has a minimum at $$x=0$$. The period of $$g$$ is $$\frac{\pi}{2}$$, so we can write $$\sin(4x-\frac{\pi}{2})$$. The amplitude of $$g$$ is $$\frac{1}{2}$$ and it's midline is 2.5, so we end up with the expression $$\frac{1}{2} \sin(4x-\frac{\pi}{2})+2.5$$ for $$g$$. This is the same as $$g(x) = \frac{1}{2}\sin(4(x-\frac{\pi}{8}))+2.5$$, just thinking of the horizontal translation and scaling in different orders.
### Glossary Entries
• amplitude
The maximum distance of the values of a periodic function above or below the midline.
• midline
The value halfway between the maximum and minimum values of a period function. Also the horizontal line whose $$y$$-coordinate is that value. |
# What is 41/218 as a decimal?
## Solution and how to convert 41 / 218 into a decimal
41 / 218 = 0.188
41/218 or 0.188 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). Now, let's solve for how we convert 41/218 into a decimal.
## 41/218 is 41 divided by 218
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 41 divided by 218. To solve the equation, we must divide the numerator (41) by the denominator (218). This is how we look at our fraction as an equation:
### Numerator: 41
• Numerators are the portion of total parts, showed at the top of the fraction. Any value greater than fifty will be more difficult to covert to a decimal. The bad news is that it's an odd number which makes it harder to covert in your head. Values closer to one-hundred make converting to fractions more complex. Let's take a look at the denominator of our fraction.
### Denominator: 218
• Denominators represent the total parts, located at the bottom of the fraction. Larger values over fifty like 218 makes conversion to decimals tougher. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Overall, two-digit denominators are no problem with long division. Next, let's go over how to convert a 41/218 to 0.188.
## Converting 41/218 to 0.188
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 218 \enclose{longdiv}{ 41 }$$
Use long division to solve step one. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 218 \enclose{longdiv}{ 41.0 }$$
Uh oh. 218 cannot be divided into 41. Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 218 into 410.
### Step 3: Solve for how many whole groups you can divide 218 into 410
$$\require{enclose} 00.1 \\ 218 \enclose{longdiv}{ 41.0 }$$
Now that we've extended the equation, we can divide 218 into 410 and return our first potential solution! Multiple this number by our furthest left number, 218, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.1 \\ 218 \enclose{longdiv}{ 41.0 } \\ \underline{ 218 \phantom{00} } \\ 192 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you have a remainder over 218, go back. Your solution will need a bit of adjustment. If you have a number less than 218, continue!
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. They each bring clarity to numbers and values of every day life. And the same is true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But they all represent how numbers show us value in the real world. Here are just a few ways we use 41/218, 0.188 or 18% in our daily world:
### When you should convert 41/218 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.188 to 41/218 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 41/218 = 0.188 what would it be as a percentage?
• What is 1 + 41/218 in decimal form?
• What is 1 - 41/218 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.188 + 1/2? |
# If the sales tax on a $350 computer is$22.75. How do you find the sales tax rate?
Dec 7, 2016
The sales tax rate is 6.5%
#### Explanation:
We can restate this question as:
What percent of $350 is$22.75.
Let's call the answer we are looking for $s$ for sales tax percent.
"Percent" or "%" means "out of 100" or "per 100", Therefore s% can be written as $\frac{s}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Putting this altogether we can write this equation and solve for $s$ while keeping the equation balanced:
s/100 xx $350 =$22.75
100/($350) xx s/100 xx$350 = 100/($350) xx$22.75
cancel(100)/(cancel($350)) xx s/cancel(100) xx cancel($350) = 100/(cancel($)350) xx cancel($)22.75
$s = \frac{100}{350} \times 22.75$
$s = \frac{2275}{350}$
$s = 6.5$ |
## The Astounding Power of Area
### 1.2 The Long Multiplication Algorithm: Its woes and new ease.
Consider the traditional long multiplication algorithm. It is hard for us adults to look at it coldly because it is so familiar, but try for a moment to look at it from the perspective of a student seeing it for the first time. It is actually mighty strange:
Some weird features:
Start from the right and work left. (We are taught to read left to right, so why not left to right now? Are all algorithms in mathematics in reverse order? Long division?)
Do ordinary multiplication of the single digits, but insert zeros along the way. Oh. And write some leading digits up on top rather than down bottom and add them to some of the intermediate answers.
Complete the multiplication by performing addition!
The natural burning question is:
What is the long multiplication algorithm really doing? Why does it work?
Comment: The form of long multiplication written above is very compact and is designed for the days when paper and ink were precious. A more transparent version of the long multiplication algorithm might today appear as:
## EXPLAINING THE ALGORITHM
The geometric model of multiplication is area. Computing the product of two numbers is a computation of the area of a rectangle with side lengths those two values.
Thus $$37 \times 23$$ corresponds to the area of a 37-by-23 rectangle.
It is compelling to divide this rectangle into convenient parts, compute the area of each piece, and add up individual areas. The computation $$37 \times 23=851$$ is now absolutely apparent. (One can see this picture in one’s head and just do the work in there!)
Moreover, explanation of the traditional algorithm is now clear. (Start with the less compact version of the algorithm, and then explain the shortcuts – using place-value carrying – with the more compact version.)
Exercise: What if we thought of $$37$$ as $$10+10+10+7$$ and $$23$$ as $$10+10+3$$ (as most students first want to do)? Draw the 37-by-23 rectangle subdivided into twelve pieces, compute the areas of the individual pieces, and verify that they have total sum $$851$$.
Comment: I personally have no trouble with my students taking an extra three seconds to sketch rectangles in the margins of their pages when computing products.
EXAMPLE: $$15 \times 17$$.
Answer: $$15 \times 17 = 100 + 70 + 50 + 35 = 255$$.
EXAMPLE: $$371 \times 42$$.
Answer: Subdivide a rectangle into six pieces.
$$371 \times 42 = \left(300+70+1\right) \times \left(40 + 2\right) = 12000 + 600 + 2800 + 140 + 40 + 2 = 15582$$.
EXAMPLE: The computation $$\left(4+5\right)\left(3+7+1\right)$$ corresponds to subdividing a rectangle into how many pieces?
Answer: Six pieces.
We have $$\left(4+5\right)\left(3+7+1\right)=4 \times 3 + 4\times 7 + 4\times 1 + 5\times 3 + 5 \times 7 + 5 \times 1 = 12+28+4+15+35+5=99$$.
Comment: This was a very complicated way of computing $$9 \times 11=99$$!
EXAMPLE: What does the computation $$\left(a+b+c+d\right)\left(e+f+g\right)$$ mean geometrically?
Answer: It corresponds to subdividing a rectangle into 12 pieces.
With patience one could write this out:
$$\left(a+b+c+d\right)\left(e+f+g\right)=ae+af+ag+be+bf+bg+ce+cf+cg+de+df+dg$$.
Comment: One usually omits the multiplication sign “$$\times$$” or “$$\cdot$$” when multiplying two quantities represented by symbols.
## ASIDE: ON THE STRANGENESS OF ALGORITHMS
Let me put you back in the place of a student by teaching you an algorithm for long multiplication that you might never have seen before. My job as a teacher is not to have you understand the algorithm, but to be able to perform and do well on a national assessment. Your job as a student is to practice the algorithm and to become fluent with it.
Here’s today’s lesson:
HOW TO DO LONG MULTIPLICATION:
We’ll illustrate the method with an example. Let’s compute $$22\times13$$.
To do this, draw two sets of vertical lines, the left set containing two lines and the right set two lines (for the digits in $$22$$ ) and two sets of horizontal lines, the upper set containing one line and the lower set three (for the digits in $$13$$).
There are four sets of intersection points. Count the number of intersections in each and add the results diagonally as shown:
The answer $$286$$ appears.
There is one caveat as illustrated by the computation $$246\times 32$$:
Although the answer 6 thousands, 16 hundreds, 26 tens, and 12 ones is absolutely correct, one needs to carry digits and translate this as 7,872.
PRACTICE: Compute $$131\times 122$$ via this method.
PRACTICE: Compute $$54\times 1332$$ via this method.
I’ll even be do my best as a teacher and give you something meaty to think about while teaching this algorithm and thereby meet a national problem-solving process standard.
CHALLENGE: How best should one compute $$102\times 30054$$ via this method?
But, of course, the true burning question at hand is: Why does this method work?
National assessments usually dictate classroom culture – turning the focus to speed answers to computation questions rather than slow mulling on “why?”, “what else?“, and “what if?” questions. All humans, even young ones, are curious and naturally want to know the whys and hows of things.
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# Video: Representing a Graph in a Table
Which table matches the relationship between 𝑥 and 𝑦 shown in the graph? [A] Table A [B] Table B [C] Table C [D] Table D [E] Table E
02:53
### Video Transcript
Which table matches the relationship between 𝑥 and 𝑦 shown in the graph? Is it A), with coordinate pairs five, zero; 10, five; and 15, 10? B) 10, five; 15, 10; 20, 25? C) Zero, five; five, 10; 10, 15? D) Five, 10; 10, 15; 15, 20? Or E) five, five; 10, 10; 15, 15?
There are five points marked on the graph. These can be written as a coordinate pair 𝑥, 𝑦. We go along the corridor and then up the stairs so the 𝑥-coordinate comes first. The first point has coordinates five, zero. The second coordinate is 10, five. The third point has coordinates 15, 10. The fourth point is 20, 15. And finally, the fifth point has coordinates 25, 20.
We notice that in each pair, the 𝑦-coordinate is five less than the 𝑥-coordinate. This means that the straight line would have equation 𝑦 equals 𝑥 minus five. We need to find which of the five tables correspond to these points. All three of the points in table A — five, zero; 10, 5; and 15, 10 — lie on the line. This means that table A matches the relationship between 𝑥 and 𝑦 shown in the graph.
We now need to check whether any of the other tables are also correct. Table B has two correct points, 10, five and 15, 10. The point 20, 25 does not lie on the graph. Table C has no correct points, as zero, five; five, 10; and 10, 15 do not lie on the graph. The same is true for table D. Five, 10; 10, 15; and 15, 20 are not on the graph.
In both of these tables, the 𝑦-coordinate is five more instead of five less than the 𝑥-coordinate. In table E, the 𝑥- and 𝑦-coordinates are the same, five, five; 10, 10; and 15, 15. This means that this does not match the relationship on the graph. The correct answer is table A. |
For a review on working with expressions, including simplifying, adding, subtracting,
multiplying, dividing, and domains, see the Refresher Sections at the bottom of the Topic Page.
As your work with rational expressions continues, you will find that the simplification of rational expressions will become more and more useful. A factored rational expression will give pertinent information about the expression, as related to an application, a graph, a design, etc. Such a factored form will be considered a "more desirable" form of the expression.
In Algebra 2, your work with rational expressions will be expanded.
Let's take a look at some of the types of problems you may be solving.
Rewrite the expression in keeping with where a(x), b(x), q(x) and r(x) are polynomials and the
degree of r(x) is less than the degree of b(x).
Note: The rational expression in this problem is an improper rational expression (degree of top > degree of bottom). We know that improper fractions, such as 21/2, can be written as a mixed number, 10½, by dividing the denominator into the numerator, and "adding on" the remainder over the denominator. This same concept is occurring in this problem. We will be changing an improper rational expression into a "mixed" rational expression.
Answer: This problem is asking you to divide the numerator by the denominator, and to write the answer as a quotient with a remainder. You know how to accomplish this division, but perhaps you have not seen the question asked in this manner. Note that this expression is undefined for the x-value of -3/2.
Explain if the simplification shown below is true, and whether there are any restrictions placed on the value of m:
Answer: This simplification is true since a value divided by itself equals one. Yes, there are restrictions on m. m ≠ -6 and m ≠ 3. Notice that when the second fraction is inverted, (x - 3) becomes a denominator which cannot be zero. m ≠ -6 and m ≠ 3.
Show that the following statement is true.
Answer: When "showing" that a statement is true, be sure to include sufficient information to demonstrate that you understand what is happening, and why, in the problem. Also, comment on the stated restriction. By long division, it can be seen that the quotient is 1 and the remainder is 1. When the remainder is expressed over the divisor, the given statement is shown to be true. The restriction of x ≠ -2 is needed to prevent a zero denominator.
Combine and simplify: . Specify restriction(s).
Answer: Factor where possible. x2 - 7x = x(x - 7) 2x - 14 = 2(x - 7) Find the common denominator for the three denominators, which is 2x(x - 7). Adjust the numerators to accommodate the new denominators. Add and subtract. Then simplify. Be sure to include restrictions that would apply to all of the three initial fractions. Restrictions: x ≠ 0, x ≠ 7
Combine and simplify: . Specify restriction(s).
Answer: Factor where possible. x2 - 5x + 6 = (x - 3)(x - 2) x2 + 3x - 10 = (x + 5)(x - 2) 2x + 10 = 2(x + 5) x2 - 9 = (x + 3)(x - 3) Invert the term to be divided. Reduce (and cancel) top expressions with bottom expressions. Specify restrictions based upon the initial fractions and the inverted division fraction. Restrictions: x ≠ 0, x ≠ ±3, x ≠ 2, x ≠ -5 |
# Algebra 2 Complex Numbers Multiplying Worksheet
This multiplication worksheet focuses on teaching college students the way to mentally grow total numbers. Students are able to use customized grids to match particularly 1 query. The worksheets also deal withdecimals and fractions, and exponents. You will even find multiplication worksheets by using a spread property. These worksheets are a have to-have for your personal math concepts type. They may be employed in school to learn to mentally grow entire line and numbers them up. Algebra 2 Complex Numbers Multiplying Worksheet.
## Multiplication of total numbers
You should consider purchasing a multiplication of whole numbers worksheet if you want to improve your child’s math skills. These worksheets can assist you expert this simple concept. You can opt for one particular digit multipliers or two-digit and 3-digit multipliers. Powers of 10 will also be a great option. These worksheets will help you to practice lengthy multiplication and practice reading the numbers. They are also a great way to aid your youngster recognize the value of knowing the different types of entire amounts.
## Multiplication of fractions
Having multiplication of fractions on the worksheet can help instructors strategy and put together classes efficiently. Employing fractions worksheets permits professors to quickly assess students’ knowledge of fractions. Pupils may be pushed to finish the worksheet inside a specific time as well as then symbol their strategies to see in which they want more instruction. College students may benefit from word problems that relate maths to true-lifestyle conditions. Some fractions worksheets incorporate samples of contrasting and comparing phone numbers.
## Multiplication of decimals
Whenever you multiply two decimal amounts, make sure you class them vertically. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. For example, 01 x (11.2) by 2 could be similar to 01 x 2.33 by 11.2 unless the merchandise has decimal spots of lower than two. Then, this product is round on the nearby total number.
## Multiplication of exponents
A math concepts worksheet for Multiplication of exponents will assist you to training multiplying and dividing phone numbers with exponents. This worksheet will even offer problems that will require individuals to multiply two various exponents. By selecting the “All Positive” version, you will be able to view other versions of the worksheet. Aside from, also you can enter in particular directions on the worksheet alone. When you’re concluded, it is possible to click “Generate” and also the worksheet will probably be saved.
## Division of exponents
The basic guideline for division of exponents when multiplying numbers is usually to deduct the exponent from the denominator from the exponent from the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. For example, \$23 split by 4 will equal 27. This method is not always accurate, however. This method can cause frustration when multiplying figures which can be too large or not big enough.
## Linear features
If you’ve ever rented a car, you’ve probably noticed that the cost was \$320 x 10 days. So the total rent would be \$470. A linear purpose of this particular type provides the kind f(by), where ‘x’ is the quantity of time the auto was booked. Furthermore, it has the shape f(x) = ax b, in which ‘b’ and ‘a’ are actual numbers. |
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# What are Irregular Polygons in Geometry?
## Introduction to Polygons
Polygons are two-dimensional shapes with at least three straight sides and angles. Regular polygons have sides that are all the same length and angles that are all the same size. All sides and angles are congruent. Irregular polygons, on the other hand, have sides and angles that are not all congruent. While the terms regular and irregular may sound like they have something to do with the shape of the polygon, they actually have to do with the length of the sides and the size of the angles.
## What is an Irregular Polygon?
An irregular polygon is a polygon that has sides and angles that are not all congruent. Irregular polygons can have any number of sides and angles, and they can be of any shape. Irregular polygons do not have to be symmetrical or regular in shape. The most common example of an irregular polygon is the triangle. A triangle can have three sides of different lengths and three angles of different sizes.
## Properties of Irregular Polygons
Irregular polygons have several interesting properties that can help in solving geometry problems. Some of these properties include:
• The sum of the interior angles is always 180 degrees.
• The sum of the exterior angles is always 360 degrees.
• The sum of the side lengths is always the same, regardless of the shape of the polygon.
## Using Irregular Polygons
Irregular polygons are useful in solving geometry problems involving area, perimeter, and angles. For example, you can use an irregular polygon to calculate the area of a room or the perimeter of a garden. You can also use an irregular polygon to calculate the interior and exterior angles of a shape.
## Practice Problems
Here are some practice problems to help you understand how to use irregular polygons in geometry:
1. What is the sum of the interior angles of an irregular pentagon?
Answer: The sum of the interior angles of an irregular pentagon is 540 degrees.
2. What is the perimeter of an irregular quadrilateral with side lengths of 5, 6, 7, and 8?
Answer: The perimeter of an irregular quadrilateral with side lengths of 5, 6, 7, and 8 is 26.
3. What is the area of an irregular triangle with side lengths of 4, 5, and 6?
Answer: The area of an irregular triangle with side lengths of 4, 5, and 6 is 9.92.
4. What is the sum of the exterior angles of an irregular octagon?
Answer: The sum of the exterior angles of an irregular octagon is 1080 degrees.
5. What is the sum of the side lengths of an irregular heptagon?
Answer: The sum of the side lengths of an irregular heptagon is the same as any other polygon, regardless of the shape.
## Conclusion
In conclusion, irregular polygons are two-dimensional shapes with sides and angles that are not all congruent. Irregular polygons have several useful properties, such as the sum of the interior angles being 180 degrees and the sum of the exterior angles being 360 degrees. Irregular polygons can be used to calculate area, perimeter, and angles in geometry problems.
The key points to remember about irregular polygons in geometry are that they have sides and angles that are not all congruent, the sum of the interior angles is 180 degrees, the sum of the exterior angles is 360 degrees, and they can be used to calculate area, perimeter, and angles.
With practice, you can become a master of irregular polygons and use them to solve geometry problems with ease.
## FAQ
### What is a irregular polygon in geometry?
An irregular polygon is a polygon that does not have all sides and angles equal. It can have any number of sides and angles that are different sizes and shapes.
### What is the best description of an irregular polygon?
An irregular polygon is a polygon with sides and angles that are of different sizes and shapes.
### What is the definition of regular and irregular polygons?
A regular polygon is a polygon with all sides and angles equal, while an irregular polygon is a polygon with sides and angles that are of different sizes and shapes.
## Problem Solver
This tool combines the power of mathematical computation engine that excels at solving mathematical formulas with the power of GPT large language models to parse and generate natural language. This creates math problem solver thats more accurate than ChatGPT, more flexible than a calculator, and faster answers than a human tutor. Learn More. |
# Chapter 1 Linear Equations and Graphs
Chapter 2 Limits and the Derivative Section 4 The Derivative Copyright 2015, 2011, and 2008 Pearson Education, Inc. 1 Objectives for Section 2.4 The Derivative The student will be able to calculate rate of change. The student will be able to calculate slope of the tangent
line. The student will be able to interpret the meaning of the derivative. The student will be able to identify the nonexistence of the derivative. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 2 2 The Rate of Change For y = f (x), the average rate of change from x = a to x = a + h is
f ( a h) f ( a ) , h 0 h The above expression is also called a difference quotient. It can be interpreted as the slope of a secant. See the picture on the next slide for illustration. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 3 3 Visual Interpretation Q
f (a + h) f (a) f ( a h) f ( a ) h slope Average rate of change = slope of the secant line through P and Q P h Copyright 2015, 2011, and 2008 Pearson Education, Inc. 4 4
Example The revenue generated by producing and selling widgets is given by R(x) = x (75 3x) for 0 x 20. What is the change in revenue if production changes from 9 to 12? Copyright 2015, 2011, and 2008 Pearson Education, Inc. 5 5 Example The revenue generated by producing and selling widgets is given by R(x) = x (75 3x) for 0 x 20. What is the change in revenue if production changes from 9
to 12? R(12) R(9) = \$468 \$432 = \$36. Increasing production from 9 to 12 will increase revenue by \$36. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 6 6 Example (continued) The revenue is R(x) = x (75 3x) for 0 x 20. What is the average rate of change in revenue (per unit change in x) if production changes from 9 to 12? Copyright 2015, 2011, and 2008 Pearson Education, Inc.
7 7 Example (continued) The revenue is R(x) = x (75 3x) for 0 x 20. What is the average rate of change in revenue (per unit change in x) if production changes from 9 to 12? To find the average rate of change we divide the change in revenue by the change in production: R (12) R (9) 36 12 12 9 3 Thus the average change in revenue is \$12 when production is increased from 9 to 12.
Copyright 2015, 2011, and 2008 Pearson Education, Inc. 8 8 The Instantaneous Rate of Change Consider the function y = f (x) only near the point P = (a, f (a)). f ( a h) f ( a ) The difference quotient , h 0 h gives the average rate of change of f over the interval [a, a+h]. If we make h smaller and smaller, in the limit we obtain the instantaneous rate of change of the function at the point P: lim
h 0 f ( a h) f ( a ) h Copyright 2015, 2011, and 2008 Pearson Education, Inc. 9 9 Visual Interpretation Q Tangent Slope of tangent = instantaneous rate of change.
f (a + h) f (a) P lim f ( a h) f ( a ) h 0 h h Copyright 2015, 2011, and 2008 Pearson Education, Inc. Let h approach 0 10
10 Instantaneous Rate of Change Given y = f (x), the instantaneous rate of change at x = a is f ( a h) f ( a ) lim h 0 h provided that the limit exists. It can be interpreted as the slope of the tangent at the point (a, f (a)). See illustration on previous slide. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 11
11 The Derivative For y = f (x), we define the derivative of f at x, denoted f (x), to be f (x) lim h 0 f (x h) f (x) h if the limit exists. If f (a) exists, we call f differentiable at a. If f (x) exist for each x in the open interval (a, b), then f is said to be differentiable over (a, b).
Copyright 2015, 2011, and 2008 Pearson Education, Inc. 12 12 Interpretations of the Derivative If f is a function, then f is a new function with the following interpretations: For each x in the domain of f , f (x) is the slope of the line tangent to the graph of f at the point (x, f (x)). For each x in the domain of f , f (x) is the instantaneous rate of change of y = f (x) with respect to x. If f (x) is the position of a moving object at time x, then v = f (x) is the velocity of the object at that time.
Copyright 2015, 2011, and 2008 Pearson Education, Inc. 13 13 Finding the Derivative To find f (x), we use a four-step process: Step 1. Find f (x + h) Step 2. Find f (x + h) f (x) Step 3. Find f ( x h) f ( x) h Step 4. Find hlim 0 f ( x h) f ( x ) h
Copyright 2015, 2011, and 2008 Pearson Education, Inc. 14 14 Example Find the derivative of f (x) = x 2 3x. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 15 15 Example Find the derivative of f (x) = x 2 3x.
Step 1. f (x + h) = (x + h)2 3(x + h) = x2 + 2xh + h2 3x 3h Step 2. Find f (x + h) f (x) = 2xh + h2 3h 2 f ( x h ) f ( x ) 2 xh h
3h Step 3. Find 2 x h 3 h h Step 4. Find lim h 0 f ( x h) f ( x ) lim 2 x h 3 2 x 3 h 0 h
Copyright 2015, 2011, and 2008 Pearson Education, Inc. 16 16 Example Find the slope of the tangent to the graph of f (x) = x 2 3x at x = 0, x = 2, and x = 3. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 17 17 Example Find the slope of the tangent to the graph of f (x) = x 2 3x
at x = 0, x = 2, and x = 3. Solution: In example 2 we found the derivative of this function at x to be f (x) = 2x 3 Hence f (0) = 3 f (2) = 1, and f (3) = 3 Copyright 2015, 2011, and 2008 Pearson Education, Inc. 18 18 Graphing Calculators Most graphing calculators have a built-in numerical differentiation routine that will
approximate numerically the values of f (x) for any given value of x. Some graphing calculators have a built-in symbolic differentiation routine that will find an algebraic formula for the derivative, and then evaluate this formula at indicated values of x. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 19 19 Example We know that the derivative of f (x) = x 2 3x is f (x) = 2x 3. Verify this for x = 2 using a graphing calculator. Copyright 2015, 2011, and 2008 Pearson Education, Inc.
20 20 Example We know that the derivative of f (x) = x 2 3x is f (x) = 2x 3. Verify this for x = 2 using a graphing calculator. Using dy/dx under the calc menu. Using tangent under the draw menu. slope tangent equation Copyright 2015, 2011, and 2008 Pearson Education, Inc.
21 21 Example Find the derivative of f (x) = 2x 3x2 using a graphing calculator with a symbolic differentiation routine. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 22 22 Example Find the derivative of f (x) = 2x 3x2 using a graphing calculator with a symbolic differentiation routine.
Using algebraic differentiation under the home calc menu. derivative Copyright 2015, 2011, and 2008 Pearson Education, Inc. 23 23 Example Find the derivative of f (x) = 2x 3x2 using the four-step process. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 24
24 Example Find the derivative of f (x) = 2x 3x2 using the four-step process. Step 1. f (x + h) = 2(x + h) 3(x + h)2 Step 2. f (x + h) f (x) = 2h 6xh 3h2 Step 3. Step 4. f ( x h) f ( x) 2 x 6 xh 3h 2 2 6 x 3h h h lim 2 6 x 3h 2 6 x
h 0 Copyright 2015, 2011, and 2008 Pearson Education, Inc. 25 25 Nonexistence of the Derivative The existence of a derivative at x = a depends on the existence of the limit f (a) lim h 0 f (a h) f (a) h
If the limit does not exist, we say that the function is nondifferentiable at x = a, or f (a) does not exist. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 26 26 Nonexistence of the Derivative (continued) Some of the reasons why the derivative of a function may not exist at x = a are The graph of f has a hole or break at x = a, or The graph of f has a sharp corner at x = a, or The graph of f has a vertical tangent at x = a.
Copyright 2015, 2011, and 2008 Pearson Education, Inc. 27 27 Summary For y = f (x), we defined the derivative of f at x, denoted f (x), to be f (x) lim h 0 f (x h) f (x) h
if the limit exists. We have seen how to find the derivative algebraically, using the four-step process. Copyright 2015, 2011, and 2008 Pearson Education, Inc. 28 28
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# Interpreting Information from Fitted Lines I
Lesson
Many statistical experiments compare measurements of pairs of variables. The aim is to see whether the two variables are related.
Examples of pairs of variables that might be studied include:
• time: number of fish in a river
• percentage of children vaccinated: number of disease cases recorded
• age of cars sold: prices obtained
Data collected as pairs of measurements can be displayed in a scatterplot. One variable is taken to be the independent or explanatory variable and its levels are measured along the horizontal axis. The other variable is called the dependent or response variable and its levels are measured along the vertical axis. The data points plotted on the plane defined by the axes may or may not show a trend or a relationship between the variables.
If the points are scattered without any clear pattern we conclude that no relationship has been detected between the variables.
If the points appear to lie close to a line, we conclude that a relationship probably exists. We insert the line that best fits the data and use it to make predictions about possible further observations.
The scatterplot above indicates a strong negative correlation between the variables. This means the dependent variable tends to decrease in a predictable way with increases in the independent variable.
In a well-designed experiment, a researcher is careful not to use the fitted line to make predictions about the response that would be observed to values of the independent variable that are outside the range of the values used in the experiment. For example, if in the experiment the smallest value of the independent variable was 10 and the largest 85, then it would be unwise to try to predict what the response would be when the independent variable was smaller than 10 or larger than 85.
To make such predictions beyond the range of the data is called extrapolation and is considered unsafe.
The usual mathematical method for fitting a line to a data set accurately is called least squares regression. A spreadsheet or a statistical application will do the calculations for this automatically. It is always possible to fit a line to a scatterplot, even when there is no genuine relationship between the variables. However, when making a judgment about whether or not a relationship exists we consider how close the data points are to the fitted line.
#### Example
The data points illustrated in the graph above turned out to be based on the sale price of an item of goods measured against the age of the item.
The regression line can be used to predict that at 23 months the value of the goods will be approximately $\$1250$$1250. Considering the amounts by which the data points are above and below the regression line, it could easily happen that the estimated value of the goods at 23 months is \100$$100 too low or too high.
#### Worked examples
##### Question 1
A dam used to supply water to the neighboring town had the following data recorded for its volume over a number of months.
Month Volume (billions of litres) $1$1 $2$2 $3$3 $4$4 $120$120 $108$108 $106$106 $86$86
1. Given the data in the table, plot the points on the number plane.
2. Which of the following lines best fits the data?
A
B
C
D
A
B
C
D
##### Question 2
The number of fish in a river is measured over a five year period.
The results are shown in the following table and plotted below.
Time in years ($t$t) Number of fish ($F$F) $0$0 $1$1 $2$2 $3$3 $4$4 $5$5 $1903$1903 $1994$1994 $1995$1995 $1602$1602 $1695$1695 $1311$1311
1. Which line best fits the data?
A
B
C
D
A
B
C
D
2. Predict the number of years until there are no fish left in the river.
3. Predict the number of fish remaining in the river after $7$7 years.
4. According to the line of best fit, how many years are there until there are $900$900 fish left in the river?
##### Question 3
One litre of gas is raised to various temperatures and its pressure is measured.
The data has been graphed below with a line of best fit.
Temperature (K) Pressure (Pa) Temperature (K) Pressure (Pa) $300$300 $302$302 $304$304 $308$308 $310$310 $2400$2400 $2416$2416 $2434$2434 $2462$2462 $2478$2478 $312$312 $314$314 $316$316 $318$318 $320$320 $2496$2496 $2512$2512 $2526$2526 $2546$2546 $2562$2562
1. The pressure was not recorded when the temperature was $306$306 K.
Is it reasonable to use the line of best fit to predict the pressure?
