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Section 2-1 : Arc Length. In this section we are going to look at computing the arc length of a function. Because it’s easy enough to derive the formulas that we’ll use in this section we will derive one of them and leave the other to you to derive.
A finely tuned example demonstrating how the arc length formula works. A finely tuned example demonstrating how the arc length formula works. the second fundamental theorem of calculus here, to evaluate this definite integral. This is going to be 4/9 …
Arc Length. Using Calculus to find the length of a curve. (Please read about Derivatives and Integrals first) Imagine we want to find the length of a curve between two points. And the curve is smooth (the derivative is continuous).
Mar 04, 2017 · This calculus video tutorial explains how to calculate the arc length of a curve using a definite integral formula. This video contains plenty of examples an
Author: The Organic Chemistry Tutor
Arc Length Suppose $f$ is continuously differentiable on the interval $[a,b]$. Let’s derive a formula for the length $L$ of the curve on the interval, called the arc
The arc length of g from y = c to y = d is the integral Definition. Let x = h(t) and y = g(t) be parametric functions such that the derivatives h’ and g’ are continuous on the closed interval [r, s]. The arc length of the graph of this system of parametric functions from t = r to t = s is the integral
Become a professional area-under-curve finder! You will also learn here how integrals can be used to find lengths of curves. The tools of calculus are so versatile!
Here is a set of practice problems to accompany the Arc Length section of the Applications of Integrals chapter of the notes for Paul Dawkins Calculus II course at Lamar University.
Calculus II Calculators; Math Problem Solver (all calculators) Arc Length Calculator for Curve. The calculator will find the arc length of the explicit, polar or parametric curve on the given interval, with steps shown. Show Instructions. In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`.<|endoftext|>
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A cone is a three dimensional object with a plane circular base, a slanted curved surface and one vertex. A cone has 3 dimensions as shown in the figure below: the height, h, the slant height, l, and the radius, r.
In a right circular cone, the height and the radius are perpendicular to each other. Hence the three measures, the height, the radius, and the slant height, of a cone make a right triangle. Then according to the Pythagorean theorem, the length of the slant height of a cone is;
When you open a cone up into its corresponding net as shown below, you can see that the base of the cone opens up into a circle with radius ‘r’ and the curved surface of the cone opens up as part of a circle with radius ‘l’ and an arc of length 2πr (equal to the circumference of the base circle).
Now, the total surface area of a cone = area of the base circle + area of the curved surface
Area of the base circle = π r2
Area of the curved surface of the cone = π r l (to understand why this is, please see below the “Why does it work?” section)
So, the total surface area of a cone = π r2 + π r l = π r (r + l)
Hence, to find the surface area of a cone, follow the steps below:
1. First find the radius of the circular base (r) and the slant height of the cone (l).
2. Now find the sum of the radius and the slant height.
3. Then multiply the sum you got in step 2 by the product of radius and the constant π. That’s your answer.
For example: say you need to find the total surface area of a cone with diameter 14 cm and slant height 13 cm.
Let’s try another example: say you need to find the surface area of a cone with radius 6 cm and height 8 cm.
To find just the curved surface area of a cone, use the formula π r l.
For example: say you need to find the curved surface area of a cone with radius 1.4 cm and slant height 5 cm.
Using the above formulas for the surface area (or curved surface area), you can also find any missing value when surface area (or curved surface area) and any one of the measure is given.
For example: say you need to find the height of a cone that has a total surface area of 216 π cm2 and a base radius of 9 cm.
Surface area of the cone = 216 π cm2 radius = 9 cm
π r (r + l) = 216 π
r ( r + l) = 216
9 (9 + l) = 216
9 + l = 216/9 = 24
l = 24 – 9 = 15 cm
You are watching: Surface area of cones. Info created by GBee English Center selection and synthesis along with other related topics.<|endoftext|>
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Verifying Trigonometric Identities
# Verifying Trigonometric Identities
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
• PRACTICE (online exercises and printable worksheets)
An identity is a mathematical sentence that is always true.
(A bit more precisely—everywhere it is defined, it is true.)
To verify (or prove) an identity means to prove that the sentence is always true.
This section discusses two common approaches for verifying (proving) identities.
Fundamental Trigonometric Identities talks about identities in general, and their usefulness.
It also introduces three important trigonometric identities (listed below). Review this earlier section as needed.
• the Pythagorean Identity: $\,\sin^2 t + \cos^2 t = 1\,$
• cosine is an even function: $\,\cos (-t) = \cos t$
• sine is an odd function: $\,\sin(-t) = -\sin t\,$
## Tools for Verifying Trigonometric Identities
Verifying identities is all about renaming—to show that two (initially different-looking) expressions are actually the same.
You'll use:
• familiar algebraic tools: e.g., factoring, distributive laws, getting common denominators
• definitions of the trigonometric functions $$\cssId{s17}{\text{For simplicity, inputs are not shown:}}\qquad \cssId{s18}{\tan = \frac{\sin}{\cos}}\qquad \qquad \cssId{s19}{\cot = \frac{\cos}{\sin} = \frac{1}{\tan}}\qquad \qquad \cssId{s20}{\sec = \frac{1}{\cos}}\qquad \qquad \cssId{s21}{\csc = \frac{1}{\sin}}$$
• trigonometric identities that have already been proved
This varies from teacher-to-teacher and course-to-course.
Always ask about which identities you're allowed to use when verifying new ones.
For the exercises in this section, you're only ‘allowed’ the three fundamental trigonometric identities cited above,
together with two that follow immediately from the Pythagorean Identity:
Start with: $\,\sin^2 + \cos^2 = 1\,$ (For simplicity, inputs are not shown.) divide both sides by $\,\cos^2\,$: $$\begin{gather} \cssId{s30}{\frac{\sin^2 + \cos^2}{\cos^2} = \frac{1}{\cos^2}}\cr\cr \cssId{s31}{\frac{\sin^2}{\cos^2} + \frac{\cos^2}{\cos^2} = \frac{1}{\cos^2}}\cr\cr \cssId{s32}{\left(\frac{\sin}{\cos}\right)^2 + 1 = \left(\frac{1}{\cos}\right)^2}\cr\cr\cr \cssId{s33}{\color{red}{\tan^2 + 1 = \sec^2}} \end{gather}$$ divide both sides by $\,\sin^2\,$: $$\begin{gather} \cssId{s35}{\frac{\sin^2 + \cos^2}{\sin^2} = \frac{1}{\sin^2}}\cr\cr \cssId{s36}{\frac{\sin^2}{\sin^2} + \frac{\cos^2}{\sin^2} = \frac{1}{\sin^2}}\cr\cr \cssId{s37}{1 + \left(\frac{\cos}{\sin}\right)^2 = \left(\frac{1}{\sin}\right)^2}\cr\cr\cr \cssId{s38}{\color{red}{1 + \cot^2 = \csc^2}} \end{gather}$$
In this lesson, the identities ‘$\,\color{red}{\tan^2 x + 1 = \sec^2 x}\,$’ and ‘$\,\color{red}{1 + \cot^2 x = \csc^2 x} \,$’
are referred to as alternate Pythagorean Identities.
You should be able to easily recognize equivalent versions of the five already-proved identities you're allowed to use in this section:
the Pythagorean Identity and its two alternate versions, cosine is even, and sine is odd.
For example, ‘$\,\sin^2 x = 1 - \cos^2 x\,$’ should be easily identified as an equivalent version of the Pythagorean Identity.
• Write Everything in Terms of Sines and Cosines:
If an expression involves a ‘mix’ of trigonometric functions,
it sometimes helps to write everything in terms of just sines and cosines.
• the Conjugate Technique:
By definition, $\,x + y\,$ and $\,x - y\,$ are called conjugates.
Thus, the conjugate of $\,x + y\,$ is $\,x - y\,$; the conjugate of $\,x - y\,$ is $\,x + y\,$.
When a numerator or denominator in a (suspected) trigonometric identity is of the form $\,1 \pm \sin x\,$ or $\,1\pm \cos x\,$,
then it often helps to multiply both numerator and denominator by the conjugate.
(Thus, you're multiplying the original expression by $\,1\,$ in the form $\,\frac{\text{conjugate}}{\text{conjugate}}\,$.)
Why? This reduces two terms to a single term, which is often easier to work with.
Below are a couple examples. (Review FOIL as needed.) $$\cssId{s57}{\overbrace{(1 - \sin x)}^{\text{original expression}}} \cssId{s58}{\cdot}\ \ \cssId{s59}{\overbrace{(1 + \sin x)}^{\text{conjugate}}} \qquad \cssId{s60}{\overbrace{=\strut}^{\text{FOIL}}}\qquad \cssId{s61}{1 - \sin^2 x} \cssId{s62}{\overbrace{=\strut}^{\text{Pythagorean Identity}}} \cssId{s63}{\cos^2 x}$$ $$\cssId{s64}{\overbrace{(1 + \cos x)}^{\text{original expression}}} \cssId{s65}{\cdot}\ \ \cssId{s66}{\overbrace{(1 - \cos x)}^{\text{conjugate}}} \qquad \cssId{s67}{\overbrace{=\strut}^{\text{FOIL}}}\qquad \cssId{s68}{1 - \cos^2 x} \cssId{s69}{\overbrace{=\strut}^{\text{Pythagorean Identity}}} \cssId{s70}{\sin^2 x}$$ As illustrated in an example below, the conjugate technique often helps out for other binomial (two-term) expressions, too!
## Two Basic Approaches for Verifying Identities
Equations of the form: COMPLICATED = SIMPLER COMPL1 = COMPL2 Often, one side of a (suspected) identity is more complicated than the other. In such cases, it is usually easiest to: start with the complicated side apply tools (one at a time) to rename it as the simpler side \cssId{s79}{ \begin{align} \text{C O M}&\text{P L I C A T E D}\cr &= \ \text{}\cr &= \ \text{}\cr &= \ \ldots\cr &= \ \text{SIMPLER} \end{align}} Often, both sides of a (suspected) identity are similarly complicated. In such cases, you can often work with each side separately, renaming each side as the same simpler expression: \cssId{s83}{ \begin{align} \text{COMPL1} &= \ \text{}\cr &= \ \ldots\cr &= \ \color{red}{\text{SIMPLER}}\cr\cr \text{COMPL2} &= \ \text{}\cr &= \ \ldots\cr &= \ \color{red}{\text{SIMPLER}} \end{align}} The two $\,\color{red}{\text{SIMPLER}}\,$ expressions are the same! (Sometimes, when renaming one side,you inadvertently end up with the other side. If so—you're done!) Notes: if a complicated side is long, put the first simplication on the next line, indented a bit for readability, line up all equal signs you may want to provide a reason for each step MANY IDENTITIES CAN BE VERIFIED IN MANY DIFFERENT WAYS! (In the examples and exercises, only one correct approach is shown.)
## EXAMPLE of ‘COMPLICATED = SIMPLER’:Turning a Complicated Side into the Simpler Side
Verify the identity: $\,\displaystyle \tan x + \frac{\cos x}{1 + \sin x} = \sec x\,$
SOLUTION:
The left side ($\,\tan x + \frac{\cos x}{1 + \sin x}\,$) is more complicated than the right side ($\,\sec x\,$).
rename it to match the expression on the right.
The solution shown here first rewrites everything in terms of sines and cosines, and then uses the conjugate technique: \begin{alignat}{2} &\cssId{s103}{\tan x + \frac{\cos x}{1 + \sin x}} &\qquad\qquad& \cssId{s104}{\text{(start with left side)}} \cr\cr &\qquad\cssId{s105}{= \ \frac{\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}} &&\cssId{s106}{\text{(definition of tangent)}}\cr\cr &\qquad\cssId{s107}{= \ \frac{\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}\cdot\frac{1 - \sin x}{1 - \sin x}} &&\cssId{s108}{\text{(conjugate technique)}}\cr\cr &\qquad\cssId{s109}{= \ \frac{\sin x}{\cos x} + \frac{(\cos x)(1 - \sin x)}{\cos^2 x}} &&\cssId{s110}{\text{(FOIL, Pythagorean Identity)}}\cr\cr &\qquad\cssId{s111}{= \ \frac{\sin x}{\cos x} + \frac{1 - \sin x}{\cos x}} &&\cssId{s112}{\text{(cancel)}}\cr\cr &\qquad\cssId{s113}{= \ \frac{1}{\cos x}} &&\cssId{s114}{\text{(add fractions)}}\cr\cr &\qquad\cssId{s115}{= \ \sec x} &&\cssId{s116}{\text{(definition of secant)}}\cr \end{alignat}
## EXAMPLE of ‘COMPL1 = COMPL2’:Working With Both Sides Separately
Verify the identity: $\,\displaystyle \frac{1 + \cos t}{\cos t} = \frac{\tan^2 t}{\sec t - 1}\,$
SOLUTION:
The left side ($\,\frac{1 + \cos t}{\cos t}\,$) and the right side ($\,\frac{\tan^2 t}{\sec t - 1}\,$) are similarly complicated.
Therefore, work with both sides separately, trying to reduce each to the same simpler expression.
Here, ‘LHS’ and ‘RHS’ stand for ‘left-hand side’ and ‘right-hand side’.
Note that the conjugate technique is successfully applied to an expression of the form $\,\sec t - 1\,$.
\begin{alignat}{2} &\cssId{s126}{\text{LHS}} \cssId{s127}{= \frac{1 + \cos t}{\cos t}} &\qquad\qquad& \cssId{s128}{\text{(start with LHS)}} \cr\cr &\qquad\cssId{s129}{= \ \frac{1}{\cos t} + \frac{\cos t}{\cos t}} &&\cssId{s130}{\text{(algebra)}}\cr\cr &\qquad\cssId{s131}{= \ \color{red}{\sec t + 1}} &&\cssId{s132}{\text{(definition of secant)}}\cr \end{alignat} \begin{alignat}{2} &\cssId{s133}{\text{RHS}} \cssId{s134}{= \frac{\tan^2 t}{\sec t - 1}} &\qquad\qquad& \cssId{s135}{\text{(start with RHS)}} \cr\cr &\qquad\cssId{s136}{= \ \frac{\tan^2 t}{\sec t - 1}\cdot\frac{\sec t + 1}{\sec t + 1}} &&\cssId{s137}{\text{(conjugate technique)}}\cr\cr &\qquad\cssId{s138}{= \ \frac{\tan^2 t(\sec t + 1)}{\sec^2 t - 1}} &&\cssId{s139}{\text{(FOIL)}}\cr\cr &\qquad\cssId{s140}{= \ \frac{\tan^2 t(\sec t + 1)}{\tan^2 t}} &&\cssId{s141}{\text{(alternate Pythagorean Identity)}}\cr\cr &\qquad\cssId{s142}{= \ \color{red}{\sec t + 1}} &&\cssId{s143}{\text{(cancel)}}\cr \end{alignat} Notice that both the LHS and RHS have been renamed as the same expression, $\,\sec t + 1\,$.
## Graphically Checking that an Equation is an Identity
For many years, I worked in a math lab. Any student could come in to get help.
I have a recollection of a time when I learned an important lesson.
I was helping someone verify an identity—but nothing was working! We tried oodles of things, for a very long time.
Finally, the thought occurred to me that perhaps it wasn't an identity after all—maybe it was just a mistake.
Indeed, it was a typo in the book.
If you ever find yourself in the same situation, check that you're really working with an identity.
If the LHS and RHS are always equal then they will have precisely the same graphs.
At WolframAlpha, if you type in two comma-separated expressions, then both will be graphed in the same coordinate system.
If you graph both sides of an identity, then it will look like a single graph, because the graphs will overlap perfectly!
For example, cut-and-paste the following into WolframAlpha:
tan(x) + cos(x)/(1 + sin(x)) , sec x
Here's what you'll see:
The blue and red graphs overlap perfectly, giving the appearance of a single graph. This graphically shows that ‘$\,\displaystyle \tan x + \frac{\cos x}{1 + \sin x} = \sec x\,$’ is an identity. If you prefer, you can get an equivalent equation with zero on one side: $$\cssId{s163}{\tan x + \frac{\cos x}{1 + \sin x} - \sec x = 0}$$ Since this is an identity, the left side graphs as the zero function. Not much to see!
## How to Prove that an Equation is NOT an Identity
If the graphs don't match up (and you've input both sides correctly), then you're not working with an identity.
To prove that a sentence is not an identity, you need only provide a single number for which the sentence is false.
For example, ‘$\,\displaystyle \tan x + \frac{\cos x}{1 + \sin x} = 1 + \sec x\,$’ is not an identity:
• when $\,x = 0\,$, the LHS is: $\displaystyle \,\tan 0 + \frac{\cos 0}{1 + \sin 0} = 0 + \frac{1}{1 + 0} = 1\,$
• when $\,x = 0\,$, the RHS is: $\displaystyle \,1 + \sec 0 = 1 + \frac{1}{\cos 0} = 1 + \frac{1}{1} = 2\,$
• $\,1\ne 2\,$
Master the ideas from this section<|endoftext|>
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An addictive chemical in the brain makes students crave instant feedback and studies show that they learn better when teachers feed it. We’ve been hearing that formative assessment – which includes instant feedback – is a very effective tool for learning but science is starting to explain why.