Yes
A
No
B
Yes
A
No
B
2. Predict the pressure when the temperature is $306$306 K.
3. Within which range of temperatures is it reasonable to use the line of best fit to predict pressure?
$\left[300,320\right]$[300,320]
A
$\left[300,600\right]$[300,600]
B
$\left[0,320\right]$[0,320]
C
$\left[280,340\right]$[280,340]
D
$\left[300,320\right]$[300,320]
A
$\left[300,600\right]$[300,600]
B
$\left[0,320\right]$[0,320]
C
$\left[280,340\right]$[280,340]
D
### Outcomes
#### S7-2
S7-2 Make inferences from surveys and experiments: A making informal predictions, interpolations, and extrapolations B using sample statistics to make point estimates of population parameters C recognising the effect of sample size on the variability of an estimate
#### 91264
Use statistical methods to make an inference |
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Hypothesis Testing | CK-12 Foundation
# 2.8: Hypothesis Testing
Created by: CK-12
## Hypothesis Testing and the P-Value
Extend: A Null and Alternative Hypotheses for Any Situation
The nature of the situations that is being tested determines if a right-tailed, left-tailed, or two-tailed hypothesis test is used. These exercises allow the student to consider these types of tests in a different, creative way. They will look at the types of test available, and think of situations where each can be used.
Directions: For each set of null and alternative hypotheses chose a number for the variable and describe a situation where this type of test would be appropriate.
Example:
1. $H_0: \mu = a\\H_a: \mu < a$
Solution: Let $a = 12 \;\mathrm{oz}$. This null and alternative hypothesis can be used for testing the mean amount of soda found in all soft-drink cans produced in a specific factory. The consumer is interested in finding out if they are receiving the full amount of soda that they paid for, or if the soda producers are cheating them.
Exercises:
1. $H_0: \mu = a\\Ha: \mu \neq a$
2. $H_0: \mu = a\\ H_a: \mu > a$
3. $H_0: \mu \le a\\H_a: \mu > a$
4. $H_0: \mu \ge a\\Ha: \mu < a$
Note: The null hypothesis must have the “equal” part of the two choices.
## Testing a Proportion Hypothesis
Students will put their new knowledge to work by applying it to their high school. They will also review, and put to use, sampling techniques learned previously in the class. This project will be a fun way to make statistics real, and to make the material relevant, active, and long lasting.
Objective: Test, and construct a confidence interval for, a hypothesis about a proportion that describes the population of your school.
Procedure:
1. Write a null and alternative hypothesis with a proportion about the students at your school.
2. Take a sample. Be sure to use proper sampling techniques to get a sample that represents the population. Review chapter six if necessary.
3. Calculate the sample proportion.
4. Test the hypothesis at probability level $0.05.$
5. Construct the $95 \%$ confidence interval for the population proportion using the sample proportion that you found.
Analysis and Conclusion:
Write a report and/or prepare a presentation for the class that explains your null and alternative hypotheses, your sampling method, and the results of your test.
Analyze the results: Were you correct? Why or why not?
How confident are you in the method you used to gather your sample?
If you were to do this over again, what would you change?
How could you improve your accuracy?
Is the hypothesis test or the confidence interval more useful for your purposes?
Is there anything else to take into consideration when interpreting your results?
## Testing a Mean Hypothesis
Extension: Comparing a Subgroup to the Whole
The text has addressed two uses for hypothesis tests. The first is to test a claim. A statement is made about the mean of some measurement made on a population, and then that claim is tested by comparing it to a sample from that population. The second is to see if a significant difference can be found between the mean of a subgroup and the mean of the group as a whole. In this second application, both means are known, and are compared with the hypothesis test. In this assignment, students will explore the latter case by writing exercises, and by doing research so they can perform this test with real data.
Part One: Writing Exercises
1. Think of three situations where it would be useful to compare a subgroup to the whole with a hypothesis test. Each exercise should be from a different context. Think about science, politics, education, social justice, sports, and other areas. Be creative.
2. Write out an exercise, like what would be presented to a student, for each situation. Choose reasonable numbers for the problem. Be sure to include all the necessary information to complete the test.
3. Provide complete solutions to these three exercises.
Part Two: Research and Apply
1. Choose a situation where you can find real data to compare a subgroup to the whole with a hypothesis test. You may have to research a few possibilities to find one where you can get all the necessary data. Make a list of the quantities you will need.
2. Write out the null and alterative hypotheses. Conduct the test, and interpret the results.
3. Create a report and/or presentation for the class. Be sure to cite the source(s) of your data.
Include the following:
How did you choose the significance level?
Were these the results you were expecting?
How reliable is the data you collected?
What other tests could be made to expand on what you discovered?
## Testing a Hypothesis for Dependent and Independent Samples
Project: Test for Significance in Experiment Results
For this project, students will perform an experiment, thereby actively reviewing proper experimental technique, and analyze the results with a hypothesis test. Because they are working with data that they produced, from a topic that is of interest to them, the learning will be deep and long lasting.
Objective: Conduct a controlled experiment and determine if the results are statistically significant.
Procedure:
1. Plan, and execute a controlled experiment. Review chapter six to ensure that your experiment meets all the guideline of a clinical trial, except for repetition.
2. Conduct the proper hypothesis test at the $0.05$ significance level to determine if the results of the experiment are statistically significant.
Analysis: Include the following in a written report and/or presentation.
1. Describe your experiment and how it meets the standards of a clinical trial (without repetition). Was your experiment blind or double blind? Did you use a placebo?
2. Are the results of your experiment in the form of a mean or a proportion?
3. Are the two groups dependent or independent?
4. Did you be use the t distribution or the normal distribution? Why? What is the critical value?
5. State the null and alternative hypotheses.
6. Calculate the standard error of the difference between the two groups. Show clear, organized work. Carefully chose the appropriate method.
7. Calculate the test statistic. Show clear, organized work.
8. Will you reject or fail to reject the null hypothesis? Why?
9. Interpret the results of the test in the context of the experiment. What have you learned from the experiment and test?
## Date Created:
Feb 23, 2012
Apr 29, 2014
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## Sketch the graph of the derivative of this function f '( x), Mathematics
Assignment Help:
Below is the sketch of a function f ( x ) . Sketch the graph of the derivative of this function f ′ ( x ) .
Solution : At first glance it seems to an all however impossible task. Though, if you have some fundamental knowledge of the interpretations of the derivative you can get sketch of the derivative. It will not be perfect sketch for the most of the part, but you have to be able to get most basic features of the derivative in the sketch.
Let's begin with the given sketch of the function with a couple of additions.
Notice as well that at x = -3 , x = -1 ,x = 2 and x = 4 the tangent line to the function is horizontal.
It means that the slope of the tangent line has to be zero. Now, we know that the slope of the tangent line at a specific point is also the value of the derivative of function at that point.
Thus, now we know that,
f ′ ( -3) = 0 f ′ ( -1) = 0 f ′ ( 2) = 0 f ′ ( 4) = 0
It is a good beginning point for us. It gives us few points on the graph of the derivative. It also breaks the domain of function up into regions where the function is increasing & decreasing. We know that if the function is increasing at point then the derivative has to be positive at that point. Similarly, we know that if the function is decreasing at point then the derivative has to be negative at that point.
Now we can give the following information regarding the derivative.
x < -3 f ′ ( x ) < 0
-3 < x < -1 f ′ ( x ) > 0
-1 < x < 2 f ′ ( x ) < 0
2 < x < 4 f ′ ( x ) > 0
x > 4 f ′ ( x ) < 0
Recall that we are giving the signs of the derivatives here & these are exclusively a function of whether the function is increasing / decreasing. The sign of function itself is totally immaterial here & will not in any way effect the sign of the derivative.
It may still seem like we don't have sufficient information to get a sketch, however we can get a little more information regarding the derivative from the graph of the function. In the range x < -3 .we know that the derivative has to be negative, though we can also see that the derivative has to be increasing in this range. Here it is -ve until we reach x = -3 and at this point the derivative has to be zero. The only way for the derivative to be -ve to the left of
x = -3 and zero at x = -3 is for the derivative to enhance as we increase x towards x = -3 .
Now, in the range -3 < x < -1 we know that the derivative has to be zero at the endpoints and positive between the two endpoints. Directly to the right of x = -3 the derivative has to also be increasing (since it starts at zero and then goes positive - thus it has to be increasing). Thus, the derivative in this range has to start out increasing and has to eventually get back to zero at x = -1 . Thus, at some point in this interval the derivative has to start decreasing before it attain x = -1 . Now, we ought to be careful here since this is just general behavior here at the two endpoints. We won't know where the derivative goes from increasing through decreasing & it may well change among increasing & decreasing many times before we reach x = -1 . All we can actually say is that immediately to the right of x =-3 the derivative will be increasing and instantly to the left of x = -1 the derivative will be decreasing.
Next, for the ranges -1 < x < 2 & 2 < x < 4 we know derivative will be zero at the endpoints & negative in between. Also, following the kind of reasoning given above we can illustrates in each of these ranges that the derivative will be decreasing only to the right of left hand endpoint & increasing only to the left of the right hand endpoint.
At last, in the last region x > 4 we know that the derivative is zero at x = 4 & positive to the right of x = 4. Once again, following the reasoning above, the derivative has to also be increasing in this range.
Putting all material together (& always taking the simplest option for increasing and/or decreasing information) gives us the given sketch for the derivative.
Note as well that it was done with the actual derivative and hence is in fact accurate. Any sketch you do will possibly not look quite the similar. The "humps" in each region may be at distinct places and/or different heights for instance. Also, note as well that we left off the vertical scale since given the information which we've got at this point there was no real way to know this information.
However, that doesn't mean that we can't get some ideas of particular points on the derivative other than where we know the derivative to be zero. In order to see this let's check out the following graph of the function (not the derivative, however the function).
At x = -2 & x = 3 we've sketched in a couple of tangent lines. We can utilize the basic rise/run slope concept to estimate the value of the derivative at these points.
Let's begin at x = 3. Here we've got two points on the line. We can see that each appear to be about one-quarter of the way off the grid line. Thus, taking that into account and the fact that we go through one complete grid we can illustrates that the slope of the tangent line, and therefore the derivative, is approximately -1.5.
At x = -2 this looks like (with some heavy estimation) that the second point is around 6.5 grids above the first point and hence the slope of the tangent line here, and therefore the derivative, is around 6.5.
Following is the sketch of the derivative with the vertical scale involved and from this we can illustrate that actually our estimates are pretty close to reality.
Note as well that this idea of estimating values of derivatives can be a complicated process and does need a fair amount of (possible bad) approximations thus whereas it can be utilized, you have to be careful with it.
#### Theorems, pythagoras theorem
pythagoras theorem
#### Find out the radius of convergence, Example: Find out the radius of conver...
Example: Find out the radius of convergence for the following power series. Solution : Therefore, in this case we have, a n = ((-3) n )/(n7 n+1 ) a n+1 = (
#### Math problem, integral from 0 to pi of dx/(a+b*cos(x)
integral from 0 to pi of dx/(a+b*cos(x)
#### Equation of the line which passes through the two points, Example: Write do...
Example: Write down the equation of the line which passes through the two points (-2, 4) and (3, -5). Solution At first glance it might not appear which we'll be capable to
#### Word problems, A baseball card was worth \$5.00 in 1940. It doubled in value...
A baseball card was worth \$5.00 in 1940. It doubled in value every decade. How much was it worth in 2000?
#### Listing method, how will you explain the listing method?
how will you explain the listing method?
#### Probability, An unbiased die is tossed twice .Find the probability of getti...
An unbiased die is tossed twice .Find the probability of getting a 4,5,6 on the first toss and a 1,2,3,4 on the second toss
#### Logarithmic form and exponential form, Logarithmic form and exponential for...
Logarithmic form and exponential form ; We'll begin with b = 0 , b ≠ 1. Then we have y= log b x is equivalent to x= b y The first one is called
#### Calculate one-sided limits, Calculate the value of the following limits. ...
Calculate the value of the following limits. Solution From the graph of this function illustrated below, We can illustrate that both of the one-sided limits suffer
#### Complex number, a ,b,c are complex numbers such that a/1-b=b/1-c=c-1-a=k.fi...
a ,b,c are complex numbers such that a/1-b=b/1-c=c-1-a=k.find the value of k |
# Difference between revisions of "2015 AMC 8 Problems/Problem 25"
One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?
$\mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 12\frac{1}{2}\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 15\frac{1}{2}\qquad \mathrm{(E) \ } 17$
$[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy]$
### SOLUTION 1
Lets draw a diagram. $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$ Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the 4 big triangles by AA. Let the height of a big triangle be $x$ then $\dfrac{x}{x-1}=\dfrac{5-x}{1}$. $$x=-x^2+6x-5$$ $$x^2-5x+5=0$$ $$x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}$$ $$x=\dfrac{5\pm \sqrt{5}}{2}$$ Which means $x=\dfrac{5-\sqrt{5}}{2}$ This means the area of each triangle is $\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}$ This the area of the square is $25-(4*\dfrac{5}{2})=\boxed{C,~15}$ |
# Divisibility by 6: How to find out if 6 divides in cleanly
Image via Wikipedia
So far we’ve learned fun & easy divisibility tricks for the numbers 3 and by 4. Learning these tricks helps us reduce fractions with serious speed, and it helps us perform other math operations with a lot more ease. So let’s keep the learning
process going.
[Note: If this is the first of these divisibility blogs that you have seen, search this blog for posts about divisibility by 3 and by 4; that way you’ll get caught up with the flow of these posts.]
The trick for 5 is incredibly simple: 5 goes into any number with a ones digit of 5 or 0. That is all you need to know. Not much else to say about 5.
And here is the trick for 6: 6 divides into any number that is divisible by BOTH 2 and 3. In other words, for the number in question, check to see if both 2 and 3 go in evenly. If they do, then 6 must also go in evenly. But if EITHER 2 or 3 does NOT go into the number, then 6 definitely will NOT go in. So you need divisibility by BOTH 2 AND 3 … in order for the trick to work.
Here’s an alternative way to say this trick, a way some kids find easier to grasp: “6 goes into all even numbers that are divisible by 3.”
EXAMPLE 1: 74 — 2 goes in, but 3 does not, so 6 does NOT go in evenly.
EXAMPLE 2: 75 — 3 goes in, but 2 does not, so 6 does NOT go in evenly.
EXAMPLE 3: 78 — 2 and 3 BOTH go in evenly, so 6 DOES go in evenly.
Notice that since the tricks for 2 and 3 are quite simple, this trick for 6 is really quite simple too. It is NOT hard to use this trick even on numbers with a bunch of digits.
EXAMPLE 4: 783,612 — 2 goes in, and so does 3, so 6 DOES go in evenly. [checking for 3, note that you need to add only the digits 7 & 8. 7 + 8 = 15, a multiple of 3, so this large number IS divisible by 3.]
Now give this a try yourself with these numbers. For each number tell whether
or not 2, 3 and 6 will divide in evenly.
PROBLEMS:
a) 84
b) 112
c) 141
d) 266
e) 552
f) 714
g) 936
h) 994
i) 1,245
j) 54,936
a) 84: 2 yes; 3 yes; 6 yes
b) 112: 2 yes; 3 no; 6 no
c) 141: 2 no; 3 yes; 6 no
d) 266: 2 yes; 3 no; 6 no
e) 552: 2 yes; 3 yes; 6 yes
f) 714: 2 yes; 3 yes; 6 yes
g) 936: 2 yes; 3 yes; 6 yes
h) 994: 2 yes; 3 no; 6 no
i) 1,245: 2 no; 3 yes; 6 no
j) 54,936: 2 yes; 3 yes; 6 yes |
# Skip Counting
Skip counting is one of the first math skills that your child will need to master before learning how to multiply.
It sets the foundation for understanding how multiplication works and breaks down multiplying into simpler tasks.
By adding a number to itself over and over again, your child sees and memorizes the counting pattern.
He or she will be closer to understanding what you mean when you start showing simple multiplication problems and in later grades, when students are asked to memorize the multiplication tables.
## Skip Counting Worksheets
Learning multiplication often begins in Kindergarten and First Grade. While children in these classes aren't likely to be working out multiplication problems, they do start learning this important skill that builds up to later problem-solving.
These drills are a helpful way to learn multiplication tables. Instead of counting by ones, you count by multiples (such as twos, threes, fours, fives, etc...).
Once a child has learned how to skip count, you can begin to introduce multiplication (which can be seen as another way to skip count).
For example, counting by twos:
2,4, 6, 8, 10
naturally progresses to:
2x1=2
2x2=4
2x3=6
2x4=8
2x5=10
Our son didn't begin solving multiplication problems until 3rd Grade. Earlier grades touched on it a little bit, but we never went into memorizing multiplication facts.
He has done pretty well learning his facts from 0 to 12 and I'd like to think that's partly due to the time we spent learning how to skip count.
Counting by twos, fives, and tens can be done easily with common items. You can count the eggs in an egg carton by twos. Count fingers and toes by fives or count the entire family's digits by tens.
Other numbers can be a bit harder to come up with everyday items to count. You might find the worksheets below helpful for counting those numbers.
I've prepared activity sheets for counting by two's and all the numbers up to ten.
Click on any number to find practice sheets for counting by that number.
### Teaching Sequence
This form of counting is generally introduced in first grade, and it is best to start first with the tens and then go on to the fives and twos.
As you work on counting by ten, you might also want to introduce some lessons on the tens and ones place value.
Try to get your child to memorize the counting pattern up to 100. The other digits will follow once the fives, tens and twos are mastered.
This printable hundreds chart is a useful tool to help your child visualize what skipping means. When you visit the individual counting pages, you'll find some worksheets that are based off of the hundreds chart as well as some other activities and videos to help you teach this skill.
More Counting Fun:
› Skip Counting |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
7.1: Addition and Subtraction of Polynomials
Difficulty Level: Advanced Created by: CK-12
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Practice Addition and Subtraction of Polynomials
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You are going to build a rectangular garden in your back yard. The garden is 2 m more than 1.5 times as long as it is wide. Write an expression to show the area of the garden.
Guidance
The word polynomial comes from the Greek word poly meaning “many”. Polynomials are made up of one or more terms and each term must have an exponent that is 0 or a whole number. This means that 3x2+2x+1\begin{align*}3x^2+2x+1\end{align*} is a polynomial, but 3x0.5+2x2+1\begin{align*}3x^{0.5}+2x^{-2}+1\end{align*} is not a polynomial. Some common polynomials have special names based on how many terms they have:
• A monomial is a polynomial with just one term. Examples of monomials are 3x\begin{align*}3x\end{align*}, 2x2\begin{align*}2x^2\end{align*} and 7\begin{align*}7\end{align*}.
• A binomial is a polynomial with two terms. Examples of binomials are 2x+1\begin{align*}2x+1\end{align*}, 3x25x\begin{align*}3x^2-5x\end{align*} and x5\begin{align*}x-5\end{align*}.
• A trinomial is a polynomial with three terms. An example of a trinomial is 2x2+3x4\begin{align*}2x^2+3x-4\end{align*}.
To add and subtract polynomials you will go through two steps.
1. Use the distributive property to remove parentheses. Remember that when there is no number in front of the parentheses, it is like there is a 1 in front of the parentheses. Pay attention to whether or not the sign in front of the parentheses is +\begin{align*}+\end{align*} or \begin{align*}-\end{align*}, because this will tell you if the number you need to distribute is +1\begin{align*}+1\end{align*} or 1\begin{align*}-1\end{align*}.
2. Combine similar terms. This means, combine the x2\begin{align*}x^2\end{align*} terms with the x2\begin{align*}x^2\end{align*} terms, the x\begin{align*}x\end{align*} terms with the x\begin{align*}x\end{align*} terms, etc.
Example A
Find the sum: (3x2+2x7)+(5x23x+3)\begin{align*}(3x^2+2x-7)+(5x^2-3x+3)\end{align*}.
Solution: First you want to remove the parentheses. Because this is an addition problem, it is like there is a +1\begin{align*}+1\end{align*} in front of each set of parentheses. When you distribute a +1\begin{align*}+1\end{align*}, none of the terms will change.
1(3x2+2x7)+1(5x23x+3)=3x2+2x7+5x23x+3\begin{align*}1(3x^2+2x-7)+1(5x^2-3x+3)=3x^2+2x-7+5x^2-3x+3\end{align*}
Next, combine the similar terms. Sometimes it can help to first reorder the expression to put the similar terms next to one another. Remember to keep the signs with the correct terms. For example, in this problem the 7 is negative and the 3x is negative.
3x2+2x7+5x23x+3=3x2+5x2+2x3x7+3=8x2x4\begin{align*}3x^2+2x-7+5x^2-3x+3&=3x^2+5x^2+2x-3x-7+3\\ &=8x^2-x-4\end{align*}
Example B
Find the difference: (5x2+8x+6)(4x2+5x+4)\begin{align*}(5x^2+8x+6)-(4x^2+5x+4)\end{align*}.
Solution: First you want to remove the parentheses. Because this is a subtraction problem, it is like there is a 1\begin{align*}-1\end{align*} in front of the second set of parentheses. When you distribute a 1\begin{align*}-1\end{align*}, each term inside that set of parentheses will change its sign.
1(5x2+8x+6)1(4x2+5x+4)=5x2+8x+64x25x4\begin{align*}1(5x^2+8x+6)-1(4x^2+5x+4)=5x^2+8x+6-4x^2-5x-4\end{align*}
Next, combine the similar terms. Remember to keep the signs with the correct terms.
5x2+8x+64x25x4=5x24x2+8x5x+64=x2+3x+2\begin{align*}5x^2+8x+6-4x^2-5x-4&=5x^2-4x^2+8x-5x+6-4\\ &=x^2+3x+2\end{align*}
Example C
Find the difference: (3x3+6x27x+5)(4x2+3x8)\begin{align*}(3x^3+6x^2-7x+5)-(4x^2+3x-8)\end{align*}
Solution: First you want to remove the parentheses. Because this is a subtraction problem, it is like there is a 1\begin{align*}-1\end{align*} in front of the second set of parentheses. When you distribute a 1\begin{align*}-1\end{align*}, each term inside that set of parentheses will change its sign..
1(3x3+6x27x+5)1(4x2+3x8)=3x3+6x27x+54x23x+8\begin{align*}1(3x^3+6x^2-7x+5)-1(4x^2+3x-8)=3x^3+6x^2-7x+5-4x^2-3x+8\end{align*}
Next, combine the similar terms. Remember to keep the signs with the correct terms.
3x3+6x27x+54x23x+8=3x3+6x24x27x3x+5+8=3x3+2x210x+13\begin{align*}3x^3+6x^2-7x+5-4x^2-3x+8&=3x^3+6x^2-4x^2-7x-3x+5+8\\ &=3x^3+2x^2-10x+13\end{align*}
Concept Problem Revisited
Remember that the area of a rectangle is length times width.
AreaAreaArea=l×w=(1.5x+2)x=1.5x2+2x\begin{align*}Area &= l \times w \\ Area &= (1.5x + 2) x \\ Area &= 1.5x^2 + 2x\end{align*}
Vocabulary
Binomial
A binomial has two terms that are added or subtracted from each other. Each of the terms of a binomial is a variable (x)\begin{align*}(x)\end{align*}, a product of a number and a variable (4x)\begin{align*}(4x)\end{align*}, or the product of multiple variables with or without a number (4x2y+3)\begin{align*}(4x^2y + 3)\end{align*}. One of the terms in the binomial can be a number.
Monomial
A monomial can be a number or a variable (like x\begin{align*}x\end{align*}) or can be the product of a number and a variable (like 3x\begin{align*}3x\end{align*} or 3x2\begin{align*}3x^2\end{align*}). A monomial has only one term.
Polynomial
A polynomial, by definition, is also a monomial or the sum of a number of monomials. So 3x2\begin{align*}3x^2\end{align*} can be considered a polynomial, 2x+3\begin{align*}2x+3\end{align*} can be considered a polynomial, and 2x2+3x4\begin{align*}2x^2+3x-4\end{align*} can be considered a polynomial.
Trinomial
A trinomial has three terms (4x2+3x7)\begin{align*}(4x^2+3x-7)\end{align*}. The terms of a trinomial can be a variable (x)\begin{align*}(x)\end{align*}, a product of a number and a variable (3x)\begin{align*}(3x)\end{align*}, or the product of multiple variables with or without a number (4x2)\begin{align*}(4x^2)\end{align*}. One of the terms in the trinomial can be a number \begin{align*}(-7)\end{align*}.
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient.
Guided Practice
1. Find the sum: \begin{align*}(2x^2+4x+3) + (x^2-3x-2)\end{align*}.
2. Find the difference: \begin{align*}(5x^2-9x+7) - (3x^2-5x+6)\end{align*}.
3. Find the sum: \begin{align*}(8x^3+5x^2-4x+2) + (4x^3+7x-5)\end{align*}.
1. \begin{align*}(2x^2+4x+3) + (x^2-3x-2)=2x^2+4x+3+x^2-3x-2=3x^2+x+1\end{align*}
2. \begin{align*}(5x^2-9x+7) - (3x^2-5x+6)=5x^2-9x+7-3x^2+5x-6=2x^2-4x+1\end{align*}
3. \begin{align*}(8x^3+5x^2-4x+2) + (4x^3+7x-5)=8x^3+5x^2-4x+2+4x^3+7x-5=12x^3+5x^2+3x-3\end{align*}
Practice
For each problem, find the sum or difference.
1. \begin{align*}(x^2+4x+5) + (2x^2+3x+7)\end{align*}
2. \begin{align*}(2r^2+6r+7) - (3r^2+5r+8)\end{align*}
3. \begin{align*}(3t^2-2t+4) + (2t^2+5t-3)\end{align*}
4. \begin{align*}(4s^2-2s-3) - (5s^2+7s-6)\end{align*}
5. \begin{align*}(5y^2+7y-3) + (-2y^2-5y+6)\end{align*}
6. \begin{align*}(6x^2+36x+13) - (4x^2+13x+33)\end{align*}
7. \begin{align*}(12a^2+13a+7) + (9a^2+15a+8)\end{align*}
8. \begin{align*}(9y^2-17y-12) + (5y^2+12y+4)\end{align*}
9. \begin{align*}(11b^2+7b-12) - (15b^2-19b-21)\end{align*}
10. \begin{align*}(25x^2+17x-23) - (-14x^3-14x-11)\end{align*}
11. \begin{align*}(-3y^2+10y-5) - (5y^2+5y+8)\end{align*}
12. \begin{align*}(-7x^2-5x+11) + (5x^2+4x-9)\end{align*}
13. \begin{align*}(9a^3-2a^2+7) + (3a^2+8a-4)\end{align*}
14. \begin{align*}(3x^2-2x+4) - (x^2+x-6)\end{align*}
15. \begin{align*}(4s^3+4s^2-5s-2) - (-2s^2-5s+6)\end{align*}
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English
distributive property
The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
Polynomial
A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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# Area of a five pointed star
A 5 pointed star is inscribed in a circle of radius $r$.
Prove that the area of the star is $$\frac{10 \tan\left(\tfrac{\pi}{10}\right)}{3-\tan^2\left(\tfrac{\pi}{10}\right)} r^2$$
Clearly the area of the 5-star is $10$ times the area of the orange-shaded triangle $\triangle OAP$ which, in turn, equals half the height times the base: $$A\left(\triangle OAP\right) = \tfrac{1}{2} \bar{AC} \cdot \bar{OP} = \tfrac{1}{2} \bar{AC} \cdot r$$ To find $\bar{AC}$ notice that it is a joint side of two right triangles, $\triangle OAC$ and $\triangle PAC$, hence $$\bar{OC} = \frac{\bar{AC}}{\tan\left(\angle AOC\right)} \qquad \bar{PC} = \frac{\bar{AC}}{\tan\left(\angle APC\right)} = r - \bar{OC} \tag{1}$$
By the symmetry $\angle EOF = \tfrac{2}{5} \pi$, and $\angle AOB = \tfrac{2}{5} \pi$, and hence $\angle AOC = \tfrac{1}{2} \angle AOB = \tfrac{1}{5} \pi$. By the inscribed angle theorem applied to angle $\angle EPF$, $$\angle APF \equiv \angle EPF = \frac{1}{2} \angle EOF = \frac{\pi}{5} \quad \therefore \quad \angle APC = \tfrac{1}{2} \angle APF = \frac{\pi}{10}$$ Using these angles in eq. $(1)$: $$r = \frac{\bar{AC}}{\tan\left(\tfrac{1}{5} \pi \right)} + \frac{\bar{AC}}{\tan\left(\tfrac{1}{10} \pi \right)}$$ and solving for $\bar{AC}$ we get $$A\left(\triangle OAP\right) = r^2 \cdot \frac{1}{2} \frac{1}{\frac{1}{\tan\left(\tfrac{1}{5} \pi \right)} + \frac{1}{\tan\left(\tfrac{1}{10} \pi \right)}} = r^2 \frac{ \tan\left(\frac{\pi}{5}\right) \tan\left(\frac{\pi}{10}\right)}{2 \left(\tan\left(\frac{\pi}{5}\right) + \left(\frac{\pi}{10}\right)\right)} \tag{2}$$ Furthermore, using angle doubling formula for tangent in the previous eq. $(2)$: $$\tan\left(\frac{\pi}{5}\right) = \frac{2 \tan\left(\frac{\pi}{10}\right)}{1 - \tan^2\left(\frac{\pi}{10}\right)}$$ we have $$A\left(\triangle OAP\right) = r^2 \frac{\tan\left(\frac{\pi}{10}\right)}{3- \tan^2\left(\frac{\pi}{10}\right)}$$ We finish by recalling the area of the 5-star equals 10 times the area of $\triangle AOP$: $$A\left({\huge \star}\right) = \frac{10 \cdot \tan\left(\frac{\pi}{10}\right)}{3- \tan^2\left(\frac{\pi}{10}\right)} r^2$$ |
Singular Matrix – Explanation & Examples
A singular matrix is a very simple matrix. Though simple, it has immense importance in linear transformations and higher-order differential equations. But for this topic, we will look at it from a much lower level of mathematics.
Let’s check the formal definition of a singular matrix:
A matrix whose determinant is $0$ and thus is non-invertible is known as a singular matrix.
In this lesson, we will discover what singular matrices are, how to tell if a matrix is singular, understand some properties of singular matrices, and the determinant of a singular matrix. Let’s start!
What is a Singular Matrix?
Simply put, a singular matrix is a matrix whose determinant is $0$. Since the determinant is $0$, we can’t find the inverses of such matrices. Thus, it’s a non-invertible matrix. A singular matrix is also known as a degenerate.
Matrices whose determinant is $0$ are called singular matrices, and matrices whose determinant is non-zero are called non-singular matrices.
Determinant of a Singular Matrix
From the definition of a singular matrix, we know that a singular matrix’s determinant is ZERO!
Singular matrices are square matrices whose determinant is $0$. Consider Matrix $A$ shown below:
$A = \begin{bmatrix} { a } & { b } \\ { c } & { d } \end {bmatrix}$
It is a $2 \times 2$ square matrix. The determinant formula for a $2 \times 2$ matrix is:
$det( A ) = | A | = ad – bc$
Now, let’s take a matrix with values. Consider Matrix $B$ shown below:
$B = \begin{bmatrix} { 1 } & { 2 } \\ { 4 } & { 8 } \end {bmatrix}$
What is the determinant of this matrix?
Let’s calculate the determinant of Matrix $B$ by using the formula:
$det( B) = | B | = ad – bc$
$| B | = ( 1 )( 8 ) – ( 2 )( 4 )$
$| B | = 0$
The determinant of matrix $B$ is $0$. Thus, this is a singular matrix!
Let’s consider the $3 \times 3$ matrix shown below:
$B = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end {bmatrix}$
It is a $3 \times 3$ square matrix. The determinant formula for a $3 \times 3$ matrix is:
$det( B ) = | B | = a \begin{vmatrix} { e } & { f } \\ { h } & { i } \end{vmatrix} – b \begin{vmatrix} { d } & { f } \\ { g } & { i } \end{vmatrix} + c \begin{vmatrix} { d } & { e } \\ { g } & { h } \end{vmatrix}$
$| B | = a ( e i – f h ) – b (d i – f g ) + c (d h – e g )$
Now, let’s take a matrix with values. Consider Matrix $C$ shown below:
$C = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ { 0 } & { – 1 } & 1 \end {bmatrix}$
What is the determinant of this matrix?
Let’s calculate the determinant of Matrix $C$ by using the formula:
$det( C ) = | C | = a ( e i – f h ) – b (d i – f g ) + c (d h – e g )$
$| C | = 1 [ ( 3 )( 1 ) – ( 1 )( – 1 ) ] – 1 [ ( 2 )( 1 ) – ( 1 )( 0 ) ] + 1 [ ( 2 )( – 1 ) – ( 3 )( 0 ) ]$
$| C | = 1 [ 3 + 1 ] – 1 [ 2 ] + 1 [ – 2 ]$
$|C | = 4 – 2 – 2$
$| C | = 0$
The determinant of matrix $C$ is $0$. Thus, this is a singular matrix!
How to tell if a Matrix is Singular
Sometimes we are working with a problem that tells us to find the inverse of a square matrix. The first step we should do is to find the determinant. Then we plug the value into the matrix inversion formula and find our inverse matrix.
What if the determinant is $0$?
Because the matrix inversion formula has $\frac{ 1 }{ determinant }$ term, we can’t find the inverse of the determinant is $0$ because we will have division by $0$! This is undefined!
So, we then conclude that the matrix we are working with is a non-invertible matrix. It doesn’t have an inverse.
This specific type of square matrices is known as singular matrices!
To find if a matrix is singular or non-singular, we find the value of the determinant.
• If the determinant is equal to $0$, the matrix is singular
• If the determinant is non-zero, the matrix is non-singular
Of course, we will find the determinant using the determinant formula depending on the square matrix’s order.
For a $2 \times 2$ matrix:
Given,
$A = \begin{bmatrix} { a } & { b } \\ { c } & { d } \end {bmatrix}$
The determinant is $| A | = ad – bc$.
Matrix $A$ is singular if and only if $| A | = ad – bc = 0$.
For a $3 \times 3$ matrix:
Given,
$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end {bmatrix}$
The determinant is $| A | = a ( e i – f h ) – b (d i – f g ) + c (d h – e g )$.
Matrix $A$ is singular if and only if $| A | = a ( e i – f h ) – b (d i – f g ) + c (d h – e g ) = 0$.
Singular Matrix Properties
There are a few properties we are going to state for singular matrices. They are given below:
1. The determinant of a singular matrix is equal to $0$. If we have Singular Matrix $A$, then $det(A) = 0$.
2. A non-invertible matrix (a matrix whose inverse doesn’t exist) is referred to as a singular matrix.
3. Singular Matrices are only defined for square matrices.
4. Singular matrices don’t have multiplicative inverses.
Let us check out some examples to enhance our understanding further. There are a few practice problems for you as well.
Example 1
Which of the following statements about singular matrices is false?
1. The matrix has no inverse.
2. The matrix’s determinant is $0$.
3. The matrix is invertible.
4. The matrix can’t be multiplied with other matrices to achieve the compatible identity matrix ($I$).
Solution
If you look at the properties of singular matrices, you will figure out that the determinant of a singular matrix is $0$, which makes the matrix non-invertible. So, Option C is false.
Example 2
For the $2$ matrices shown below, check whether they are singular matrices or not.
$A = \begin{bmatrix} { 6 } & { -3 } \\ { 4 } & { -2 } \end {bmatrix}$
$B = \begin{bmatrix} 0 & 1 & -1 \\ 1 & 4 & 0 \\ 3 & -1 & 1 \end {bmatrix}$
Solution
Calculating the determinant will tell us whether Matrix $A$ and $B$ are singular or not. Let’s calculate the determinant of Matrix $A$:
$| A | = ad – bc$
$= (6)(-2) – (-3)(4)$
$= -12+12$
$= 0$
Matrix $A$ is a singular matrix.
Now, let’s calculate the determinant of Matrix $B$:
$|B| = a ( e i – f h ) – b (d i – f g ) + c (d h – e g )$
$| B | = 0 [ ( 4 )( 1 ) – ( 0 )( – 1 ) ] – 1 [ ( 1 )( 1 ) – ( 0 )( 3 ) ] + (-1) [ ( 1 )( – 1 ) – ( 3 )( 4 ) ]$
$| B | = -1(1) -1(-13)$
$|B | = -1 + 13$
$| B | = 12$
Since the determinant isn’t $0$, Matrix $B$ is a non-singular matrix.
Practice Questions
1. For the $2$ matrices shown below, comment whether they are singular or not.
$A = \begin{bmatrix} { -1 } & { 2 } \\ { -2 } & { 4 } \end {bmatrix}$
$B = \begin{bmatrix} { -6 } & { -15 } \\ { -4 } & { 10 } \end {bmatrix}$
2. What value of $y$ will make the matrix $C = \begin{bmatrix} { 10 } & { -2 } \\ { y } & { -1 } \end {bmatrix}$ singular?
1. We check the determinant of each matrix. If it’s $0$, then it is a singular matrix. Finding the determinant of Matrix $A$:
$| A | = ad – bc$
$= (-1)(4) – (2)(-2)$
$= – 4 + 4$
$= 0$
Matrix $A$ is singular.
Now, calculating the determinant for Matrix $B$:
$| B | = ad – bc$
$= (-6)(10) – (-15)(-4)$
$= -60 – 60$
$= – 120$
Matrix $B$ is not singular.
2. For Matrix $C$ to be singular, we have to equate the determinant equation to $0$. Then use some algebra to solve for $y$. The process is shown below:
$| C | = 0$
$ad – bc = 0$
$(10)(-1) – (-2)(y) = 0$
$-10 + 2y = 0$
$2y = 10$
$y = \frac{ 10 }{ 2 }$
$y = 5$ |
# A Systematic Approach to Factoring Step 1
```A Systematic Approach to Factoring
Step 1
Count the number of terms.
(Remember****Knowing the number of terms will allow you to eliminate unnecessary
tools.)
Step 2
Is there a greatest common factor?
Tool 1
(Page 3)
(Remember****This is the most important tool. It is often used to hide other tools. When
you factor the greatest common factor, it allows you to recognize other tools.)
Step 3
Can you group the expression?
(Grouping) Tool 2
(page 6)
(Remember****Grouping requires 4 terms. If there are four terms present, the object is to
group the first two and factor them. You will then have to factor the second two terms and try
to make them look the first two terms.)
Step 4
Can you use the “AC” method?
Tool 3
(page 7)
(Remember****The “A times C” method requires the problem to have three terms. Then
number that we call “A” when multiplied by the number called “C” must have a sum that
equals the middle term. We call the middle term “B”. We say that “A times C” must equal
B.)
Ax2 + Bx+ C
Elementary & Intermediate Algebra
1
Step 5
Can the problem be factored as the difference of two perfect squares.
Tool 4
(Remember****This tool requires two terms. Both the first and second terms must be perfect
squares. A minus sign must separate the two terms. A plus “+” sign cannot separate the
difference of two squares.)
Step 6
Is the problem the sum/difference of perfect cubes?
Tool 5
(Remember****Does the first and second terms have a cube root? This tool requires that a
polynomial to have exactly two terms. The problem can either have the sum or the difference
of two perfect cubes. Variables that are perfect cubes have exponents that are divisible by
three.)
Elementary & Intermediate Algebra
2
Polynomial Factoring
Factoring polynomials is one of the most important topics covered in Elementary Algebra.
We will divide the concept into five tools.
Tool 1
Factoring the greatest common monomial:
(The most important tool!!!!)
A monomial is a one termed polynomial. For example: 3x, 5x2t5, 6, -3v
The term “factor” in this context means to divide. We are going to divide the largest factor
found in each term of a polynomial from that polynomial. This is the most important of the
five tools that we will discuss.
Important Terms:
Factor – The original definition of the factor is an algebraic expression that is multiplied.
Examples: 12 has factors 3 x 4 and 2 x 6
3x2y4z has factors 3, x,2 y,4 z
When factors are variables, they are simply
algebraic factors.
We say in words that this expression is 3 times x2 times y4 times z. The key is that these
expressions (algebraic expressions) are separated by multiplication. ***Remember, factors
are multiplied and factoring is dividing.***
Factoring – The changing of an addition/subtraction problem into a multiplication problem.
Term – A term is an algebraic expression separated by either a plus sign or a minus sign.
Example: 3x2 + 5y3z – 8m + 7
Constant – A number that does not have a variable associated with it.
Combine – To add or subtract. (Which ever is implied in the problem.)
Numerical Coefficient (coefficient) – The number that is multiplied by the variable.
Elementary & Intermediate Algebra
3
3x2 + 5y3z – 8m + 7
This algebraic expression has four terms. The first term 3x2 has two factors. This implies that
a factor is not a terms but factors make up a term. The second term in the expression is 5y3z.
This term has three factors. The factors are 5, y,3 and z. These factors are separated by an
understood multiplication symbol. ***Remember, factors are multiplied.***
Factoring the greatest common monomial
3x2yz3 – 6x3y4z2 + 12xy2z5
The term “common” implies that eh factors that you are looking for have something in
common. You must first understand that there are three terms in this expression. 3x2yz3 is the
first term. – 6x3y4z2 is the second term, and 12xy2z5 is the third term. You are looking for the
largest factor that is common to all three terms. In the end, you will divide each of these
terms by this greatest common factor, (greatest common monomial) that you are trying to
create.
You are looking for common factors in each term.
3x2yz3m – 6x3y4z2 + 12xy2z5m
The numerical coefficients 3, 6, and 12 have a greatest
common factor of 3. This 3 becomes a part of the
greatest common factor.
3x2yz3m – 6x3y4z2 + 12xy2z5m
The variable x2, x3, and x are common variables in that
they are all x’s. The smallest variable {x} is the only
one that can divide evenly into each term. This means
that {x} is part of the greatest common factor.
3x2yz3m – 6x3y4z2 + 12xy2z5m
The variables y, y4, and y2 are all common in that each
term has at least one variable {y}. The smallest number
of y’s present is y1 or simply y. this y becomes part of
the greatest common factor.
3x2yz3m – 6x3y4z2 + 12xy2z5m
The variables z3, z2, and z5 are all common in that each
term has at least one variable {z}. The smallest number
of z’s present is z2. This variable z2 becomes the final
part of the greatest common factor.
The last factor in the first term is m. The variable m cannot be a part of the greatest common
factor because there is not at least one m in each of the terms.
Elementary & Intermediate Algebra
4
The greatest common factor (GCF) is 3xyz2. You will divide each term by this GCF.
***Remember when you divide polynomials, you actually divide the numerical coefficients,
but you simply subtract the exponents of the variables.*** The result is xzm from the first
term, 2x2y3 from the second term, and 4yz3m from the third term. The answer should be
written as follows: (The signs stay consistent.)
3xyz2 ( xzm - 2x2y3 + 4yz3m ).
You should recognize that this problem has been factored because it is a multiplication
problem. (****Remember, Factoring is the changing of an addition/subtraction problem into
a multiplication problem. This rule carries over into any type of polynomial. When you are
looking for a GCF this is what should be done. Again, it must be said that this is the most
important of the five tools that we are going to discuss. It can be a part of each of the
remaining four tools.
Simplify the following.
5x2(3m + 1) – 7x5(3m + 1) + 4x3(3m + 1)
5x2(3m + 1) – 7x5(3m + 1) + 4x3(3m + 1)
This polynomial has three terms. The minus sign separates the first two terms ad the plus
sign separates the last two terms. The first term has three factors. The first factor is 5, the
second is x2, and the third factor is the expression (3m + 1).
You should notice right away that the factors 5, 7, and 4, are all numerical coefficients, but
they do not share a GCF. This implies that my overall GCF will not have a numerical
coefficient. Each term has {x} to some degree. The first term has x2 which implies that there
are two x’s. The second term has five x’s and the third term has three x’s. When we ask,
“What is the largest number of x’s present in each term?” You can see right away that the
number in question is not going to be the largest number of x’s present overall. The largest
number of x’s present is x5, but you can only divide each term by x2 because you cannot
subtract 5 from 2 and have a positive exponent. Obviously, the expression (3m + 1) is
common in all three terms. ****Remember, the expression (3m + 1) is a factor just the same
as any other factor; you must treat it as you would any other factor.****
Solution:
x2(3m + 1) (5 - 7 x3 + 4x)
Elementary & Intermediate Algebra
5
Tool 2
Grouping
Grouping – Grouping requires four or more terms. However, there must be an even number
of terms. When using Tool 2, you must guarantee that you cannot use Tool 1 first.
Procedure:
6x2 – 4x – 3xa +2a
Group the first two terms. (6x2 – 4x) Then group the second two terms. (– 3xa +2a)
Find the GCF for the first two terms. Then find the GCF for the second two terms.
2x(3x – 2) - a(3x – 2)
In this case you must factor {-a}. Only factoring {-a} will make the two expressions have
something in common. Notice, factoring {-a} produces expressions which look exactly the
same.
You are left with a two termed expression:
2x(3x – 2) - a(3x – 2)
You can now factor the expression (3x – 2) from both terms.
(3x – 2) (2x – a)
That completes Tool 2.
Elementary & Intermediate Algebra
6
Tool 3
The AC Method
Tool 3 is called the AC method. This is short for the A * C method. The AC method
requires that a polynomial have exactly three terms. When we use Tool 3, we are simply
taking a three termed polynomial and changing it into a four termed polynomial so we can
group factor it by grouping (Tool 2).
We will introduce the quadratic equation in standard form:
Ax2 + Bx + C
6x2 - x - 1
You will notice that the number that represents {A} is 6 and the number that represents {C}
is -1. So
A = 6 and C = -1
When we say A * C, what we are really saying is 6 * -1. In this case A * C = -6.
Now we must find factors of {-6}:
Factors:
1*6
2*3
You must look for the factors of {-6} whose sum produces the coefficient of the middle term.
The middle term’s coefficient is {-1}. The only way to get {-1} is by adding the factors in a
very specific way. The factors {1, 6} cannot be combined (added/subtracted) to equal -1.
However, the factors {2, 3} can be added to result in a -1 only when the combination -3 + 2
is present. So only when 3 is negative and 2 is positive will we obtain –1 as the difference.
The rule says that the largest factor must have the same sign as the middle term. You
must then replace the –x (the middle term) with its equivalent expression.
6x2 - x - 1
- x = - 3x + 2x
6x2 - 3x + 2x - 1
You will notice that you have a nice three termed polynomial into a four termed polynomial.
This changed was made so that we can now use the grouping method (Tool 2).
(6x2 - 3x) + (2x – 1)
3x(2x - 1) +1(2x – 1)
Elementary & Intermediate Algebra
7
(3x + 1) (2x – 1) Factored!
****Remember**** Grouping requires that the problem to have four terms and an {A * C}
has three terms. Logic says that if you want to group, you must make this three termed
polynomial into a four termed polynomial. Replacing the middle term with two terms
accomplishes this requirement.
Examples:
Algebraic
Expression
A*C
Replacement
Factors for Middle
Term
Results
12x2 - 17x + 5
12 * 5
- 12x – 5x
12x2 - 12x – 5x + 5
6x2 + 41xy + 7y2
6*7
42xy - xy
6x2 + 42xy - xy + 7y2
2x2 - 11x - 40
2 * 40
- 16x + 5x
2x2 - 16x + 5x - 40
4y2 – 15y + 9
4*9
- 12y - 3y
4y2 - 12y - 3y + 9
Tool 4
The Difference of Two Perfect Squares
The difference of two perfect squares is Tool 4. Understanding this tool requires an
understanding of the words “perfect square” and “square root”.
Perfect Square – A perfect square is a number that is created by multiplying a number times
itself.
Example: 49
64
81
100
2*2=4
4 is a perfect square because it is created by multiplying the
number two times itself.
3*3=9
9 is a perfect square because it is created by multiplying the
number three times itself.
Elementary & Intermediate Algebra
8
Perfect square numbers are created by multiplying a number times itself. Perfect square
variables simply have exponents that are even.
Square Root – A square root is the number that is multiplied times itself to produce a perfect
square.
Number
…is a square root
because…
Multiply
Perfect Square
7
…is a square root
because…
7*7
49
5
…is a square root
because…
5*5
25
x
…is a square root
because…
x*x
x2
x3
…is a square root
because…
x3 * x3
x6
***Remember*** To find the square root of a variable, simply divide the exponent by two.
This implies that the exponents must be even. Only even numbers are divisible by two.
Examples:
x6 25x2y4 9z4 x4
Number …is a square root because… Multiply
7
…is a square root because…
7*7
5
…is a square root because…
5*5
x
…is a square root because…
x*x
x3
…is a square root because…
x3 * x3
Elementary & Intermediate Algebra
9
When we find the difference of two perfect squares, we must first recognize a perfect square.
Recognition is the key!!!
Example:
9x2 – y2
25 – z2
36x4z2 – 49
36(x + 7y)2 - 49(x + 7y)2
All problems that qualify to be called the difference of two perfect squares, (difference of
squares, for short) must have only two terms. These terms must be separated by a minus
sign. The sum of squares cannot be factored. When we see that the problem is difference
of two perfect squares, we must treat them all exactly the same way.
We make two sets of parenthesis:
Step 1
(
parenthesis. The order does not
matter.
Step 2
(
We find the square root of the
first term and write it as the first
term in both parenthesis.
Step 3
( 5x +
We then write the square root of
The second term as the second
term in both parenthesis.
Step 4
( 5x + y4 ) ( 5x - y4 )
Example:
)(
+
)(
)
-
)
25x2 – y4
) ( 5x -
)
Examples:
25y2 – 36 = ( 5y + 6 ) ( 5y - 6 )
16y2 z2 – 49 = ( 4yz + 7 ) ( 4yz - 7 )
Tool 4
The Sum & Difference of Two Perfect Cubes
Perfect cube – A perfect cube is a number that is obtained by multiplying a number times
itself three times.
Elementary & Intermediate Algebra
10
Example:
1
8
27
64
125
216
343
Cube root: A cube root is the number that is multiplied times itself three times itself to
create a perfect cube.
Examples:
The numbers:
2 is the cube root of 8.
3 is the cube root of 27.
3 * 3 * 3 = 27
4 * 4 * 4 = 64
so the cube root of 27 is 3.
so the cube root of 64 is 4.
Like the exponents of the variables of perfect squares must be divisible by two, the exponents
of the variables of perfect cubes must be divisible by three. This implies that the following
variables are perfect cubes.
Examples:
x3
x15
y3z9
The following terms are perfect cubes:
Examples:
27x3
8y3v9
64x3m6
***Remember*** The numerical coefficients (the number that is multiplied by the variable),
must meet a different requirement than a variable to be called a perfect cube. The numerical
coefficient must have a cube root and all variables must have exponents that are divisible by
three.
When we factor the difference of two perfect cubes, we must make two sets of parenthesis.
They must be of different sizes because a two termed polynomial (binomial) will go in the
first set and the second set will house a three termed polynomial (trinomial).
8x3 - 27
Example:
Because of the minus sign and the fact that both terms are perfect cube, this is going to be
the difference of two perfect cubes. Recognition is the key!!!
Step 1
Two sets of parenthesis
(
)(
)
The parenthesis must not be of equal size. One must house a binomial—a two termed
polynomial, and the other must house trinomial—a three termed polynomial.
Step 2
Because this is the difference of two perfect cubes, the signs that must go in
the parenthesis must be very specific. The terms of the binomial are separated
by a minus sign and the terms of the trinomial are separated by two plus signs.
(
-
)(
+
Elementary & Intermediate Algebra
11
+
)
Step 3
8x3 - 27 = ( 2x - 3 ) (
Cube root of the first term^
+
+
)
^Cube root of the second term
The rule says that you must write the cube root of the first as the first and the cube root of the
second as the second. Since the cube root of the first term is 2x, is must be written as the first
term of the binomial. The cube root of the second term, which is 3, must be written as the
second term in the binomial.
To fill the three spaces of the trinomial, you will not use the original problem. You must
ignore the expression 8x3 - 27. Instead, you must take the first term 2x and square it.
Squaring it means squaring each factor in the term. 2x then becomes 4x2. 4x2 becomes the
first term in the trinomial.
Step 4
8x3 - 27 = ( 2x - 3 ) ( 4x2 +
First term^
+
)
^First term squared
To get the last term in the trinomial, you must square the last term in the binomial. Since the
last term in the binomial is 3, when you square it, it becomes 9. This implies that 9 becomes
the last term in the trinomial.
8x3 - 27 = ( 2x - 3 ) ( 4x2 +
+9)
Second term^
^Second term squared
Finally, to get the middle term of the trinomial, you must ignore the signs in the middle two
terms in the binomial and just multiply them together. 2x times 3 is 6x.
Thus,
8x3 - 27 = ( 2x - 3 ) ( 4x2 + 6x + 9 )
^The product of the first & last terms in the binomial
***Remember*** Since the signs were already set at the beginning of the problem, you will
ignore the signs of the terms in the binomial and just multiply the terms. We say that we just
take their absolute values and then multiply them.
Formulas:
Difference of Two Perfect Cubes
( F3 – L3 ) = (F2 + FL + L2 )
Sum of Two Perfect Cubes
( F3 + L3 ) = (F2 - FL + L2 )
Elementary & Intermediate Algebra
12
Elementary & Intermediate Algebra
13
``` |
GreeneMath.com - Graphing Linear Inequalities in two Variables Practice Set
# In this Section:
In this section, we review how to graph a linear inequality in two variables. A linear inequality in two variables is of the form: ax + by < c, where a, b, and c are real numbers, a and b are not both zero, and < could be: >, ≥, or ≤. To graph a linear inequality in two variables, we solve the inequality for y. We then replace the inequality symbol with an equality symbol and graph the resulting equation. This gives us our boundary line. The boundary line separates the solution region from the non-solution region. The boundary line is dashed for a strict inequality and solid for a non-strict inequality. If our inequality is strict, the boundary line is not part of the solution. Therefore a dashed line shows this line is excluded from the solution region. If the inequality is non-strict, we draw a solid line to show the line is included. Once the boundary line is drawn, we shade below the line for a less than and above the line for a greater than. Note this only works when we have solved the inequality for y. If we don’t solve the inequality for y, we use a test point. The test point method tells us to choose a point on either side of the line. If that point works as a solution to the inequality, the test point lies in the solution region and that region should be shaded. If the point does not work, the test point lies in the non-solution region and we must shade the other region.
Sections:
# In this Section:
In this section, we review how to graph a linear inequality in two variables. A linear inequality in two variables is of the form: ax + by < c, where a, b, and c are real numbers, a and b are not both zero, and < could be: >, ≥, or ≤. To graph a linear inequality in two variables, we solve the inequality for y. We then replace the inequality symbol with an equality symbol and graph the resulting equation. This gives us our boundary line. The boundary line separates the solution region from the non-solution region. The boundary line is dashed for a strict inequality and solid for a non-strict inequality. If our inequality is strict, the boundary line is not part of the solution. Therefore a dashed line shows this line is excluded from the solution region. If the inequality is non-strict, we draw a solid line to show the line is included. Once the boundary line is drawn, we shade below the line for a less than and above the line for a greater than. Note this only works when we have solved the inequality for y. If we don’t solve the inequality for y, we use a test point. The test point method tells us to choose a point on either side of the line. If that point works as a solution to the inequality, the test point lies in the solution region and that region should be shaded. If the point does not work, the test point lies in the non-solution region and we must shade the other region. |
# Multiplication
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Four bags of three marbles gives twelve marbles (4 × 3 = 12).
Multiplication can also be thought of as scaling. In the above animation, we see 2 being multiplied by 3, giving 6 as a result
4 × 5 = 20, the rectangle is composed of 20 squares, having dimensions of 4 by 5.
Area of a cloth 4.5m × 2.5m = 11.25m2; 4½ × 2½ = 11¼
Multiplication (often denoted by the cross symbol "×", or by the absence of symbol) is one of the four elementary, mathematical operations of arithmetic; with the others being addition, subtraction and division.
The multiplication of two whole numbers is equivalent to the addition of one of them with itself as many times as the value of the other one; for example, 3 multiplied by 4 (often said as "3 times 4") can be calculated by adding 3 copies of 4 together:
${\displaystyle 3\times 4=4+4+4=12}$
Here 3 and 4 are the "factors" and 12 is the "product".
One of the main properties of multiplication is that the result does not depend on the place of the factor that is repeatedly added to itself (commutative property). 3 multiplied by 4 can also be calculated by adding 4 copies of 3 together:
${\displaystyle 3\times 4=3+3+3+3=12}$
The multiplication of integers (including negative numbers), rational numbers (fractions) and real numbers is defined by a systematic generalization of this basic definition.
Multiplication can also be visualized as counting objects arranged in a rectangle (for whole numbers) or as finding the area of a rectangle whose sides have given lengths. The area of a rectangle does not depend on which side is measured first, which illustrates the commutative property.
In general, multiplying two measurements gives a new type, depending on the measurements. For instance:
${\displaystyle 2.5{\mbox{ meters}}\times 4.5{\mbox{ meters}}=11.25{\mbox{ square meters}}}$
${\displaystyle 11{\mbox{ meters/second}}\times 9{\mbox{ seconds}}=99{\mbox{ meters}}}$
The inverse operation of the multiplication is the division. For example, since 4 multiplied by 3 equals 12, then 12 divided by 3 equals 4. Multiplication by 3, followed by division by 3, yields the original number (since the division of a number other than 0 by itself equals 1).
Multiplication is also defined for other types of numbers, such as complex numbers, and more abstract constructs, like matrices. For these more abstract constructs, the order that the operands are multiplied sometimes does matter.
## Notation and terminology
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The multiplication sign ×
(HTML entity is ×)
In arithmetics, multiplication is often written using the sign "×" between the terms; that is, in infix notation. For example,
${\displaystyle 2\times 3=6}$ (verbally, "two times three equals six")
${\displaystyle 3\times 4=12}$
${\displaystyle 2\times 3\times 5=6\times 5=30}$
${\displaystyle 2\times 2\times 2\times 2\times 2=32}$
The sign is encoded in Unicode at Template:Unichar.
There are other mathematical notations for multiplication:
• Multiplication is sometimes denoted by dot signs, either a middle-position dot or a period:
${\displaystyle 5\cdot 2\quad {\text{or}}\quad 5\,.\,2}$
The middle dot notation, encoded in Unicode as Template:Unichar, is standard in the United States, the United Kingdom, and other countries where the period is used as a decimal point. When the dot operator character is not accessible, the interpunct (·) is used. In other countries that use a comma as a decimal point, either the period or a middle dot is used for multiplication.{{ safesubst:#invoke:Unsubst||date=__DATE__ |\$B=
{{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }}
• In algebra, multiplication involving variables is often written as a juxtaposition (e.g., xy for x times y or 5x for five times x). This notation can also be used for quantities that are surrounded by parentheses (e.g., 5(2) or (5)(2) for five times two).
In computer programming, the asterisk (as in 5*2) is the standard notation: it belongs to most character sets and appears on every keyboard. This usage originated in the FORTRAN programming language.
The numbers to be multiplied are generally called the "factors" or "multiplicands". When thinking of multiplication as repeated addition, the number to be multiplied is called the "multiplicand", while the number of addends is called the "multiplier". In algebra, a number that is the multiplier of a variable or expression (e.g., the 3 in 3xy2) is called a coefficient.
The result of a multiplication is called a product. A product of integers is a multiple of each factor. For example, 15 is the product of 3 and 5, and is both a multiple of 3 and a multiple of 5.
## Computation
The common methods for multiplying numbers using pencil and paper require a multiplication table of memorized or consulted products of small numbers (typically any two numbers from 0 to 9), however one method, the peasant multiplication algorithm, does not.
Multiplying numbers to more than a couple of decimal places by hand is tedious and error prone. Common logarithms were invented to simplify such calculations. The slide rule allowed numbers to be quickly multiplied to about three places of accuracy. Beginning in the early twentieth century, mechanical calculators, such as the Marchant, automated multiplication of up to 10 digit numbers. Modern electronic computers and calculators have greatly reduced the need for multiplication by hand.
### Historical algorithms
Methods of multiplication were documented in the Egyptian, Greek, Indian and Chinese civilizations.
The Ishango bone, dated to about 18,000 to 20,000 BC, hints at a knowledge of multiplication in the Upper Paleolithic era in Central Africa.
#### Egyptians
{{#invoke:main|main}} The Egyptian method of multiplication of integers and fractions, documented in the Ahmes Papyrus, was by successive additions and doubling. For instance, to find the product of 13 and 21 one had to double 21 three times, obtaining 1 × 21 = 21, 4 × 21 = 84, 8 × 21 = 168. The full product could then be found by adding the appropriate terms found in the doubling sequence:
13 × 21 = (1 + 4 + 8) × 21 = (1 × 21) + (4 × 21) + (8 × 21) = 21 + 84 + 168 = 273.