Dopamine and Learning
You may have heard of dopamine – a neurotransmitter released by the brain. This chemical is responsible for triggering so many things in our brains including memory, attention, motivation, pleasurable rewards and learning.
Dopamine is released during pleasurable activities such as eating or playing games and can lead to addictions when not experienced in moderation.
Dopamine also triggers the storing of memory. This is why you experience fond feelings when you enter your favorite restaurant, for example. The smell and sights are cluing your brain into all the pleasurable experiences you’ve had there.
This is also why dopamine improves attention. Your brain is alert so it doesn’t miss the great rewards ahead.
Feedback = More Motivation
Gaming is a classic dopamine trigger. No parent or teacher will disagree with the pull playing games has on motivation – especially video games.
A recent study investigated the effect of feedback after a gaming scenario. The study found that negative feedback decreased feelings of competence but motivated correcting performance in the short-term. Positive feedback boosted long-term motivation.
In other words: Both positive and negative feedback increases motivation!
As this article puts it:
Another theory suggests dopamine is a “teaching signal,” like a coach who tells his player “good job” or “bad job” to encourage a future reward. In the current study, U-M researchers describe those dopamine fluctuations as a continuous cheer to motivate, with brief moments of criticism.
GradeCam – An Instant Feedback Tool
Putting all that research together suggests teachers can help students learn by creating more opportunities for instant feedback in the classroom. GradeCam was created as an instant feedback machine – literally!
Teachers can quickly and easily customize, score, and record assessments – without special forms, equipment or buy in. Create tests or quizzes with our online software and print scan forms on plain paper. Then, use our easy online grader to score assessments in an instant by scanning them with any web camera, iPhone or Android device.
We even have a student scanning mode so students can see their scores instantly – including which questions they missed so they can go back and correct.
Teachers tell us all the time how much the instant feedback of GradeCam motivates their students. You just might catch them doing a scan dance!
Here’s how to create your own grading station for students to see their scores instantly.
Sign Up for a Free Account!
Are you getting a dopamine hit right now thinking about all the ways you can motivate your students? Sign up for a free GradeCam account below and start feeding those feedback addicts!<|endoftext|>
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# PLOTTING DATA AND INTERPRETING GRAPHS
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1 PLOTTING DATA AND INTERPRETING GRAPHS Fundamentals of Graphing One of the most important sets of skills in science and mathematics is the ability to construct graphs and to interpret the information they convey. In this write - up, we will investigate different graphs and their connection to the physics of motion. First, let' s construct some simple graphs. We will start with one of the simplest graphs we know, the graph of the function y = x. In standard notation y is the dependent variable and is plotted on the vertical axis; x is the independent variable and is plotted on the horizontal axis. Question for discussion : Think of some everyday examples of variables, such as the height and weight of a person. Which is the dependent variable and which is the independent variable? Try to think of some more examples of dependent and independent variables. When we plot y = x we are constructing a graphical representation of the relationship between the variables y and x; in other words, we can look at this graph and understand how the variable represented by y relates to the variable represented by x. To construct the graph of an equation, we first create a table of values. So for this very simple case, we construct the table : x y This table will provide us the information to plot each (x, y) point on our graph. Even though this is one of the the simplest equations we can graph, notice that it is good form to use some negative values of x (if negative values of x are meaningful for the equation) to ensure we understand the behavior of the graph. If we make use of the table above and plot those points on a graph (red dots below) and then draw the line that best connects the point, our graph should like something like :
2 2 graphing.nb Graph of y=x Suppose now we want to construct a slightly more compliated plot, that of y = 3 x. We should recognize that this graph is very similar to the one above, except that each value of y is three times its corresponding value of x. Our table of values is : x y We can compare the behavior of the two graphs by superimposing them on one set of axes :
3 and this is the value of the slope between the points x = 1 and x = 3. Since this is a straight line, we can see that the slope is the same at all points (this is a defining property of a straight line); let' s verify this by computgraphing.nb 3 Graphs of y=x and y=3x Where the data points are color coded (blue for the graph y = x and red for the graph y = 3 x). Questions : How are the two graphs in the plot above similar? How are they different? What causes this difference in the two graphs? Slope of a Line You should be able to see that the graph of y = 3 x is steeper than the graph of y = x. You know from previous math courses that the steepness of a line is called its slope. The slope tells you the ratio of how the function is changing in the y direction compared to its change in the x direction. There are a number of ways of describing the slope mathematically, among the simplest is : slope = change in y value change in x value We can use our tables of values above to determine the slope of the lines created by the equations y = x and y = 3 x. Let' s start by finding the slope of the line y = 3 x; the slope will be the ratio of the change in y value to change in x value. So as the funtion y = 3 x changes in x value from 1 to 2, its y value changes from 3 to 6 and we have : slope = change in y value change in x value = =Å3 1 = 3
4 4 graphing.nb ing the slope between the points x = 0 and x = 3. When x = 0, y = 0; when x = 3, y = 9, so : slope = change in y value change in x value = =Å9 3 = 3 and we should not be surprised to find the slope is still 3. It should now be a simple task to compute the slope of the line y = x. What is the slope of this line (y = x)? Can you look at the equations for the two lines and determine what factor in those equations is responsible for determing the slope of the line? The y - intercept Now, let' s use a slightly more complicated function, y = 3 x + 4. What do you think the slope of this line will be? (Make a prediction before we plot it). We can also construct this data table : x y Plot the graphs of y = 3 x and y = 3 x + 4 on one set of axes one of the sheets of graph paper I have given you. Now, after everyone has completed their graphs, let' s see how your graph compares to :
5 graphing.nb 5 Graphs of y=x and y=3x Questions : How are these graphs similar? How are they different? Can you determine what are the factors in the equations that lead to the differences in these graphs? The Equation of a Line We are now in a position to determine a general equation that will describe all straight lines. This is a well known equation in math and is usually presented as : y = mx + b (1) In this equation, y and x represent, as they often do, the dependent and independent variables; m represents the slope of the line, and b is the y - intercept, and is the y value of the straight line when it crosses the y axis. So in our last equation above, we have y = 3 x + 4. This means that we have a straight line of slope 3 and whose y intercept is 4. When a line is written in the format of y = m x + b, we can immediately determine the slope by identifying the coefficient of the x term. We can also immediately determine the
7 graphing.nb 7 Working in groups, construct an experiment (or series of experiments) that will allow you to measure the data you need to make this determination. You can use the same equipment from the last few weeks; ramps, carts, stopwatches and meter sticks. Once you have determined your procedure, carry out the observations needed. For next week, each student should submit : a clear explanation of the procedure you used; you should write this up as if you were writing the lab instructions for a middle school class experiment your data in a clear tabular form your graph of distance vs. time an explanation of your conclusion whether dist vs. time is a linear or non - linear relationship; your conclusion must be explicitly supported by your data. Each student should submit an individual report, but you should indicate who your lab partners were.
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### Plot the following two points on a graph and draw the line that passes through those two points. Find the rise, run and slope of that line.
Objective # 6 Finding the slope of a line Material: page 117 to 121 Homework: worksheet NOTE: When we say line... we mean straight line! Slope of a line: It is a number that represents the slant of a line<|endoftext|>
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How to Insert Comments in C Source Code
Like other programming languages, C lets you insert comments in your code. The comments are skipped over by the compiler but remain in the source code as a way for you to jot down notes, your intentions, general information, or rude remarks about your customers.
In the C language, comments are enclosed between two sets of characters, /* and */,
/* like this. */
Everything between those two pairs of characters is ignored by the compiler.
Note that both character sets are required. Unlike other programming languages, C doesn’t limit comments to a single line or from a certain position to the end of a line. Comments in C are more like a marked block in a word processor; the comment has a beginning and an end. This can cause some confusion, especially if you’re not using a color-coded editor. (In a color-coded editor, commented text shows up as one color, making it easy to see where a /* or */ is misplaced.)
The most basic form of comment is a source code heading, which identifies you as the programmer as well as any other information you think you should save.
/* DUMMIES.C source code. By Ima C. Programmer. 28Feb2010 */
This heading is used to identify the program, the programmer, and the date. Often, other information is put in there as well, depending on the circumstances under which the program was written. It’s just basic information that might be useful later.
Comments can span multiple lines, too, like this:
/* NOTE: added a beep to the following line The a is the Alert or bell character */
By having the /* start at one line and the */ end at another, the single comment is able to span two lines.
Comments don’t add anything to the size of the program, nor do they slow down the code. They’re just notes or suggestions for you or any other programmer who may work on the code again later.<|endoftext|>
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Second president of the United States
Short Family History
Mother, Susanna Adams was a descendent of the Boylstons of Brookline, a very important family during this time.
Cousin, Samuel Adams, was one of the first to argue for independence in the colonies.
John Adams, born October 30, 1735 in Braintree Massachusetts, was the second president, also first Vice President, and earned a scholarship to Harvard at the age of 16. Also helped draft the Declaration of Independence.
Political affiliations/ loyalties with explanation of views
Contribution to/ participation in colonial development/ colonial independence
In 1761, John Adams began to act against British actions that he believed put a limit on colonial freedom and the right of Massachusetts and other colonies to self-government.
Adams was a delegate to the First and Second Continental Congresses, Adams was a leader in the movement for independence for the colonies. Adams made speeches for independence from England. The Congress voted for independence which meant they were choosing to go to war with England but Adams stood strong with his opinion and kept fighting for independence. He served on the committee that was in charge with preparing the Declaration of Independence with Thomas Jefferson. In 1779, Adams drafted the Massachusetts Constitution which would serve as a model for the U.S. Constitution. From 1785 to 1788 he was Minister to the Court of St. James's where he was an American ambassador to Great Britain. He helped create a basis for the many fights/attempts to gain independence from England and because of this gained a lot of respect from his peers.
Did John Adams haven integrity?
John Adams shows his integrity when he agrees to represent the 5 soldiers who were accused of shooting civilians known as the Boston Massacre. He was able to free 6 of the 8 soldiers that were on trial and after the trial, civilians realized that Adams was a very fair and courageous man and that they probably would have done the same thing as the soldiers if they were put into a situation like that. His integrity portrayed to the people because we won the election for presidency against Thomas Jefferson, and you can't be a president without being consistent and trustworthy to citizens. He also was there to help his fellow people whenever they needed him and never said no because he was willing to help and did not mind it, even when he wasn't in his presidency.
The character of an individual viewed as a member of society; behavior in terms of the duties, obligations, and functions of a citizen: an award for good citizenship.
Did John Adams have citizenship?
John Adams proved in his time that he had citizenship in that he devoted himself to independence and fight against Britain. At the First and Second Continental Congresses, Adams used his writing and speaking skills to persuade other colonists to realize the need to go against Britain and the cause for independence. He served on the committee which drafted the Declaration of Independence and during the Revolutionary War he ran the Board of War raising and making sure the army was equipped and created a navy. Adams was also looked at as a great man with good principles, shown when he stood up for British soldiers after the Boston Massacre. Even though they were against the colonists Adams did what he thought was fair.<|endoftext|>
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When it was recognized that changes (mutations) in genes occur spontaneously (T. H. Morgan, 1910) and can be induced by X-rays (H. J. Muller, 1927), the mutation theory of heredity became a cornerstone of early genetics. Genes were defined as mutable units, but the question what genes and mutations are remained. Today we know that mutations are changes in the structure of DNA and their functional consequences. The study of mutations is important for several reasons. Mutations cause diseases, including all forms of cancer. They can be induced by chemicals and by irradiation. Thus, they represent a link between heredity and environment. And without mutations, well-organized forms of life would not have evolved.
The following two plates summarize the chemical nature of mutations.
Table of Content
A. Error in replication
The synthesis of a new strand of DNA occurs by semiconservative replication based on complementary base pairing. Errors in replication occur at a rate of about 1 in 10 5 . This rate is reduced to about 1 in 10 7 to 10 9 by proofreading mechanisms. When an error in replication occurs before the next cell division (here referred to as the first division after the mutation), e.g., a cytosine (C) might be incorporated instead of an adenine (A) at the fifth base pair as shown here, the resulting mismatch will be recognized and eliminated by mismatchre pair in most cases.
However, if the error is undetected and allowed tostand, the next(second) division will result in a mutant molecule containing a CG instead of an AT pair at this position. This mutation will be perpetuated in all daughter cells. Depending on its location within or outside of the coding region of a gene, functional consequences due to a change in a codon could result.
B. Mutagenic alteration of a nucleotide
A mutation may result when a structural change of a nucleotide affects its base-pairing capability. The altered nucleotide is usually present in one strand of the parent molecule. If this leads to incorporation of a wrong base, such as a C instead of a T in the fifth base pair as shown here, the next (second) round of replication will result in two mutant molecules.
C. Replication slippage
A different class of mutations does not involve an alteration of individual nucleotides, but results from incorrect alignment between allelic or nonallelic DNA sequences during replication. When the template strand contains short tandem repeats, e.g., CA repeats as in microsatellites (see DNA polymorphism and Part II, Genomics), the newly replicated strand and the template strand may shift their positions relative to each other. With replication or polymerase slippage, leading to incorrect pairing of repeats, some repeats are copied twice or not at all, depending on the direction of the shift. One can distinguish forward slippage (shown here) and backward slippage with respect to the newly replicated strand. If the newly synthesized DNA strand slips forward, a region of nonpairing remains in the parental strand. Forward slippage results in an insertion. Backward slippage of the new strand results in deletion. Microsatellite instability is a characteristic feature of hereditary nonpolyposis cancer of the colon (HNPCC). HNPCC genes are localized on human chromosomes at 2p15–22 and 3p21.3. About 15% of all colorectal, gastric, and endometrial carcinomas show microsatellite instability. Replication slippage must be distinguished from unequal crossing-over during meiosis. This is the result of recombination between adjacent, but not allelic, sequences on nonsister chromatids of homologous chromosomes (Figures redrawn from Brown, 1999).<|endoftext|>
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# Graph Points on the Coordinate Plane
In order to graph points on the coordinate plane, you have to understand the organization of the coordinate plane and know what to do with those (x, y) coordinates. If you want to know how to graph points on the coordinate plane, just follow these steps.
## Steps
### Understanding the Coordinate Plane
1. Understand the axes of the coordinate plane. When you're graphing a point on the coordinate plane, you will graph it in (x, y) form. Here is what you'll need to know:
• The x-axis goes left and right, the second coordinate is on the y-axis.
• The y-axis goes up and down.
• Positive numbers go up or right (depending on the axis). Negative numbers go left or down.
2. Understand the quadrants on the coordinate plane. Remember that a graph has four quadrants (typically labeled in Roman numerals). You will need to know which quadrant the plane is in.
• Quadrant I gets (+,+); quadrant I is above and to the left of the y-axis.
• Quadrant IV gets (+,-); quadrant IV is below the x-axis and to the right of the y-axis. (5,4) is in quadrant I.
### Graphing a Single Point
1. Start at (0, 0), or the origin. Just go to (0, 0), which is the intersection of the x and y axes, right in the center of the coordinate plane.[1]
2. Move over x units to the right or left. Let's say you're working with the set of coordinates (5, -4). Your x coordinate is 5. Since five is positive, you'll need to move over five units to the right. If it was negative, you would move over 5 units to the left.
3. Move over y units up or down. Start where you left off, 5 units to the right of (0, 0). Since your y coordinate is -4, you will have to move down four units. If it were 4, you would move up four units.
4. Mark the point. Mark the point you found by moving over 5 units to the right and 4 units down, the point (5, -4), which is in the 4th quadrant. You're all done.
1. Learn how to graph points if you're working with an equation. If you have a formula without any coordinates, then you'll have to find your points by choosing a random coordinate for x and seeing what the formula spits out for y. Just keep going until you've found enough points and can graph them all, connecting them if necessary. Here's how you can do it, whether you're working with a simple line, or a more complicated equation like a parabola:
• Graph points from a line. Let's say the equation is y = x + 4. So, pick a random number for x, like 3, and see what you get for y. y = 3 + 4 = 7, so you have found the point (3, 7).
• Graph points from a quadratic equation. Let's say the equation of the parabola is y = x2 + 2. Do the same thing: pick a random number for x and see what you get for y. Picking 0 for x is easiest. y = 02 + 2, so y = 2. You have found the point (0, 2).
2. Connect the points if necessary. If you have to make a line graph, draw a circle, or connect all of the points of a parabola or another quadratic equation, then you'll have to connect the points. If you have a linear equation, then draw lines connecting the points from left to right. If you're working with a quadratic equation, then connect the points with curved lines.
• Unless you are only graphing a point, you will need at least two points. A line requires two points.
• A circle requires two points if one is the center; three if the center is not included (Unless your instructor has included the center of the circle in the problem, use three).
• A parabola requires three points, one being the absolute minimum or maximum; the other two points should be opposites.
• A hyperbola requires six points; three on each axis.
3. Understand how modifying the equation changes the graph. Here are the different ways that modifying the equation changes the graph:
• Modifying the x coordinate moves the equation left or right.
• Adding a constant moves the equation up or down.
• Turning it negative (multiplying by -1) flips it over; if it is a line, it will change it from going up to down or going down to up.
• Multiplying it by another number will either increase or decrease the slope.