#### Babylonians
The Babylonians used a sexagesimal positional number system, analogous to the modern day decimal system. Thus, Babylonian multiplication was very similar to modern decimal multiplication. Because of the relative difficulty of remembering 60 × 60 different products, Babylonian mathematicians employed multiplication tables. These tables consisted of a list of the first twenty multiples of a certain principal number n: n, 2n, ..., 20n; followed by the multiples of 10n: 30n 40n, and 50n. Then to compute any sexagesimal product, say 53n, one only needed to add 50n and 3n computed from the table.
#### Chinese
38 × 76 = 2888
In the mathematical text Zhou Bi Suan Jing, dated prior to 300 BC, and the Nine Chapters on the Mathematical Art, multiplication calculations were written out in words, although the early Chinese mathematicians employed Rod calculus involving place value addition, subtraction, multiplication and division. These place value decimal arithmetic algorithms were introduced by Al Khwarizmi to Arab countries in the early 9th century.
### Modern method
Product of 45 and 256. Note the order of the numerals in 45 is reversed down the left column. The carry step of the multiplication can be performed at the final stage of the calculation (in bold), returning the final product of 45 × 256 = 11520.
The modern method of multiplication based on the Hindu–Arabic numeral system was first described by Brahmagupta. Brahmagupta gave rules for addition, subtraction, multiplication and division. Henry Burchard Fine, then professor of Mathematics at Princeton University, wrote the following:
The Indians are the inventors not only of the positional decimal system itself, but of most of the processes involved in elementary reckoning with the system. Addition and subtraction they performed quite as they are performed nowadays; multiplication they effected in many ways, ours among them, but division they did cumbrously.[1]
### Computer algorithms
{{#invoke:main|main}}
The standard method of multiplying two n-digit numbers requires n2 simple multiplications. Multiplication algorithms have been designed that reduce the computation time considerably when multiplying large numbers. In particular for very large numbers methods based on the Discrete Fourier Transform can reduce the number of simple multiplications to the order of n log2(n) log2 log2(n).
## Products of measurements
{{#invoke:main|main}}
When two measurements are multiplied together the product is of a type depending on the types of the measurements. The general theory is given by dimensional analysis. This analysis is routinely applied in physics but has also found applications in finance. One can only meaningfully add or subtract quantities of the same type but can multiply or divide quantities of different types.
A common example is multiplying speed by time gives distance, so
50 kilometers per hour × 3 hours = 150 kilometers.
## Products of sequences
### Capital Pi notation
The product of a sequence of terms can be written with the product symbol, which derives from the capital letter Π (Pi) in the Greek alphabet. Unicode position U+220F (∏) contains a glyph for denoting such a product, distinct from U+03A0 (Π), the letter. The meaning of this notation is given by:
${\displaystyle \prod _{i=1}^{4}i=1\cdot 2\cdot 3\cdot 4,}$
that is
${\displaystyle \prod _{i=1}^{4}i=24.}$
The subscript gives the symbol for a dummy variable (i in this case), called the "index of multiplication" together with its lower bound (1), whereas the superscript (here 4) gives its upper bound. The lower and upper bound are expressions denoting integers. The factors of the product are obtained by taking the expression following the product operator, with successive integer values substituted for the index of multiplication, starting from the lower bound and incremented by 1 up to and including the upper bound. So, for example:
${\displaystyle \prod _{i=1}^{6}i=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6=720}$
More generally, the notation is defined as
${\displaystyle \prod _{i=m}^{n}x_{i}=x_{m}\cdot x_{m+1}\cdot x_{m+2}\cdot \,\,\cdots \,\,\cdot x_{n-1}\cdot x_{n},}$
where m and n are integers or expressions that evaluate to integers. In case m = n, the value of the product is the same as that of the single factor xm. If m > n, the product is the empty product, with the value 1.
### Infinite products
{{#invoke:main|main}}
One may also consider products of infinitely many terms; these are called infinite products. Notationally, we would replace n above by the lemniscate ∞. The product of such a series is defined as the limit of the product of the first n terms, as n grows without bound. That is, by definition,
${\displaystyle \prod _{i=m}^{\infty }x_{i}=\lim _{n\to \infty }\prod _{i=m}^{n}x_{i}.}$
One can similarly replace m with negative infinity, and define:
${\displaystyle \prod _{i=-\infty }^{\infty }x_{i}=\left(\lim _{m\to -\infty }\prod _{i=m}^{0}x_{i}\right)\cdot \left(\lim _{n\to \infty }\prod _{i=1}^{n}x_{i}\right),}$
provided both limits exist.
## Properties
Multiplication of numbers 0-10. Line labels = multiplicand. X axis = multiplier. Y axis = product.
For the real and complex numbers, which includes for example natural numbers, integers and fractions, multiplication has certain properties:
Commutative property
The order in which two numbers are multiplied does not matter:
${\displaystyle x\cdot y=y\cdot x}$.
Associative property
Expressions solely involving multiplication or addition are invariant with respect to order of operations:
${\displaystyle (x\cdot y)\cdot z=x\cdot (y\cdot z)}$
Distributive property
Holds with respect to multiplication over addition. This identity is of prime importance in simplifying algebraic expressions:
${\displaystyle x\cdot (y+z)=x\cdot y+x\cdot z}$
Identity element
The multiplicative identity is 1; anything multiplied by one is itself. This is known as the identity property:
${\displaystyle x\cdot 1=x}$
Zero element
Any number multiplied by zero is zero. This is known as the zero property of multiplication:
${\displaystyle x\cdot 0=0}$
Zero is sometimes not included amongst the natural numbers.
There are a number of further properties of multiplication not satisfied by all types of numbers.
Negation
Negative one times any number is equal to the additive inverse of that number.
${\displaystyle (-1)\cdot x=(-x)}$
Negative one times negative one is positive one.
${\displaystyle (-1)\cdot (-1)=1}$
The natural numbers do not include negative numbers.
Inverse element
Every number x, except zero, has a multiplicative inverse, ${\displaystyle {\frac {1}{x}}}$, such that ${\displaystyle x\cdot \left({\frac {1}{x}}\right)=1}$.
Order preservation
Multiplication by a positive number preserves order:
if a > 0, then if b > c then ab > ac.
Multiplication by a negative number reverses order:
if a < 0 and b > c then ab < ac.
The complex numbers do not have an order predicate.
Other mathematical systems that include a multiplication operation may not have all these properties. For example, multiplication is not, in general, commutative for matrices and quaternions.
## Axioms
{{#invoke:main|main}}
In the book Arithmetices principia, nova methodo exposita, Giuseppe Peano proposed axioms for arithmetic based on his axioms for natural numbers.[2] Peano arithmetic has two axioms for multiplication:
${\displaystyle x\times 0=0}$
${\displaystyle x\times S(y)=(x\times y)+x}$
Here S(y) represents the successor of y, or the natural number that follows y. The various properties like associativity can be proved from these and the other axioms of Peano arithmetic including induction. For instance S(0). denoted by 1, is a multiplicative identity because
${\displaystyle x\times 1=x\times S(0)=(x\times 0)+x=0+x=x}$
The axioms for integers typically define them as equivalence classes of ordered pairs of natural numbers. The model is based on treating (x,y) as equivalent to xy when x and y are treated as integers. Thus both (0,1) and (1,2) are equivalent to −1. The multiplication axiom for integers defined this way is
${\displaystyle (x_{p},\,x_{m})\times (y_{p},\,y_{m})=(x_{p}\times y_{p}+x_{m}\times y_{m},\;x_{p}\times y_{m}+x_{m}\times y_{p})}$
The rule that −1 × −1 = 1 can then be deduced from
${\displaystyle (0,1)\times (0,1)=(0\times 0+1\times 1,\,0\times 1+1\times 0)=(1,0)}$
Multiplication is extended in a similar way to rational numbers and then to real numbers.
## Multiplication with set theory
It is possible, though difficult, to create a recursive definition of multiplication with set theory. Such a system usually relies on the Peano definition of multiplication.
### Cartesian product
The definition of multiplication as repeated addition provides a way to arrive at a set-theoretic interpretation of multiplication of cardinal numbers. In the expression
${\displaystyle \displaystyle n\cdot a=\underbrace {a+\cdots +a} _{n},}$
if the n copies of a are to be combined in disjoint union then clearly they must be made disjoint; an obvious way to do this is to use either a or n as the indexing set for the other. Then, the members of ${\displaystyle n\cdot a\,}$ are exactly those of the Cartesian product ${\displaystyle n\times a\,}$. The properties of the multiplicative operation as applying to natural numbers then follow trivially from the corresponding properties of the Cartesian product.
## Multiplication in group theory
There are many sets that, under the operation of multiplication, satisfy the axioms that define group structure. These axioms are closure, associativity, and the inclusion of an identity element and inverses.
A simple example is the set of non-zero rational numbers. Here we have identity 1, as opposed to groups under addition where the identity is typically 0. Note that with the rationals, we must exclude zero because, under multiplication, it does not have an inverse: there is no rational number that can be multiplied by zero to result in 1. In this example we have an abelian group, but that is not always the case.
To see this, look at the set of invertible square matrices of a given dimension, over a given field. Now it is straightforward to verify closure, associativity, and inclusion of identity (the identity matrix) and inverses. However, matrix multiplication is not commutative, therefore this group is nonabelian.
Another fact of note is that the integers under multiplication is not a group, even if we exclude zero. This is easily seen by the nonexistence of an inverse for all elements other than 1 and -1.
Multiplication in group theory is typically notated either by a dot, or by juxtaposition (the omission of an operation symbol between elements). So multiplying element a by element b could be notated a ${\displaystyle \cdot }$ b or ab. When referring to a group via the indication of the set and operation, the dot is used, e.g., our first example could be indicated by ${\displaystyle \left(\mathbb {Q} \smallsetminus \{0\},\cdot \right)}$
## Multiplication of different kinds of numbers
Numbers can count (3 apples), order (the 3rd apple), or measure (3.5 feet high); as the history of mathematics has progressed from counting on our fingers to modelling quantum mechanics, multiplication has been generalized to more complicated and abstract types of numbers, and to things that are not numbers (such as matrices) or do not look much like numbers (such as quaternions).
Integers
${\displaystyle N\times M}$ is the sum of M copies of N when N and M are positive whole numbers. This gives the number of things in an array N wide and M high. Generalization to negative numbers can be done by
${\displaystyle N\times (-M)=(-N)\times M=-(N\times M)}$ and
${\displaystyle (-N)\times (-M)=N\times M}$
The same sign rules apply to rational and real numbers.
Rational numbers
Generalization to fractions ${\displaystyle {\frac {A}{B}}\times {\frac {C}{D}}}$ is by multiplying the numerators and denominators respectively: ${\displaystyle {\frac {A}{B}}\times {\frac {C}{D}}={\frac {(A\times C)}{(B\times D)}}}$. This gives the area of a rectangle ${\displaystyle {\frac {A}{B}}}$ high and ${\displaystyle {\frac {C}{D}}}$ wide, and is the same as the number of things in an array when the rational numbers happen to be whole numbers.
Real numbers
${\displaystyle (x)(y)}$ is the limit of the products of the corresponding terms in certain sequences of rationals that converge to x and y, respectively, and is significant in calculus. This gives the area of a rectangle x high and y wide. See Products of sequences, above.
Complex numbers
Considering complex numbers ${\displaystyle z_{1}}$ and ${\displaystyle z_{2}}$ as ordered pairs of real numbers ${\displaystyle (a_{1},b_{1})}$ and ${\displaystyle (a_{2},b_{2})}$, the product ${\displaystyle z_{1}\times z_{2}}$ is ${\displaystyle (a_{1}\times a_{2}-b_{1}\times b_{2},a_{1}\times b_{2}+a_{2}\times b_{1})}$. This is the same as for reals, ${\displaystyle a_{1}\times a_{2}}$, when the imaginary parts ${\displaystyle b_{1}}$ and ${\displaystyle b_{2}}$ are zero.
Further generalizations
See Multiplication in group theory, above, and Multiplicative Group, which for example includes matrix multiplication. A very general, and abstract, concept of multiplication is as the "multiplicatively denoted" (second) binary operation in a ring. An example of a ring that is not any of the above number systems is a polynomial ring (you can add and multiply polynomials, but polynomials are not numbers in any usual sense.)
Division
Often division, ${\displaystyle {\frac {x}{y}}}$, is the same as multiplication by an inverse, ${\displaystyle x\left({\frac {1}{y}}\right)}$. Multiplication for some types of "numbers" may have corresponding division, without inverses; in an integral domain x may have no inverse "${\displaystyle {\frac {1}{x}}}$" but ${\displaystyle {\frac {x}{y}}}$ may be defined. In a division ring there are inverses, but ${\displaystyle {\frac {x}{y}}}$ may be ambiguous in non-commutative rings since ${\displaystyle x\left({\frac {1}{y}}\right)}$ need not the same as ${\displaystyle \left({\frac {1}{y}}\right)x}$.
## Exponentiation
{{#invoke:main|main}} When multiplication is repeated, the resulting operation is known as exponentiation. For instance, the product of three factors of two (2×2×2) is "two raised to the third power", and is denoted by 23, a two with a superscript three. In this example, the number two is the base, and three is the exponent. In general, the exponent (or superscript) indicates how many times to multiply base by itself, so that the expression
${\displaystyle a^{n}=\underbrace {a\times a\times \cdots \times a} _{n}}$
indicates that the base a to be multiplied by itself n times. |
# Area of a rectangle
Here you will learn about finding the area of a rectangle, including finding the area of rectilinear figures, finding missing side lengths, solving area problems involving unit conversion, and solving real-world word problems involving the area of a rectangle.
Students will first learn about area of a rectangle as part of measurement and data in 3rd grade.
## What is the area of a rectangle?
The area of a rectangle is the amount of space inside the rectangle. It is measured in units squared, or square units. ( cm^2, m^2, in^2, etc.)
In order to find the area, you need to use the area formula:
\text { Area }=\text { length} \times \text { width}
For example,
\begin{aligned} \text { Area } & =\text { length } \times \text { width } \\\\ & =7 \times 4 \\\\ & =28 \mathrm{~m}^2 \end{aligned}
You can also use this area formula:
\text { Area }=\text { base} \times \text { height}
Notice it multiplies the same parts of the rectangle as the formula A=l \times w, but uses the terms base and height instead of length and width. Either of these formulas can be used to find the area of a rectangle.
A rectangle is a quadrilateral ( 4 sided shape) where every angle is a right angle (90^{\circ}). Opposite sides of the rectangle are equal length.
The area of a rectangle is calculated by multiplying the length of the rectangle by the width of the rectangle.
The final answer must be in square units. For example, square centimeters (cm^2), square meters (m^2), square feet (ft^2), square inches (in^2), etc.
## Common Core State Standards
How does this relate to 3rd grade math and 4th grade math?
• Grade 3 – Measurement and Data (3.MD.7)
Relate area to the operations of multiplication and addition.
a. Find the area of a rectangle with whole-number side lengths by tiling it, and
show that the area is the same as would be found by multiplying the side
lengths.
b. Multiply side lengths to find areas of rectangles with whole number side
lengths in the context of solving real world and mathematical problems,
and represent whole-number products as rectangular areas in mathematical
reasoning.
c. Use tiling to show in a concrete case that the area of a rectangle with whole-
number side lengths a and b + c is the sum of a \times b and a \times c. Use area
models to represent the distributive property in mathematical reasoning.
d. Recognize area as additive. Find areas of rectilinear figures by decomposing
them into non-overlapping rectangles and adding the areas of the non-
overlapping parts, applying this technique to solve real world problems.
• Grade 4 – Measurement and Data (4.MD.3)
Apply the area and perimeter formulas for rectangles in real world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor.
## How to find the area of a rectangle
In order to find the area of a rectangle:
1. Identify the length and width of the rectangle.
2. Write down the formula for the area of a rectangle.
3. Substitute the given values and calculate.
## Area of a rectangle examples
### Example 1: finding area given the length and width
Find the area of the rectangle below.
1. Identify the length and width of the rectangle.
Length = 6 \, m
Width = 4 \, m
2Write down the formula for the area of a rectangle.
\text { Area }=\text { length } \times \text { width }
3Substitute the given values and calculate.
\begin{aligned} \text { Area }&=\text { length } \times \text { width } \\\\ \text { Area }&=6 \times 4 \\\\ & =24 \end{aligned}
In this case, you are working with meters so your final answer must be in square meters.
\text { Area }=24 \mathrm{~m}^{2}
### Example 2: finding the area of a rectangle requiring converting units
Find the area of the rectangle below.
Identify the length and width of the rectangle.
Write down the formula for the area of a rectangle.
Substitute the given values and calculate.
### Example 3: real-world word problem
Ms. Crawely is tiling her bathroom floor. The dimensions of the floor are 6 \, m by 4 \, m. Each tile is 50 \, cm by 50 \, cm. How many tiles will she need to cover the bathroom floor?
Identify the length and width of the rectangle.
Write down the formula for the area of a rectangle.
Substitute the given values and calculate.
### Example 4: calculating side length given the area
Find the width of the rectangle below.
Identify the length and width of the rectangle.
Write down the formula for the area of a rectangle.
Substitute the given values and calculate.
### Example 5: area of a rectilinear figure
Find the area of the field below:
Identify the length and width of the rectangle.
Write down the formula for the area of a rectangle.
Substitute the given values and calculate.
### Example 6: word problem
Rectangle ABCD, shown below, has a perimeter of 22 \, m and a side length of 8 \, m. Find the area of the rectangle.
Identify the length and width of the rectangle.
Write down the formula for the area of a rectangle.
Substitute the given values and calculate.
### Teaching tips for area of a rectangle
• Before learning the area of a rectangle formula, allow students to explore area by tiling with square tiles. These concrete manipulatives act as unit squares and will help students build an understanding of area before they begin calculating it.
• Rather than having students practice finding the area of a rectangle on multiple skill worksheets, provide them with a variety of practice problems, activities, and/or projects that have a real-world context. This will deepen their understanding of this skill.
### Easy mistakes to make
• Using incorrect units for the answer
A common error is to forget to include square units in your answer when finding the area of a rectangle.
• Forgetting to convert measures to a common unit
Before using the formula for the area of a rectangle, you need to ensure that the units are the same. If different units are given (for example, length = 4 \, m and width = 3 \, cm ), then you must convert them either both to centimeters or both to meters.
### Practice area of a rectangle questions
1. Find the area of the rectangle below.
60 \, mm
17 \, mm^2
60 \, mm^2
17 \, mm
Multiply the length and width together to get the area of the rectangle.
12 \mathrm{~mm} \times 5 \mathrm{~mm}=60 \mathrm{~mm}^2
2. Find the area of the rectangle below.
480 \, m^2
480 \, cm^2
4.8 \, m^2 or 48 \, 000 \, cm^2
124 \, m^2
Prior to multiplying the length times the width, convert the length and width to a common unit (both length and width to meters or both to centimeters).
Remember there are 100 centimeters in 1 meter.
3. Mr. Measure is tiling his kitchen floor. The dimensions of the floor are 7 \, m by 5 \, m. Each tile is 20 \, cm by 20 \, cm. How many tiles will he need to cover the bathroom floor?
35 tiles
875 tiles
400 tiles
11 tiles
Find the area of the floor by multiplying the length of the rectangle times the width.
Prior to multiplying the length and width for the tile, convert the length and width to a common unit (since you calculated the floor in meters here, you converted tiles to meters).
Remember there are 100 centimeters in 1 meter.
Take the area of the tiles and divide it into the area of the floor to get the number of tiles needed to cover the floor.
4. Find the width of the rectangle below.
4 \, m
19600 \, m
19 \, 600 \, m^2
4 \, m^2
Using the formula for the area of a rectangle, substitute your given values.
As you are trying to find the width of the rectangle, you need to rearrange the formula dividing the length into the area.
5. Below is a blueprint for a new flower bed. Find the area of the flower bed.
57 \, m^2
22 \, m^2
45 \, m^2
30 \, m^2
Split the rectilinear figure into 3 rectangles.
Find the missing side length of rectangle C by subtracting 3 from the total length which is 5 \, m.
Add up all the individual areas to get the total area of the rectilinear figure.
Rectangle A
\begin{aligned} \text { Area }&=\text { length } \times \text { width } \\\\ &=3 \times 2 \\\\ &=6 \\\\ \text { Area }&=6 \, m^2 \end{aligned}
Rectangle B
\begin{aligned} \text { Area }&=\text { length } \times \text { width } \\\\ &=3 \times 2 \\\\ &=6 \\\\ \text { Area }&=6 \, m^2 \end{aligned}
Rectangle C
\begin{aligned} \text { Area }&=\text { length } \times \text { width } \\\\ &=9 \times 2 \\\\ &=18 \\\\ \text { Area }&=18 \, m^2 \end{aligned}
\begin{aligned} \text { Total Area }&= 6 \, m^2+6 \, m^2+18 \, m^2 \\\\ & =30 \, m^2 \end{aligned}
6. The perimeter of rectangle WXYZ is 26 \, m. It has a side length of 11 \, m. Find the area of the rectangle.
2 \, m
2 \, m^2
22 \, m^2
22 \, m
Since side XY is 11 \, m, side WZ is also 11 \, m. That means the sum of the remaining side lengths must equal 4 \, m.
So side WX and side YZ are each 2 \, m. Then, you multiply 11 \, m by 2 \, m \; (length \times width) to find the area.
## Area of a rectangle FAQs
How do you find the area of a rectangle?
Multiply the length of a rectangle by the width of a rectangle to find its area.
What is the difference between the area of a rectangle and the perimeter of a rectangle?
The area of a rectangle is the amount of space inside the rectangle. It is measured in square units. The perimeter of a rectangle is the distance around the outside of a rectangle, or the sum of the sides of a rectangle. It is measured in any unit of length.
What is the area of a rectangle formula?
The formula for area of a rectangle is Area = length \times width.
## Still stuck?
At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.
Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. |
Question 6a5a1
Aug 31, 2017
$70$
Explanation:
The Least Common Denominator (L.C.D) as you said, can be gotten from the Lowest Common Multiple (L.C.M) from of the two digits..
Hence we have for $10 = 2 \times 5$ and $14 = 2 \times 7$
Since $2$ is both common between them, we should take of of the $2 ' s$ instead of both..
Hence the L.C.M for both digits are $\to 2 \times 5 \times 7 = 70$
Hence the L.C.D of $10$ and $14$ is $70$
Though the L.C.D, is best explained using fractions, but what is given is just whole number..
To make it more crystal..
An example, Find the L.C.D of 1/2 + 1/4 = ?#
Like i said it has to do with the Denominators, like making the denominators common..
$\frac{1}{\textcolor{b l u e}{2}} + \frac{1}{\textcolor{b l u e}{4}}$
We have $2 \mathmr{and} 4$ as the respective denominators, so in other to make it common, we have to look for the L.C.M
L.C.M of $2 = \textcolor{red}{2}$
L.C.M of $4 = \textcolor{red}{2} \times 2$
The one in red color is common in both digits, so we will just take one $2$ out of them both..
$\therefore$ L.C.M of $2 \mathmr{and} 4 = \textcolor{red}{2} \times 2 = 4$
$\Rightarrow L . C . D = 4$
$\Rightarrow \frac{1}{4} + \frac{1}{4}$
Hope this helps! |
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• Level: GCSE
• Subject: Maths
• Word count: 1180
# Stair shape maths GCSE coursework
Extracts from this document...
Introduction
page
Coursework – Number Stairs
In order to investigate the stair shapes, I will look at the relationship among the total of 3 step stair shapes and position of stair shapes, on a 10 × 10 number grid. The shaded area, in the grid, is a 3 stair shape at position 25. The position is shown on the bottom left of every stairs. I will investigate different positions and their totals, on the 10 × 10 number grid, with the 3 stair shape (to begin with). With these results, I will be able to create a formula to show the total of 3 step stair shape, at any position, on a 10 × 10 number grid.
I will investigate further, by looking at the relationship between the different sized stair shapes, and the different sized number grids. 3 stair shape I will begin my investigation by obtaining enough results to formulate an a equation, for a 3 step stair shape I will began my investigation by obtaining enough results to formulate an equation, for a 3 step stair shape.
First I will draw a 3 stair shape at the bottom left with the number 1.
Middle
being the pattern.
1. The difference when moving one square upwards
18 The difference when moving one square side ways
I have also found a formula to get the total of each stairs: 6n + 44
I got this formula via:
The 6n comes from the six numbers in the stair shape i.e.:
The Formula = 6n +44, how I get the 6n is mentioned above now I am going to show how I get the 44.
This shows the six n and numbers. 20+10+11+2+1= 44
This is how I get 6n +44 which will work on finding any 3 step stairs at least that’s what I think.
This is how I worked out my formula:
T = n + (n+1) n + 2 + n + 10 + n +11+ n + 20 = 6n+44
6n + 44 will work with any number in this 10 × 10 grid and with any 3 step stairshape.
Now for part 2 I am going to investigate the further relationship between the stair totals and other step stairs on an other number grid. For this I am going to use 9× 9 and a 3 step stair shape.
Grid: 9 Steps: 3 @1 1+2+3+10+11+19 = 46
Grid: 9 Steps: 3 @4 4+5+6+13+14+22 = 64
Grid: 9 Steps: 3 @7 7+8+9+16+17+25 = 82
Conclusion
I have also found a formula to get the total of each stairs: 6n + 40
I got this formula via:
The 6n comes from the six numbers in the stair shape i.e.:
The Formula = 6n +40, how I get the 6n is mentioned above now I am going to show how I get the 40.
This shows the six n and numbers. 18+9+10+1+2= 40
This is how I get 6n +40 which will work on finding any 3 step stairs at least that’s what I think.
This is how I worked out my formula:
T = n + (n+1) n + 2 + n + 9 + n +10 + n + 18 = 6n+40
6n + 40 will work with any number in this 9 × 9 grid and with any 3 step stairshape.
The difference between a 10× 10 grid and a 9 × 9 is that to find a total of 3 step stair in a 10× 10 grid is that the formula is different 6n +44 from a 9 × 9 grid which is 6n + 40 this is the result of my investigation.
I am going to investigate the further relationship between the stair totals and other step stairs on an other number grid. For this I am going to use 8× 8 and a 3 step stair shape.
Grid: 8 Steps: 3 @41 41+42+43+49+50+57 = 282
Grid: 8 Steps: 3 @44 44+45+46+52+53+60 = 300
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. ## GCSE Maths Sequences Coursework
Stage (N) 1 2 3 Sequence 3 6 9 1st difference +3 +3 2nd difference Shaded I can see a clear pattern here, the sequence for shaded is going up regularly in 3's, and therefore this sequence is a linear sequence and has an Nth term. Nth term = 3N+?
2. ## Number Grid Coursework
The lengths will range from 2 to 5. With these boxes, in 5 different random locations on the width 10 grid, the differences of the two products will be calculated. Again, I believe 5 calculations are enough to display any patterns.
1. ## Number Grids Investigation Coursework
these numbers together, then multiplying the product by 10, I get the difference. I can put this into an algebraic formula, like so: D = 10 (m - 1) (n - 1) This also fits with my formula for squares as, in squares the length and the width are equal, (m - 1)
2. ## What the 'L' - L shape investigation.
by 5 grid I have found: 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 I will display my results from these calculations into a table format as follows: Number In Sequence 1 2
1. ## Investigation of diagonal difference.
I will now analyse the cutout and see if G can be applied. n n + 2 N + 10 n + 12 From analysing the cutout I have noticed that G can be applied as the bottom left corner is n + 10 and 10 is the length of the grid.
2. ## Number Grid Maths Coursework.
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 Rectangle Difference 1,3,11,13 (3x2) 20 56,57,76,77 (2x3) 20 7,10,17,20 (4x2) 30 32,35,52,55 (4x3) 60 48,50,88,90 (3x5)
1. ## For other 3-step stairs, investigate the relationship between the stair total and the position ...
then the square above would be x - 10 = 11 (31 - 10 = 21) and so on, Using the algebra equation and adding the values we get is -90 11 x 11 Grid 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
2. ## Maths Grids Totals
going to prove it algebraically: "x" is the number in the top-left corner. The number to the right of it is "x+ (n-1)" because it is "x" added to the length of the square (take away one). Because the square is on a 9 x 9 grid, the formula for
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# 4.3 Parallel and Perpendicular Lines
## Presentation on theme: "4.3 Parallel and Perpendicular Lines"— Presentation transcript:
4.3 Parallel and Perpendicular Lines
Algebra 2/Geometry High School
Parallel and Perpendicular Lines
Teacher Page Content: Write the equation of a parallel or perpendicular line to a given line, in slope intercept form. Grade Level: Algebra 2/Geometry Creator: Don Sutcliffe Other – Print slides as a handout for students to take notes. MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Teacher Page Curriculum Objectives: Algebra 2 – II.A – The learner will find the slope of a linear equation and the relationships of the slopes of parallel and perpendicular lines. MAP Objectives: Missouri Show-Me Standard – Math – MA1, MA2, MA4, MA5 Missouri Show-Me Standard – Goal – G1.6, G3.4, G3.5, G4.1 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Parallel Lines Two non-vertical lines are parallel if and only if their slopes are equal. If l1║l2, then m1= m2. If m1= m2, then l1║l2. m1 m2 l1 l2 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (3,6) and is parallel to y = 2/3x+2. m = 2/3 and the point is (3,6) y = mx+b 6 = 2/3(3)+b 6 = 2+b 4 = b y = 2/3x+4 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (4,-5) and is parallel to y = -2x-4. m = -2 and the point is (4,-5) y = mx+b -5 = -2(4)+b -5 = -8+b 3 = b y = -2x+3 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (-6,4) and is parallel to y=1/3x-1. m=1/3 and the point is (-6,4) y =1/3x+6 MAP TAP Parallel and Perpendicular Lines
Perpendicular Lines Two non-vertical lines are perpendicular if and only if the product of their slopes is -1. If l1┴l2, then m1 ● m2= -1. If m1● m2 = -1, then l1┴l2. m1 l1 m2 Slopes are negative reciprocals l2 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (6,-5) and is perpendicular to y = 2x+3. m = -1/2 and the point is (6,-5) y = mx+b -5 = -1/2(6)+b -5 = -3+b -2 = b y = -1/2x-2 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (6,-7) and is perpendicular to y = 2/3x+1. m = -3/2 and the point is (6,-7) y = mx+b -7 = -3/2(6)+b -7 = -9+b 2 = b y = -3/2x+2 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (-4,-3) and is perpendicular to y = x+6. m = -1 and the point is (-4,-3) y = -x-7 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
End Show MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Parallel Lines Two non-vertical lines are parallel if and only if their slopes are equal. If l1║l2, then m1= m2. If m1= m2, then l1║l2. m1 m2 l1 l2 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (3,6) and is parallel to y = 2/3x+2. m = _____ and the point is (__,__) y = mx+b MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (4,-5) and is parallel to y = -2x-4. m = ____ and the point is (__,__) MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (-6,4) and is parallel to y=1/3x-1. m=___ and the point is (__,__) MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Two non-vertical lines are perpendicular if and only if the product of their slopes is -1. If l1┴l2, then m1 ● m2= -1. If m1● m2 = -1, then l1┴l2. m1 l1 m2 l2 MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (6,-5) and is perpendicular to y = 2x+3. m = ___ and the point is (__,-__) MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (6,-7) and is perpendicular to y = 2/3x+1. m = ____ and the point is (__, __) MAP TAP Parallel and Perpendicular Lines
Parallel and Perpendicular Lines
Write the equation of the line that passes through (-4,-3) and is perpendicular to y = x+6. m = ___ and the point is (__,__) MAP TAP Parallel and Perpendicular Lines |
## Thursday, November 12, 2015
### Determining the Determinant of a 3x3 matrix
Introduction
Determinant is a unique value representing a square matrix. It is useful in determining the inverse of a matrix. The determinant of a 1x1 matrix is the entry in the matrix itself whereas in the case of a 2x2 matrix, it is the upper left into lower right minus the upper right into the lower left. The determinant thus obtained is called the minor of the 2x2 matrix. Conventionally, the determinant of a higher order matrix is calculated by a recursive method with the help of the minors. Let us now see a method of rotations to find the determinant.
Let us consider the 2x2 matrix given on the left. Its determinant as per the rule given above is AD-BC.
However, look at the diagram given right. It can be explained as follows.
Rotation Method
Rotate left the second row to one position and then multiply the entries in the respective columns. Subtract the second column product from that of the first.
This process can be extended into higher orders also. This is done by carefully completing all the rotations depending on the number of rows. Let us see the case of a 3x3 matrix.
At first, rotate left the second row to one position and third row to two positions. The matrix thus obtained is given on the right.
Multiply the values in each column separately and then add the three products.
Thus we get A= (A1.B2.C3)+(A2.B3.C1)+(A3.B1.C2).
Now we rotate left the second row to one more position and the third row to two more positions. The matrix thus obtained is as follows.
Multiply the values in each column separately and then add the three products. Thus we get B=(A1.B3.C2)+(A2.B1.C3)+(A3.B2.C1).
The determinant of the 3x3 matrix is then
A-B = (A1.B2.C3+A2.B3.C1+A3.B1.C2)-( A1.B3.C2+A2.B1.C3+A3.B2.C1).
Cylindrical Rotation Method
This process is better understood if we can represent the matrix cylindrically. Consider a cylinder that has three horizontal sections which can be rotated freely with respect to a central vertical axis. Entries in the matrix are given on the exterior of the cylinder.
After the first set of two types of left rotations, the cylinder looks like what is on the left.
Multiplying the entries in the columns separately we get
A=(A1.B2.C3)+(A2.B3.C1)+(A3.B1.C2).
After the second set of two types of left rotations, the cylinder looks like what is on the right.
Multiplying the entries in the columns separately we get
B = (A1.B3.C2)+(A2.B1.C3)+(A3.B2.C1).
The determinant of the 3x3 matrix is then
A-B = (A1.B2.C3+A2.B3.C1+A3.B1.C2)-( A1.B3.C2+A2.B1.C3+A3.B2.C1).
Palm Method
This can be visualized in yet another way also.
Let us use the three central fingers on the left palm to represent a 3x3 matrix.
Instead of the first set of rotations, multiply the entries from left as indicated by the dark lines, starting from the diagonal to get the value A.
To get the value B, we can multiply the entries from right starting from the antidiagonal. This can be done also by turning the palm and then multiplying the entries from left as given below from the diagonal.
Conclusion
An advantage of the above mentioned process is the elimination of the repeated use of plus(+) and minus(–) which is sometimes disturbing for beginners and non-Mathematics students. Can this method be extended to higher order matrices?
### Develop a skeleton for each day
One has to develop one's own habit of writing. One important characteristic of habit is that one may not realize what one does in a hab... |
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This question tests the student's ability to identify the factors of some composite numbers and the highest common factors of two numbers.
#### a)
\n
i)
\n
$\\var{fourfac}$ has four factors: $1$, $3$, $\\var{fourfac/3}$ and $\\var{fourfac}$.
\n
It is possible to pair the factors up to prove that they are factors.
\n
\\\begin{align} 1\\times\\var{fourfac}&=\\var{fourfac}\\text{.}\\\\ 3\\times\\var{fourfac/3}&=\\var{fourfac}\\text{.}\\\\ \\end{align} \
\n
ii)
\n
$\\var{sixfac}$ has six factors: $1$, $2$, $3$, $\\var{sixfac/3}$, $\\var{sixfac/2}$ and $\\var{sixfac}$.
\n
Again, it is possible to pair the factors up to prove that they are factors.
\n
\\\begin{align} 1\\times\\var{sixfac}&=\\var{sixfac}\\text{.}\\\\ 2\\times\\var{sixfac/2}&=\\var{sixfac}\\text{.}\\\\ 3\\times\\var{sixfac/3}&=\\var{sixfac}\\text{.}\\\\ \\end{align} \
\n
\n\n
#### b)
\n
We now look for common factors between the two lists of factors, and the highest common factor will be the largest of these.
\n
\n
For $\\var{fourfac}$ and $\\var{sixfac}$, the highest common factor is $\\var{hc}$.
\n
#### c)
\n
Dividing both the numerator and denominator by the highest common factor gives:
\n
\$\\frac{\\var{sixfac}}{\\var{fourfac}} = \\frac{\\frac{\\var{sixfac}}{\\var{hc}}}{\\frac{\\var{fourfac}}{\\var{hc}}} = \\frac{\\var{sixfac/hc}}{\\var{fourfac/hc}}\\text{.}\$
\n
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Simplify $\\var{sixfac}/\\var{fourfac}$ by finding the highest common factor.
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Identify the factors of the following numbers in ascending order.
\n
i) Factors of $\\var{fourfac}$: $1$, [[4]], [[5]], $\\var{fourfac}$
\n
\n
ii) Factors of $\\var{sixfac}$: [[0]], $2$, [[1]], [[2]], $\\var{sixfac/2}$, [[3]]
What is the heighest common factor of $\\var{fourfac}$ and $\\var{sixfac}$?
\n
The heighest common factor is [[0]]
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If the same number appears in a list of factors for two numbers, it is a common factor. The largest of these common factors is the highest common factor.
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Use the result above to reduce the following fraction to its simplest form.
\n
$\\displaystyle \\frac{\\var{sixfac}}{\\var{fourfac}} =$ [[0]] [[1]]
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To simplify the fraction, divide both the numerator and denominator by the highest common factor.
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## Math Expressions Common Core Grade 3 Unit 1 Lesson 5 Answer Key Multiply and Divide with 2
Math Expressions Grade 3 Unit 1 Lesson 5 Homework
Use this chart to practice your 2s count-bys, multiplications, and divisions. Then have your Homework Helper test you.
Multiply and Divide with 2 Grade 5 Answer Key Math Expressions Question 1.
2 × 4 = ___
By multiplying 2 with 4 we get 8
Multiply and Divide with 2 Grade 3 Answers Key Math Expressions Question 2.
20 ÷ 5 = ___
By dividing 20 with 5 we get 4
Lesson 5 Answer Key Multiply and Divide with 2 Grade 3 Unit 1 Question 3.
6 * 2 = ___
By multiplying 6 with 2 we get 12
Multiply and Divide with 2 Grade 5 Answer Key Unit 1 Question 4.
45/5 = ___
By dividing 45 with 5 we get 9
Multiply and Divide with 2 Answer Key Grade 3 Unit 1 Math Expressions Question 5.
2 • 10 = ___
By multiplying 2 with 10 we get 20
Unit 1 Lesson 5 Math Expressions Grade 3 Question 6.
$$\frac{20}{2}$$ = ___
By dividing 20 with 2 we get 10
Lesson 5 Multiply Math Expressions Grade 3 Unit 1 Question 7.
5 × 10 = ___
by multiplying 5 with 10 we get 50.
Multiply and Divide with 2 Lesson 5 Answer Key Math Expressions Question 8.
16 ÷ 2 = ___
By dividing 16 with 2 we get 8.
Math Expressions Grade 3 Answer Key Unit 1 Lesson 5 Question 9.
6 × 5 = ___
By multiplying 6 with 5 we get 30.
Question 10.
30/5 = ___
By dividing 30 with 5 we get 6.
Question 11.
5 • 7 = ___
By multiplying 5 with 7 we get 35
Question 12.
By dividing 18 with 2 we get 9
Question 13.
8 * 2 = ___
By multiplying 8 with 2 we get 16.
Question 14.
$$\frac{25}{5}$$ = ___
By dividing 25 with 5 we get 5.
Question 15.
5 • 4 = ___
By multiplying 5 with 4 we get 20.
Question 16.
16 / 2 = ___
By dividing 16 with 2 we get 8.
Question 17.
By dividing 10 with 2 we get 5.
Question 18.
2 * 7 = ___
By multiplying 2 with 7 we get 14.
Question 19.
5 × 5 = ___
By multiplying 5 with 5 we get 25.
Question 20.
14 ÷ 2 = ___
By dividing 14 with 2 we get 7.
Question 21.
By dividing 35 with 5 we get 7
Write an equation and solve the problem.
Question 1.
Tanya had 14 cups to fill with juice. She put them in 2 equal rows. How many cups were in each row?
Explanation:
Given
Tanya had 14 cups to fill with juice. She put them in 2 equal rows.
To find how many cups in each row
we have to divide 14 with 2 we get 7 cups in each row.
Question 2.
Rebecca has 3 pairs of running shoes. She bought new shoelaces for each pair. How many shoelaces did she buy?
Given
Rebecca has 3 pairs of running shoes. She bought new shoelaces for each pair.
each pair had two shoe lays
by multiplying 3 with 2 we get 6 shoelaces
Question 3.
Jason served his family dinner. He put 5 carrots on each of the 4 plates. How many carrots did Jason serve in all?
Given
Jason served his family dinner. He put 5 carrots on each of the 4 plates.
by multiplying 5 carrots with 4 plates we get 20 carrots.
Question 4.
Olivia filled 8 vases with flowers. She put 5 flowers in each vase. How many flowers did she put in the vases?
Explanation:
Given
Olivia filled 8 vases with flowers. She put 5 flowers in each vase
To find How many flowers did she put in the vases
we have to multiple 8 with 5 we get 40.
Question 5.
Devon has 30 model airplanes. He put the same number on each of the 5 shelves of his bookcase. How many model airplanes did Devon put on each shelf?
Explanation:
Given
Devon has 30 model airplanes. He put the same number on each of the 5 shelves of his bookcase
To find How many model airplanes did Devon put on each shelf
we have to divide 30 with 5 we get 6 airplanes in each shelf.
Question 6.
There are 12 eggs in a carton. They are arranged in 2 rows with the same number of eggs in each row. How many eggs are in each row?
Explanation:
Given
There are 12 eggs in a carton. They are arranged in 2 rows with the same number of eggs in each row.
To find How many eggs are in each row
by dividing 12 with 2 we get 6 eggs in each row.
Math Expressions Grade 3 Unit 1 Lesson 5 Remembering
Make a math drawing for the problem and label it with a multiplication equation. Then write the answer to the problem.
Question 1.
Kishore has 4 stacks with 3 books in each stack. How many books are there in all?
Explanation:
Given
Kishore has 4 stacks with 3 books in each stack
To find How many books are there in all
we have to multiple 4 with 3 we get 12 books
Question 2.
Cindy had 6 envelopes. She put 2 stamps on each one. How many stamps did she use?
Given,
Cindy had 6 envelopes. She put 2 stamps on each one.
6 × 2 = 12
Write a multiplication equation for the array.
Question 3.
How many dots?
By observing picture we get
there are 6 rows and 3 columns
there fore by multiplying 6 with 3 we get 18 dots
Multiply or divide to find the unknown numbers.
Question 4.
7 * 5 = ___
By multiplying 7 with 5 we get 35.
Question 5.
45 ÷ 5 = ___
By dividing 45 with 5 we get 9.
Question 6.
__ × 5 = 50
By multiplying 10 with 5 we get 50
Question 7.
8 * 5 = ___
By multiplying 8 with 5 we get 40.
Question 8.
5 • __ = 25
By multiplying 5 with 5 we get 25.
Question 9.
$$\frac{10}{5}$$ = ___ |
$$\require{cancel}$$
# 02. Calculating the Rotational Inertia
[ "article:topic", "authorname:dalessandrisp" ]
### Calculating the Rotational Inertia
To fully utilize Newton's second law in rotational form, we must be able to set up and evaluate the integral that determines the rotational inertia. (To be honest, this is a lie. For the vast majority of common shapes, and many quite uncommon shapes, these integrals have already been evaluated. A table of selected results is at the end of this section.) To test our understanding of the relationship for rotational inertia for a thin disk about an axis passing through its center of mass and perpendicular to its circular face.
pic
1. Choose the chunks of mass, dm, to be ring-shaped. This is because you must multiply each dm by the distance of the chunk from the rotation axis squared. To facilitate doing this, it's crucial that every point in the chunk be the same distance from the axis, i.e., have the same r. if the ring-shaped chunk is thin enough, for example dr (infinitesimally) thick, then this is true.
2. Realize that the mass of the little chunk is directly proportional to its volume, assuming the disk has a constant density. If it does have a constant density, the ratio of the chunk's mass to its volume must be the same as the ratio of the total mass of the disk to its volume.
pic
where R is the radius of the disk and T is its thickness. The volume of the ring-shaped chunk, dV, is equal to the product of the circumference of the ring (2pr), the thickness of the disk (T), and the thickness of the ring (dr). Thus,
pic
3. Plug the expression for dm into
pic
To include all the chunks of mass, the integral must go from r = 0 m up r = R.
pic
Thus, the rotational inertia of a thin disk about an axis through its CM is the product of one-half the total mass of the disk and the square of its radius. Notice that the thickness of the disk does not effect its rotational inertia. A consequence of this fact is that a cyclinder has the same rotational inertia as a disk, when rotated about an axis through its CM and perpendicular to its circular face.
### The Parallel-Axis Theorem
Now let's imagine we need to calculate the rotational inertia of a thin disk about an axis perpendicular to its circular face and along the edge of the disk. It would be convenient if we could determine the rotational inertia about an axis along the edge using the rotational inertia about an axis through the CM (which we've already calculated). In fact, there is a very convenient method to determine the rotational inertia about any axis parallel to an axis through the CM if we know the rotational inertia about an axis through the CM.
Imagine you want to determine the rotational inertia of an arbitrarily shaped object about an arbitrary axis. The solid circle denotes an axis of interest. The two axes are parallel.
Notice that
pic
with
pic
and
pic
Thus,
pic
The rotational inertia about the axis of interest is given by:
pic
Note that xCM, yCM, and rCM are constants that depend only on the distance between the two axes. Thus, xCM, yCM, and rCM can by brought outside of the integral.
pic
Now comes the key observation in the derivation. Examine the term x dm. Remember that x is the horizontal distance from the CM. If this distance from the CM. If this distance is integrated over all the chunks of mass, dm, throughtout the entire object, this integral must equal zero because the CM is defined to be in exactly the spot where a mass-weighted average over distance is equal to zero. x dm and y dm are equal to zero by the definition of CM! (Pretty cool, huh?)
Thus,
pic
Noting that dm is the total mass of the object, M, and r2 dm is the rotational inertia about the CM, ICM, then
pic
This result states that the rotational inertia about an axis parallel to an axis through the CM, I, is equal to the rotational inertia about an axis through the CM, ICM, plus the product of the total mass of the object and the distance between the axes, rCM, squared.
To answer the original question, let's determine the rotational inertia of a thin disk about an axis perpendicular to its circular face and along the egde of the disk using the parallel-axis theorem.
We know that the rotational inertia for a thin disk about an axis passing through its center of mass and perpendicular to its circular face is 1/2 MR2 and the axis of interest is R. Thus,
pic |
## Basic Algebra
### Introduction
The letter 'x' is a very common symbol used in algebra to represent an unknown quantity. For this reason it is wise to avoid use of the similar multiplication sign. Usually the multiplication sign is 'understood', e.g
5x
actually represents
5 × x
in the 'old' representation.
If it is really necessary to use a specific sign to denote multiplication, then the use of the 'full-stop' or brackets is to be recommended, i.e. instead of
5 × 3
use
5 . 3
(but this is only where no confusion would arise with a decimal point!)
or
(5)(3)
Question
(a) A cup of tea costs 80 pence.
Write down an expression, in terms of x, for the cost, in pence, of x cups of tea.
(b)A cup of coffee costs 95 pence.
Write down an expression, in terms of y, for the cost, in pence, of y cups of coffee.
. (c) Write down an expression, in terms of x and y, for the total cost, in pence, of x cups of tea and y cups of coffee.
(a) 80x
(b) 95y
(c) 80x + 95y
### Substitution
Proceed by example
Example What is the value of
### 3x - 4y
when x=3 and y=2?
### = 1
Example What is the value of
when p=2 and q = -7? |
# EXPLORING THE CONCEPT OF SLOPE
Document Sample
``` EXPLORING THE CONCEPT OF SLOPE
Tools for Teaching Algebra for All Workshop
OBJECTIVE
The student understands the meaning of the slope and intercepts of linear functions and interprets
and describes the effects of changes in parameters of linear functions in real-world and
mathematical situations.
SET-UP
Students should work in groups of 3-4.
MATERIALS
Activity sheets, 2 meter sticks for each for each group, l level for demonstration, graphing
calculators
PREREQUISITES
Use of meter stick to measure, use of graphing calculator to graph lines, conversions of ratios to
equivalent decimal and percent forms
PROCEDURE
Demonstrate how to measure the steepness of objects using meter sticks and levels. The first
demonstration is clearer if the horizontal measurement of the object is 100 cm. The ratio,
decimal equivalent, and percent will be dealing with parts of 100. The second demonstration
should not have horizontal measurement of 100 cm. Calculator use should be encouraged.
Example 1.
Vertical Horizontal Ratio Decimal Percent
8 100 8/100 0.08 8%
Example 2.
Vertical Horizontal Ratio Decimal Percent
30 75 30/75 0.40 40%
ACTIVITY I.
Using meter sticks, have students measure and record the vertical and horizontal distances of at
least three geographic areas with objects that have steepness, i.e., steps, handrails, sidewalks,
the students to sketch diagrams of the objects they measured and to show the placement of the
meter stick. If students are measuring sidewalks or ramps, the best procedure is to find the
vertical distance from the ground at a point where the horizontal measurement is 100 cm as
shown in Example 1.
After the measurements for each object are recorded, the participants are to complete the table.
Questions for Discussion
Have groups compare ratios for the same object (ratios will most likely vary). Are the
ratios approximately the same for each object?
What is the steepness of the roof of the school (if the roof is flat) or what is the steepness
of a flat roof? (Ans: zero)
What is the steepness of a wall in the classroom? (Ans: undefined slope because the
horizontal distance is zero)
Predict the steepness of the roof of your home? How would you determine the steepness
of the roof of your home?
What conclusions can be drawn about the ratio of the vertical distance of an object to its
horizontal distance?
Which object is more steep—one in which the vertical measure is greater than the
horizontal measure or one in which the vertical measure is less than the horizontal
measure? (Ans: one in which the vertical measure is greater than the horizontal
measure)
ACTIVITY II.
Have students sketch stair steps with the given steepness. Allow time for students to compare
and contrast the sketches with members in their group.
2
Questions for Discussion
Are all the sketches exactly alike?
What is different about some of the sketches?
Do some of the stairs seem to be going up whereas others are going down?
What happens to the stairs when the numerator of the steepness is less than the
denominator?
What happens to the stairs when the numerator of the steepness is greater than the
denominator?
What happens to the stairs as the values of the numerator and denominator get farther and
farther apart?
What happens to the stairs as the values of the numerator and denominator approach the
same number?
What happens to the stairs when the numerator and denominator are the same number?
ACTIVITY III.
This third activity (Identifying Steepness or Slope) is a summarizing activity at the
pictorial/graphical representational level. The activity provides static pictorial/graphical
representations of the concept and requires the students to reverse their thinking process. Rather
than being given the steepness and asked to sketch the corresponding stairs, students are given
sketches of stairs and asked to determine the corresponding steepness represented. Students
must be able to approach a concept from either direction before they reach understanding.
ACTIVITY IV.
Allow students to explore functions in the form y = mx. Ask students to summarize their
findings for each set of linear functions. Emphasize that this activity begins to connect the
concept of slope at the algebraic representational level with the graphical level.
1. As the slope increases from 1 upward, the line becomes steeper.
2. As the slope decreases from 1 toward 0, the line becomes less steep.
3. As the slope increases from 0 upward, the line becomes steeper.
4. A negative slope reflects the line across the y-axis. Its steepness remains the same.
5. As the | m | > 1, the line becomes steeper. The lines with positive slope travel upward to
the right; the lines with negative slope travel upward to the left.
6. As | m | > 0, the line becomes steeper. The lines with positive slope travel upward to the
right; the lines with negative slope travel upward to the left.
3
ACTIVITY I. MEASURING STEEPNESS
DIRECTIONS: Measure the vertical and horizontal distances of the objects
space provided. Then, write the measurements as a ratio (vertical
measure/horizontal measure), an equivalent decimal rounded to two places,
and a percent.
4
ACTIVITY II. EXPLORING STEEPNESS
Make a sketch of stair steps with the given steepness. The steepness (ratio) is
vertical measure compared to horizontal measure.
5
ACTIVITY III. IDENTIFYING STEEPNESS OR SLOPE
Match each stair step diagram with the ratio of measures (vertical/horizontal)
that best describes the diagram.
(a) 3/3 (b) 3/5 (c) 5/5 (d) 5/3 (e) 1/3 (f) 3/1
6
ACTIVITY IV. EXPLORING THE CONCEPT OF SLOPE
WITH THE GRAPHING CALCULATOR
Select a standard viewing window on your calculator: (-10, 10, 1, -10, 10, 1).
Graph each of the following sets of linear functions and look for patterns.
Summarize your findings beside each set of linear functions.
Summary:
Summary:
7
Summary:
Summary:
Summary:
Summary:
8
```
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# What is calculus 1?
Wiki User
2012-01-14 18:15:30
Traditionally, and in my learning experiences, calculus is taught in three stages, often referred to as Calculus I, Calculus II, and Calculus III (often shortened to Calc I, Calc II, Calc III). You are asking about Calculus I only, but it is easy to explain all three.
Calc I usually covers only derivative calculus, Calc II covers integral calculus and infinite series, and Calc III covers both derivative and integral calculus, but in multiple variables instead of only one independent variable ( xyz = x+y+z as opposed to y = x). This is a traditional collegiate leveling of calculus. This is often changed around in secondary education (in the United States at least). Programs such as AP Calculus often change around this order. AP Calculus AB covers Calc I and introduces Calc II, while AP Calculus BC covers the remainder of Calc II.
Now that you know the subject matter, what does it mean? Derivative calculus is a generalized category meant to encompass the computation and application of only derivatives, which are basically rates of change of a mathematical function. A basic mathematical function such as y = x + 2 describes a mathematical relationship: for every additional independent variable "x", a dependent variable "y" will have a value of (x + 2). But, how do you describe how quickly the value of "y" changes for each additional "x"? This is where derivatives come from. The derivative of the function y = x + 2, as you would learn in Calc I, is y' = 1. This means that y changes at a constant rate (called y') of "1" for each additional x. In more familiar terms, this is the slope of this function's graph.
However, not all functions have constant slopes. What about a parabola, or any other "curvy" graph? The "slopes" of these graphs would be different for any given value of a dependent variable "x". A function such as y = x2 + 2 would have a derivative, as you would learn in Calc I, of y' = 2x, meaning that the original value of "y" will change at a rate of two times the value of "x" (2x), for each additional increment of "x".
You can continue into further derivatives, called second, third, fourth (and so on) derivatives, which are derivatives of derivatives. This is essentially asking "At what rate does a derivative change?".
The beginning of Calc I is concerned with introducing what a derivative is, ways to describe the behavior of mathematical functions, and how to compute derivatives. After this introduction is complete, you will begin to apply derivatives to mathematical problems. The description of how derivatives are used to solve these problems is not worth going into, because it would be better for you to connect derivatives to their applications on your own, but you can use derivatives to answer such questions as:
What is the maximum/minimum value of a mathematical function on a given interval or on its entire domain?
This kind of knowledge can be applied like so: Suppose a mathematical function is found that describes the volume of a box. Knowing that you can use the derivative of this function to find its maximum value, you can then find what value of a certain variable will yield the maximum volume of the box.
Another type of application is called a "related rates" problem, in which a known mathematical relationship is used with some given information to describe another property. A question of this type could be: Suppose you have a cylindrical tank of water with a small hole in the bottom, and you measure that the water is flowing out at 2 gallons per minute. At what rate is the height of the water in the tank changing? (This is a simple related rates problem).
A full description of integral calculus (Calc II and a basis of Calc III), would take far too long to explain, and it would be easier to explain once you have taken Calc I. Calc III takes the same idea as Calc I and Calc II, but instead of one independent variable "x" changing one dependent variable "y", there are several variables, although in most applications you will only see three, "x", "y", and "z", although the ideas you will learn in the class will apply to potentially infinite variables. The basic ideas of derivatives and integrals will hold here, but the mathematical methods needed and applications possible with multiple variables require additional learning.
Wiki User
2012-01-14 18:15:30
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Other Types of Graphs
As well as straight line coordinate graphs there are many functions whose graphs have different shapes.
These functions will become more important in future years.
Below are some examples of the different types of functions and their graphs are sketched by plotting points.
Parabolas
A parabola is a curved graph produced by a quadratic function, one which contains a "squared" x-term.
y = x2
x y Coordinates -3 (-3)2 = 9 (-3, 9) -2 (-2)2 = 4 (-2, 4) -1 (-1)2 =1 (-1, 1) 0 (0)2 = 0 (0, 0) 1 (1)2 = 1 (1, 1) 2 (2)2 = 4 (2, 4) 3 (3)2 = 9 (3, 9)
This shape is called a parabola.
Hyperbolas
The equation of a hyperbola contains an x-term and a y-term multiplied together. The general equation is xy = c, where c is a constant.
xy = 6
x y Coordinates -3 -2 (-3, 2) -2 -3 (-2, -3) -1 -6 (-1, -6) 0 no number possible undefined 1 6 (1, 6) 2 3 (2, 3) 3 2 (3, 2)
x multiplied by y has to always be 6.
The lines x = 0 and y = 0 on the graph are called asymptotes.
They are not crossed by the curves.
This shape is called a hyperbola.
Growth Curves
The equation of a growth curve has a power of x. The general equation is y = ax, where a is a constant.
y = 2x
x y Coordinates -3 0.125 (-3, 0.125) -2 0.25 (-2, 0.25) -1 0.5 (-1, 0.5) 0 1 (0, 1) 1 2 (1, 2) 2 4 (2, 4) 3 8 (3, 8)
Check these results using the button on a calculator.
The x-axis is an asymptote.
This shape is called a growth curve or an exponential function.
Circles
The equation of a circle contains an x2 and a y2 term. The general equation is x2 + y2 = r2, where r is the radius.
x2 + y2 = 4
x x2 y2 y Coordinates -2 4 0 0 (-2, 0) -1 1 3 -√3 or √3 (-1, √3) and (-1, √3) 0 0 4 2 or -2 (0, -2) and (0, 2) 1 1 3 -√3 or √3 (1, √3) and (1,√3) 2 4 0 0 (2, 0)
The middle terms above, in red, have to always add to 4.
This shape is called a circle. |
1 / 6
# Math III - PowerPoint PPT Presentation
Math III. October 22 nd , 2012. Zeroes and their Multiplicity. Multiplicity refers to the number of times a factor appears in a polynomial. Example 1: (x - 3)(x + 2). Zeroes : each occur only once in the polynomial, so they have a multiplicity of __________.
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### Math III
October 22nd, 2012
• Multiplicity refers to the number of times a factor appears in a polynomial.
• Zeroes:
• each occur only once in the polynomial, so they have a multiplicity of __________
Example 2: (x – 3)4(x + 2)7
• Zeroes: __________________
• (x – 3) occurs ________________
• (x + 2) occurs ________________
• So, the root x = _________ has a multiplicity of _______________
and x = __________ has a multiplicity of ________________
• If the multiplicity is:
• Even graph DOES NOT cross the x-axis
• Odd graph DOES cross the x-axis
Example 3: The following graph shows a fifth-degree polynomial. List the polynomial’s zeroes with their multiplicities.
Zeroes:
Crosses the x-axis @
Touches the x-axis @
If you add their minimum multiplicities, 1+1+1+1+1 |
# NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
(Last Updated On: March 29, 2023)
Arithmetic Progressions Introduction : NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression (AP) is a sequence of numbers in which each term is obtained by adding a fixed number to the preceding term, except for the first term. This fixed number is called the common difference (d) of the AP. The first term of an AP is denoted by ‘a’ and the nth term by ‘an’. The general formula for nth term of an AP is given as:
an = a + (n-1)d
Where, an = nth term of the AP a = first term of the AP d = common difference of the AP n = the number of terms in the AP
For example, consider the sequence 2, 5, 8, 11, 14, …. This is an arithmetic progression with first term (a) = 2 and common difference (d) = 3. To find the nth term of this AP, we can use the formula:
an = a + (n-1)d
So, the 10th term (n = 10) of this AP would be:
a10 = 2 + (10-1)3 = 2 + 27 = 29
APs are used in a variety of mathematical and real-life applications, such as in calculating the interest rate of a loan, in finding the distance covered by a moving object, in designing musical scales, and in many other fields.
## NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
Arithmetic Progressions Class 10 has total of 4 exercises consists of 49 Problems. find the nth terms and the sum of n consecutive terms are important topics in this chapter 5.