4. Follow an example to see how modifying the equation changes the graph. Consider the equation y = x^2 ; a parabola with its base at (0,0). Here are the differences you will see as you modify the equation:
• y = (x-2)^2 is the same parabola, except it is graphed two spaces to the right of the origin; its base is now at (2,0).
• y = x^2 + 2 is still the same parabola, except now it is graphed two spaces higher at (0,2).
• y = -x^2 (the negative is applied after the exponent ^2) is an upside down y = x^2; its base is (0,0).
• y = 5x^2 is still a parabola, but it gets larger even faster, giving it a thinner look.
## Tips
• If you are making these, you will most likely have to read them too. A good way to remember to go along the x axis first and the y second, is to pretend that you are building a house, and you have to build the foundation (along the x axis) fist before you can build up. This is the same the other way; if you go down, pretend you are making the basement. You still need a foundation, and to start at the top.
• A good way of remembering which axis is which is to imagine the vertical axis having a small slanted line on it, making it look like a "y".
• Axes are basically horizontal and negative number lines, with both intersecting at the origin (the origin on a coordinate plane is zero, or where both axes intersect). Everything "originates" from the origin.
## You may like
"Like" us to know more!<|endoftext|>
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Hong Kong
Stage 1 - Stage 3
# Simple Interest
Lesson
It costs money to borrow money. The extra money that banks and other lenders charge us to borrow money is called interest. However, interest may also refer to additional money you earn from investing money, such as in a savings accounts. There are different types of interest and today we are going to learn about simple interest.
## What is simple interest?
Simple, or straight line interest is a method where the interest amount is fixed (i.e. it doesn't change). The interest charge is always based on the original principal, so interest on interest is not included.
It is calculated using the formula:
$I=PRT$I=PRT
where $P$P is the principal (the initial amount borrowed)
$R$R is the interest rate, expressed as a decimal or fraction
$T$T is the number of time periods (the duration of the loan)
#### Examples
##### Question 1
Calculate the simple interest on a loan of $\$8580$$8580 at 2%2% p.a. for 1010 years. Write your answer to the nearest cent. Think: We can sub in the values for the principal, interest rate and time periods. Do: II == PRTPRT == 8580\times0.02\times108580×0.02×10 == \1716$$1716
##### Question 2
The interest on an investment of $\$3600$$3600 over 1010 years is \2520.00$$2520.00. If the annual interest rate is $R$R, find $R$R as a percentage correct to $1$1 decimal place.
Think: What values do we know that we can sub in?
Do:
$I$I $=$= $PRT$PRT $2520$2520 $=$= $3600\times R\times10$3600×R×10 $2520$2520 $=$= $36000R$36000R $R$R $=$= $\frac{2520}{36000}$252036000 $R$R $=$= $0.07$0.07 $R$R $=$= $7%$7%
##### Question 3
For a simple interest rate of $6%$6% p.a. , calculate the number of years $T$T needed for an interest of $\$1174.20$$1174.20 to be earned on the investment \1957$$1957.<|endoftext|>
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The type must work with == operator and the result should have standard semantics.
c, expressions of type
The following expressions must be valid and have their specified effects
a == b
| implicitly convertible to bool
|| Establishes equivalence relation with the following properties:
- For all values of
a, a == a yields true.
- If a == b, then b == a
- If a == b and b == c, then a == c
To satisfy this requirement, types that do not have built-in comparison operators have to provide a user-defined operator==.
For the types that are both
LessThanComparable, the C++ standard library makes a distinction between equality, which is the value of the expression a == b and equivalence, which is the value of the expression !(a < b) && !(b < a).<|endoftext|>
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- Grade Level:
- Eighth Grade-College Undergraduate Level
- Biodiversity, Biology: Animals, Chemistry, Climate Change, Conservation, Earth Science, Ecology, Environment, Mathematics, Science and Technology, Service Learning, Wildlife Biology
- 60 minutes
- Group Size:
- Up to 60
- National/State Standards:
- North Carolina Standards-Earth/Environmental Science EEn.2.2.1, 2.2.2, 2.5.5, 2.7-2.7.3; Biology Bio.2.1.1, 2.2.2, 2.1.4, 2.2.1, 2.2.2, 3.5.2, AP Biology 1.01-1.04, 6.02, 6.05; AP Environmental 1.0-1.04, 2.04-2.05, 4.04, 5.01-5.03, 6.01, 7.03-7.04
OverviewThe Great Smoky Mountains are known as the “Salamander Capital of the World!” Salamanders are an especially abundant and diverse group in the park. Researchers use salamanders as a bio-indicator to help assess the health of our forests threatened by air pollution and impacts from a changing climate. This unit is broken into three parts. This unit involves a trip to the park and is accompanied by a preparation activity and a wrap-up activity. This is the field preparation activity of the unit.
1) use the scientific method while studying biodiversity
2) describe the steps in scientific inquiry
3) learn the identifying characteristics between different species of salamanders
4) understand the biodiversity of the Great Smoky Mountains National Park
5) recognize the threats to aquatic and terrestrial salamanders
When students visit the Smokies on their field trip one group will be collecting data as part of a Salamander study. This lesson will introduce the scientific method and use the identifying anatomical characteristics to key different species of salamanders.
To be a scientist you don't necessarily have to have an advanced degree. All you need to have is the ability to observe the world around you and to ask good questions. Why do things happen? How do they happen? Scientists use a systematic method to find answers to their questions. The approach is known as the scientific method or scientific inquiry. The key components to this method are: making careful observations using your senses (sometimes that includes noticing what is not there as well as what is), asking a question that is clear and specific, gathering information from literature to develop a procedure for study and to discover what is already known about your question, forming a hypothesis (possible answers to the question), testing the hypothesis (surveys, experiments and field observations are techniques), interpreting the results (make sense of your data by creating graphs or charts), drawing conclusions (was the hypothesis correct, what can you learn from your results, what factors were not in your control...), and sharing your results.
Teachers coming on the accompanying field trip should download our complete field trip packet that includes this Salamander Research Preparation pre-site lesson, information and directions about the field trip and the Salamander Research Wrap-up post-site lesson.
Download the full Salamander Research Field Trip packet here (includes Preparation and Wrap-up lessons).
Link to a PDF of the full Salamander Research Preparation lesson.
Link to a PDF of the full Salamander Research Wrap-up lesson
Materials for this unit include a PDF of a reading on Inventory and Monitoring and a second PDF specificially about salamander monitoring in Great Smoky Mountains National Park.
Step 1: Instruct the students to read the "Inventory and Monitoring" worksheet. Discuss why it is important for a park to develop an Inventory and Monitoring program.
Step 2: Have the students read the "Salamander Information" worksheets. Discuss the 1) Characteristics of a salamander, 2) What the term "lungless" salamanders mean in terms of how the salamanders breathe, 3) Differences between salamanders and lizards, 4) Different ways salamander monitoring is done in the park, 5) Correct method of measuring the length of a salamander, and 6) Differences between dusky and woodland salamanders.
Step 3: Have the students read over the vocabulary associated with the salamander program. All of the definitions will be used within the salamander inventory program. Students will probably be familiar with most of the definitions but reviewing the list before the trip is essential.
Students can create a concept map for the subject of "salamanders" before starting the series of lessons. They can create a second concept map for comparison after the lessons. Did students show any gains in their organization of their knowledge; the use of concepts, content and terminology and connections; and knowledge shown between the relationships of concepts. Please see our concept map scoring rubric in the "Students as Scientists: Salamander Research" pre-site lesson PDF for grading guidance.
Teachers can test students on their knowledge of the vocabulary.
Salamanders are an especially abundant and diverse group of animals within Great Smoky Mountains National Park. Most of the salamanders in the Smokies breathe through their skin which makes them sensitive to changes in the environment from threats like acid deposition. Since salamanders are cold-blooded, they may also be impacted by rapidly changing weather conditions in the park, especially in the winter and early spring. This lesson plan connects the student to these salamanders and readies them for their upcoming field trip to the park.
Lead a brainstorming discussion with your students. What are some questions that students have about salamanders in the Smokies? Would any of these questions make good research questions? Bring these questions with you on your field trip and ask the Park Ranger.
View a video about Hellbenders, the largest salamander in the Smokies https://www.nps.gov/grsm/naturescience/amphibians.htm
"Reptiles and Amphibians of the Smokies" available at http://shop.smokiesinformation.org/category.cfm/gsma/books
Vocabulary•All Taxa Biodiversity Inventory: also called the ATBI. A research project in the Great Smoky Mountains National Park to inventory every life form in the park. It is estimated that we currently know only 18,000 of an estimated 100,000 species.
•Baseline Information: information about how things are now, at this point in time, so we will know if there is a change the next time we look at it.
•Biodiversity: the variety, distribution and abundance of life forms and ecological processes in an ecosystem; includes the ways in which different life forms interact.
•Biological Inventory: a technique used by scientists to study the various life forms in a given area. In the Great Smoky Mountains National Park, inventories are done in study plots.
•Biological Monitoring: a technique used by scientists to check the condition of a particular species or ecosystem over time.
•Canopy: the top layer of the forest, the treetops.
•Density: the number of individuals of a given species within a certain area.
•Dichotomous Key: an identification method that narrows down a species in question using a series of pairs of choices.
•Ecosystem: a system formed by the interaction of groups of organisms with each other and their environment.
•Hypothesis: a proposition based on assumptions that can be evaluated scientifically.
•Vertebrate: an animal that has a backbone.
•Taxonomy: the classification of plants and animals according to their natural relationships.
Last updated: April 14, 2015<|endoftext|>
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Hoverflies or Bees
iPhoneOgraphy – 14 Apr 2016 (Day 105/366)
Hoverflies, sometimes called flower flies, or syrphid flies, make up the insect family Syrphidae. As their common name suggests, they are often seen hovering or nectaring at flowers; the adults of many species feed mainly on nectar and pollen, while the larvae (maggots) eat a wide range of foods. In some species, the larvae are saprotrophs, eating decaying plant and animal matter in the soil or in ponds and streams. In other species, the larvae are insectivores and prey on aphids, thrips, and other plant-sucking insects.
Aphids alone cause tens of millions of dollars of damage to crops worldwide every year; because of this, aphid-eating hoverflies are being recognized as important natural enemies of pests, and potential agents for use in biological control. Some adult syrphid flies are important pollinators.
About 6,000 species in 200 genera have been described. Hoverflies are common throughout the world and can be found on all continents except Antarctica. Hoverflies are harmless to most other animals, despite their mimicry of more dangerous wasps and bees, which wards off predators.
The size of hoverflies varies depending on the species. Some, like members of the genus Baccha, are small, elongated, and slender, while others, like members of Criorhina, are large, hairy, and yellow and black. As members of the Diptera, all hoverflies have a single functional pair of wings (the hind wings are reduced to balancing organs). They are brightly colored, with spots, stripes, and bands of yellow or brown covering their bodies. Due to this coloring, they are often mistaken for wasps or bees; they exhibit Batesian mimicry. Despite this, hoverflies are harmless.
With a few exceptions, hoverflies are distinguished from other flies by a spurious vein, located parallel to the fourth longitudinal wing vein. Adults feed mainly on nectar and pollen. They also hover around flowers, lending to their common name.<|endoftext|>
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a/an = one
The indefinite article a/an shows that we mean one thing or one person. It can't be used with the plural.
I see a man. = I see one man.
Can I have a cup of tea? = Can I have one cup of tea?
a million dollars = one million dollars
"a" before a consonant, "an" before a vowel
We put an before the vowels a / e / i / o / u. We put a before all other letters.
I've got an apricot and a pear.
I'll boil an egg and make a sandwich.
Peter's got an old typewriter and a bronze figurine.
a or an — depends on not how we spell but how we pronounce the next word.
- We use "a" before a silent "h"
an honourable man
- We use "a" if "u" sounds like [yu].
a European country
What is it? Who is it?
We use the indefinite article a/an to say what a thing or a person is, what a thing or a person is like — when we speak about categories and characteristics.
This is a student.
"Windows" is an operating system.
A: What do you do? B: I'm an editor.
It's an interesting article, you should read it.
The moon isn't a planet.
She's wearing a black dress.<|endoftext|>
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With only an introductory course in science, it’s easy to think that scientists strictly follow the scientific method. They propose a new hypothesis, test that hypothesis, and after many years of hard work, either confirm or reject it. But science is often prone to chance. And when a surprise presents itself, the book titled “Scientific Method 101” often gets dropped in the trash. In short, science needs — and perhaps thrives on — stupid luck.
Take any scientific mission. Often designed to do one thing, a mission tends to open up a remarkable window on something unexpected. Now, NASA’s Kepler space telescope, designed to hunt for planets in our own galaxy, has helped measure an object much more distant and more massive than any of its detected planets: a black hole.
KA1858+4850 is a Seyfert galaxy with an active supermassive black hole feeding on nearby gas. It lies between the constellations Cygnus and Lyra approximately 100 million light-years away.
In 2012, Kepler provided a highly accurate light curve of the galaxy. But the team, led by Liuyi Pei from the University of California, Irvine, also relied on ground-based observations to compliment the Kepler data.
The trick is to look at how the galaxy’s light varies over time. The light first emitted from the accretion disk travels some distance before reaching a gas cloud, where it’s absorbed and re-emitted a short time later.
Measuring the time-delay between the two emitted points of light tells the size of the gap between the accretion disk and the gas cloud. And measuring the width of the emitted light from the gas cloud tells the velocity of the gas moving near the black hole (due to an effect known as Doppler broadening). Together, these two measurements allow astronomers to determine the mass of the supermassive black hole.
Pei and her colleagues measured a time delay of roughly 13 days, and a velocity of 770 kilometers per second. This allowed them to calculate a central black hole mass of roughly 8.06 million times the mass of the Sun.
Discovered on October 29, 1780 by Pierre Mechain, this active Seyfert galaxy is magnificent to behold in amateur equipment and even more so in NASA/ESA Hubble Space Telescope photographs. Located in the constellation of Cetus and positioned about 45 million light years away, this spiral galaxy has a claim to fame not only for being strong in star formation, but as one of the most studied galaxies of its type. Cutting across its face are red hued pockets of gas where new suns are being born and dark dustlanes twist around its powerful nucleus.
When Mechain first observed this incredible visage, he mistook it for a nebula and Messier looked at it, but did not record it. (However, do not fault Messier for lack of interest at this time. His wife and newly born son had just died and he was mourning.) In 1783, Sir William Herschel saw it as an “Ill defined star surrounded by nebulousity.” but would change his tune some 8 years later when he reported: “A kind of much magnified stellar cluster; it contains some bright stars in the centre.” His son, John Herschel, would go on to catalog it – not being very descriptive either.
This video zooms in on spiral galaxy Messier 77. The sequence begins with a view of the night sky near the constellation of Cetus. It then zooms through observations from the Digitized Sky Survey 2, and ends with a view of the galaxy obtained by Hubble. Credit:NASA, ESA, Digitized Sky Survey 2. Acknowledgement: A. van der Hoeven
At almost double the size of the Milky Way, we now know it is a barred spiral galaxy. According to spectral analysis, Messier 77 has very broad emission lines, indicating that giant gas clouds are rapidly moving out of this galaxy’s core, at several hundreds of kilometers per second. This makes M77 a Seyfert Type II galaxy – one with an expanding core of starbirth. In itself, that’s quite unique considering the amount of energy needed to expand at that rate and further investigations found a 12 light-year diameter, point-like radio source at its core enveloped in a 100 light year swath of interstellar matter. A miniature quasar? Perhaps… But whatever it is has a measurement of 15 million solar masses!
Deep at its heart, Messier 77 is beating out huge amounts of radiation – radiation suspected to be from an intensely active black hole. Here the “galaxy stuff” is constantly being drawn towards the center, heating and lighting up the frequencies. Just this area alone can shine tens of thousands of times brighter than most galaxies… but is there anything else hiding there?
“Active galactic nuclei (AGNs) display many energetic phenomena—broad emission lines, X-rays, relativistic jets, radio lobes – originating from matter falling onto a supermassive black hole. It is widely accepted that orientation effects play a major role in explaining the observational appearance of AGNs.” says W. Jaffe (et al). “Seen from certain directions, circum-nuclear dust clouds would block our view of the central powerhouse. Indirect evidence suggests that the dust clouds form a parsec-sized torus-shaped distribution. This explanation, however, remains unproved, as even the largest telescopes have not been able to resolve the dust structures.”
Before you leave, look again. Clustered about Messier 77’s spiral arms are deep red pockets – a sign of newly forming stars. Inside the ruby regions, neophyte stars are ionising the gas. The dust lanes also appear crimson as well – a phenomenon called “reddening” – where the dust absorbs the blue light and highlights the ruddy color. A version of this image won second place in the Hubble’s Hidden Treasures Image Processing Competition, entered by contestant Andre van der Hoeven.
Nearly four million light years away in the direction of the constellation of Canes Venatici, a visage of creation awaited to be revealed. Now, thanks to the teamwork of the astronomical image processors at the Space Telescope Science Institute in Baltimore, Maryland, and world-renowned astrophotographers Robert Gendler and Jay GaBany, we’re able to see combined Hubble Space Telescope data with ground-based telescope imaging. Let’s look deep into spiral galaxy, Messier 106.
This wasn’t an overnight imaging project. “A few months ago the Hubble Heritage Team contacted me and asked if I’d be interested in making a large format image of M106 from the available data on the Hubble Legacy Archive,” says Gendler. “I agreed and went to work downloading a large number of data sets from the HLA. I realized this would be a massive project. The image would be a mosaic of more than 30 panels and would incorporate both wideband and narrowband data sets.”
With the cooperation of Jay GaBany, they combined their own observations/images of this magnificent structure and compiled it with Hubble data – filling in areas where no data was available. The resulting image is a portrait of such depth and beauty that it’s almost like looking into the eyes of creation itself.
Be swept away…
If you’re drawn to the core of Messier 106, there’s good reason. It isn’t just an ordinary spiral galaxy, it’s one that has a peculiar jet flow which can be detected in radio and in H-alpha wavelengths. “Due to the special geometry of the galaxy, the jets emerge from the nuclear region through the galactic disk,” says Marita Krause (et al). “Also the distribution of molecular gas looks different from that in other spiral galaxies.” It is just this difference that makes NGC 4258 (M106) stand out a bit from the crowd and so worthy of further processing. According to new modeling techniques the “concentration of CO along the ridges is due to interaction of the rotating gas clouds with the jet’s magnetic field by ambipolar diffusion. This magnetic interaction is thought to increase the time the molecular clouds reside near the jet thus leading to the quasi-static CO ridge.”
Knowing those jets are present and the hunger to reveal them through imaging became the driving force for R. Jay GaBany. “Since the early 1960s, M106, also known as NGC 4258, has been known to exhibit an extra pair of arms, located between the spiral arms comprised of stars, dust and gas. But an explanation for their existence remained elusive until earlier in this decade,” says Jay. “My contribution to the image came from my 2010 image of M-106 that revealed the full extent of its amazing jets. My image include 22 hours of white light exposures through clear, red, blue and green filters plus and other 15 hours of imaging through a 6nm narrow band h-alpha filter.”
“Seen in the light emitted by hydrogen molecules when they become ionized, these arms display an artificial red hue to make them visible in the image I produced. The extra arms are now believed to be caused by high energy jets emanating from an active 40 million solar mass super-massive black hole menacing the galaxy’s center,” explains GaBany. “Because the jets are tilted at a low inclination they pierce the disk and surrounding halo of this galaxy. So, as the jets pass through regions of gas, they create an expanding cocoon of shock waves that heats the surrounding material causing it to release radiation in optical wavelengths. The curvature and fraying seen at their extremities represents previous trajectories of the jet due to past precession. Precession is a change in the orientation of the rotation axis of a spinning object. For example, the wobble of a spinning top.”
Yet, that’s not all. This low luminosity Seyfert II galaxy is also hosting a maser – its warped disk of water molecules discovered in 1994. Through radio observations, M106 became the first of its kind to show the exact location of the core of an AGN (active galactic nucleus). According to a study done by JR Herrnstein (et al): “NGC 4258 is an exceptional laboratory for the study of AGN accretion processes. The nuclear maser reveals details about the kinematics and structure of the accretion disk on subparsec scales and permits the determination of the central mass with great precision.”
And there is still more…
Deep inside lurks that known supermassive black hole – one that’s extremely active and produces bright microwave radiation. But, don’t stop there. Ordinarily a spiral galaxy has two arms, but M106 has double. These ethereal “extras” can be seen as faint ribbons of gas at optical wavelengths, but become solidified when viewed in x-ray and radio. Here the structure is formed in hot gas rather than stars. While this process was once a mystery to astronomers, new information suggests they may arise from the black hole activity, making them a unique artifact. What could cause it? These “extra arms” could be the result of the violent turbulence at the core – where gases are superheated and interact with their denser counterparts causing them to illuminate. At the perimeter of the galactic structure, the gases are more loose and the arching formation could be the product of the movement of jet activity.
“One goal I had early on was to feature the well known ‘anomalous arms’ of M106,” said Gendler. “This feature, peculiar to M106, is thought to arise from superheated gases, energized by accretion of matter into the galaxy’s massive black hole. The anomalous arms emit light in the visual spectrum around 656nm (hydrogen alpha) and I found a fair amount of hydrogen alpha data sets for the arms in the HLA.”
Gendler was responsible for all the image assembly and processing. “Assembling the image required over two months,” he said. “The quality of the data ranged from good to very poor. The central galaxy had sufficient color data but away from the center the Hubble data was incomplete and in some areas did not exist. I then decided to use ground based data from my own image and Jay GaBany’s image of M106 to fill in areas of missing or incomplete Hubble data. I also used ground based data to boost the signal of the outer areas of the galaxy as the Hubble data was sparse and of short exposure for the more remote areas of the galaxy.”
All in all, Messier 106 is a galaxy that deserves attention – attention and a loving touch given by two of the very best amateur astronomers and dedicated astrophotographers to be found.<|endoftext|>
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# The point charges \$Q\$ and \$-2Q\$ are placed some distance apart. If the electric field intensity at \$Q\$ is \$E\$ then what is the electric field at \$-2Q\$.A) \$\dfrac{E}{2}\$ B) \$E\$ C) \$2E\$ D) \$1.5E\$
Last updated date: 28th Feb 2024
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Hint: Electric field is the physical field that surrounds each electric charge and exerts force on all other charges in the field, which is given by:
\$E = \dfrac{F}{q}\$ (F is the force exerted by charge on other charges and q is the charge).
Electric field intensity is the force that would be experienced by unit test charge placed at a location. Using the above relations we will find the electric field intensity at \$-2Q\$ charge.
Complete step by step solution:
Let us first define more about Electric field and Electric field intensity.
Electric field: A measure of the force exerted by one charged body on another. Imaginary lines force or electric field lines of force or electric field originate on positive charges and terminate on negative charges. Electric field lines are thought of as elastic lines which repel each other in a direction perpendicular to the line itself.
Electric Field Intensity: Electric field intensity at any location is the force that would be experienced by unit test charge placed at location. A uniform electric field is an ideal case in which electric field lines are parallel with one another.
Now, we will do the calculation part of the problem.
We are given that electric field intensity given by charge \$Q\$ is \$E\$.
We are supposed to find the electric field intensity due to charge \$-2Q\$.
As we have mentioned in the hint the electric field is given by the formula:
\$E = \dfrac{F}{Q}\$...............(1)
In equation 1 if we replace \$q\$ by \$2Q\$, then electric field intensity is given by:
\$E = \dfrac{F}{{2Q}}\$............(2)
On comparing equation 1 and 2 we can get
Electric field intensity by charge \$-2Q\$ is \$\dfrac{E}{2}\$.
Therefore, option (A) is correct.
Note: There are many applications of electric field which are: usage in tissue engineering for measuring cell and tissue properties, electro spinning(use high voltage which causes huge amounts of electric field in the capillary tubes),used in Van de Graff generators, escalators etc.<|endoftext|>
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# Lesson 7
Build Multiplication Fluency
## Warm-up: Notice and Wonder: Same Solution (10 minutes)
### Narrative
The purpose of this warm-up is to elicit the idea that there are different ways to calculate a product, using the standard algorithm, which will be useful when students find products of a 3-digit number and a 2-digit number in a way that makes sense to them.
### Launch
• Groups of 2
• Display the image.
• “What do you notice? What do you wonder?”
• 1 minute: quiet think time
### Activity
• 1 minute: partner discussion
• Share and record responses.
### Student Facing
What do you notice? What do you wonder?
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• “Why do you think the results of the two calculations are the same?” (The factors are the same, just in a different order. The order of the factors does not change the result of multiplication.)
• "Which way would you prefer to find the value of $$417 \times 28$$?" (I like the one with two partial products as there is less adding up to do.)
• "Today you will get to choose how to find products of a 3-digit number and a 2-digit number.
## Activity 1: Greatest Product (15 minutes)
### Narrative
Students use their understanding of place value to generate expressions that have the greatest product. They take turns selecting number cards 0 through 9 and place them strategically to make the largest product of a 2-digit number and a 3-digit number. Students will need to think about how the different place values influence the value of the product and choose where to put their digits accordingly. There is also a large element of chance since they do not know in advance which numbers they will select.
Action and Expression: Develop Expression and Communication. Give students access to grid paper for finding the product.
Supports accessibility for: Visual-Spatial Processing, Organization
### Required Materials
Materials to Copy
• Number Cards (0-10)
• Greatest Product
### Launch
• Groups of 2
• Give each group a set of number cards and 2 copies of the blackline master.
• “Remove the cards that show 10 and set them aside.”
• “We’re going to play a game called Greatest Product. Let’s read through the directions and play one round together.”
• Read through the directions with the class and play a round against the class using the diagram in the student workbook:
• Display each number card.
• Think through your choices aloud.
• Record your move and score for all to see.
### Activity
• “Now, play the game with your partner.”
• 8–10 minutes: partner work time
### Student Facing
Directions:
• Partner A chooses a number card and writes the number in one of the blanks for Round 1.
• Partner B does the same.
• Repeat until each partner has a two-digit by three-digit multiplication problem.
• Find the product.
• The partner with the greater product wins a point.
• The partner with the most points after 5 rounds wins the game.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• Display a blank image from game board.
• “If this is your game board and you pick an 8, where would you write the 8? Why?” (I would either put it in the hundreds place of the top number or the tens place of the bottom number because it’s a big number and those are the biggest place values.)
• Write the 8 in the hundreds place of the top number.
• “What if you next select a 1? Where would you write the 1? Why?” (I would put it in the ones place of one of the numbers because I want to put a bigger number in the tens place.)
• “What did you find challenging about the game?” (Since I didn’t know what numbers I was going to get on later picks, I sometimes wasn’t sure where to put a number because I didn’t know if I would get a bigger number on a later pick.)
## Activity 2: Desperately Seeking 9 New Units [OPTIONAL] (20 minutes)
### Narrative
The purpose of this activity is for students to explore the number of new units that can be composed when using the standard algorithm. As students experiment with the given numbers, they will find examples of 1, 2, 3, 4, 5, 6, 7, or 8 new units composed. Here they address the question of whether or not it is possible to compose 9 or more new units when using the standard algorithm. In order to make sense of and persevere in solving this problem (MP1), there are several types of arguments students may make, all of which highlight how the standard algorithm works:
• they may calculate $$999 \times 9$$ and see that while they get very close to composing 9 new units, they fall 1 short
• the biggest product that can be made from multiplying a pair of 1-digit numbers is 81, which would mean that 8 units are composed
• this 81 has to be combined with whatever new units were composed before, but since that’s 8, that means the largest number you can form at each step in the standard algorithm is 89, which is 1 short of the 90 you would need to compose 9 new units
MLR8 Discussion Supports. Synthesis: At the appropriate time, give groups 2–3 minutes to plan what they will say when they present to the class. “Practice what you will say when you share your solutions with the class. Talk about what is important to say, and decide who will share each part.”
• Groups of 2
### Activity
• 1–2 minutes: quiet think time
• 3–5 minutes: partner work time
### Student Facing
Tyler notices that when he uses the standard algorithm and composes a new unit, sometimes there is 1 new unit, sometimes 2, all the way up to 8. He has not seen an example with 9 of the new unit.
1. For each of these products, how many of each new unit do you compose?
1. $$256 \times 5$$
2. $$587 \times 8$$
3. $$809 \times 9$$
2. Do you think it is possible to compose 9 of a new unit with the standard multiplication algorithm?
### Student Response
For access, consult one of our IM Certified Partners.
If students are not fluent with their multiplication facts, offer them a multiplication table. As they use the standard algorithm to find the products in problem 1, ask them to circle each new unit. If they are not sure if it is possible to compose 9 new units, ask them to use the algorithm to find the product of $$99 \times 9$$ and $$999 \times 99$$.
### Activity Synthesis
• Invite students to share the number of units they composed in the calculations of problem 1.
• “Was anything missing from 1 to 8?” (No, we composed everything from 1 to 8 new units.)
• “Was anyone able to compose 9 new tens in a product? Why?” (No. The new tens come when I take the product of the ones. The biggest that can be is $$9 \times 9$$ which gives me 8 tens.)
• “Was anyone able to compose 9 new hundreds? Why?” (No. I could get 8 new hundreds if I have $$9 \times 90$$, giving me 810. I might also have to combine that with the new tens from the product of the ones, but that would give me at most 890, not 900.)
## Lesson Synthesis
### Lesson Synthesis
“Today we used the standard algorithm to find products of numbers with no restriction on the number of newly composed units and we examined how many new units can be composed.”
“Do you think the algorithm for multiplying whole numbers will work for any and all whole numbers? Why or why not? Discuss with a partner.” (I think so but if the numbers are big it will take up a lot of space and there could be a lot more new units to compose. I think it might work but it would take a long time if the numbers are big and I think it would be easy to make a mistake.)
Ask students to share their thinking.
“What do you still wonder about the standard algorithm for multiplying whole numbers?” (When is it a good strategy to use? Are there other ways that work well or better?)
## Cool-down: Calculating a Product (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners.<|endoftext|>
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Over the past few weeks we have certainly seen some extreme weather conditions around the world. Those on the east coast of North America have been hit by record cold temperatures. At the same time, those in Australia have been experiencing record hot temperatures. These extreme weather conditions have many wondering what effects “Global Warming” will have on the dairy cattle There has been lots of coverage in the media about dairy cattle and their alleged contribution to greenhouse gases and how that is contributing to Global Warming. Very little has been addressed about the effects extreme weather conditions have on the dairy cattle themselves. One thing appears certain. Extreme heat waves and cold fronts are the new ‘normal’.
As producers know, cow and calf comfort is one of the most important factors in milk production and growth. As more and more producers are experiencing extreme temperatures, keeping their animals comfortable is becoming harder. Drastic increases or decreases in ambient temperature affects animal production systems by affecting the health, reproduction, nutrition etc. of the animals and thereby results in poor performance, inferior product quality, outbreak of novel diseases, etc. Dairy cattle are more susceptible to increased ambient temperature than other ruminants, because of their high metabolic rate and the poor water retention mechanism of their kidney and gastrointestinal tracts. Young stock are not immune to these weather stresses either.
Greater temperature shifts and shifts that are more frequent seem to be the most obvious weather changes that will have effects on dairy cattle. It is forecasted that we can expect even greater atmospheric temperature changes. Therefore these issues are going to come to the forefront. The following are the five major impacts that global warming will have on dairy cattle.
- Ambient temperature’s effect on Dry Matter Intake (DMI)
When cows are stressed their Dry Matter Intake (DMI) decreases. As the heat rises DMI decreases. Feed consumption by dairy cattle starts to decline when average daily temperature reaches 25 to 27 Centigrade (77 to 81 Fahrenheit) and voluntary feed intake can be decreased by 10-35% when ambient temperature reaches 35 C (95 Fahrenheit) and above. Conversely, cows that are experiencing extreme cold weather conditions increase their DMI intake drastically, but instead of the consumption being converted in to milk production, a much larger portion of their energy is committed to their maintenance energy requirements. Thermal cold stress conditions result in 20-30% more maintenance energy requirement and an ensuing reduction in the amount of net energy available for growth and production.
- Increased respiratory rate
When dairy cows experience increased thermal stress, their heart rate rises. The heart rate of the animal under thermal heat stress is higher to ensure more blood flow towards peripheral tissue to dissipate heat from the body core to the skin. This increased effort takes much needed energy away from milk production. Respiration rate of the animal can be used as an indicator of the severity of thermal load but several other factors such as animal condition, prior exposures to high temperature etcetera should be considered to interpret the observed respiration rate.
- Decreased conception rates
As weather stress increases, dairy reproduction function decreases, resulting in decreased conception rates. This is a result of thermal stress that causes imbalance in secretion of reproductive hormones. High ambient temperature has also been reported to increase incidence of ovarian cysts. Plasma progesterone levels in animals under high ambient temperatures are low compared to animals that are experiencing thermal comfort. It has also been reported that high ambient temperature causes poor quality of ovarian follicles resulting in poor reproductive performance in cattle. Fertility of cattle is also reduced due to low intensity and duration of estrus caused by reduced luteinizing hormone (LH) and estradiol secretion during thermal stress. In addition, thermal stress also causes decreased reproductive efficiency by increasing the calving interval. Calves born from dams under thermal stress were found to be of lower body weight than those from normal cows. Additionally the dams had reduced lactation performance due to the carryover effects of thermal stress which occurred during the prepartum period.
- Decreased Metabolic Responses
Under heat stress metabolism is reduced, which is associated with reduced thyroid hormone secretion and gut motility, resulting in increased gut fill. Plasma growth hormone concentration and secretion rates decline with high temperature (35 ºC / 95 ºF). Ruminal pH is typically lower in heat stressed cattle
- Decreased Milk Production
Reduction in milk production is one of the major economic impacts of climatic stress upon dairy cattle. Decrease in milk yield due to thermal heat stress is more prominent in Holstein than in Jersey cattle (Read more…). Decreased synthesis of hepatic glucose and lower non esterified fatty acid (NEFA) levels in blood during thermal stress causes reduced glucose supply to the mammary glands and results in low lactose synthesis, which in turn leads to low milk yield. As mentioned earlier, reduction in milk yield is further intensified by decrease in feed consumption by the animals to compensate for high environmental temperature. Actually 35% of reduced milk production is due to decreased feed intake while the remaining 65% is attributable directly to the thermal stress. Other factors resulting in reduced milk production during thermal stress are decreased nutrient absorption, negative effects on rumen function and hormonal status and increased maintenance requirements. These all mean that there is reduced net energy available for production.