Topics and Sub Topics in NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions:
Section Name Topic Name 5 Arithmetic Progressions 5.1 Introduction 5.2 Arithmetic Progressions 5.3 Nth Term Of An AP 5.4 Sum Of First N Terms Of An AP 5.5 Summary
## Arithmetic Progressions formulas
Here are some of the important formulas related to Arithmetic Progressions (AP):
1. nth term of an AP: The nth term (an) of an AP with first term (a) and common difference (d) can be found using the formula:
an = a + (n-1)d
1. Sum of n terms of an AP: The sum (Sn) of the first n terms of an AP with first term (a) and common difference (d) can be found using the formula:
Sn = n/2[2a + (n-1)d]
1. Sum of first n natural numbers: The sum of the first n natural numbers is given by the formula:
1 + 2 + 3 + … + n = n(n+1)/2
1. Sum of squares of first n natural numbers: The sum of the squares of the first n natural numbers is given by the formula:
1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6
1. Number of terms in an AP: The number of terms (n) in an AP with first term (a), last term (l) and common difference (d) can be found using the formula:
n = (l-a)/d + 1
These formulas can be used to solve problems related to APs and find the values of various parameters.
## What is Arithmetic Progressions
Arithmetic Progressions (AP) is a sequence of numbers in which each term is obtained by adding a fixed number (called the common difference) to the preceding term, except for the first term. In other words, an arithmetic progression is a sequence of numbers in which each term after the first is obtained by adding a constant value to the previous term.
For example, the sequence 2, 5, 8, 11, 14, … is an arithmetic progression with a first term of 2 and a common difference of 3. Each term in this sequence is obtained by adding 3 to the previous term.
APs are used in a variety of mathematical and real-life applications, such as in calculating the interest rate of a loan, in finding the distance covered by a moving object, in designing musical scales, and in many other fields. The study of APs is an important part of mathematics and is often included in school curricula.
## NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.1 is part of the NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.1.
Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter 5 Chapter Name Arithmetic Progressions Exercise Ex 5.1 Number of Questions Solved 4 Category NCERT Solutions
## Use of Arithmetic Progressions
Arithmetic Progressions (AP) have a wide range of applications in mathematics and real-life situations. Some of the important uses of APs are:
1. Financial calculations: APs are used to calculate the interest rates on loans, investments, and mortgages. For example, if the interest rate on a loan is an AP with a known common difference, then the total interest paid can be easily calculated.
2. Distance and time calculations: APs are used to calculate the distance and time traveled by a moving object with uniform acceleration or deceleration. For example, the distance traveled by a car in a given time interval can be calculated using an AP formula.
3. Music theory: APs are used in the construction of musical scales and chords. The frequencies of musical notes are arranged in an AP, with the common difference being a constant multiple of 2^(1/12).
4. Statistics: APs are used in statistics to analyze and organize data. For example, the ages of a group of people can be arranged in an AP to study the distribution of ages.
5. Mathematics: APs are used in various mathematical fields such as algebra, geometry, and calculus. They are used to model many real-world situations and to solve complex problems.
Thus, Arithmetic Progressions have a wide range of uses in various fields and are an important tool in mathematics and science.
## Methods of Arithmetic Progressions
There are several methods for working with Arithmetic Progressions (AP), including:
1. Using the nth term formula: The nth term formula is a general formula used to find any term of an AP. It is given by:
an = a + (n-1)d
where a is the first term of the AP, d is the common difference, and n is the term number.
1. Using the sum of n terms formula: The sum of n terms formula is used to find the sum of the first n terms of an AP. It is given by:
Sn = n/2 [2a + (n-1)d]
where a is the first term of the AP, d is the common difference, and n is the number of terms.
1. Using the middle term formula: The middle term formula is used to find the middle term of an AP. It is given by:
am = a + (n/2-1)d
where a is the first term of the AP, d is the common difference, n is the number of terms, and am is the middle term.
1. Using the common difference formula: The common difference formula is used to find the common difference of an AP when two terms and their positions are given. It is given by:
d = (an – am)/(n – m)
where a is the first term of the AP, d is the common difference, n is the position of the second term, m is the position of the first term, an is the nth term of the AP, and am is the mth term of the AP.
1. Using the sum of two APs formula: The sum of two APs formula is used to find the sum of the terms of two APs with the same number of terms and common difference. It is given by:
S = n/2 [(a1 + b1) + (an + bn)]
where a1 and b1 are the first terms of the two APs, an and bn are the last terms of the two APs, n is the number of terms, and S is the sum of the terms.
These methods can be used to solve problems related to APs and find the values of various parameters.
## NCERT Solutions for Class 10 Maths Chapter 5 Ex 5.1 Arithmetic Progressions Questions
In which of the following situations, does the list of numbers involved make an arithmetic progression and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 14 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every meter of digging, when it costs ₹ 150 for the first meter and rises by ₹ 50 for each subsequent meter.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum. Solution:
(i) The list of numbers involved does not make an arithmetic progression. The fare increases by ₹ 8 after each additional km, but this increase is not a fixed constant. In an arithmetic progression, the difference between consecutive terms should be a constant, but here the difference between consecutive terms varies.
(ii) The list of numbers involved makes an arithmetic progression. When a vacuum pump removes 1/14 of the remaining air, it leaves 13/14 of the original air in the cylinder. So, the amount of air remaining after each cycle is obtained by multiplying the previous amount of air by 13/14. Thus, the list of numbers involved forms a geometric progression, and its logarithm will form an arithmetic progression.
(iii) The list of numbers involved makes an arithmetic progression. The cost of digging the well increases by ₹ 50 for each subsequent meter, which is a fixed constant. So, the difference between consecutive terms is constant, and hence, the list of numbers forms an arithmetic progression.
(iv) The list of numbers involved does not make an arithmetic progression. The amount of money in the account after each year is not obtained by adding a fixed constant to the previous amount. It is obtained by compounding the interest on the previous amount, which leads to an exponential growth rather than a linear one. So, the list of numbers involved forms a geometric progression.
### Ex 5.1 Class 10 Maths Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = 12
(v) a = -1.25, d = -0.25
We can use the formula for nth term of an arithmetic progression to find the first four terms of the given APs:
(i) a = 10, d = 10 The nth term formula is: an = a + (n – 1)d Using this formula, we get:
• first term, a1 = 10
• second term, a2 = 10 + (2 – 1)10 = 20
• third term, a3 = 10 + (3 – 1)10 = 30
• fourth term, a4 = 10 + (4 – 1)10 = 40
So, the first four terms of this AP are 10, 20, 30, 40.
(ii) a = -2, d = 0 The nth term formula is: an = a + (n – 1)d Using this formula, we get:
• first term, a1 = -2
• second term, a2 = -2 + (2 – 1)0 = -2
• third term, a3 = -2 + (3 – 1)0 = -2
• fourth term, a4 = -2 + (4 – 1)0 = -2
So, the first four terms of this AP are -2, -2, -2, -2.
(iii) a = 4, d = -3 The nth term formula is: an = a + (n – 1)d Using this formula, we get:
• first term, a1 = 4
• second term, a2 = 4 + (2 – 1)(-3) = 1
• third term, a3 = 4 + (3 – 1)(-3) = -2
• fourth term, a4 = 4 + (4 – 1)(-3) = -5
So, the first four terms of this AP are 4, 1, -2, -5.
(iv) a = -1, d = 12 The nth term formula is: an = a + (n – 1)d Using this formula, we get:
• first term, a1 = -1
• second term, a2 = -1 + (2 – 1)12 = 11
• third term, a3 = -1 + (3 – 1)12 = 23
• fourth term, a4 = -1 + (4 – 1)12 = 35
So, the first four terms of this AP are -1, 11, 23, 35.
(v) a = -1.25, d = -0.25 The nth term formula is: an = a + (n – 1)d Using this formula, we get:
• first term, a1 = -1.25
• second term, a2 = -1.25 + (2 – 1)(-0.25) = -1.5
• third term, a3 = -1.25 + (3 – 1)(-0.25) = -1.75
• fourth term, a4 = -1.25 + (4 – 1)(-0.25) = -2
So, the first four terms of this AP are -1.25, -1.5, -1.75, -2.
### Ex 5.1 Class 10 Maths Question 3.
For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3, ……
(ii) -5, -1, 3, 7, ……
(iii) 13 , 53 , 93133 , ……..
(iv) 0.6, 1.7, 2.8, 3.9, …….
We can observe the given sequence of numbers to find the common difference and the first term of the corresponding arithmetic progressions (APs).
(i) 3, 1, -1, -3, …… The common difference is d = -2 (subtract 2 from each term to get the next term). The first term is a1 = 3.
(ii) -5, -1, 3, 7, …… The common difference is d = 4 (add 4 to each term to get the next term). The first term is a1 = -5.
(iii) 13 , 53 , 93, 133 , …….. The common difference is d = 40 (add 40 to each term to get the next term). The first term is a1 = 13.
(iv) 0.6, 1.7, 2.8, 3.9, ……. The common difference is d = 1.1 (add 1.1 to each term to get the next term). The first term is a1 = 0.6.
### Ex 5.1 Class 10 Maths Question 4.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, …….
(ii) 2, 52 , 3, 72 , …….
(iii) -1.2, -3.2, -5.2, -7.2, ……
(iv) -10, -6, -2,2, …..
(v) 3, 3 + 2–√, 3 + 22–√, 3 + 32–√, …..
(vi) 0.2, 0.22, 0.222, 0.2222, ……
(vii) 0, -4, -8, -12, …..
(viii) 12 , 12 , 12 , 12 , …….
(ix) 1, 3, 9, 27, …….
(x) a, 2a, 3a, 4a, …….
(xi) a, a2, a3, a4, …….
(xii) 2–√8–√18−−√32−−√, …..
(xiii) 3–√6–√9–√12−−√, …..
(xiv) 12, 32, 52, 72, ……
(xv) 12, 52, 72, 73, ……
(i) 2, 4, 8, 16, ……. This is an AP with a common difference of d = 4. Three more terms are 32, 64, 128.
(ii) 2, 52 , 3, 72 , ……. This is not an AP as the common difference is not constant.
(iii) -1.2, -3.2, -5.2, -7.2, …… This is an AP with a common difference of d = -2. Three more terms are -9.2, -11.2, -13.2.
(iv) -10, -6, -2, 2, ….. This is an AP with a common difference of d = 4. Three more terms are 6, 10, 14.
(v) 3, 3 + 2–√, 3 + 22–√, 3 + 32–√, ….. This is not an AP as the common difference is not constant.
(vi) 0.2, 0.22, 0.222, 0.2222, …… This is not an AP as the common difference is not constant.
(vii) 0, -4, -8, -12, ….. This is an AP with a common difference of d = -4. Three more terms are -16, -20, -24.
(viii) −12 , −12 , −12 , −12 , ……. This is an AP with a common difference of d = 0. Three more terms are -12, -12, -12.
(ix) 1, 3, 9, 27, ……. This is not an AP as the common difference is not constant.
(x) a, 2a, 3a, 4a, ……. This is an AP with a common difference of d = a. Three more terms are 5a, 6a, 7a.
(xi) a, a2, a3, a4, ……. This is not an AP as the common difference is not constant.
(xii) 2–√, 8–√, 18−−√, 32−−√, ….. This is not an AP as the common difference is not constant.
(xiii) 3–√, 6–√, 9–√, 12−−√, ….. This is an AP with a common difference of d = 3√. Three more terms are 15−√, 18−√, 21−√.
(xiv) 12, 32, 52, 72, …… This is an AP with a common difference of d = 20. Three more terms are 92, 112, 132.
(xv) 12, 52, 72, 73, …… This is not an AP as the common difference is not constant.
## Arithmetic Progression (AP)
Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is a constant, known as the common difference. This constant difference is denoted by ‘d’. The first term of the AP is denoted by ‘a’, and the nth term of the AP is denoted by ‘an’.
The general formula to find the nth term of an AP is given by:
an = a + (n-1)d
Where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the nth term.
The sum of the first ‘n’ terms of an AP can be found using the following formula:
Sn = (n/2) [2a + (n-1)d]
Where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the number of terms.
APs are used in many mathematical and real-world applications, such as in calculating interest rates, in determining the position and velocity of moving objects, and in solving various problems related to arithmetic and algebraic operations.
### Arithmetic Progression (AP) some examples
Here are some examples of Arithmetic Progressions (APs):
Example 1: 2, 4, 6, 8, 10, …
In this AP, the first term is ‘2’, and the common difference is ‘2’, as we add 2 to each term to get the next one.
Example 2: -3, -1, 1, 3, 5, …
In this AP, the first term is ‘-3’, and the common difference is ‘2’, as we add 2 to each term to get the next one.
Example 3: 5.5, 8, 10.5, 13, 15.5, …
In this AP, the first term is ‘5.5’, and the common difference is ‘2.5’, as we add 2.5 to each term to get the next one.
Example 4: -10, -5, 0, 5, 10, …
In this AP, the first term is ‘-10’, and the common difference is ‘5’, as we add 5 to each term to get the next one.
Example 5: 1/2, 1, 3/2, 2, 5/2, …
In this AP, the first term is ‘1/2’, and the common difference is ‘1/2’, as we add 1/2 to each term to get the next one.
These are just a few examples of Arithmetic Progressions. There are many more APs in mathematics and the real world.
### Formula for common Difference (d)
The formula for the common difference (d) in an arithmetic progression (AP) is:
d = (a_n – a_1) / (n – 1)
where d is the common difference, a_n is the nth term of the AP, a_1 is the first term of the AP, and n is the total number of terms in the AP.
### nth Term (or General Term) of an Arithmetic Progressions
The nth term (or general term) of an arithmetic progression (AP) can be found using the following formula:
a_n = a_1 + (n-1)d
where a_n is the nth term of the AP, a_1 is the first term of the AP, n is the number of terms, and d is the common difference.
This formula can be derived by noticing that in an AP, the difference between consecutive terms is constant. Therefore, to find the nth term, we can add the common difference (d) to the (n-1)th term. Since the first term is a_1, we can express the nth term as a_1 + (n-1)d.
### Sum of the FIRST ‘n’ Terms of an A.P.
The sum of the first n terms of an arithmetic progression (AP) can be found using the following formula:
S_n = (n/2)(a_1 + a_n)
where S_n is the sum of the first n terms, a_1 is the first term of the AP, a_n is the nth term of the AP, and n is the number of terms in the AP.
This formula can be derived by using the formula for the nth term of an AP, which is a_n = a_1 + (n-1)d. By substituting this expression for a_n in the sum of the first n terms, we get:
S_n = a_1 + (a_1 + d) + (a_1 + 2d) + … + (a_1 + (n-1)d)
S_n = n(a_1) + d(1 + 2 + … + (n-1))
The sum of the first (n-1) positive integers can be found using the formula n(n-1)/2. Therefore, we get:
S_n = n(a_1) + d(n(n-1)/2)
Simplifying this expression, we get:
S_n = (n/2)(2a_1 + (n-1)d)
which is the same as the formula given above.
### Arithmetic Mean Between Two Numbers
The arithmetic mean between two numbers is the number that is exactly midway between them.
Let the two numbers be a and b, then the arithmetic mean between them is given by:
Arithmetic Mean = (a + b)/2
For example, the arithmetic mean between 3 and 7 is:
Arithmetic Mean = (3 + 7)/2 = 5
So, the arithmetic mean between 3 and 7 is 5.
## FAQs About NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression (AP)
1. What is an arithmetic progression (AP)? An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference.
2. What is the formula for the nth term of an arithmetic progression? The formula for the nth term of an arithmetic progression is: an = a1 + (n-1)d where an is the nth term, a1 is the first term, n is the term number, and d is the common difference.
3. What is the formula for the sum of the first n terms of an arithmetic progression? The formula for the sum of the first n terms of an arithmetic progression is: Sn = (n/2)(a1 + an) where Sn is the sum of the first n terms, a1 is the first term, an is the nth term, and n is the number of terms.
4. What is the formula for the common difference of an arithmetic progression? The formula for the common difference of an arithmetic progression is: d = (an – a1)/(n-1) where d is the common difference, an is the nth term, a1 is the first term, and n is the number of terms.
5. How do you find the arithmetic mean between two numbers? The arithmetic mean between two numbers a and b is given by: Arithmetic Mean = (a + b)/2
6. How do you identify an arithmetic progression? To identify an arithmetic progression, you need to check if the difference between any two consecutive terms is constant. If the difference is the same for all pairs of consecutive terms, then the sequence is an arithmetic progression.
7. What is the difference between an arithmetic progression and a geometric progression? An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant. A geometric progression, on the other hand, is a sequence of numbers where each term is obtained by multiplying the previous term by a constant factor
Q: Are the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions free to download?
Q: Are the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions accurate?
A: Yes, the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions are accurate. They have been prepared by subject matter experts and are based on the latest CBSE syllabus.
Q: Are the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions helpful for board exams?
A: Yes, the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions are helpful for board exams. They provide a comprehensive understanding of the chapter and help students to solve problems related to Arithmetic Progressions with ease.
Q: Are the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions useful for competitive exams?
A: Yes, the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions are useful for competitive exams. They cover all the important topics related to Arithmetic Progressions and provide a strong foundation for solving problems related to AP in various competitive exams like JEE, NEET, etc.
Q: Can I rely on NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions as my only source of preparation?
A: While the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions provide a comprehensive understanding of the chapter, it is always recommended to refer to other reference books and study materials for additional practice and better preparation
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# Determine the values of $r$ for which the differential equation $y'+8y=0$ has solutions of the form $y=e^{rt}$
Determine the values of $r$ for which the differential equation $y'+8y=0$ has solutions of the form $y=e^{rt}$.
We have never done a problem like this in class. We've only done direction lines and separable differential equations.
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If you plug $y = e^{rt}$ into the given differential equation, you get $$re^{rt} + 8e^{rt} = 0,$$ or $$(r + 8)e^{rt} = 0. \tag{\ast}$$ Now $e^{rt} \neq 0$ for all $t$, so you can divide both sides of $(\ast)$ by $e^{rt}$ to get $$r + 8 = 0.$$ This tells us that $r = -8$.
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I think this is the most appropriate answer to the question. – Tunococ Jan 25 '13 at 3:37
This equation is separable:
$$\frac{dy}{dt} = 8 y \implies \frac{dy}{y} = -8 \, dt$$
Now integrate both sides:
$$\int \frac{dy}{y} = -8 t + C$$
where $C$ is a constant. The integral on the left is a natural log:
$$\log{y} = -8 t + C \implies y = e^{-8 t+C} = A e^{-8 t}$$
so $r=-8$.
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$$y'/y=(\log(y))'\implies y'/y=-8 = (\log(y))'.$$ Integrate both sides to get $$-8x + C=\log(y)\implies y(x)=e^{-8x+C}=C_0 e^{-8x}.$$ Setting $C_0=1,$ we see that $r=-8.$
You can easily turn this into a separable equation and avoid the trick $y'/y=(\log(y))'.$ Here's your equation: $$\frac{\mathrm dy}{\mathrm dx}=-8y\\\implies\int\frac{\mathrm dy}{y}=-\int 8x\,\mathrm dx$$ and I think you know what to do next: find the solution of the from $C_0 e^{kt}$ and compare it to $e^{rt}.$
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Hint: $y^\prime+8y=0$ $$\frac{y'}{y}=-8$$ Integrate both sides to get $$\ln y=-8t+c$$ $y=\exp(c)\exp(-8t)$
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# How do you find all zeros with multiplicities of f(x)=3x^3+3x^2-11x-10?
Apr 10, 2017
The zeros are $- 2$, $\frac{1}{2} + \frac{\sqrt{69}}{6}$, $\frac{1}{2} - \frac{\sqrt{69}}{6}$ all with multiplicity $1$.
#### Explanation:
Given:
$f \left(x\right) = 3 {x}^{3} + 3 {x}^{2} - 11 x - 10$
By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $- 10$ and $q$ a divisor of the coefficient $3$ of the leading term.
That means that the only possible rational zeros are:
$\pm \frac{1}{3} , \pm \frac{2}{3} , \pm 1 , \pm \frac{5}{3} , \pm 2 , \pm \frac{10}{3} , \pm 5 , \pm 10$
Trying each in turn, we eventually find:
$f \left(- 2\right) = 3 {\left(\textcolor{b l u e}{- 2}\right)}^{3} + 3 {\left(\textcolor{b l u e}{- 2}\right)}^{2} - 11 \left(\textcolor{b l u e}{- 2}\right) - 10$
$\textcolor{w h i t e}{f \left(- 2\right)} = - 24 + 12 + 22 - 10$
$\textcolor{w h i t e}{f \left(- 2\right)} = 0$
So $- 2$ is a zero and $\left(x + 2\right)$ a factor:
$3 {x}^{3} + 3 {x}^{2} - 11 x - 10 = \left(x + 2\right) \left(3 {x}^{2} - 3 x - 5\right)$
The remaining quadratic factor is in the form:
$a {x}^{2} + b x + c$
with $a = 3$, $b = - 3$ and $c = - 5$
We can find its zeros using the quadratic formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$\textcolor{w h i t e}{x} = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(3\right) \left(- 5\right)}}{2 \cdot 3}$
$\textcolor{w h i t e}{x} = \frac{3 \pm \sqrt{9 + 60}}{2 \cdot 3}$
$\textcolor{w h i t e}{x} = \frac{1}{2} \pm \frac{\sqrt{69}}{6}$ |
# How do you solve the following linear system: y = - 3x - 3, 4x-5y=18 ?
Nov 15, 2015
Solution Set$= \left(x , y\right) = \left(\frac{3}{19} , - \frac{66}{19}\right)$
#### Explanation:
Adding $3 x$ to both sides of 1st equation
$3 x + y = - 3$
Multiplying both sides of above equation by $5$
$15 x + 5 y = - 15$
Adding above equation to 2nd equation of question
$19 x = 3$
$x = \frac{3}{19}$
Subsituting value of x in 1st equation of question
$y = - 3 \left(\frac{3}{19}\right) - 3$
$y = - \frac{9}{19} - 3$
$y = \frac{- 9 - 57}{19}$
$y = - \frac{66}{19}$ |
# Equations with Indices – Corbettmaths
Hi welcome to score math video on equations in this video, we’re gonna, look at how to solve equations which involve indices or roots! So let’s have a look at our first question. Our first question says: solve x to the power of 1/2 equals 8 now to solve an equation.
We want to do the inverse operations to the both sides until we’re left with x equals a certain number. So here we’ve got x to the power of 1/2. Now the power of 1/2 means two square roots, so we would want to do the opposite of the inverse operation so, instead of square root, n we’re gonna square both sides of this equation, so squaring the left-hand side. Well, just leave us with x and squaring the right-hand side will give us what 8 squared is 64. So our answer would be x, equals 64. At the check our answer, we could just substitute the 64 in so 64 to the power of 1/2! What’s the square root of 64, which is it so that’s right. So if you’ve got an equation where it’s x to the power of 1/2 to solve it, you would just square both sides of the equation if it was x to the power of 1/3. Well, that means the cube root of x, but then we would cube both sides of the equation if it, if it was x to the power of 1/4 i mean so for fruits, so we would do to the power 4 to both sides of the equation. So if you’ve got an equation, where you’ve got x to the power of 1 over something, you would just take that power. The denominator in you did that power to both sides of the equation and you a solve your equation?
Ok, let’s have a look at a question that, where it’s not a 1 on the numerator, so our next question solve x to the power 3/4 equals 8 now x to the part of 3/4, that’s the same as saying the fourth root of x and then cubing it. So whenever we want to solve this equation, we monitor the inverse.
So, instead of doing the cubing, we’re gonna cube root! Both sides of the equation and then i said of the fourth root: we’re gonna do the power 4 to both sides of the equation! So, first of all, let’s do the cube root the opposite of this cubing! So taking the cube root, look at the numerator, it’s a phrase. We’re gonna, take the cube root of both sides of the equation and so taking the cube root of the left-hand side, but i would just be x to the power of 1/4, we’re doing it to get rid of the 3 on the numerator and the cube root of the left-hand side. Well, the cube root of 8 is equal to 2. Now it’s just with this part of the equation is just like our last question? If we wanted to solve this, we wanna do the inverse operation of the four fruits, so that would be to the power of 4. So that would leave us with x? + 2 to the power 4 to to the par 4 is equal to 2 times 2 times 2 times, 2 well 2 times, 2 is 4 times. 2 is 8 times, 2 is 16, so answer would be x, equals 16 and again we can check our answer! 16 by the 4th root of 16 is equal to 2 and cubed is equal to it and that’s it alright.
So our next question it says: solve x to the power of negative 2 equals 9. So this questions got a negative power, so we’re going to have to be careful whenever we’re solving this so x to the power negative 2! What’s the same as 1 over x squared when you’ve got a negative power, you can write it as 1 over or the reciprocal. Now we’ll come back to that later. On the question to show you how you can sort of to this question using a slightly different approach, so we’ve got 1 over x, squared equals 9! Now we’re going to this we’re going to multiply both sides to the equation by x squared so they give us one equals and 9 x squared and then we’re gonna divide both sides of the equation by 9 to get 1/9 equals x. Squared now we’re going to be careful here, because what we’re going to do is we’re going to square both sides, but we’ve got to be careful because we’ve got x, squared equals 1/9! Now a positive squared is a positive, but also negative! Squared is equal to a positive. So whenever we square both sides, we need to remember that we could have the positive or the negative solution. Whenever you got x, squared equals a number a positive number and your square root in it. You could have the positive or negative roots, so we have got the square root of 1/9, and so the square root of the 9 for our square root of 1 is 1 and square root of 9 is equal to 3.
So that means that x will be equal to positive or negative 1/3 i’m, just gonna write out and fill the x equals, 1/3 or x equals negative 1/3, and we can check our answers so we had 1/3 and we done to the part of negative 2. That would be the reciprocal of that?
Well, the reciprocal of 1/3 is equal to 3 and squared is 9 and that’s our solution there or if we had negative preferred well, the reciprocal of that would be negative? 3 and squared would be 9? So there’s our two solutions. Now there was a slightly different approach.
We could have taken on that question rather than starting off by writing 1 over x squared! So what we could have done was we could have taken the reciprocal of both sides, speaking with to get rid of the negative son, because the negative sign means 1 over. If we get rid of that, we can just say well x, squared that’s taking the reciprocal of it is equal to 1, and that could mean that we could just go straight to this part here and then we could square root both sides, remembering it’s the positive or negative solutions. Okay, our next question: our next question is to solve the cube root of 9x. Minus 1 equals 4!
So again we want to do the inverse operations to both sides? This is the cube root, so we’re going to cube both sides of this equation, so keeping the left hand side. You just leave us with 9x minus 1 and cubing the right hand.
Side would be 64! Now we need to find what 9x is so we’re gonna, add 1 to both sides, so 9x equals 65 and then dividing by 9 will give us x, equals 65 over 9 and so x equals 65 over 9 and that doesn’t simplify. So that’s our solution and that’s it! Okay? Next question: our next question is a little different because we have x to the power three-halfs, but then we’ve got a mixed number here of 4 and 17 over 27. What we’re gonna do is we’re gonna write this as a top-heavy fraction to begin with, and that might help us, so we’ve got x to the power of 3 over 2 all right. This is a top-heavy fraction about 4 times, 27 well 4 times, 27 is equal to 108, plus 17 is equal to 125 and then that’s gonna be over. The denominator stays the same? So it’s 27 now to solve this.
Remember we’re going to take whatever it’s on the numerator! We’re gonna take that rich of both sides? So we’ve got a cubed here, so we need to do the inverse, which is the cube root of both sides, and that’s quite nice, because both of the numbers are cube numbers. So it’s gonna, give us x to the 1/2. Getting rid of the 3 will equal and taking the cube root of both of these numbers would give us well.
The cube root of 125 is 5 and the cube root of 27 is equal to 3 and then finally, we’ve got x to the power of 1/2, which is the square root. So we need to square both sides of the equation, so squaring both sides will give us x equals 25 over 9, and that’s it right. So our last two question is a little bit different than the ones we’ve done so far because they involve the laws of indices, so you’ve got our first question says: solve 3 to the power of 4x equals 27 to the power of 5 minus x!
Now, in this question, we’ve got our free and our 27.
Now these numbers aren’t chosen at random and they’re chosen because 27 is equal to 3 cubed. So what we’re going to do is we’re going to rewrite the 27 as free cubed, so we’ve got 3 ^ 4 x equals 3, cubed ^, 5 minus x!
Now what we’re going to do this we’re going to use the laws of indices here, so we’ve got a power of a power so whenever you’ve got x to the power of a all to the power of b, that would be the same as x to the a b? So you multiply the end, the b! So if we multiply the 3 this cubed by the 5 – x, then we can write it as 3 ^, 15, -, 3 x, so will give us, on the left hand, side 3 to the power of 4x equals and multiplying the 3 here, the cubed by the 5 and the minus x? Where goes free to the power of 15 – 3 x, now we’ve got 3 to the power of something equals 3 to the power of something submitted! The – something’s must be the same, so in other words, that must give us 4x equals 15, – 3 x never solve this. We can just add 3x to both sides of the equation, so that’ll be 7x, equals 15 and dividing by 7 gives us x equals 15 over 7! That’s it! so if we’ve got a question where you’ve got the 3 and the 27, you can write, the 27 are 3 cubes and then you use the laws of indices of silver!
Let’s have a look at a similar question, i so similar, but a little bit different. We’ve got our question where it says solve it: to the power of 4 plus x, over 4 to the power of 5.
– x equals no point 5. Now? The first thing i’m going to do is i’m going to rewrite the null point!
5 is 1/2, so we’re gonna, write x or it’s the power of 4 plus x over 4 to the power 5 – x equals 1/2. Now all of these, the it the 4 and 1/2 are all numbers which can be written as powers of 2!
It is the same as 2 cubed 4 is obviously 2 squared and 1/2.
Well, that’s the same as 2 to the minus 1. So i’m going to rewrite this as 2 cubed to the power of 4 plus x, over 2 squared to the power of 5 – x, equals two to the negative one. So then, what i’m not going to do is well! Are you gonna make a bit of space and then we’re going to go to do is i’m going to use the laws of indices here, because these are powers over power, so i’m going to multiply these two powers together, so it’ll be 2 to the power off 3 times! 4 is 12 and 3 times.
X is plus 3 x over and then x in the powers together here will give us 2 to the power of or 2 times, 5 is equal to 10 and 2 times minus x. We minus 2 x that equals 2 to the negative 1! Now here we’ve got 2 to the power of something divided by 2 to the power of something again using the laws of indices? If you had x to the power of a divided by x to the power of b, that’s the same as x to the power of a minus b.