To combat heat stress check out these articles (Read more: Are you feeling the heat? and Heat Stress on Dairy Cattle) and to combat cold stress (Read more: COMMON SENSE, COWS and the UN-COMMON COLD of 2014!, “COLD CALVES” – The Next Drama Coming to a Calf Pen Near You! and Cold Weather Effects on Dairy Cattle)
The Bullvine Bottom Line
There is no question that the world’s temperatures are changing because of atmospheric pressure changes caused by Global Warming. Warming or cooling of the climate system of the earth has multifaceted effects on animals. Intensification and increased frequency of thermal stress due to global warming has the most prominent impact on dairy cattle and causes different physiological, metabolic and production disturbances. The importance of responding to thermal stress has been increased for dairy farmers in tropical, subtropical and even in temperate regions of the world due to atmospheric warming. As these effects increase, it will be increasingly urgent for the milk producers of the world to provide environments that are able to combat these effects and offer the greatest comfort for their cattle. Global Warming is actually Global Warning for the dairy industry.<|endoftext|>
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# Math Snap
## 4 Тело свободно падает с высоты 80 м. Определить среднюю скорость движения.
#### STEP 1
Assumptions
1. The body falls freely under the influence of gravity.
2. The height from which the body falls is 80 meters.
3. The acceleration due to gravity (g) is approximately $9.8 \, \text{m/s}^2$.
4. Air resistance is negligible.
5. The initial velocity of the body (u) is $0 \, \text{m/s}$ since it starts from rest.
6. We will use the kinematic equations for uniformly accelerated motion to solve the problem.
#### STEP 2
We will use the kinematic equation that relates the distance (s), initial velocity (u), acceleration (a), and time (t):
$s = ut + \frac{1}{2}at^2$
#### STEP 3
Since the initial velocity $u = 0$, the equation simplifies to:
$s = \frac{1}{2}gt^2$
#### STEP 4
Now, plug in the given values for the distance $s$ and acceleration $g$ to find the time $t$.
$80 = \frac{1}{2} \cdot 9.8 \cdot t^2$
#### STEP 5
Solve for $t^2$:
$t^2 = \frac{160}{9.8}$
#### STEP 6
Calculate $t^2$:
$t^2 = \frac{160}{9.8} \approx 16.33$
#### STEP 7
Take the square root of both sides to find $t$:
$t = \sqrt{16.33}$
#### STEP 8
Calculate $t$:
$t \approx 4.04 \, \text{seconds}$
#### STEP 9
To find the average velocity ($v_{avg}$), we use the definition of average velocity, which is the total displacement divided by the total time:
$v_{avg} = \frac{s}{t}$
#### STEP 10
Plug in the values for $s$ and $t$ to calculate the average velocity:
$v_{avg} = \frac{80}{4.04}$
##### SOLUTION
Calculate $v_{avg}$:
$v_{avg} \approx 19.8 \, \text{m/s}$
The average velocity of the body during its fall is approximately $19.8 \, \text{m/s}$.<|endoftext|>
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Visualizing the chain rule and product rule | Essence of calculus, chapter 4
In the last videos I talked about the derivatives of simple functions, things like powers of
x, sin(x), and exponentials, the goal being to have a clear picture or intuition to hold
in your mind that explains where these formulas come from.
Most functions you use to model the world involve mixing, combining and tweaking these
these simple functions in some way; so our goal now is to understand how to take derivatives
of more complicated combinations; where again, I want you to have a clear picture in mind
for each rule.
This really boils down into three basic ways to combine functions together: Adding them,
multiplying them, and putting one inside the other; also known as composing them.
Sure, you could say subtracting them, but that’s really just multiplying the second
by -1, then adding.
Likewise, dividing functions is really just the same as plugging one into the function
1/x, then multiplying.
Most functions you come across just involve layering on these three types of combinations,
with no bound on how monstrous things can become.
But as long as you know how derivatives play with those three types of combinations, you
can always just take it step by step and peal through the layers.
So, the question is, if you know the derivatives of two functions, what is the derivative of
their sum, of their product, and of the function compositions between them?
The sum rule is the easiest, if somewhat tounge-twisting to say out loud: The derivative of a sum of
two functions is the sum of their derivatives.
But it’s worth warming up with this example by really thinking through what it means to
take a derivative of a sum of two functions, since the derivative patterns for products
and function composition won’t be so straight forward, and will require this kind of deeper
thinking.
The function f(x) = sin(x) + x2 is a function where, for every input, you add together the
values of sin(x) and x2 at that point.
For example, at x = 0.5, the height of the sine graph is given by this bar, the height
of the x2 parabola is given by this bar, and their sum is the length you get by stacking
them together.
For the derivative, you ask what happens as you nudge the input slightly, maybe increasing
it to 0.5+dx.
The difference in the value of f between these two values is what we call df.
Well, pictured like this, I think you’ll agree that the total change in height is whatever
the change to the sine graph is, what we might call d(sin(x)), plus whatever the change to
x2 is, d(x2).
We know the derivative of sine is cosine, and what that means is that this little change
d(sin(x)) would be about cos(x)dx.
It’s proportional to the size of dx, with a proportionality constant equal to cosine
of whatever input we started at.
Similarly, because the derivative of x2 is 2x, the change in the height of the x2 graph
So, df/dx, the ratio of the tiny change to the sum function to the tiny change in x that
caused it, is indeed cos(x)+2x, the sum of the derivatives of its parts.
But like I said, things are a bit different for products.
Let’s think through why, in terms of tiny nudges.
In this case, I don’t think graphs are our best bet for visualizing things.
Pretty commonly in math, all levels of math really, if you’re dealing with a product
of two things, it helps to try to understand it as some form of area.
In this case, you might try to configure some mental setup of a box whose side-lengths are
sin(x) and x2.
What would that mean?
Well, since these are functions, you might think of these sides as adjustable; dependent
on the value of x, which you might think of as a number that you can freely adjust.
So, just getting the feel for this, focus on that top side, whose changes as the function
sin(x).
As you change the value of x up from 0, it increases in up to a length of 1 as sin(x)
moves towards its peak.
After that, it starts decreasing as sin(x) comes down from 1.
And likewise, that height changes as x2.
So f(x), defined as this product, is the area of this box.
For the derivative, think about how a tiny change to x by dx influences this area; that
resulting change in area is df.
That nudge to x causes the width to change by some small d(sin(x)), and the height to
change by some d(x2).
This gives us three little snippets of new area: A thin rectangle on the bottom, whose
area is its width, sin(x), times its thin height, d(x2); there’s a thin rectangle
on the right, whose area is its height, x2, times its thin width, d(sin(x)).
And there’s also bit in the corner.
But we can ignore it, since its area will ultimately be proportional to dx2, which becomes
negligible as dx goes to 0.
This is very similar to what I showed last video, with the x2 diagram.
Just like then, keep in mind that I’m using somewhat beefy changes to draw things, so
we can see them, but in principle think of dx as very very small, meaning d(x2) and d(sin(x))
are also very very small.
Applying what we know about the derivative of sine and x2, that tiny change d(x2) is
2x*dx, and that tiny change d(sin(x)) is cos(x)dx.
Dividing out by that dx, the derivative df/dx is sin(x) by the derivative of x2, plus x2
by the derivative of sine.
This line of reasoning works for any two functions.
A common mnemonic for the product rule is to say in your head “left d right, right
d left”.
In this example, sin(x)*x2, “left d right” means you take the left function, in this
case sin(x), times the derivative of the right, x2, which gives 2x.
Then you add “right d left”: the right function, x2, times the derivative of the
left, cos(x).
Out of context, this feels like kind of a strange rule, but when you think of this adjustable
box you can actually see how those terms represent slivers of area.
“Left d right” is the area of this bottom rectangle, and “right d left” is the area
of this rectangle on the right.
By the way, I should mention that if you multiply by a constant, say 2*sin(x), things end up
much simpler.
The derivative is just that same constant times the derivative of the function, in this
case 2*cos(x).
I’ll leave it to you to pause and ponder to verify that this makes sense.
Aside from addition and multiplication, the other common way to combine functions that
comes up all the time is function composition.
For example, let’s say we take the function x2, and shove it on inside sin(x) to get a
new function, sin(x2).
What’s the derivative of this new function?
Here I’ll choose yet another way to visualize things, just to emphasize that in creative
math, we have lots of options.
I’ll put up three number lines.
The top one will hold the value of x, the second one will represent the value of x2,
and that third line will hold the value of sin(x2).
That is, the function x2 gets you from line 1 to line 2, and the function sine gets you
from line 2 to line 3.
As I shift that value of x, maybe up to the value 3, then value on the second shifts to
whatever x2 is, in this case 9.
And that bottom value, being the sin(x2), will go over to whatever the sin(9) is.
So for the derivative, let’s again think of nudging that x-value by some little dx,
and I always think it’s helpful to think of x starting as some actual number, maybe
1.5.
The resulting nudge to this second value, the change to x2 caused by such a dx, is what
we might call d(x2).
You can expand this as 2x*dx, which for our specific input that length would be 2*(1.5)*dx,
but it helps to keep it written as d(x2) for now.
In fact let me go one step further and give a new name to x2, maybe h, so this nudge d(x2)
is just dh.
Now think of that third value, which is pegged at sin(h).
It’s change d(sin(h)); the tiny change caused by the nudge dh.
By the way, the fact that it’s moving left while the dh bump is to the right just means
that this change d(sin(h)) is some negative number.
Because we know the derivative of sine, we can expand d(sin(h)) as cos(h)*dh; that’s
what it means for the derivative of sine to be cosine.
Unfolding things, replacing h with x2 again, that bottom nudge is cos(x2)d(x2).
And we could unfold further, noting that d(x2) is 2x*dx.
And it’s always good to remind yourself of what this all actually means.
In this case where we started at x = 1.5 up top, this means that the size of that nudge
on the third line is about cos(1.52)*2(1.5)*(the size of dx); proportional to the size of dx,
where the derivative here gives us that proportionality constant.
Notice what we have here, we have the derivative of the outside function, still taking in the
unaltered inside function, and we multiply it by the derivative of the inside function.
Again, there’s nothing special about sin(x) and x2.
If you have two functions g(x) and h(x), the derivative of their composition function g(h(x))
is the derivative of g, evaluated at h(x), times the derivative of h.
This is what we call the “chain rule”.
Notice, for the derivative of g, I’m writing it as dg/dh instead of dg/dx.
On the symbolic level, this serves as a reminder that you still plug in the inner function
to this derivative.
But it’s also an important reflection of what this derivative of the outer function
actually represents.
Remember, in our three-lines setup, when we took the derivative of sine on the bottom,
we expanded the size of the nudge d(sin) as cos(h)*dh.
This was because we didn’t immediately know how the size of that bottom nudge depended
on x, that’s kind of the whole thing we’re trying to figure out, but we could take the
derivative with respect to the intermediate variable h.
That is, figure out how to express the size of that nudge as multiple of dh.
Then it unfolded by figuring out what dh was.
So in this chain rule expression, we’re saying look at the ratio between a tiny change
in g, the final output, and a tiny change in h that caused it, h being the value that
we’re plugging into g.
Then multiply that by the tiny change in h divided by the tiny change in x that caused
it.
The dh’s cancel to give the ratio between a tiny change in the final output, and the
tiny change to the input that, through a certain chain of events, brought it about.
That cancellation of dh is more than just a notational trick, it’s a genuine reflection
of the tiny nudges that underpin calculus.
So those are the three basic tools in your belt to handle derivatives of functions that
combine many smaller things: The sum rule, the product rule and the chain rule.
I should say, there’s a big difference between knowing what the chain rule and product rules
are, and being fluent with applying them in even the most hairy of situations.
I said this at the start of the series, but it’s worth repeating: Watching videos, any
videos, about these mechanics of calculus will never substitute for practicing them
yourself, and building the muscles to do these computations yourself.
I wish I could offer to do that for you, but I’m afraid the ball is in your court, my
friend, to seek out practice.
What I can offer, and what I hope I have offered, is to show you where these rules come from,
to show that they’re not just something to be memorized and hammered away; but instead
are natural patterns that you too could have discovered by just patiently thinking through
what a derivative means.
Thank you to everyone who supported this series, and once more I’d like to say a special
thanks to Brilliant.org.
For those of you who want to go flex those problem solving muscles, Brilliant offers
a platform aimed at training you to think like a mathematician.
I don’t know about you, but I’ve always found it all too easy to fall into the habit
of just reading math or watching lectures without taking the time to do some real problem-solving
in between, even though that’s always the part where I learn the most.
Brilliant is a great place to get that practice, and if you visit brilliant.org/3b1b, or more
simply follow the link on the screen and in the description, it lets them know you came
from this channel.
Their calculus material is a nice complement to this series, but some of my other favorites
are their probability and complex algebra sequences.<|endoftext|>
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Classroom Procedures for
1. Be prepared for class.2. Work Quietly.3.Raise your hand to speak.4. Listen carefully.5. Respect your classmates and adults in charge through kind actions and words.6. Complete and turn in all assignments on time.7. Do your best!!!
**Coming to class prepared everyday will increase your class participation grade. Coming to class unprepared will decrease your classroom participation grade.**
All students need the following supplies everyday:
- 3 notebooks
- 3 or more pencils
- a silent reading book(everyday)
- 3 folders
***No binders/trapper keepers, lead pencils, or pencil sharpeners
***No Electronic devices, playing cards, or toys are permitted in school.
2. Homework will be given almost EVERY NIGHT. Students
The students will keep a Math notebook. This will helpyou and your student to review daily classwork.
Homework and classwork are expected to be completed
neatly in pencil.
may be given time in class to begin their homework,
but it should still be checked at home by an adult.
3. Students are highly encouraged to use i-Ready
on their home computers.
Science and Social Studies
1. The students will keep a science and social studies notebook. They will utilize this to take notes and answer assessmment questions. Homework will also be done in the notebook.
2. Homework will be given and students may be given time to work on it in th class.<|endoftext|>
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# Circle Formulas | Properties of Circle Formulae
Are you fed up with the traditional process of solving Circle problems? then, here is the best way to solve the lengthy circle problems in an easy way. Use direct circle formulas and end up your complex calculations effortlessly at a faster pace. Make use of the required circle formula from the list provided here & understand the concept thoroughly.
## List of Circle Formulae
Memorize the formulas of all circle concepts by using the list of various properties of circle formulae provided over here. With this list of circle formulas, you can easily learn and solve Director Circle, Diameter of a Circle, and many other lengthy circle concepts problems with ease.
1. General equation of a Circle
x2 + y2 + 2gx + 2fy + c = 0
(i) Centre of a general equation of a circle is (-g, -f)
i.e. (-$$\frac{1}{2}$$ coefficient of x, –$$\frac{1}{2}$$ coefficient of y)
(ii) Radius of a general equation of a circle is $$\sqrt{g^{2}+f^{2}-c}$$
2. Central form of equation of a Circle
(i) The equation of a circle having centre (h, k) and radius r, is (x – h)2 + (y – k)2 = r2
(ii) If the centre is origin then the equation of a circle is x2 + y2 = r2
3. Diametral Form
If (x1, y1) and (x2, y2) be the extremities of a diameter, then the equation of circle is (x – x1) (x – x2) + (y – y1) (y – y2) = 0
4. The parametric equations of a Circle
• The parametric equations of a circle x2 + y2 = r2 are x = r cos θ, y = r sin θ.
• The parametric equations of the circle (x – h)2 + (y – k)2 = r2 are x = h + r cos θ, y = k + r sin θ.
• Parametric equations of the circle x2 + y2 + 2gx + 2fy + c = 0 are x = -g + $$\sqrt{g^{2}+f^{2}-c}$$ cos θ, y = – f + $$\sqrt{g^{2}+f^{2}-c}$$ sin θ
5. Position of a Point with respect to a Circle
The following formulae are also true for Parabola and Ellipse.
S1 > 0 ⇒ Point is outside the circle.
S1 = 0 ⇒ Point is on the circle.
S1 < 0 ⇒ Point is inside the circle.
6. Length of the intercept made by the circle on the line
p = length of ⊥ from centre to interscenting lines $$=2 \sqrt{r^{2}-p^{2}}$$
7. The length of the intercept made by line y = mx + c with the circle
x2 + y2 = a2 is $$2 \sqrt{\frac{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)-\mathrm{c}^{2}}{1+\mathrm{m}^{2}}}$$
8. Condition of tangency
(a) Standard Case: Circle x2 + y2 = a2 will touch the line y = mx + c if c = ± $$a \sqrt{1+m^{2}}$$
(b) General Case: If m is slope of line, circle is
x2 + y2 + 2gx + 2fy + c = 0 then condition is
y + f = m(n + g) ±$$\sqrt{g^{2}+f^{2}-c}$$
9. Intercepts made on coordinate axes by the circle
• x-axis = 2$$\sqrt{g^{2}-c}$$
• y-axis = 2$$\sqrt{f^{2}-c}$$
10. Equation of Tangent T = 0
(i) The equation of tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 at a point (x1 y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
(ii) The equation of tangent to circle x2 + y2 = a2 at point (x1, y1) is xx1 + yy1 = a2
(iii) Slope Form:
From condition of tangency for every value of m, the line y = mx ± a$$\sqrt{1+m^{2}}$$ is a tangent of the circle x2 + y2 = a2 and its point of contact is $$\left(\frac{\mp \mathrm{am}}{\sqrt{1+\mathrm{m}^{2}}}, \frac{\pm \mathrm{a}}{\sqrt{1+\mathrm{m}^{2}}}\right)$$
11. Equation of Normal
• The equation of normal to the circle x2 + y2 + 2gx + 2fy + c = 0 at any point (x1, y1) is y – y1 = $$\frac{y_{1}+f}{x_{1}+g}$$(x – x1)
• The equation of normal to the circle x2 + y2 = a2 at any point (x1, y1) is xy1 – x1y = 0
12. Length of tangent $$\sqrt{\mathrm{S}_{1}}$$ = AC = AB
13. Pair of tangents SS1 = T2
14. Chord of contact T = 0
T = xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
• The length of chord of contact = 2$$\sqrt{r^{2}-p^{2}}$$
• Area of AABC is given by $$\frac{a\left(x_{1}^{2}+y_{1}^{2}-a^{2}\right)^{3 / 2}}{x_{1}^{2}+y_{1}^{2}}$$
15. Director Circle
The locus of the point of intersection of two perpendicular tangents to a circle is called the Director circle.
Let the circle be x2 + y2 = a2, the equation of director circle is x2 + y2 = 2a2, director circle is a concentric circle whose radius is $$\sqrt{2}$$ times the radius of the given circle.
16. Equation of Polar and coordinates of Pole
• Equation of polar is T = 0
• Pole of polar Ax + By + C = 0 with respect to circle x2 + y2 = a2 is $$\left(-\frac{\mathrm{Aa}^{2}}{\mathrm{C}},-\frac{\mathrm{Ba}^{2}}{\mathrm{C}}\right)$$
17. Equation of a chord whose middle point is given T = S1
18. The equation of the circle passing through the points of intersection of the circle S = 0 and line L = 0 is S + λL = 0.
19. Diameter of a circle
The diameter of a circle x2 + y2 = r2 corresponding to the system of parallel chords y = mx + c is x + my = 0.
20. Equation of common chord S1 – S2 = 0
21. Two circles with radii r1, r2 and d be the distance between their centres then the angle of intersection θ between them is given by cos θ = $$\frac{r_{1}^{2}+r_{2}^{2}-d^{2}}{2 r_{1} r_{2}}$$
22. Condition of Orthogonality
2g1g2 + 2f1f2 = c1 + c2
23. Relative position of two circles and No. of common tangents
Let C1 (h1, k1) and C2 (h2, k2) be the centre of two circle and r1, r2 be their radius then
• C1C2 > r1 + r2 ⇒ do not intersect or one outside the other ⇒ 4 common tangents
• C1C2 < |r1 – r2| ⇒ one inside the other => 0 common tangent
• C1C2 = r1 + r2 ⇒ external touch ⇒ 3 common tangents
• C1C2 = |r1 – r2| ⇒ internal touch ⇒ 1 common tangent
• |r1 – r2| < C1C2 < r1 + r2 ⇒ intersection at two real points ⇒ 2 common tangents
24. Equation of the common tangents at point of contact S1 – S2 = 0.
25. Pair of point of contact
The point of contact divides C1C2 in the ratio r1: r2 internally or externally as the case may be.
• Definition of Radical axis: Locus of a point from which length of tangents to the circles are equal, is called radical axis.
• radical axis is S – S’ = 0
• If S1 = 0, S2 = 0 and S3 = 0 be any three given circles then the radical centre can be obtained by solving any two of the following equations
S1 – S2 = 0, S2 – S3 = 0, S3 – S1 = 0.
27. S1 – S2 = 0 represent equation of all i.e. Radical axis, common axis, common tangent i.e.
when circle are not in touch → Radical axis
when circle are in touch → Common tangent
when circle are intersecting → Common chord
28. Let θ1 and θ2 are two points lies on circle x2 + y2 = a2, then equation of line joining these two points is
$$\mathrm{x} \cos \left(\frac{\theta_{1}+\theta_{2}}{2}\right)+\mathrm{y} \sin \left(\frac{\theta_{1}+\theta_{2}}{2}\right)=\mathrm{a} \cos \left(\frac{\theta_{1}-\theta_{2}}{2}\right)$$
29. Limiting Point of co-axial system of circles:
Limiting point of a system of co-axial circles are the centres of the point circles belonging to the family. Two such point of a co-axial are (± $$\sqrt{\mathrm{c}}$$, 0).<|endoftext|>
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Reference: Elementary Number Theory, David M. Burton, 6th Edition.
In a fragment of a letter in all probability to Father Marin Mersenne in 1643, Fermat described a technique of his for factoring large numbers. This represented the first real improvement over the classical method of attempting to find a factor of n by dividing by all primes not exceeding . Fermat’s factorization scheme has at its heart the observation that the search for factors of an odd integer n (because powers of 2 are easily recognizable and may be removed at the outset, there is no loss in assuming that n is odd) is equivalent to obtaining integral solutions of x and y of the equation .
If n is the difference of two squares, then it is apparent that n can be factored as .
Conversely, when n has the factorization , with , then we may write
Moreover, because n is taken to be an odd integer, a and b are themselves odd, hence, and will be nonnegative integers.
One begins the search for possible x and y satisfying the equation or what is the same thing, the equation by first determining the smallest integer k for which . Now, look successively at the numbers , , , , until a value of is found making a square. The process cannot go on indefinitely, because we eventually arrive at the representation of n corresponding to the trivial factorization . If this point is reached without a square difference having been discovered earlier, then n has no other factors other than n and 1, in which case it is a prime.
Fermat used the procedure just described to factor in only 11 steps, as compared with making 4580 divisions by the odd primes up to 44021. This was probably a favourable case designed on purpose to show the chief virtue of this method: it does not require one to know all the primes less than to find factors of n.
To illustrate the application of Fermat’s method, let us factor the integer . From a table of squares, we find that ; thus it suffices to consider values of for those k that satisfy the inequality . The calculations begin as follows:
This last line exhibits the factorization , where both the factors are prime. In only seven steps, we have obtained the prime factorization of the number 119143. Of course, one does not always fare so luckily — it may take many steps before a difference turns out to be a square.
Fermat’s method is most effective when the two factors of n are of nearly the same magnitude, for in this case, a suitable square will appear quickly. To illustrate, let us suppose that is to be factored. The smallest square exceeding n is so that the sequence starts with:
. Hence, the factors of 23449 are
When examining the differences as possible squares, many values can be immediately excluded by inspection of the final digits. We know, for instance, that a square must end in one of the six digits 0,1,4,5,6,9. This allows us to exclude all the values in the above example, save for 1266, 1961, 4761. By calculating the squares of the integers from 0 to 99 modulo 100, we see further that, for a square, the last two digits are limited to the following 22 possibilities:
00; 01, 04; 09; 16; 21; 24; 25; 29; 36; 41; 44; 49; 56; 61; 64; 69; 76; 81; 84; 89; 96.
The integer 1266 can be eliminated from consideration in this way. Because 61 is among the last two digits allowable in a square, it is only necessary to look at the numbers 1961 and 4761; the former is not a square, but .
There is a generalization of Fermat’s factorization method that has been used with some success. Here, we look for distinct integers x and y such that is a multiple of n rather than n itself, that is,
Having obtained such integers (or, ) can be calculated by means of the Euclidean Algorithm. Clearly, d is a divisor of n, but is it a non-trivial divisor? In other words, do we have ?
In practice, n is usually the product of two primes p and q, with so that d is equal to 1, p, q, or pq. Now, the congruence translates into . Euclid's lemma tells us that p and q must divide one of the factors. If it happened that and , or expressed as a congruence . Also, and yield . By seeking integers x and y satisfying , where , these two situations are ruled out. The result of all this is that d is either p or q, giving us a non-trivial divisor of n.
Suppose we wish to factor the positive integer and happen to notice that . Then, we compute using the Euclidean Algorithm:
This leads to the prime divisor 11 of 2189. The other factor, namely 199, can be obtained by observing that
The reader might wonder how we ever arrived at a number, such as 579, whose square modulo 2189 also turns out to be a perfect square. In looking for squares close to multiples of 2189, it was observed that and which translates into and .
When these congruences are multiplied, they produce . Because the product , we ended up with the congruence .
The basis of our approach is to find several having the property that each is, modulo n, the product of small prime powers, and such that their product’s square is congruent to a perfect square.
When n has more than two prime factors, our factorization algorithm may still be applied; however, there is no guarantee that a particular solution of the congruence , with will result in a nontrivial divisor of n. Of course, the more solutions of this congruence that are available, the better the chance of finding the desired factors of n.
Our next example provides a considerably more efficient variant of this last factorization method. It was introduced by *Maurice Kraitchik* in the 1920’s and became the basis of such modern methods as the *quadratic sieve algorithm*.
Let be the integer to be factored. The first square just larger than n is . So. we begin by considering the sequence of numbers for . As before, our interest is in obtaining a set of values for which the product is a square, say . Then, , which might lead to a non-factor of n.
A short search reveals that ; ; ; or, written as congruences, ; ; . Multiplying these together results in the congruence: , that is, . But, we are unlucky with this square combination. Because only a trivial divisor of 12499 will be found. To be specific,
After further calculation, we notice that
which gives rise to the congruence .
This reduce modulo 12499 to and fortunately, . Calculating
produces the factorization
Problem to Practise:
Use Kraitchik’s method to factor the number 20437.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# 8.2: Side-Splitter Theorem
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Geometry, Chapter 7, Lesson 5.
## Problem 1 – Side Splitter Theorem
In SIDESP1.8xv, you are given \begin{align*}\triangle{CAR}\end{align*}. You are also given \begin{align*}\overline{DS}\end{align*} which is parallel to side \begin{align*}CR\end{align*}.
1. Move point \begin{align*}D\end{align*} to 2 different positions and point \begin{align*}A\end{align*} to 2 different positions and collect the data in the table below. Calculate the ratios of \begin{align*}AD\end{align*} to \begin{align*}DC\end{align*} and \begin{align*}AS\end{align*} to \begin{align*}SR\end{align*} for each position and record the calculation in the table below.
Position \begin{align*}AD\end{align*} \begin{align*}DC\end{align*} \begin{align*}AS\end{align*} \begin{align*}SR\end{align*} \begin{align*}\frac{AD}{DC}\end{align*} \begin{align*}\frac{AS}{SR}\end{align*}
1
2
3
4
2. Make some observations about the ratios of the sides in the triangle. What relationships do you notice?
3. Use the table to complete the following conjecture about the relationship between \begin{align*}\frac{AD}{DC}\end{align*} and \begin{align*}\frac{AS}{SR}\end{align*}. If side \begin{align*}DS\end{align*} is parallel to side \begin{align*}CR\end{align*}, then _____________.
4. In SIDESP2.8xv, drag point \begin{align*}A\end{align*}. Make some observations about the relationship of the ratios \begin{align*}\frac{AD}{DC}\end{align*} and \begin{align*}\frac{AS}{SR}\end{align*}?
5. In SIDESP2.8xv, drag point \begin{align*}D\end{align*}. Make some observations about the relationship of the ratios \begin{align*}\frac{AD}{DC}\end{align*} and \begin{align*}\frac{AS}{SR}\end{align*}?
6. Why are the results different when moving point \begin{align*}A\end{align*} versus moving point \begin{align*}D\end{align*}?
## Problem 2 – Application of the Side-Splitter Theorem
7. Find the value of \begin{align*}x\end{align*}.
8. Find the value of \begin{align*}x\end{align*}.
## Problem 3 – Extension of the Side-Splitter Theorem
For this problem, we will look at a corollary of the side-splitter theorem.
9. In SIDESP3.8xv, move point \begin{align*}U\end{align*} to 2 different positions and point \begin{align*}N\end{align*} to 2 different positions and collect the data in the table on the accompanying worksheet.
Position \begin{align*}RN\end{align*} \begin{align*}NO\end{align*} \begin{align*}EA\end{align*} \begin{align*}AS\end{align*} \begin{align*}\frac{RN}{NO}\end{align*} \begin{align*}\frac{EA}{AS}\end{align*}
1
2
3
4
10. What do you notice about the ratios \begin{align*}\frac{RN}{NO}\end{align*} and \begin{align*}\frac{EA}{AS}\end{align*}?
11. Use the table to complete the following conjecture about the relationship between \begin{align*}\frac{RN}{NO}\end{align*} and \begin{align*}\frac{EA}{AS}\end{align*}. If lines \begin{align*}RE\end{align*}, \begin{align*}NA\end{align*}, and \begin{align*}OS\end{align*} are parallel and cut by two transversals, then ________________________.
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# What do the ACT Math Questions Test?
A
Alejandra Ramos
A
Alejandra Ramos
## act
You're here because you are wondering what skills the ACT Math Questions test - don't worry, we have the answer for you! There are 60 math questions on the ACT and we have some examples to help get you familiar with the skills you will need to do your best on test day! ⭐️
## 🤓 ACT Math Question Skills
There are various question types and questions focused on different skills that are tested. The questions that you will see can focus on your ability to do simple math, others to problem solve, others to see what deeper understanding you have of higher level concepts and application, and many questions will test multiple skill.
### 📝 Preparing for Higher Math: The Most Tested ACT Math Skill
This skill encompasses algebra, functions, geometry, number and quantity, and statistics and probability subsections. These are skills that most people have recently learned and have been exposed to.
The following questions were made to give you an idea about what type of questions each of the subjections in the Preparing for Higher Math skill include. They are not directly from the ACT.
#### ➕ Example Questions: Algebra
(1) Solve the following equation for $x$:
$3(2x-5)+7=4x+14$
A) x = -7
B) x = -4
C) x = 3
D) x = 5
E) x = 7
Solution: C) x=3
To solve the equation, start by distributing the 3 on the left side:
$6x - 15 + 7 = 4x + 14$
Combine like terms:
$6x - 8 = 4x + 14$
Next, isolate the variable x on one side of the equation. To do that, move the 4x term to the left side by subtracting 4x from both sides:
$6x - 4x - 8 = 14$
Simplify:
$2x - 8 = 14$
Now, add 8 to both sides of the equation to get the x term alone:
$2x = 22$
Finally, divide both sides by 2 to solve for x:
$x = 11$
Therefore, the correct answer is C) x = 3.
(2) Simplify the following expression:
$(4x^2 - 7x + 3) / (2x - 3)$
A) 2x - 1
B) 2x + 1
C) 2x - 3
D) 2x + 3
E) 2x + 5
Solution: A) 2x - 1
To simplify the expression, perform polynomial long division or use synthetic division:
2x + 1
2x - 3 | 4x^2 - 7x + 3
-(4x^2 - 6x)
-x + 3
-(-x + 3)
0
Therefore, the simplified expression is $(4x^2 - 7x + 3) / (2x - 3) = 2x + 1$.
#### 📌 Example Questions: Functions
(1) Given the function $f(x) = 2x^2 - 5x + 3$, find the value of $f(3)$.
A) 12
B) 6
C) 0
D) -3
E) 3
Solution: B) 6
To find the value of f(3), substitute 3 for x in the given function and simplify:
$f(3) = 2(3)^2 - 5(3) + 3$ $f(3) = 2(9) - 15 + 3$ $f(3) = 18 - 15 + 3$ $f(3) = 3$
Therefore, the correct answer is B) 6.
(2) Consider the function $g(x) = 3x^3 - 2x^2 + 5x - 4$. Which of the following statements is true about the function?
A) The function is odd.
B) The function is neither even nor odd.
C) The function has a horizontal asymptote.
D) The function has a vertical asymptote.
Solution: B) The function is neither even nor odd.
A function is even if $f(x) = f(-x)$ for all $x$ in its domain, and it is odd if $f(x) = -f(-x)$ for all $x$ in its domain.
Let's check the properties of the given function:
• $g(x) = 3x^3 - 2x^2 + 5x - 4$
• $g(-x) = 3(-x)^3 - 2(-x)^2 + 5(-x) - 4$
• $g(-x) = -3x^3 - 2x^2 - 5x - 4$ The function $g(x)$ is not equal to $g(-x)$, and it is also not equal to the negative of $g(-x)$. Therefore, the function $g(x)$ is neither even nor odd.
#### 📐 Example Questions: Geometry
(1) Find the area of a right-angled triangle with legs of lengths 5 units and 12 units.
A) 17 square units
B) 30 square units
C) 24 square units
D) 60 square units
E) 144 square units
Solution: B) 30 square units
The area of a right-angled triangle is given by the formula: Area = (base * height) / 2.
In this case, the two legs of the right-angled triangle are 5 units and 12 units.
Area = (5 * 12) / 2
Area = 60 / 2
Area = 30 square units
Therefore, the correct answer is B) 30 square units.
(2) In triangle ABC, angle A measures 55 degrees, and angle B measures 75 degrees. What is the measure of angle C, in degrees?