You take away the powers so here if we have 2 to the power of 12 plus 3x over 2 to the power of 10 minus 2x. If we take away the powers, we can write it as 2 to the power of something. So let’s take away the powers and see what we get. So if we had 12 plus 3x and we subtract from that 10 minus 2x or that will give us well 12, take away. 10 is 2 +, 3, x, -, minus 2x. Will that be plus 5x? somebody said the left hand?
Side would have become 2 to the power of 2 plus 5 x and that equals 2 to the negative 1? Now, if you got 2 to the power of something equals 2 to the power of something well, that makes it 2 for something’s must be equal to each other. That means a 2 + 5 x must be the same as minus 1. So let’s write that down.
2 plus 5x is equal to negative 1, so a desk just ticked away from both sides of the equation so to be 5 x, equals a negative 3 and dividing by 5 would give us x equals negative 3/5. That’s it. ? |
Difference between revisions of "2019 AMC 10A Problems/Problem 13"
Problem
Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$. Construct the circle with diameter $\overline{BC}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$. What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$
Solution 1
$[asy] unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("A",(1,0),SE);label("C",(0,2.75),N);label("B",(-1,0),SW);label("E",(0,0),S);label("D",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));[/asy]$
Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find $\angle ABC=70^{\circ}$. We can find $\angle ECB=20^{\circ}$ and $\angle DBC=50^{\circ}$ by the triangle angle sum on $\triangle ECB$ and $\triangle DBC$.
$$\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}$$
$$\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}$$
Then, we take triangle $BFC$, and find $\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}.$
Solution 2
Alternatively, we could have used similar triangles. We start similarly to Solution 1.
Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Therefore, $$\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.$$
So, $\triangle BEF \sim BDA$ by AA Similarity, since $\angle EBF = \angle DBA$ and $\angle BEC = 90^{\circ} = \angle BDA$. Thus, we know $$\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.$$
Finally, we deduce $$\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110^{\circ}}.$$
Solution 3 (outside angles)
Through the property of angles formed by intersecting chords, we find that $$m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}$$
Through the Outside Angles Theorem, we find that $$m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}$$
Adding the two equations gives us $$m\angle BFC + m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB$$
Since $\overarc{BC}$ is the diameter, $m\overarc{BC}=180^{\circ}$, and because $\triangle ABC$ is isosceles and $m\angle ACB=40^{\circ}$, we have $m\angle CAB=70^{\circ}$. Thus $$m\angle BFC=180^{\circ}-70^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}$$
Solution 4
Notice that if $\angle BEC = 90^{\circ}$, then $\angle BCE$ and $\angle ACE$ must be $20^{\circ}$. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that $\angle EBD \cong \angle ECD = 20^{\circ}$. Thus $\angle CBF = 70 - 20 = 50^{\circ}$, and so $\angle BFC = 180 - 20 - 50 = 110^{\circ}$, which is $\boxed{\textbf{(D)}}$.
Solution 5
$\triangle{ABC}$ is isosceles so $\angle{CAB}=70^{\circ}$. Since $CB$ is a diameter, $\angle{CDB}=\angle{CEB}=90^{\circ}$. Quadrilateral $ADFE$ is cyclic since $\angle{ADF}+\angle{AEF}=180^{\circ}$. Therefore $\angle{BFC}=\angle{DFE}=180^{\circ}-\angle{CAB}=\boxed{110^{\circ}}$
Video Solution
Education, the Study of Everything
See Also
2019 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# The Ultimate List of Math Hacks, Tricks, and Tips
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Regardless of the age, grade or degree, mathematics could be challenging at times no matter if you’re learning arithmetic, geometry, fractions, algebra or children learning times tables. However, the important thing is the subject has its hacks and tricks to hijack, helpful in tackling tough problems and managing time in the assessment.
Mathematics’ tricks are techniques, most of them are methods of quick calculations through which complex mathematical problems are solved easily and mathematical skills are improved. Most of the hacks are devised for the basic operations of addition, subtraction, multiplication, division percentage, fractions etc.
## Arithmetic Operations
Through the principles of ‘tens and unit places’, it becomes relatively easy to perform calculations rather than using finger count or calculator.
• The two-digit addition of numbers becomes easy and quick to perform such as:
• Say, 33 + 23
• Split the second number into tens and units, such that: 23= 20 + 3
• Finish the addition of ten’s, such that: 33 + 20 = 53
• Then add the rest with unit place digit, such that: 53 + 3 = 56.
1. ### Subtraction
• Consider two numbers say, 1000 and 546
• Subtract 1 from each number, we get: 999 and 545
• Then subtract, the both numbers, we get: 454
• And, 1000 – 546 = 454
1. ### Multiplication
• We have numbers say, 20 and 12
• Splitting 20, we get 4 x 5
• Multiply 5 with 12, we get 60
• Multiply 60 x 4, we get: 240.
1. ### Multiply 2 and 4
When a number is multiple of 2 and 4, then the last digit of the output value will always be an even number.
• 19 x 4 = 76
• 19 x 6 = 114
1. ### Division
The numbers which are ‘evenly divided by certain numbers’:
• Divisible by 2: The even numbers ending at 0, 2, 4, 6 or 8.
• Divisible by 3: If the sum of the digits is divisible by 3.
• Divisible by 4: If the last two digits of the number are divisible by 4.
• Divisible by 5: If the last digit of the number is 0 or 5.
• Divisible by 6: If a number is divisible by 2 or 3.
• Divisible by 8: If the last three digits are divisible by 8.
• Divisible by 9: If the sum of the digits is divisible by 9.
• Divisible by 10: If the last digit of a number is 0.
1. ### Finding Percentage
• Let’s say: we need to find 5% of the number 255.
• Move the decimal point over one place, say 255 becomes 25.5.
• Divide the number 25.5 by 2, we get:12.75
• 75 is the solution.
1. ### Squares Ending at 5
• Consider the number 125 to find its square.
• Start writing the last two-digit number that is 25.
• Take the first number i.e., 12 and the one that follows i.e., 13.
• Multiple 12 x 13, to get 156.
• Use it as prefix with 25 i.e., 15625.
• 1252 = 15625
1. ### Memorizing Table of 9
• Let’s focus on the pattern of the complete table of 9.
• 09, 18, 27, 36, 45, 54, 63, 72, 81, 90
• The numbers at ten’s place are increasing by 1, whereas those at unit’s place are decreasing by 1.
1. ### Distance Formula
In order to calculate the total time for a trip, a simple hack is the distance formula which is usually remember by all: Distance = Rate x Time or Time = Distance/ Rate, if one intends to calculate the total time. For instance, for 5 miles of distance, at the rate of 2 miles per hour, the time would be calculated as: 5/2 = 2.5 hours.
1. ### Fraction into Percentages
Everyone has the understanding about percentages and how to measure them, but most of the time it is not understand that what fraction of a pie makes a certain percentage? Mostly, easy fractions are known to all such that 1/2 makes 50%, 1/4 makes 25%. Let us memorize a few hard approximate values to use them arbitrarily when questioned.
• 1/6th makes 16% of the pie.
• 1/7th makes 14% of the pie.
• 1/8th makes 12-12.5% of the pie.
• 1/9th makes 11% of the pie
• 1/11th makes 9% of the pie.
• 1/12th makes 8% of the pie.
1. ### Butterfly Method
This method is used for the quick addition or subtraction of fractions.
1. ### Mnemonics
In Mathematics, mnemonics are taught to the students which help in memorizing the formulas better, for their quick and effective calculations.
• “Some People Have Curly Brown Hair Through Proper Brushing”.
This mnemonic is useful in learning the formulas of trigonometric rations where every first letter starts with:
• Sin θ = Perpendicular/ Hypotenuse
• Cos θ = Base/ Hypotenuse
• Tan θ = Perpendicular/Base
• King Henry Died Drinking Chocolate Milk”
This mnemonic is useful in learning the metric systems with the order of values where every first letter starts with:
• Kilo, Hecto, Deca, Deci, Centi, Milli.
• Please Excuse My Dear Aunt Sally”
The mnemonic is used for the order of arithmetic operations to be solved first, where every first letter starts with:
• Parenthesis, Exponents, Multiply, Divide, Addition, then Subtraction.
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# Multiplication Properties Worksheets Printable PDF [Grade 3]
Multiplication Properties Worksheets-Multiplication properties can be learned in a relatively short amount of time with practice. Worksheets can be helpful in reinforcing what has been learned and aiding in the memory of multiplication facts.
Math can be a challenging subject, but it can also be fun and exciting when you are able to practice and master the basics. One way to do this is by using Properties of Multiplication Worksheets with Answers. These sheets allow students to practice their multiplication facts without having to worry about mistakes. By working through these sheets, students will have a better foundation upon which they can build more complex multiplication problems.
## Multiplication Properties Worksheets
Math is a subject that can be very confusing for students. One way to make the subject easier to understand is by using multiplication properties worksheets. These help students learn the multiplication property, and how to use it in real-world problems.
PDF
One example of a problem that can be solved using multiplication properties is finding the product of two numbers. To do this, divide one number by the other and find the answer in parentheses.
For example, if someone tells you that 5 multiplied by 3 equals 12, you would take 5 (divided by 3) and move the decimal two places to the right, which equals 1.2 (in parentheses).
This process can be repeated for any other combination of numbers. This type of worksheet can be helpful for both math beginners and veterans alike.
PDF
Free Properties of Multiplication Worksheet PDF can be a helpful tool for students with difficulty. Additionally, by practising these properties regularly, students will develop a deeper understanding of them and will be able to solve problems faster in future encounters.
### What Are the Properties of Multiplication?
Multiplication is a mathematical operation that is used to calculate the result of multiplying two numbers.
PDF
The properties of multiplication can be summarized as follows:
1. Multiplication is associative: The order of operations does not matter when multiplying two numbers.
2. Multiplication is commutative: The order of operations does not matter when multiplying two numbers.
3. Multiplication is distributive: The order of operations does not matter when distributing the work between multiple multipliers.
4. Multiplication is always associative, commutative, and distributive over addition and subtraction: These three properties known as the law of multiplication.
5. The product of two numbers cannot be negative: This property follows from the fact that multiplication is always associative and commutative (and therefore identity).
In learning multiplication, it is important to have a solid foundation. One way to do this is by providing students with Printable Multiplication Properties Worksheet for Grade 3 that covers all the basics. This printable multiplication properties worksheet for grade 3 is one such resource.
Multiplication Properties of Exponents Worksheet with Answers given to check your progress.
PDF
Do you find yourself multiplying two-digit numbers quickly and accurately? If so, you’re in luck! This worksheet provides a multiplication property for every one of the natural numbers from 2 through to 20. Simply fill in the number in the corresponding box and see what happens!
For example, if you filled in 5 in the first box and 9 in the second box, the answer would be 25. Multiplication property worksheets can be a great way to review multiplication facts or to simply improve your speed and accuracy when multiplying two-digit numbers. |
# SAT Math : How to find slope of a line
## Example Questions
← Previous 1 3
### Example Question #1 : Slope And Line Equations
Based on the table below, when x = 5, y will equal
x y -1 3 0 1 1 -1 2 -3
11
–10
–9
–11
–9
Explanation:
Use 2 points from the chart to find the equation of the line.
Example: (–1, 3) and (1, –1)
Using the formula for the slope, we find the slope to be –2. Putting that into our equation for a line we get y = –2x + b. Plug in one of the points for x and y into this equation in order to find b. b = 1.
The equation then will be: y = –2x + 1.
Plug in 5 for x in order to find y.
y = –2(5) + 1
y = –9
### Example Question #1 : Slope And Line Equations
What is the slope of a line that runs through points: (-2, 5) and (1, 7)?
5/7
3/2
2
2/3
2/3
Explanation:
The slope of a line is defined as a change in the y coordinates over a change in the x coordinates (rise over run).
To calculate the slope of a line, use the following formula:
### Example Question #3 : Slope And Line Equations
A line passes through the points (–3, 5) and (2, 3). What is the slope of this line?
-3/5
2/5
–2/5
–2/3
2/3
–2/5
Explanation:
The slope of the line that passes these two points are simply ∆y/∆x = (3-5)/(2+3) = -2/5
### Example Question #2 : Slope And Line Equations
Which of the following lines intersects the y-axis at a thirty degree angle?
yx((√3)/3) + 1
y = x - √2
y = x
y = x√3 + 2
y = x√2 - 2
y = x√3 + 2
Explanation:
### Example Question #5 : Lines
What is a possible slope of line y?
–2
2
–2
Explanation:
The slope is negative as it starts in quadrant 2 and ends in quadrant 4. Slope is equivlent to the change in y divided by the change in x. The change in y is greater than the change in x, which implies that the slope must be less than –1, leaving –2 as the only possible solution.
### Example Question #5 : Slope And Line Equations
What is the slope between and ?
Explanation:
Let and
so the slope becomes .
### Example Question #12 : How To Find Slope Of A Line
What is the slope of line 3 = 8y - 4x?
2
-2
0.5
-0.5
0.5
Explanation:
Solve equation for y. y=mx+b, where m is the slope
### Example Question #13 : How To Find Slope Of A Line
Find the slope of the line 6X – 2Y = 14
-3
3
12
-6
3
Explanation:
Put the equation in slope-intercept form:
y = mx + c
-2y = -6x +14
y = 3x – 7
The slope of the line is represented by M; therefore the slope of the line is 3.
### Example Question #14 : How To Find Slope Of A Line
If 2x – 4y = 10, what is the slope of the line?
–0.5
2
–2
–5/2
0.5
0.5
Explanation:
First put the equation into slope-intercept form, solving for y: 2x – 4y = 10 → –4y = –2x + 10 → y = 1/2*x – 5/2. So the slope is 1/2.
### Example Question #1 : How To Find The Slope Of A Line
What is the slope of the line with equation 4x – 16y = 24?
–1/4
1/2
1/4
1/8
–1/8
1/4
Explanation:
The equation of a line is:
y = mx + b, where m is the slope
4x – 16y = 24
–16y = –4x + 24
y = (–4x)/(–16) + 24/(–16)
y = (1/4)x – 1.5
Slope = 1/4
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# Coast Guard MST Marine Science Technician ASVAB Study Guide (2023)
Algebra Geometry FAQs Videos
### Algebra
##### Expressions498
###### Classifications
A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms. Linear expressions have no exponents. A quadratic expression contains variables that are squared (raised to the exponent of 2).
###### Operations Involving Monomials
You can only add or subtract monomials that have the same variable and the same exponent. However, you can multiply and divide monomials with unlike terms.
###### Multiplying Binomials
To multiply binomials, use the FOIL method. FOIL stands for First, Outside, Inside, Last and refers to the position of each term in the parentheses.
To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse.
##### Solving Equations292
###### One Variable
An equation is two expressions separated by an equal sign. The key to solving equations is to repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the equal sign and the answer on the other.
###### Two Variables
When solving an equation with two variables, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)
(Video) Marine Science Technician (MST) Explained
###### Two Equations
When presented with two equations with two variables, evaluate the first equation in terms of the variable you're not solving for then insert that value into the second equation. For example, if you have x and y as variables and you're solving for x, evaluate one equation in terms of y and insert that value into the second equation then solve it for x.
When solving quadratic equations, if the equation is not set equal to zero, first manipulate the equation so that it is set equal to zero: ax2 + bx + c = 0. Then, factor the quadratic and, because it's set to zero, you know that one of the factors must equal zero for the equation to equal zero. Finding the value that will make each factor, i.e. (x + ?), equal to zero will give you the possible value(s) of x.
###### Inequalities
Solving equations with an inequality (<, >) uses the same process as solving equations with an equal sign. Isolate the variable that you're solving for on one wide of the equation and put everything else on the other side. The only difference is that your answer will be expressed as an inequality (x > 5) and not as an equality (x = 5).
### Geometry
##### Lines & Angles5214
###### Line Segment
A line segment is a portion of a line with a measurable length. The midpoint of a line segment is the point exactly halfway between the endpoints. The midpoint bisects (cuts in half) the line segment.
###### Right Angle
A right angle measures 90 degrees and is the intersection of two perpendicular lines. In diagrams, a right angle is indicated by a small box completing a square with the perpendicular lines.
###### Acute & Obtuse Angles
(Video) Coast Guard Job Tier List
An acute angle measures less than 90°. An obtuse angle measures more than 90°.
###### Angles Around Lines & Points
Angles around a line add up to 180°. Angles around a point add up to 360°. When two lines intersect, adjacent angles are supplementary (they add up to 180°) and angles across from either other are vertical (they're equal).
###### Parallel Lines
Parallel lines are lines that share the same slope (steepness) and therefore never intersect. A transversal occurs when a set of parallel lines are crossed by another line. All of the angles formed by a transversal are called interior angles and angles in the same position on different parallel lines equal each other (a° = w°, b° = x°, c° = z°, d° = y°) and are called corresponding angles. Alternate interior angles are equal (a° = z°, b° = y°, c° = w°, d° = x°) and all acute angles (a° = c° = w° = z°) and all obtuse angles (b° = d° = x° = y°) equal each other. Same-side interior angles are supplementary and add up to 180° (e.g. a° + d° = 180°, d° + c° = 180°).
##### Polygons81016
###### Triangle Geometry
A triangle is a three-sided polygon. It has three interior angles that add up to 180° (a + b + c = 180°). An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite (d = b + c). The perimeter of a triangle is equal to the sum of the lengths of its three sides, the height of a triangle is equal to the length from the base to the opposite vertex (angle) and the area equals one-half triangle base x height: a = ½ base x height.
###### Triangle Classification
An isosceles triangle has two sides of equal length. An equilateral triangle has three sides of equal length. In a right triangle, two sides meet at a right angle.
###### Pythagorean Theorem
The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, $$c = \sqrt{a + b}$$
A quadrilateral is a shape with four sides. The perimeter of a quadrilateral is the sum of the lengths of its four sides (a + b + c + d).
(Video) MST, United States Coast Guard
###### Rectangle & Square
A rectangle is a parallelogram containing four right angles. Opposite sides (a = c, b = d) are equal and the perimeter is the sum of the lengths of all sides (a + b + c + d) or, comonly, 2 x length x width. The area of a rectangle is length x width. A square is a rectangle with four equal length sides. The perimeter of a square is 4 x length of one side (4s) and the area is the length of one side squared (s2).
###### Parallelogram
A parallelogram is a quadrilateral with two sets of parallel sides. Opposite sides (a = c, b = d) and angles (red = red, blue = blue) are equal. The area of a parallelogram is base x height and the perimeter is the sum of the lengths of all sides (a + b + c + d).
###### Rhombus
A rhombus has four equal-length sides with opposite sides parallel to each other. The perimiter is the sum of the lengths of all sides (a + b + c + d) or, because all sides are the same length, 4 x length of one side (4s).
###### Trapezoid
A trapezoid is a quadrilateral with one set of parallel sides. The area of a trapezoid is one-half the sum of the lengths of the parallel sides multiplied by the height. In this diagram, that becomes ½(b + d)(h).
##### Circles528
###### Dimensions
(Video) ASVAB vs AFQT in the Coast Guard: What Score do you Need to go MST?
A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter (AC, CB, CD). A chord is a line segment that connects any two points along its perimeter (AB, AD, BD). The diameter of a circle is the length of a chord that passes through the center of the circle (AB) and equals twice the circle's radius (2r).
###### Calculations
The circumference of a circle is the distance around its perimeter and equals π (approx. 3.14159) x diameter: c = π d. The area of a circle is π x (radius)2 : a = π r2.
##### Solids245
###### Cubes
A cube is a rectangular solid box with a height (h), length (l), and width (w). The volume is h x l x w and the surface area is 2lw x 2wh + 2lh.
###### Cylinders
A cylinder is a solid figure with straight parallel sides and a circular or oval cross section with a radius (r) and a height (h). The volume of a cylinder is π r2h and the surface area is 2(π r2) + 2π rh.
##### Coordinate Geometry222
###### Coordinate Grid
The coordinate grid is composed of a horizontal x-axis and a vertical y-axis. The center of the grid, where the x-axis and y-axis meet, is called the origin.
###### Slope-Intercept Equation
A line on the coordinate grid can be defined by a slope-intercept equation: y = mx + b. For a given value of x, the value of y can be determined given the slope (m) and y-intercept (b) of the line. The slope of a line is change in y over change in x, $${\Delta y \over \Delta x}$$, and the y-intercept is the y-coordinate where the line crosses the vertical y-axis.
(Video) MST: MARINE SCIENCE TECHNICIAN RESEARCH: COAST GUARD JOBS VLOG 064
## FAQs
### Is the Coast Guard ASVAB test hard? ›
The Coast Guard is the most challenging branch of the military to enter. A minimum ASVAB score of 36 is required to enlist.
What ASVAB score do you need for Coast Guard? ›
What is the qualifying ASVAB score for the Coast Guard? In order to join the Coast Guard, you must achieve a minimum overall qualifying ASVAB score of 40 if you are a high school senior or have a high school diploma, and 50 if you are a GED holder.
How do you become a marine science technician in the Coast Guard? ›
Prospective MSTs must meet Armed Services Vocational Aptitude Battery (ASVAB) score requirements of 114 in Verbal Ability plus Arithmetic Reasoning (VE+AR), and a minimum Mathematics Knowledge (MK) score of 56. Training for the rating is accomplished through a 12-week course at USCG Training Center Yorktown, VA.
What score on the ASVAB do I need to be in the Marines? ›
The Armed Service Vocational Aptitude Battery (ASVAB) test is a multiple-choice test taken by all who aspire to become Marines. To pass the test, aspiring Marines must achieve a score of 31 or higher, and those with nontraditional degrees or a GED must score at least a 50.
How fast do you have to run a mile in the Coast Guard? ›
Coast Guard Basic PFT
EventMaleFemale
Push-ups (60 sec)2915
Sit-ups (60 sec)3832
Run (1.5 mile)12:5115.26
Sit and reach*16.50 "19.29"
2 more rows
How hard is it to get a 90 on the ASVAB? ›
Your AFQT score is very important.
This percentile is based on the number of questions that you answered correctly compared to others that have taken the ASVAB. If you receive a percentile score of 90, that simply means that you have scored as well or better than 90% of other test-takers.
Is getting a 50 on the ASVAB hard? ›
With ASVAB Standard scores, the majority of students score between 30 and 70. This means that a standard score of 50 is an average score, and a score of 60 would be an above-average score.
Is Coast Guard the hardest bootcamp? ›
While the Coast Guard might be considered the easiest military branch in terms of physical fitness, it is by far the hardest to join.
How many recruits fail Coast Guard boot camp? ›
Every wet-behind-the-ears recruit is scared of failing Coast Guard Boot Camp – and for good reason – last I heard the attrition rate is over 10% nowadays. BUT that's not what this is about. That's just a harsh fact that you have to swallow.
How long is MST a school USCG? ›
Marine Science Technician "A" (MST) Course
The course is 9 weeks long and is open to qualified active and reserve enlisted personnel. The training topics covered include Pollution Investigation, Pollution Response, Occupational Safety and Health, Facility Inspections and Vessel Boardings.
### Is marine technician a good career? ›
Is a boat mechanic a good career? Boat mechanics and service technicians can get many job roles with excellent prospects of a high salary. Along with decent wages, boat mechanics have immense growth and learning opportunities. Earn your degree in marine tech from NEIT and begin your new career path today!
What is a Marines salary? ›
While ZipRecruiter is seeing annual salaries as high as $121,500 and as low as$11,500, the majority of salaries within the Marine Corps jobs category currently range between $37,500 (25th percentile) to$73,000 (75th percentile) with top earners (90th percentile) making $104,000 annually across the United States. Which branch has the easiest ASVAB? › At the ASVAB stage, the easiest military branch to join is the Army or Air Force. At the basic training stage, the easiest military branch to join is the Air Force. What is the hardest part of the ASVAB? › According to recent research, the mathematics knowledge test is considered to be the hardest ASVAB subtest. To get a well understanding as well as tips and tricks to get the highest ASVAB Scores in this section, read more information on our free ASVAB Math study guide! What percent of recruits fail basic training? › Yes, it is possible to fail basic training. You could go through the trouble of leaving your home, job, family and friends and come back a failure. In fact, this happens to about 15% of recruits who join the military every year. Can you have your phone in Coast Guard boot camp? › Family Communication During Coast Guard Boot Camp You are allowed to bring your cell phone, but you may not receive or make personal phone calls for granted liberty on Sunday of the final two weeks of training. Any future phone calls you make while in boot camp will be at the discretion of your Company Commander. How often do coast guards come home? › TWO- “At least your spouse gets ?to come home every night.” The reality is that Coast Guard missions are often away from home. CG members do not get to go home every night. Depending on their unit, they may be gone from a couple of days to upwards of a year. Is a 70% good on the practice ASVAB? › With ASVAB Standard scores, the majority of students score between 30 and 70. This means that a standard score of 50 is an average score, and a score of 60 would be an above-average score. How far in advance should I study for the ASVAB? › To study effectively for the ASVAB, you really need to begin studying at least two months before you plan to take the test, if not more! Here are some basic steps to take: Find a comfortable, quiet area to study. Gather paper, pens and pencils, a calculator and other tools. How likely is it to fail the ASVAB? › There is no pass or fail on the ASVAB. You cannot "ace" the ASVAB or "flunk" it! Your scores reflect your own abilities! You will want to do your best so that you will be eligible for the military skill specialty that matches your ability and interests. ### What is the easiest way to pass the ASVAB test? › The easiest way to pass the ASVAB is by taking practice exams. You can use these practice exams to become more familiar with the exam in terms of content, topics, and the timing aspect. You can use the practice exams to narrow down the subjects you need the most help in. How many questions can I miss on the ASVAB? › You will not be penalized for incorrect answers, but you will be penalized for questions that are left blank, as these will count as incorrect answers. Each question is 1 point, so if you leave 15 questions blank, you will automatically reduce your ASVAB score to 85. What parts of the ASVAB matter the most? › Four of these sections- Arithmetic Reasoning, Mathematics Knowledge, Paragraph Comprehension, and Word Knowledge- make up your AFQT score. This score is the most important as it determines if you can enlist in the military at all. What is the easiest military branch? › The Air Force is reported to be the “easiest” branch when it comes to physical challenges and difficulties. Next to the Coast Guard, Air Force members are among the least likely to see combat. Deployments in the Air Force are not as common as the Army, Navy, or Marine Corps, and they are generally shorter in duration. What branch has hardest boot camp? › The military branch with the toughest basic training is the Marine Corps. The hardest military branch for non-males because of exclusivity and male dominance is the Marine Corps. Is the Navy or Coast Guard harder? › Navy probably trains harder only because Coast Guard doesn't have as much time to train. CG is involved in a ton a search and rescue as well as law enforcement. They are the basically the national fire and police of the waterways. Navy serves a different purpose as they are national defense. What happens if you cry during basic training? › With all the pressure that comes with boot camp, it's no wonder people cry. Crying isn't looked at as a bad thing, though some drill instructors give the recruit grief if they think he's just feeling sorry for himself. But in most cases, you are not considered less of a marine if you cry. Can a drill sergeant hit you? › The military's drill sergeants and instructors are prohibited from hitting their recruits. Where is MST stationed? › MSTs are stationed at Sectors, and Marine Safety Units and Detachments. Is MST secondary to PTSD? › Posttraumatic stress disorder (PTSD) is one of the most common mental health diagnoses among MST survivors. Other mental health diagnoses that are frequently related to MST include depression and other mood disorders and substance use disorders. ### What is the hardest marine job? › One of the military's most dangerous jobs is explosive ordnance disposal technician. These troops risk life and limb responding to IEDs and other explosive threats. Two seasoned Marine Corps EOD techs recently told Insider about the challenges of the job. How long does it take to be a marine technician? › How Long Does It Take to Become a Marine Mechanic? Once you graduate high school, completing a certificate course will require about one year, while an associate's degree can take up to two years. Apprenticeship programs for marine mechanics can span over four years. What does a marine science technician do in the Coast Guard? › Coast Guard: Marine Science Technician From protecting U.S. waters and the public from oil and hazardous material responses to conducting safety and security inspections, an MST plays the essential role of enforcing regulations for the safety of the marine environment and the security of the port. Are Marines paid for life? › Our current retirement system consists of a 20 year cliff vested annuity (a defined benefit plan). This means that if you enter the Marine Corps and serve for at least 20 years, you will earn a monthly retirement annuity for the remainder of your life. What rank do most Marines retire at? › It will generally take an officer in the Marine Corps16 to 22 years to rise to the rank of Lieutenant Colonel. As a result, many career officers who are eligible to retire after 20 years of active service retire at this rank. ... O-5 Lieutenant Colonel - U.S. Marine Corps Ranks. ClassField Officer Basic Pay$6,112/mo
3 more rows
Do Marines get paid more if married? ›
There is no military spouse pay or stipend, but the military offers a number of benefits to help service members and their families. Your first stop after the wedding should be the nearest military ID card issuing facility to enroll in DEERS, the Defense Enrollment Eligibility Reporting System.
Who fights more Army or Marines? ›
The Army has a force of roughly 500,00 active duty Soldiers, while the active duty Marine Corps is under 200,000. . The Army is composed of multiple subgroups, including the Infantry, Special Forces and the Army Rangers.
Is the Coast Guard hard to get into? ›
The acceptance rate at United States Coast Guard Academy is 14.2%. For every 100 applicants, only 14 are admitted. This means the school is extremely selective. Meeting their GPA requirements and SAT/ACT requirements is very important to getting past their first round of filters and proving your academic preparation.
What branch needs the highest ASVAB score? ›
To join the Air Force as an enlisted member, you must get a decent score on the Armed Services Vocational Aptitude Battery (ASVAB) test. In fact, the Air Force requires the highest score of any branch of the service for admission. Jobs in the Air Force are called Air Force Specialty Code (AFSC).
How many times can you fail the ASVAB? ›
Retest Policy
If you take the test once and don't do well, you can retake it after just one month. If you fail to perform well on this second attempt, you can retake it again after another month. Your fourth attempt, and any subsequent attempt, requires a six-month wait from the previous test.
### What is the best study material for the ASVAB? ›
1. Mometrix ASVAB Study Guide 2021-2022: ASVAB Test Prep Secrets (5th Edition) ...
2. Kaplan ASVAB Prep Plus 2022–2023. ...
3. Kaplan ASVAB Total Prep 2022–2023. ...
4. 2021/2022 ASVAB For Dummies. ...