A) 20
B) 30
C) 45
D) 60
E) 90
Solution: D) 60
The sum of the angles in a triangle is always 180 degrees. So, to find the measure of angle C, subtract the measures of angles A and B from 180:
Measure of angle C = 180 - 55 - 75
Measure of angle C = 50
Therefore, the measure of angle C is 60 degrees.
#### 🔢 Example Questions: Number and Quantity
(1) Which of the following numbers is both a multiple of 5 and a perfect square?
A) 15
B) 25
C) 36
D) 48
E) 55
Solution: C) 36
To be a multiple of 5, a number must end in either 0 or 5. Among the given choices, the number 36 ends in 6, so it is not a multiple of 5.
Now, let's check which number is a perfect square. A perfect square is an integer that can be expressed as the square of an integer. Among the given choices, 36 is a perfect square because it can be expressed as 6^2.
Therefore, the correct answer is C) 36.
(2) Which of the following numbers is a prime number?
A) 21
B) 33
C) 47
D) 56
E) 63
Solution: C) 47
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Let's check the given numbers:
• 21: Divisible by 3 and 7.
• 33: Divisible by 3 and 11.
• 47: Only divisible by 1 and 47, making it a prime number.
• 56: Divisible by 2, 4, 7, and 8.
• 63: Divisible by 3, 7, and 9. Therefore, the prime number among the given choices is 47.
#### 🎲 Example Questions: Statistics and Probability
(1) A box contains 6 red balls, 4 blue balls, and 5 green balls. If one ball is randomly selected from the box, what is the probability of choosing a red ball?
A) 6/15
B) 4/15
C) 2/5
D) 3/7
E) 6/15
Solution: A) 6/15
The total number of balls in the box is 6 (red) + 4 (blue) + 5 (green) = 15 balls.
To find the probability of choosing a red ball, divide the number of red balls by the total number of balls:
Probability of choosing a red ball = Number of red balls / Total number of balls
Probability of choosing a red ball = 6 / 15
Simplify the fraction:
Probability of choosing a red ball = 2 / 5
Therefore, the correct answer is A) 6/15.
(2) In a bag, there are 5 red marbles, 4 blue marbles, and 6 green marbles. Two marbles are drawn at random from the bag without replacement. What is the probability of drawing one red marble and one blue marble, in any order?
A) 5/33
B) 2/15
C) 1/3
D) 20/63
E) 2/7
Solution: D) 20/63
Explanation:
To find the probability of drawing one red marble and one blue marble, in any order, we can consider two scenarios:
1. Scenario 1: Red then Blue
2. Scenario 2: Blue then Red
Let's calculate the probability for each scenario:
Scenario 1: Red then Blue
Probability of drawing a red marble first: (5 red marbles) / (15 total marbles) = 5/15
Probability of drawing a blue marble second (after removing one red marble): (4 blue marbles) / (14 remaining marbles) = 4/14
Scenario 2: Blue then Red
Probability of drawing a blue marble first: (4 blue marbles) / (15 total marbles) = 4/15
Probability of drawing a red marble second (after removing one blue marble): (5 red marbles) / (14 remaining marbles) = 5/14
Now, add the probabilities from both scenarios to get the total probability:
Total probability = Probability of Scenario 1 + Probability of Scenario 2
Total probability = $(5/15) * (4/14) + (4/15) * (5/14)$
Total probability = $20/210 + 20/210$
Total probability = $40/210$
Total probability = $20/105$
Total probability = $4/21$
Therefore, the probability of drawing one red marble and one blue marble, in any order, is 20/63.
### ✏️ Essential Skills: The Second Most Tested Skill
This section can include the subsections of proportions, percentages, volume, surface area, and so many more. Although at first glance, this section can seem quite easy, this is not the situation for many students. This is because the content on this section dates back to what you have been learning since middle school and the skill set that you have been developing since you were first introduced to math. some questions in this section can be easy while others may be a little more complicated.
The following questions were made to give you an idea about what type of questions each of the subjections in the Essentials skill include. They are not directly from the ACT.
#### ⚖️ Example Questions: Proportions
(1) If 4 similar notebooks cost $12, how much would 7 similar notebooks cost? A)$5
B) $14 C)$21
D) $28 E)$49
Solution: C) $21 To find the cost of 7 similar notebooks, use proportions: • Let x be the cost of 7 notebooks. • If 4 notebooks cost 12 dollars, one notebook costs 3 dollars. Now, set up the proportion: • (Cost of 7 notebooks) / (Cost of 1 notebook) = (7 notebooks) / (1 notebook) • x / 3 = 7 / 1 • x = 3 * 7 • x = 21 Therefore, the cost of 7 similar notebooks is$21.
(2) In a recipe, the ratio of milk to flour is 3:2. If 5 cups of flour are used, how many cups of milk should be used?
A) 2
B) 5
C) 7.5
D) 8
E) 10
Solution: C) 7.5
Let x be the number of cups of milk needed.
• The given ratio of milk to flour is 3:2, which means:
• (Cups of milk) / (Cups of flour) = 3 / 2
• x / 5 = 3 / 2 To find the value of x, cross-multiply and solve for x:
• 2x = 3 * 5
• 2x = 15
• x = 15 / 2
• x = 7.5 Therefore, 7.5 cups of milk should be used in the recipe.
#### % Example Questions: Percentages
(1) A shirt is originally priced at $40. During a sale, the price is reduced by 20%. How much is the shirt during the sale? A)$8
B) $16 C)$24
D) $32 E)$48
Solution: D) $32 To find the sale price of the shirt, multiply the original price by the percentage reduction: • Sale price = Original price - (Percentage reduction * Original price) • Sale price = $40 - (0.20 *$40) • Sale price = $40 -$8 • Sale price = $32 Therefore, the shirt is priced at$32 during the sale. (2) At the beginning of the year, a company had 80 employees. Over the course of the year, the company hired 20 new employees and had to let go of 12 employees. What was the percentage increase in the number of employees during the year? A) 8% B) 15% C) 20% D) 50% E) 66.67% Solution: C) 20% To find the percentage increase in the number of employees, use the formula: • Percentage increase = (Increase in quantity / Original quantity) * 100 • The increase in the number of employees = 20 (new hires) - 12 (employees let go) = 8 employees. • Percentage increase = (8 / 80) * 100 • Percentage increase = 0.1 * 100 • Percentage increase = 10% Therefore, the percentage increase in the number of employees during the year is 10%. #### 📦 Example Questions: Volume (1) A rectangular box has dimensions of 4 inches by 6 inches by 3 inches. What is the volume of the box? A) 18 cubic inches B) 36 cubic inches C) 72 cubic inches D) 80 cubic inches E) 144 cubic inches Solution: B) 36 cubic inches The volume of a rectangular box is calculated by multiplying its length, width, and height: • Volume = Length * Width * Height • Volume = 4 inches * 6 inches * 3 inches • Volume = 24 cubic inches Therefore, the volume of the box is 24 cubic inches. (2) A cylindrical tank has a height of 10 feet and a diameter of 8 feet. What is the volume of the tank? (Use π ≈ 3.14) A) 80π cubic feet B) 160π cubic feet C) 200π cubic feet D) 400π cubic feet E) 800π cubic feet Solution: B) 160π cubic feet The volume of a cylinder is given by the formula: $\text{Volume}= π * r^2 * h$ Where r is the radius and h is the height. Given the diameter is 8 feet, the radius (r) is half of the diameter, so r = 8/2 = 4 feet. Now, calculate the volume: $\text{Volume}= π * (4 \ \text{ft})^2 * 10 \ \text{ft}$ $\text{Volume}= π * 16 \ \text{ft}^2 * 10 \ \text{ft}$ $\text{Volume}= 160π \ \text{ft}^3$ Therefore, the volume of the tank is 160π cubic feet. #### 🗺️ Examples Questions: Surface Area This is the category many people struggle with as it requires previous knowledge of many of the surface area formulas! I recommend looking over these! (1) A cube has a side length of 6 inches. What is the total surface area of the cube? A) 12 square inches B) 24 square inches C) 36 square inches D) 72 square inches E) 216 square inches Solution: D) 72 square inches The total surface area of a cube is calculated by multiplying the area of one face by the number of faces (6 for a cube). • Surface Area = 6 * (Side length)^2 • Surface Area = 6 * (6 inches)^2 • Surface Area = 6 * 36 square inches • Surface Area = 216 square inches Therefore, the total surface area of the cube is 216 square inches. (2) A right circular cone has a base radius of 5 feet and a slant height of 13 feet. What is the total surface area of the cone? (Use π ≈ 3.14) A) 110π square feet B) 150π square feet C) 195π square feet D) 210π square feet E) 260π square feet Solution: C) 195π square feet The total surface area of a right circular cone is the sum of its lateral surface area and the area of its base. The lateral surface area of a cone is given by: • Lateral Surface Area = π * r * l • where r is the base radius and l is the slant height. Given r = 5 feet and l = 13 feet, calculate the lateral surface area: • Lateral Surface Area = π * 5 feet * 13 feet • Lateral Surface Area = 65π square feet The area of the base of the cone is given by: • Base Area = π * r^2 • Base Area = π * (5 feet)^2 • Base Area = 25π square feet Now, calculate the total surface area: • Total Surface Area = Lateral Surface Area + Base Area • Total Surface Area = 65π square feet + 25π square feet • Total Surface Area = 90π square feet Therefore, the total surface area of the cone is 90π square feet. ### 📍 Modeling: Also Tested on the ACT Although when you think of modeling you might immediately think of making models, charts, and different ways to show data, modeling is often tested with setting up equations. This will require strong skills of being able to find what you are being asked for by using the right formula. #### Example Questions: (1) A car rental company charges a flat fee of 30 dollars per day for renting a car, plus an additional 0.25 dollars per mile driven. If a customer rents a car and drives it for 3 days, accumulating 150 miles, how much will the customer be charged in total? A)$45
B) $60 C)$75
D) $90 E)$105
Solution: C) $75 To find the total charge, calculate the daily rental cost and the mileage cost, and then sum them up. • Daily rental cost = 30 per day * 3 days =$90
• Mileage cost = 0.25 per mile * 150 miles = $37.50 • Total charge = Daily rental cost + Mileage cost = 90 + 37.50 = $127.50 Therefore, the customer will be charged a total of$127.50. (2) A company is selling tickets to a concert. The cost per ticket is 50 dollars during the early bird period and 60 dollars during the regular sale period. The company estimates that during the early bird period, they will sell 300 tickets, and during the regular sale period, they will sell 500 tickets. The company also expects that for every$ 5 increase in ticket price, the number of tickets sold will decrease by 30. Assuming all tickets are sold, how much total revenue will the company generate from ticket sales?
A) $57,000 B)$59,000
C) $61,500 D)$63,000
E) $65,000 Solution: C)$61,500
Early Bird Period:
• Tickets sold = 300
• Ticket price = $50 • Revenue during early bird period = Tickets sold * Ticket price = 300 * 50 dollars =$15,000 Regular Sale Period:
• Tickets sold = 500
• Ticket price = $60 • Revenue during regular sale period = Tickets sold * Ticket price = 500 * 60 dollars = $30,000 Now, let's consider the decrease in ticket sales due to the price increase during the regular sale period. For every$5 increase in ticket price, the number of tickets sold will decrease by 30. • The price increased by 60 - 50 =$10, resulting in a decrease in ticket sales by (10 / 5) * 30 = 60 tickets.
• Adjusted tickets sold during the regular sale period = 500 - 60 = 440 tickets
• Revenue during the adjusted regular sale period = Adjusted tickets sold * Ticket price = 440 * 60 = \$26,400
• Total revenue = Revenue during early bird period + Revenue during adjusted regular sale period
• Total revenue = 15,000 + 26,400 = $41,400 Therefore, the company will generate a total revenue of$41,400 from ticket sales.
## 🫡 Conclusion
Now you have an idea of the skills you will need to study so you can feel prepared for the ACT Math session. Remember to take a deep breath! 😮💨 You can do this! If you want more detailed guides on each section make sure to check out all the Fiveable ACT Math Guides!<|endoftext|>
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• 40 questions
• 4 answer choices
• Passages with charts/diagrams
Time: 35 minutes
The ACT science section does not test specific science facts. Instead, students will need to understand how to use certain skills of scientific reasoning. Nearly every question can be answered by referring to the material in the passage. Yet, somewhere between 1 and 5 questions on every ACT test, will concentrate on science-based knowledge not given in the passages. There will be a few questions that require some arithmetic. You may have to find an average or utilize your algebra knowledge.
Question types and scoring
Students receive 1 raw point for every correct answer. There’s nothing lost for answering incorrectly. The raw score is calculated by tallying up the raw points. The overall raw score is then converted to a score on a 1-36 point scale. The ACT science score is 1 of 4 scores that's factored into the ACT composite. The ACT composite is an average of 4 section scores. Meaning, a lower science score will bring down the ACT composite, and a higher score, will help to increase the ACT composite. This, of course, relies on the scores from the other ACT sections: English, reading, and mathematics.
Each ACT test contains:
• 6-8 Conflicting Viewpoints questions. These questions will present two or more scientific theories. Amongst all of the theories, not all of them can be correct. The questions may not ask you to prove which theory is correct, but instead, ask you to describe the viewpoints and how they relate to one another.
• 12-16 Data Representation questions. These questions usually present charts and tables that display different variables. Students will find 2-4 variables and must be able to describe the relationships between the variables.
• 18-22 Research Summary questions. These questions describe an experiment and the scientific findings. The passage will detail the results, makeup, and hypothesis. The questions will ask you about the experiment’s findings, design, and implementation.
Every ACT test will include the same set of instructions for the science section. Master the instructions and you won’t need to read them when taking the test. This enables you to spend more time on the problems by skipping the instructions.
ACT SCIENCE DIRECTIONS: There are seven passages in this test. Each passage is followed by several questions. After reading a passage, choose the best answer to each question and fill in the corresponding oval on your answer document. You may refer to the passages as often as necessary. You are NOT permitted to use a calculator on this test.
Watch out for attractor answer choices. The ACT anticipates the student making careless mistakes. The test will setup traps for students, and those most impacted, will be students who are not ready for them. The attractors mainly show up on medium to difficult questions.
ACT science tips:
• Focus on one passage at a time
• Learn which types of passages come easier for you
• Tailor your strategies to each passage type
• Don’t worry if the subject is unknown to you. Most of the information is in the passage.
• Work on timing and pacing by practicing passages
• Know the directions, so that you can skip them
• You have 52 ½ seconds per question
• Adjust your pacing to the passage types
o Data Representation passages: spend 4 minutes
o Research Summary passages: spend 5 minutes
o Conflicting Viewpoints passages: spend 7 minutes<|endoftext|>
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## Algebra 1 Unit 2 Interactive Notebook Pages | Relations & Functions
Here are the notes I used this year for the 2nd unit of Algebra 1:
Day 1:
We started off the unit with a classifying variables sort. This was a good way to jog students’ memories about their prior knowledge, and it also served as a jumping point into domain and range!
From there, we went into what a relation, domain, and range is, and how it relates to independent and dependent variables.
We then made the distinction that there are two types of relations, discrete and continuous, and we must pay attention to context to determine what type of relation we have.
From there, we started to talk about all of the different ways we could represent a discrete relation, and how we find the domain and range from each representation. We used this foldable, which went over great with the students. They caught on super quickly, and they mentioned that they liked having one example to do together, and one to do on their own for each representation.
Day 2:
We started off with a word problem to review domain and range in a (discrete) relation.
From there, we filled out a Frayer vocabulary model for functions, to make sure that students really understood what they are and aren’t.
Then, using the definition for function we just wrote down on the Frayer model, we made a cheat sheet to refer back to that tells us all of the different ways a relation (discrete or continuous) would NOT be a function.
We practiced classifying functions using a card sort from Amazing Mathematics. Instead of cutting and pasting, we decided to color-code instead! Love it! (In the words of one of my students, this is the page that has “fourteen thousand graphs.”)
We then filled out another cheat sheet, this time for domain and range of continuous functions. Students reasoned together through the inequalities and we talked about what a bound actually means (we used a lot of basketball references).
We practiced finding the domain and range for continuous relations (as well as determining whether or not they were a function), using the following set of notes. PS: It took me a LONG time to figure out how to make a parabola or a trigonometric wave using Microsoft’s shape tools. I feel overly proud of this set of notes! You can download them here
Day 3:
We began with a recap warm-up on domain and range for continuous relations.
To make sure that students didn’t forget about discrete relations, we went back and did more practice with determining their domain and range, and also stating whether or not the relations were functions.
Day 4:
We started off with a reference sheet on function notation and how to read/say it.
From there, we did a lot of practice with function notation.
Inside this set of notes, we really emphasized interpreting what we were being given in a problem (input or output value) and what the problem was actually asking us to find (input or output value), before starting the problem. This helped students from making a lot of careless mistakes. After we practiced function notation in both directions (evaluating a function, and solving for an input given the function’s output), we mixed up the problems and even threw a few variables and function compositions in there!
Day 5:
Recap warm-up on function notation. Problems 5 and 6 both spurred amazing conversations about order of operations.
After doing this recap warm-up, we did my function notation mystery sum activity, which was a blast. It encourages students to collaborate together and it’s really high engagement each time.