5. Newstone Test Prep ASVAB Exam Prep 2021-2022.
24 Apr 2022
What level of math is on the ASVAB? ›
What kind of math is on the ASVAB? There are two ASVAB math tests: Arithmetic Reasoning and Math Knowledge. The Arithmetic Reasoning test covers word problems that involve arithmetic. The Math Knowledge test covers high school math skills.
What if I don't score high enough on the ASVAB? ›
The Army will allow a retest only if the applicant's previous ASVAB test has expired, or the applicant failed to achieve an AFQT score high enough to qualify for enlistment, or when unusual circumstances occur, such as if an applicant, through no fault of their own, is unable to complete the test.
What job can you get with a 31 on the ASVAB? ›
To join the Army as an enlisted member, you usually must take the Armed Services Vocational Aptitude Battery (ASVAB) test and get a good score. The maximum ASVAB score is 99. For enlistment into the Army, you must get a minimum ASVAB score of 31.
Is Coast Guard the hardest to get into? ›
The Coast Guard is one of the more difficult branches to join because it accepts far fewer new recruits than the other branches of the military, and qualifying requirements are strict. In addition to basic citizenship and physical fitness prerequisites, the Coast Guard sets academic standards as will.
What is the hardest ASVAB test? ›
According to recent research, the mathematics knowledge test is considered to be the hardest ASVAB subtest. To get a well understanding as well as tips and tricks to get the highest ASVAB Scores in this section, read more information on our free ASVAB Math study guide!
Is Coast Guard basic hard? ›
As with any military service, your journey begins at basic training. Basic training is tough. You'll be challenged every day, both mentally and physically. You'll be pushed, tested and worked harder than you ever thought possible.
What is the failure rate of Coast Guard boot camp? ›
Every wet-behind-the-ears recruit is scared of failing Coast Guard Boot Camp – and for good reason – last I heard the attrition rate is over 10% nowadays.
What is the most elite Coast Guard unit? ›
Maritime Security Response Teams (Coast Guard)
The MSRTs are the Coast Guard's elite. They specialize in maritime counterterrorism and high-risk maritime law enforcement. Like Navy SEALs, they also excel at Visit, Board, Search, and Seizure (VBSS) operations and often deploy outside the US.
What branch has the easiest ASVAB? ›
At the ASVAB stage, the easiest military branch to join is the Army or Air Force. At the basic training stage, the easiest military branch to join is the Air Force.
### What's the lowest ASVAB score ever? ›
An AFQT score is a percentile score. This means a person can receive a score from 0-99, and the score means "You have scored higher than X% of test takers". The lowest score a person can get is 0 - they did not score higher than anyone on the grading curve.
What is the safest branch of the military? ›
Today's article will look at “safety” in terms of man-to-man combat and machine-to-machine accidents and give you this answer: THE SPACE FORCE is the safest military branch. But, since there are different jobs within a branch, it is also worth considering the degree of safety at the job level.
What GPA do you need for Coast Guard? ›
Average GPA: 3.81
With a GPA of 3.81, United States Coast Guard Academy requires you to be near the top of your class, and well above average. You'll need mostly A's, ideally with several AP or IB classes to help show your preparation at a college level.
What is the shortest boot camp? ›
Counting the half-week you spend in forming (in-processing), you'll spend a total of seven-and-a-half weeks in Coast Guard basic training at Cape May, (N.J.,) the shortest basic training of all the services.
## Videos
1. US Coast Guard to Civilian: Transferable Job Skills as an MST/Typical Advancement Timelines E1-E6
(Duck Scrubber, First Class)
2. A Day in the Life of an MST--3rd Class Edition (Post "A" School)
(Duck Scrubber, First Class)
3. Marine Science Technician Certificate Program
4. Boatswain's Mate (BM)
(GoCoastGuard)
5. What Can Colorblind People Do in the Coast Guard?
(Mike Saint-Antoine)
6. My Coast Guard career and the FUTURE!!
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Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
296
Use techniques from Chapter 12 to solve equations
14.1 Solve the equation: xy = 0.
According to the zero product property, a product can only equal 0 if at least
one of the factors (in this case either x or y) equals 0. Therefore, xy = 0 if and
only if x = 0 or y = 0.
14.2 Solve the equation: x(x – 3) = 0.
The product of x and (x – 3) is equal to 0. According to the zero product
property (explained in Problem 14.1), one (or both) of the factors must equal 0.
x = 0 or x – 3 = 0
Solve x – 3 = 0 by adding 3 to both sides of the equation.
x = 0 or x = 3
The solution to the equation x(x – 3) = 0 is x = 0 or x = 3.
14.3 Solve the equation: (x + 2)(2x – 9) = 0.
According to the zero product property, the product left of the equal sign only
equals 0 if either (x + 2) or (2x – 9) equals 0. Set both factors equal to 0 and
solve the equations.
The solution to the equation (x + 2)(2x – 9) = 0 is x = –2 or .
Note: Problems 14.4–14.5 demonstrate two different ways to solve the equation x
2
= 16.
14.4 Solve the equation using square roots.
Problems 13.32–13.36 demonstrate that squaring both sides of an equation that
contains a square root eliminates the root. Conversely, taking the square root
of both sides of an equation containing a perfect square eliminates the perfect
square.
You cant
multiply two
numbers and
get zero unless one
(or both) of those
numbers is 0. That’s a
property that is unique
to 0. For example, if
xy = 4, then theres
no guarantee that
either x = 4 or
y = 4. You could set
x = 2 and y = 2, or
maybe x = 8 and
y = 0.5.
There
are two
possible
solutions to |
the equation.
Use the word
or” to separate
them because
plugging either
x = 0 OR x = 3 into
the equation produces
a true statement.
Mathematically,
saying “x = 0 AND
x = 3” means x has to
equal both of those
things at the same
time, and that
doesnt make
sense.
Back in Chapter 13, you only squared
both sides AFTER you isolated the square
root on one side of the equal sign. Similarly,
only square root both sides when the perfect
square is isolated on one side of the equal sign.
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# Word Problems on Profit and Loss
Math word problems on profit and loss will help us to review worked-out examples using the formula of profit and loss as a percentage of cost price/sale price.
Word Problems on Profit and Loss
1. By selling 33 m of carpet, a man loses an amount equal to the selling price of 3 m of carpet. Find his gain or loss per cent.
Solution:
Loss = (CP of 33 m) - (SP of 33 m)
⇒ (SP of 3 m) = (CP of 33 m) - (SP of 33 m)
⇒ (SP of 33 m) + (SP of 3 m) = (CP of 33 m)
⇒ (SP of 36 m) = (CP of 33 m).
Let the CP of 1 m be $x. Then, CP of 36 m =$ 36x
SP of 36m = (CP of 33m) = $33x. Thus, CP =$ 36x and SP = $33x. Since, (CP) > (SP), there is a loss. Loss =$ (36x - 33x) = $3x. Loss% = [(loss/CP) × 100]% = [(3x/36x) × 100] % = 25/3% = 8¹/₃% 2. Ronald buys a geyser for$ 3680 and sells it at a gain of 7¹/₂%. For how much does he sell it?
Solution:
CP of the geyser = $3680. Gain % = 7¹/₂% = 15/2%. Therefore, SP of the geyser = [{(100 + gain %)/100} × CP] =$ [{(100 + ¹⁵/₂)/100} × 3680]
= ${(215/200) × 3680} =$ 3956
Hence, Ronald sells the geyser for $3956. More solved examples for eighth grade math word problems on profit and loss formula for finding cost price and selling price. Word Problems on Profit and Loss 3. Jenny buys a calculator for$ 720 and sells it at a loss of 6²/₃)%. For how much does she sell it?
Solution:
CP of the calculator = $720. Loss % = 20/3% SP of the calculator = [{(100 - Loss %)/100} × CP] =$ [{(100 - 20/3)/100} × 720]
= ${(280/300) × 720} =$ 672
Hence, Jenny sells it for $672. 4. On selling of fan for$ 810, Sam gains 8%. For how much did he purchase it?
Solution:
SP of the fan = $810, gain % = 8%. Therefore, CP of the fan = {100/(100 + gain %) × SP} =$ {100/(100 + 8) × 810}
= ${(100/108) × 810} =$ 750
Hence, Sam purchased the fan for $750. 5. On selling a table for$ 987, Ron loses 6%. For how much did he purchase it?
Solution:
SP of the table = $987, loss % = 6%. Therefore, CP of the table = {100/(100 - loss %) × SP} =$ {100/(100 - 6) × 987}
= $(100/94) × 987 =$ 1050
Hence, Ron purchased the table for $1050. Practice word problems on profit and loss will help the students to review the questions to calculate profit % and loss % before solving the worksheet on profit and loss. 6. On selling a bat for$ 371, a man gains 6%. For how much should he sell it to gain 8%?
Solution:
SP of the bat = $371, gain % = 6%. Therefore, CP of the bat = {100/(100 + gain %) × SP} =$ {100/(100 + 6) × 371}
= ${(100/106) × 371} =$ 350
Now, CP = $350 and the desired gain% = 8%. Therefore, SP = [{(100 + gain %)/100} × CP] =$ [{(100 + 8)/100} × 350]
= ${(108/100) × 350} =$ 378
Hence, the selling price to obtain the desired gain is $378. 7. By selling a Jeans for$ 432, John loses 4%. For how much should John sell it to gain 6%?
Solution:
SP of the shirt = $432. Loss = 4% Therefore, CP of the shirt = {100/(100 - loss %) × SP} =$ {100 /(100 - 4) × 432}
= ${(100/96) × 432} =$ 450
Now, CP = $450, desired gain % = 6%. Desired SP = [{(100 + gain %)/100} × CP] =$ [{(100 + 6)/100} × 450]
= ${(106/100) × 450} =$ 477.
Hence, the desired selling price is \$ 477.
Profit, Loss and Discount
Calculating Profit Percent and Loss Percent
Word Problems on Profit and Loss
Examples on Calculating Profit or Loss
Practice Test on Profit and Loss
Discount
Practice Test on Profit Loss and Discount
Profit, Loss and Discount - Worksheets
Worksheet to Find Profit and Loss
Worksheets on Profit and Loss Percentage
Worksheet on Gain and Loss Percentage
Worksheet on Discounts
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Closed Looped Gain Of A NonInverting Operational Amplifier
TECTechnics Classroom TECTechnics Overview
Closed Looped Gain Of A NonInverting Operational Amplifier
Figure 9.1 illustrates a closed looped noninverting op-amp:
(a) Find the exact expression for the voltage gain Av.
(b) Find the expression for the voltage gain, Av if the op-amp is ideal.
(c) Evaluate both expressions derived in (a) and (b) if R1 = 1 kΩ, R2 = 10 kΩ, Rd = 1 kΩ, and AOL = -104. What is the percent difference?
The strings: S7P5A51 (change - physical+).
The math:
Pj Problem of Interest is of type change (change-physical).
(a) Consider figure 9.1. By KCL, if = iin + i1 -------(1)
So, (vo - v1)/R2 = vd/Rd + v1/R1 --------(2)
So, vo/R2 - vd/Rd = v1(1/R1 + 1/R2) -------(3)
From the open loop ideal op-amp:
open loop gain, AOL = vo/vd
Where vd = vs - v1.
So, vd = vo/AOL
And v1 = vs + vd
So, substituting these values for vd and v1 in eq (3), we have:
vo[1/r2 - (R1 + R2)/(R1R2)AOL - 1/AOLRd] =vs(R1 + R2)/(R1R2)-------(4)
Multiply eq (4) by R1R2/vs, we have:
vo/vs[R1 - (R1 + R2)/AOL - (R1R2)/(AOLRd)] = (R1 + R2)
So, exact gain Av = vo/vs = (R1 + R2)/[R1 - (R1 + R2)/AOL - (R1R2)/(AOLRd)]-------(5)
(b) If op-amp is ideal, then iin = 0
So, vs equals v1 approximaly, so vd = 0
So, eq (2) becomes:
(vo - vs)/R2 = vs/R1 --------(6)
So, ideal Av = vo/vs = 1 + R2/R1 -------(7)
(c) Substituting the given values into eq (5) and eq (7) give exact Av = 10.977 and ideal Av = 11.
So, percent difference = 0.21%. So the ideal op-amp model is an ok approximation.
Blessed are they that have not seen, and yet have believed. John 20:29 |
## The Diagonal Method on Finite Binary Strings
Georg Cantor developed the diagonal method to study infinite sets, but to understand the diagonal method, it is best to begin by applying the technique to finite sets. For example, we could apply the diagonal method to all permutations of a three character binary strings. This set is as follows:
```0 000
1 001
2 010
3 011
4 100
5 101
6 110
7 111```
Anyone who has studied basic computer architecture knows that there are eight ways to order a three character binary string. We can list these permutations in any order. In this case, the first three permutations are: 000, 001, 010. The digits on the diagonal are 000. If I replace the zeros with ones and ones with zeros (the bitnot function), I get the string 111. The string 111 is a permutation of a three character binary string. But it is not in the first three positions of the set!
We can shuffle this set and, in every case, the diagonal method will produce a three character binary string that is not in the first three positions of the set. For example, the first three members of an ordering might be: 100,001,111. The characters on the diagonal are 101. The bitnot function produces the string 010.
This phenomena is not restricted to three character binary strings. We can perform the diagonal method on say a six character binary strings, as follows:
```0 010010
1 111000
2 010101
3 111101
4 100011
5 000111```
In this experiment, the digits on the diagonal are 010111. The diagonal method produces the string 101000. This string is not in the first six strings in the list.
Applying the diagonal method to an n-character binary string produces an n-character string that is not in the first n members of the set. We have just proven that the set of permutations of an n-character binary string is wider than it is deep. We have proven that 2^n > n, where n > 1.
Cantor's diagonal method is function that takes a full set as its input and produces a string of characters as its output. The length of the string is determined by the size of the set. As a computer scientist, I prefer to work in binary. The function takes one character from each member in the set to create a string. It then returns the bitnot of that string.
When looking at a finite set that contains nothing but the diagonal of itself, we get an interesting case of the liar's paradox. Imagine the set A = {dp(A)}. This set contains only one member—the diagonal product of itself. By definition dp(A) must return 0 or 1. It cannot return a 0, because the bitnot of 0 is 1. It cannot return a 1 because the bitnot of 1 is 0. This is the exact same form as the paradox: "This sentence is false." If the sentence were false, then it is true, which means that it is false ...
Whenever you try to include the diagonal product in the first n members of set of n-digit binary strings, your set returns a liar's paradox. For example, were I to define set B = {100, 101, dp(B)}, my set would return an anomaly. dp(B) returns a three digit binary string. The first two characters of the diagonal are 10. The third character cannot be 0 as the bitnot of 0 is 1. It cannot be 1 as the bitnot or 1 is 0.
In Boolean algebra, the not function is a form of negation. The bit not might be called a form of hypernegation. When performed on an infinite, it is a form of ultra-hyper negation. However, it should be clear that when performed on the set of Reals, the absurdity returned by the diagonal method is of the same form as the Liar's Paradox.
Let's say I defined a set A that simply included a
### The Set of Halves
The diagonal method is a function that operates on the first n members of a set. It records enough information about the first members to produce a number that should be part of the main set, but is not yet recorded. This leads me to wonder if it is possible to create a similar operation for the so called "denumerable" sets.
For example, lets look at the set of all halves. The set of all halves can be listed as follows:
``` 1 2 3 4 5 6 7 8 9 ...
1/2 1 3/2 2 5/2 3 7/2 4 9/2 ... ```
Note, there are 2n members of this set that are less than or equal to n. I would like to define a function on the first n members of this set that will return either the number 1/2 or the lowest valued member of the set not in the first n members of the set. The pseudo code for this function might be as follows:
```FUNCTION halver(nArray array of numbers)
RETURN Number
i, n Number;
bArray(n) Array of Boolean;
BEGIN
// n is the length of the array
rv = 1/2+ // set the initial return value to 1/2
n = nArray.length;
// Initialize the values of boolean array to false.
For i=1 to n
str(i) = False;
End Loop;
// Process the in coming array.
For i = 1 to n
If nArray(n) is an integer
bArray(i) = True;
Else
// Do nothing
End If;
If nArray(n) = rv Then
// Find the next lowest value not in list
Loop
rv ++;
Exit When bArray(rv)
End Loop;
If rv = n then
// this will never happen
print "Wow. The set of halves is the same size as the set of integers!!!"
End If;
End If;
End Loop;
RETURN rv;
END;```
Like the diagonal method, this function returns a member of the set of halves that is not yet listed. This new function proves conclusively that 2n > n, where n > 1.
It is possible to extend this function to the set of rational numbers. The rational versions of this function returns the number 1/2, or the lowest natural number not within the first n members of the set. Once again this function proves that the number rational numbers created with the first n numbers is greater than n.
### To Infinity and Beyond
Now, to the heart of the matter. Georg Cantor conjectured that, when performed on a completed infinity, the diagonal method would continue to return a valid real number, while the halver function would stop returning a valid number. This supposition is not proven by the diagonal method. However, the diagonal method gives us a starting place to make the supposition.
Perhaps the best way to describe this is to say that, if you fed the halver function an infinite set, it would return a transfinite number.
Of course, it is impossible to verify this conjecture. As you check through the code for the function, you will see that the halver function blows apart when I try to initialize an infinite array. Just like a program would explode if you tried feeding it an infinite decimal.
### Conclusion
When performed on a set of finite binary strings, the diagonal method simply demonstrates that 2^n > n, for n >1. The diagonal method simply shows that the set of real numbers is wider than it is deep. It is possible to create similar functions for so called denumerable sets.
Just as the diagonal method shows that there is a real number not yet in any given list of real numbers, we can create programs that show there is a rational number not yet in any given list of rational numbers.
Perhaps one of the greatest mistakes that a teacher can make while presenting the diagonal method is to confuse this attribute of the proof on finite sets with the much more subtle distinctions that Cantor was trying to make on infinite sets. |
12.3: Division of Polynomials
Difficulty Level: At Grade Created by: CK-12
Learning Objectives
• Divide a polynomial by a monomial.
• Divide a polynomial by a binomial.
• Rewrite and graph rational functions.
Introduction
A rational expression is formed by taking the quotient of two polynomials.
Some examples of rational expressions are
\begin{align*}\frac{2x}{x^2-1} \qquad \frac{4x^2-3x+4}{2x} \qquad \frac{9x^2+4x-5}{x^2+5x-1} \qquad \frac{2x^3}{2x+3}\end{align*}
Just as with rational numbers, the expression on the top is called the numerator and the expression on the bottom is called the denominator. In special cases we can simplify a rational expression by dividing the numerator by the denominator.
Divide a Polynomial by a Monomial
We’ll start by dividing a polynomial by a monomial. To do this, we divide each term of the polynomial by the monomial. When the numerator has more than one term, the monomial on the bottom of the fraction serves as the common denominator to all the terms in the numerator.
Example 1
Divide.
a) \begin{align*}\frac{8x^2-4x+16}{2}\end{align*}
b) \begin{align*}\frac{3x^2+6x-1}{x}\end{align*}
c) \begin{align*}\frac{-3x^2-18x+6}{9x}\end{align*}
Solution
a) \begin{align*}\frac{8x^2-4x+16}{2}=\frac{8x^2}{2}-\frac{4x}{2}+\frac{16}{2}=4x^2-2x+8\end{align*}
b) \begin{align*}\frac{3x^3+6x-1}{x} = \frac{3x^3}{x}+\frac{6x}{x}-\frac{1}{x}=3x^2+6-\frac{1}{x}\end{align*}
c) \begin{align*}\frac{-3x^2-18x+6}{9x}=-\frac{3x^2}{9x}-\frac{18x}{9x}+\frac{6}{9x}=-\frac{x}{3}-2+\frac{2}{3x}\end{align*}
A common error is to cancel the denominator with just one term in the numerator.
Consider the quotient \begin{align*}\frac{3x+4}{4}\end{align*}.
Remember that the denominator of 4 is common to both the terms in the numerator. In other words we are dividing both of the terms in the numerator by the number 4.
The correct way to simplify is:
\begin{align*}\frac{3x+4}{4}=\frac{3x}{4}+\frac{4}{4}=\frac{3x}{4}+1\end{align*}
A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just \begin{align*}3x\end{align*}. This is incorrect, because the entire numerator needs to be divided by 4.
Example 2
Divide \begin{align*}\frac{5x^3-10x^2+x-25}{-5x^2}\end{align*}.
Solution
\begin{align*}\frac{5x^3-10x^2+x-25}{-5x^2}=\frac{5x^3}{-5x^2}-\frac{10x^2}{-5x^2}+\frac{x}{-5x^2}-\frac{25}{-5x^2}\end{align*}
The negative sign in the denominator changes all the signs of the fractions:
\begin{align*}-\frac{5x^3}{5x^2}+\frac{10x^2}{5x^2}-\frac{x}{5x^2}+\frac{25}{5x^2}=-x+2-\frac{1}{5x}+\frac{5}{x^2}\end{align*}
Divide a Polynomial by a Binomial
We divide polynomials using a method that’s a lot like long division with numbers. We’ll explain the method by doing an example.
Example 3
Divide \begin{align*}\frac{x^2+4x+5}{x+3}\end{align*}.
Solution
When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor.
To start the division we rewrite the problem in the following form:
\begin{align*}& {x+3 \overline{ ) x^2+4x+5 }}\end{align*}
We start by dividing the first term in the dividend by the first term in the divisor: \begin{align*}\frac{x^2}{x}=x\end{align*}.
We place the answer on the line above the \begin{align*}x\end{align*} term:
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\end{align*}
Next, we multiply the \begin{align*}x\end{align*} term in the answer by the divisor, \begin{align*}x + 3\end{align*}, and place the result under the dividend, matching like terms. \begin{align*}x\end{align*} times \begin{align*}(x + 3)\end{align*} is \begin{align*}x^2+3x\end{align*}, so we put that under the divisor:
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \ \ x^2 + 3x\end{align*}
Now we subtract \begin{align*}x^2+3x\end{align*} from \begin{align*}x^2+4x+5\end{align*}. It is useful to change the signs of the terms of \begin{align*}x^2+3x\end{align*} to \begin{align*}-x^2-3x\end{align*} and add like terms vertically:
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x}\\ & \qquad \qquad \quad \ x\end{align*}
Now, we bring down the 5, the next term in the dividend.
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\end{align*}
And now we go through that procedure once more. First we divide the first term of \begin{align*}x + 5\end{align*} by the first term of the divisor. \begin{align*}x\end{align*} divided by \begin{align*}x\end{align*} is 1, so we place this answer on the line above the constant term of the dividend:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\end{align*}
Multiply 1 by the divisor, \begin{align*}x + 3\end{align*}, and write the answer below \begin{align*}x + 5\end{align*}, matching like terms.
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\\ & \qquad \qquad \quad \ x + 3\end{align*}
Subtract \begin{align*}x + 3\end{align*} from \begin{align*}x + 5\end{align*} by changing the signs of \begin{align*}x + 3\end{align*} to \begin{align*}-x -3\end{align*} and adding like terms:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\\ & \qquad \qquad \ \ \underline{-x - 3}\\ & \qquad \qquad \qquad \quad 2\end{align*}
Since there are no more terms from the dividend to bring down, we are done. The quotient is \begin{align*}x + 1\end{align*} and the remainder is 2.
Remember that for a division with a remainder the answer is \begin{align*}\text{quotient}+\frac{\text{remainder}}{\text{divisor}}\end{align*}. So the answer to this division problem is \begin{align*}\frac{x^2+4x+5}{x+3}=x+1+\frac{2}{x+3}\end{align*}.
Check
To check the answer to a long division problem we use the fact that
\begin{align*}(\text{divisor} \times \text{quotient}) + \text{remainder} = \text{dividend}\end{align*}
For the problem above, here’s how we apply that fact to check our solution:
\begin{align*}(x+3)(x+1)+2 & = x^2+4x+3+2\\ & = x^2+4x+5\end{align*}
To check your answers to long division problems involving polynomials, try the solver at http://calc101.com/webMathematica/long-divide.jsp. It shows the long division steps so you can tell where you may have made a mistake.
Rewrite and Graph Rational Functions
In the last section we saw how to find vertical and horizontal asymptotes. Remember, the horizontal asymptote shows the value of \begin{align*}y\end{align*} that the function approaches for large values of \begin{align*}x\end{align*}. Let’s review the method for finding horizontal asymptotes and see how it’s related to polynomial division.
When it comes to finding asymptotes, there are basically four different types of rational functions.
Case 1: The polynomial in the numerator has a lower degree than the polynomial in the denominator.
Take, for example, \begin{align*}y=\frac{2}{x-1}\end{align*}. We can’t reduce this fraction, and as \begin{align*}x\end{align*} gets larger the denominator of the fraction gets much bigger than the numerator, so the whole fraction approaches zero.
The horizontal asymptote is \begin{align*}y = 0\end{align*}.
Case 2: The polynomial in the numerator has the same degree as the polynomial in the denominator.
Take, for example, \begin{align*}y=\frac{3x+2}{x-1}\end{align*}. In this case we can divide the two polynomials:
\begin{align*}& \overset{\qquad \qquad \ 3}{x-1 \overline{ ) 3x+2 \;}}\\ & \qquad \underline{-3x+3}\\ & \qquad \qquad \quad 5\end{align*}
So the expression can be written as \begin{align*}y=3+\frac{5}{x-1}\end{align*}.
Because the denominator of the remainder is bigger than the numerator of the remainder, the remainder will approach zero for large values of \begin{align*}x\end{align*}. Adding the 3 to that 0 means the whole expression will approach 3.
The horizontal asymptote is \begin{align*}y = 3\end{align*}.
Case 3: The polynomial in the numerator has a degree that is one more than the polynomial in the denominator.
Take, for example, \begin{align*}y=\frac{4x^2+3x+2}{x-1}\end{align*}. We can do long division once again and rewrite the expression as \begin{align*}y=4x+7+\frac{9}{x-1}\end{align*}. The fraction here approaches zero for large values of \begin{align*}x\end{align*}, so the whole expression approaches \begin{align*}4x + 7\end{align*}.
When the rational function approaches a straight line for large values of \begin{align*}x\end{align*}, we say that the rational function has an oblique asymptote. In this case, then, the oblique asymptote is \begin{align*}y = 4x + 7\end{align*}.
Case 4: The polynomial in the numerator has a degree that in two or more than the degree in the denominator. For example: \begin{align*}y=\frac{x^3}{x-1}\end{align*}.
This is actually the simplest case of all: the polynomial has no horizontal or oblique asymptotes.
Example 5
Find the horizontal or oblique asymptotes of the following rational functions.
a) \begin{align*}y=\frac{3x^2}{x^2+4}\end{align*}
b) \begin{align*}y=\frac{x-1}{3x^2-6}\end{align*}
c) \begin{align*}y=\frac{x^4+1}{x-5}\end{align*}
d) \begin{align*}y=\frac{x^3-3x^2+4x-1}{x^2-2}\end{align*}
Solution
a) When we simplify the function, we get \begin{align*}y=3-\frac{12}{x^2+4}\end{align*}. There is a horizontal asymptote at \begin{align*}y = 3\end{align*}.
b) We cannot divide the two polynomials. There is a horizontal asymptote at \begin{align*}y = 0\end{align*}.
c) The power of the numerator is 3 more than the power of the denominator. There are no horizontal or oblique asymptotes.
d) When we simplify the function, we get \begin{align*}y=x-3+\frac{6x-7}{x^2-2}\end{align*}. There is an oblique asymptote at \begin{align*}y = x - 3\end{align*}.
Notice that a rational function will either have a horizontal asymptote, an oblique asymptote or neither kind. In other words, a function can’t have both; in fact, it can’t have more than one of either kind. On the other hand, a rational function can have any number of vertical asymptotes at the same time that it has horizontal or oblique asymptotes.
Review Questions
Divide the following polynomials:
1. \begin{align*}\frac{2x+4}{2}\end{align*}
2. \begin{align*}\frac{x-4}{x}\end{align*}
3. \begin{align*}\frac{5x-35}{5x}\end{align*}
4. \begin{align*}\frac{x^2+2x-5}{x}\end{align*}
5. \begin{align*}\frac{4x^2+12x-36}{-4x}\end{align*}
6. \begin{align*}\frac{2x^2+10x+7}{2x^2}\end{align*}
7. \begin{align*}\frac{x^3-x}{-2x^2}\end{align*}
8. \begin{align*}\frac{5x^4-9}{3x}\end{align*}
9. \begin{align*}\frac{x^3-12x^2+3x-4}{12x^2}\end{align*}
10. \begin{align*}\frac{3-6x+x^3}{-9x^3}\end{align*}
11. \begin{align*}\frac{x^2+3x+6}{x+1}\end{align*}
12. \begin{align*}\frac{x^2-9x+6}{x-1}\end{align*}
13. \begin{align*}\frac{x^2+5x+4}{x+4}\end{align*}
14. \begin{align*}\frac{x^2-10x+25}{x-5}\end{align*}
15. \begin{align*}\frac{x^2-20x+12}{x-3}\end{align*}
16. \begin{align*}\frac{3x^2-x+5}{x-2}\end{align*}
17. \begin{align*}\frac{9x^2+2x-8}{x+4}\end{align*}
18. \begin{align*}\frac{3x^2-4}{3x+1}\end{align*}
19. \begin{align*}\frac{5x^2+2x-9}{2x-1}\end{align*}
20. \begin{align*}\frac{x^2-6x-12}{5x^4}\end{align*}
Find all asymptotes of the following rational functions:
1. \begin{align*}\frac{x^2}{x-2}\end{align*}
2. \begin{align*}\frac{1}{x+4}\end{align*}
3. \begin{align*}\frac{x^2-1}{x^2+1}\end{align*}
4. \begin{align*}\frac{x-4}{x^2-9}\end{align*}
5. \begin{align*}\frac{x^2+2x+1}{4x-1}\end{align*}
6. \begin{align*}\frac{x^3+1}{4x-1}\end{align*}
7. \begin{align*}\frac{x-x^3}{x^2-6x-7}\end{align*}
8. \begin{align*}\frac{x^4-2x}{8x+24}\end{align*}
Graph the following rational functions. Indicate all asymptotes on the graph:
1. \begin{align*}\frac{x^2}{x+2}\end{align*}
2. \begin{align*}\frac{x^3-1}{x^2-4}\end{align*}
3. \begin{align*}\frac{x^2+1}{2x-4}\end{align*}
4. \begin{align*}\frac{x-x^2}{3x+2}\end{align*}
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