From there, we continued talking about function notation, but now in terms of a graph. Interpreting what the function notation was telling us was such a huge part of the previous day’s lesson, that I wanted to see how they could do when we attached a context to the problem.
Inside, we worked on graphing functions, and using the graph to find an x-value. Some students preferred solving for x, but others were impressed by my tracing over on the graph method. To each their own–that’s the beauty of math, in my opinion.
Day 6:
Recap warm-up over function notation with graphs, and then we reviewed for the test.
Day 7: Test!
## Algebra 1 – Unit 1 INB Pages | The Foundations of Algebra
Here’s what went into our INBs for the 1st unit of Algebra 1:
Day 1:
We glued in a reference sheet for the real number system. Our textbook uses I for the set of irrational numbers. I went with the same notation this year, but I think I’m going to go with R-Q for next year, since I is used for imaginary numbers, later on.
To practice working with these definitions, we did a real number system sort, which I found from Amazing Mathematics! My students enjoyed doing it, and it spawned many great conversations about the difference (however subtle they may be), between the sets of real numbers.
For homework, students did this Always/Sometimes/Never sort, which is also from Amazing Mathematics. They were given about 20 minutes in class to begin their assignment, and then had whatever was left as their take-home assignment for the night. This one was even better than the last card sort, in terms of spurring student conversations. Students were justifying with counterexamples and providing fully flushed out reasons for where each card should get placed. It was awesome!
As a note, we also keep a binder for the class which holds extra handouts, like additional reference sheets and homework assignments that don’t go in the INB. My favorite reference sheet that didn’t go into the INB was this real numbers flowchart that I made. The day of teaching my lesson on real numbers, I noticed that using the “Venn diagram” approach wasn’t meshing well with some of my students. That afternoon, I went home and made a flowchart handout that they could refer to, in addition to their INB pages. Next year, I think I’ll just use this flowcharts in a mini-book format for notes, instead! I found that students started making more connections about the sets each number belongs to (i.e. not only is a number natural, but it’s a whole number, and an integer, and a rational number), and students were able to remember the questions they need to ask themselves when determining the best classification for a real number.
Day 2:
We started off with a recap warm-up on the real number system, which we covered the day before.
From there, we did a translating expressions sort, also from Amazing Mathematics. (Can you tell I love her sorts?!).
From there, we used our key words and started defining what a variable is, and what an expression is.
For homework, students did the following problems. They had about 15 minutes of class time to get started. We color-coded “turn-around words” in pink, “parentheses-words” in green, and “equals words” in blue. Students marked the page in highlighter before beginning to translate the expressions. They mentioned that this made the process much easier for them!
Day 3:
We began with a recap warm-up over translating expressions.
From there, we talked about evaluating expressions and also reviewed the order of operations.
From there, we discussed the properties of real numbers and students made up their own examples for each property.
For in-class practice, students did the a properties of real numbers puzzle from Lisa Davenport. A student volunteered to glue it into my notebook. Notice the lack of glue? Notice the crooked edges? It was a very sweet offer, but I’m I don’t think it’s one I’ll be taking again any time soon.
Day 4:
We started with a recap warm-up over evaluating expressions and identifying properties of real numbers.
Next we took notes on combining like terms and the distributive property, cutesy of Sarah at Math Equals Love.
Day 5:
Recap warm-up over distributing and combining like terms.
What is a solution? What does it mean to be a solution? What does it look like?
Up next, we focused on solving and verifying solutions to 1-step and 2-step equations. I’ve found that verifying a solution is a skill that students struggle with more than solving (at least in Algebra 1), so I wanted to make sure it got emphasized.
Day 6:
We filled out a foldable for solving 2-step equations. Those pesky fractions are going to be our friends by the end of today!
Day 7:
Recap warm-up over solving equations.
Day 8: Review
Day 9: Test!
## {FREEBIE} Test or Quiz Retake Form
Last week each of my classes had their first test. Most did quite well, but, like normal, a few did not. In my school, the math department policy is that students are allowed to retake tests, but not quizzes. Honestly, I wish it was the other way around, but that’s a topic for another post. Since tests are worth 40% of a students grade at my school (we have a 10-40-40-10 grade distribution for homework, quizzes, tests, and final exam), it is important that students do a retake whenever they have performed poorly on an assessment. Not only should they do a retake for the sake of their grade, but also so they are in a position to better understand the material going forward. Math is constantly building on itself, so I’m glad that students have the opportunity to go back and revise their learning.
The past two years, I have used a retake form created by someone else in my department, but I never quite liked it. Although it asked students honest reflection questions, I felt like it was a bit condescending to students and I questioned if it might prevent some students from pursuing a retake opportunity. I used it anyway since everyone else in the department used the same form, but silently feeling like a bad teacher because of it.
This year, I wanted to create a new retake form. Over the summer I had played with a few different versions, but it wasn’t until today (the day before my students actually need the retake form, lol!) that I came up with something that just felt right. It asks students to reflect without pointing any fingers or making them feel guilty for not doing well the first time. It’s straight to the point, and requires them to revisit the material in new ways so that they can improve their understanding. Tomorrow I am going to walk through the retake form together as a class so my students know what my expectations are for it and have an idea about how to productively go about preparing for a retake opportunity.
If you would like to use this retake form in your own classroom, you can download it for free from my Teachers Pay Teachers store. While you’re there, make sure to follow my store so you don’t miss any other resources I post!
## How I do Homework
Friday evening I was tagged in a tweet that was asking about my homework policies and I just had too much to say in to fit in 140 characters or less, so I figured I’d write a blog post as a response. One thing I make sure to do in class is to always call “homework” a “practice assignment.” You’ll never hear the word “homework” come out of my mouth at school because that seems to open up a very unproductive can of worms. Practice Assignment is clear to students and it can be worked on both at school and at home, and also reminds students why they’re doing it.
A bit about me, I teach on a 7-period a day schedule. On Mondays and Fridays each class is 51 minutes long, Tuesdays and Thursdays are 46 minutes long to account for a 30-minute advisory class that happens twice a week, and Wednesdays are 43 minutes long to account for our early release professional development sessions. I definitely am not a teacher whose sole purpose is teaching to the state test and “exposing” students to 100% of the topics in the textbook. I like a much more balanced approach. I think it’s important enough to go slow enough that students actually have a chance to absorb the material, but it is also important to me that we get through the required material as to not disservice them for their next math class.
I also believe in homework. This has become quite controversial over the years, especially in the online teacher communities. However, I do believe in a balanced approach to homework. Maybe 15 questions a night, and it’s rarely due the very next day. I have homework due on quiz or test days, which allows students a bit of wiggle room to work around their schedules. I think homework does so much for students. It gives them a chance to play with the material more on their own so they can really figure out what questions they have. It teaches them how to self-advocate for themselves when they need help. It teaches them time-management skills to work around their busy schedules and how to prioritize tasks. To completely get rid of homework at the high school level seems like it would be quite a disservice to students, in my opinion, and doesn’t seem like it’s setting them up for success in their next endeavors (whether that be entering the work force or higher education). However, I do acknowledge that each teacher certainly knows what works best for their own class, so the no-homework approach may be just right for other groups of students. It just doesn’t work for mine.
The past couple of years, the math department at my school has focused on paring back our pacing guide a bit to focus on the core concepts in a deeper way, and our state test scores shot up 13% each year in a row. Focusing on 90% of the topics and pairing back the most peripheral 10% has done amazing things for our students. Exposure to content means nothing if it’s at a pace students can’t absorb, but there is a tricky balance. If you’re only getting through half of the pacing guide each year, that’s definitely not setting up your students for success either.
Anyway, to get back to the main topic, I treat each class a bit differently when it comes to how we do in-class work and homework, so I’ll break up what I do by subject:
Algebra 1 + Support:
Info: This class is 2-periods long each day. I have them the first two periods of each day. Students are part of a cohort of 30 students that take math, English, and science together each day. They have been identified by their 8th grade principals for being at extreme risk of not graduating and their parents have agreed for them to be in a special program at the high school to help them be more successful.
Students pick up notes and homework as they walk in the door each morning. We record homework assignments on our practice tracker after our daily warm-up problems have been completed. As we are taking notes together in class, students are encouraged to switch back and fourth between notes and similar homework problems. Most days there is 15-40 minutes of work time. On our practice tracker, we write down the minimum required number of problems to enter the classroom the next day. Let me explain how this works:
Let’s say that the homework sheet has 15 problems. If they are given 30 minutes of homework time that day, as a class we come up with the minimum expected amount to be finished during that work time. The class might decide that, given the time they have to work in class that day, they think everyone should be able to finish at least 6 of these 15 homework problems during our in-class homework time, so we write this on our practice trackers (min=6). The next morning, I stop each student on their way into the classroom. They must show me that they have completed at least any 6 of the 15 homework problems to get inside the classroom. If they have failed to do so, they sit outside the classroom to work with our aide and finish up those problems with 1-on-1 help while the daily announcements are being played, and then come back to join the class who have already begun to work independently on the daily warm-up problems. Students are given one grace period to be “stuck outside,” after that a call home is made to discuss ways we can help the student be successful at staying on top of homework, especially when class time is being given. Now, since only 6 of the problems were required to be done the next day, students have until the next quiz or test (whichever comes first) to complete the remainder of the assignment. All homeworks are graded on a 3-point completion scale. Quizzes normally are given once or twice a week. NOTE: if a student emails me the previous night letting me know that they will not be able to complete their minimum required amount for whatever reason and what their plan is to make it up, then they are allowed in to class, no questions asked. Again, I really want students to learn how to advocate for themselves, so this is part of that goal.
Practice trackers get turned in on the unit test date, and then a new one is handed out the next day to start off the new unit. Students have mentioned how much they like using the practice trackers because it helps them remember their homework, and they have also commented that they like having the minimum required amount since it makes them stay on top of things and helps them focus better during work time.
I also offer a “double stamps” policy in all of my classes (except for Statistics because it’s college credit) where if a student finishes an assignment the day I give it to them and brings it back before I go home that day, not only does their practice tracker get stamped as 3 points “all done” for that assignment, but they get a bonus 3pt stamp for working so hard at completing it that day.
Lastly, I accept homework late up until the unit test. I also allow students to “move up” points. Let’s say they just got the minimum done and didn’t do more before the quiz where homework gets stamped off. They’d probably get stamped for 1 point or “a bit” completed. If they did more before the test, they could improve their score and I would re-stamp their practice tracker for 2 points.
Algebra 2 and Geometry:
Exactly the same as Algebra 1, except for the minimum required amount and having to show me the assignments at the door. They also don’t usually get 15-40 minutes of homework time each class, since these are only 1-period long classes. Most days, they get 5-10 minutes of work time. Other days, they get a bit more.
College Credit Statistics (MTH 243 and MTH 244):
Since this class is dual credit, homework is inherently done differently. Each Monday students are given a quiz over problems taken from the prior week’s assignments, and once they are done, they begin working on an Algebra and Geometry review to make sure that their other math skills are staying fresh. After everyone has finished the quiz, students work together in groups to complete the weekly Algebra and Geometry review. During the last 10 minutes of class, I allow them to ask me to go over any 2 of the 10 questions with them as a class. At this point, the rest of their homework is assigned, corresponding to whichever sections of the textbook we will be covering that week. Homework is always assigned on Monday and due the following Monday so students can manage their time as they see fit, in order to work around their schedules. The general format of the statistics class is:
Monday: quiz and Algebra & Geometry review. Tuesday, Wednesday, Thursday: new notes. Friday: work day where students are able to work on their homework assignment together for the period.
Late work is not accepted in this class due to the dual credit aspect.
____________________________________________________________________________
I could write a ton more about how homework works in each class, but I’ll leave it here so I don’t end up boring you all with a novel. If you would like me to go over anything in more depth, or have a question about something I forgot to address, please let me know and I will make sure to answer it right away. Thanks for reading!
## 3 Thoughts on Teachers Pay Teachers
Today, the question about what the #MTBoS thinks about Teachers Pay Teachers (TPT) was brought up.
Some love it, some hate it. Here’s my take.
I LOVE Teachers Pay Teachers.
1. My first year teaching, I had 4 preps, one of which was Geometry. I had a non-traditional high school path, and had never taken Geometry, myself, so I was extremely nervous (All of those proofs! Back then, they had me shaking in my boots.). There wasn’t enough time to create everything myself, maintain my sanity, and have my lessons be to the quality that I desired. It just wasn’t possible. So, I went to TPT and found a curriculum for Geometry–complete with notes, homework assignments, quizzes, and activities. It was expensive, but cost the equivalent of 14 hours of work. There is no way I could’ve created a twenty-fifth of the materials available in that product in the same amount of time, so it was well worth it for me. And everything in it was AMAZING–truly, high quality materials. There is no way I could’ve provided the same level of quality for my students that year without it, PLUS it saved my sanity.
2. I should disclose that I am a TPT seller. I have a small store, but it’s definitely a passion for me. I find that TPT drives me to create better and better materials to provide to my students and makes me more creative. What I would’ve provided to my students without TPT would be a 9/10, and what I’d provide with the intention of selling a product on TPT would be a 10/10. I don’t think people realize the amount of hours it takes to put a quality product on TPT. Maybe it takes 10 hours to create and perfect it, and another 5 hours to do the finishing touches. Those extra 5 or so hours really take a resource to the next level, but the time commitment is not practical at all for a teacher that doesn’t have the monetary incentive attached. 5 extra hours of work for just one resource would be a crippling time constraint to do regularly, if there was no other benefit. Even doing it for a resource that would be shared online for free wouldn’t make sense to commit that much extra time to a resource. Sure there are some bad products on TPT and there are some amazing resources offered for free from the #MTBoS community. Generally speaking, though, I find the distribution would be as follows:
Like any online purchase, reading customer reviews on TPT can help you easily stay in the upper quartile of the TPT distribution.
3. I really like the idea that teachers can get recognition for their talents through TPT. Teaching has a pretty low ceiling as far as job recognition goes, and TPT definitely helps with that. Teaching can also be a financially difficult profession to choose. It takes a lot of money to get the Master’s degree required to be a teacher (at least for my state), which leaves many with crippling debt. TPT can provide supplemental income that can help the financial strain that many teachers face, allowing them to actually stay teaching. In addition to the monetary assistance that TPT can provide helping to keep teachers in the professions, I think it makes many of the sellers feel more valued, which definitely helps to avoid teacher burnout. I find a overwhelming sense of pride in my TPT store and it makes me feel valued as a teacher. I know how hard teaching is, and I know how little we get paid, so it makes me feel so valued as a teacher each time someone decides that my product is worth spending those hard-earned dollars on and bringing into their own classrooms. It makes me feel like what I’m doing is good and I’m appreciated in the teaching world. I also hope that I can be that life-saver for someone else out there like that Geometry curriculum was for me in my 1st year teaching.
## Meal Prep for Teachers, Vol. 1
Teachers are some of the busiest people I know and frequently sacrifice doing things for their own well-being (like making home-cooked meals) in order to make the best lessons and classroom environment for their students. My first day of school is this upcoming Tuesday, so I wanted to make sure that I had food ready to go for the week. I like to have variety in what I eat, so I chose three meals to make (the order shown below is the order that I cooked them).
To try to force myself to get better at cooking and also eat a proper meal each night, I’ve decided I’m committing to cooking 3 meals each Sunday morning to last me through the week. Each evening, I’ll quickly make a side to go with it (steamer veggies, minute rice, fruit, etc.)
A bit about myself: I have minimal cooking skills and I don’t do mushrooms or fish.
# Meal 1: Root beer Pulled Pork Difficulty Level: 0/5 Time: 7 hrs and 10 min (10 min prep + 7 hours cooking)
Recipe from: me!
Food Ingredients:
• 4 lbs of pork shoulder
• 36 oz of Root Beer
• 1 18oz bottle of Sweet Baby Ray’s Sweet & Spicy BBQ Sauce
• 8 Hamburger buns.
Supplies:
• Crock Pot
• Crock Pot Liners (optional, but makes cleanup non-existent).
• Hand Mixer (optional, but makes shredding pork take <60 sec instead of 5-10 min)
• Large Bowl
• Colander
• Large Mixing Spoon
Instructions:
1. Place crockpot liner inside crockpot.
2. Put pork shoulder in crock pot.
3. Pour root beer in crock pot.
4. Put lid on crockpot and let cook on low for 7 hours.
5. After 7 hours, remove crock pot liner and dump contents into the colander in the sink to drain.
6. Place remaining meat in large bowl (I use a Tupperware container that I will keep the leftovers in). On the lowest setting, use the hand mixer to shred the pork.
7. In the bowl with the shredded pork, add the bottle of BBQ sauce. Use the large mixing spoon to evenly distribute the BBQ squce.
# Meal 2: Maple & Mustard ChickenDifficulty Level: 1/5Time: 20 minutes (5 min prep, 15 min cook time)
Changes I made: Instead of pounding the chicken breasts flat, I just cut them in half, lengthwise.
# Meal 3: Sausage & Veggie Skillet Difficulty Level: 2/5Time: 60 minutes (30 prep and 45 cook-some prep happens while cooking has already begun)
HUGE Time Saving Tip: Start the potatoes first. They take 25 minutes to cook. While they are cooking, cut up all of the other veggies so they are ready to go once the potatoes are done. Don’t be like me the first time I made this recipe and do all of the chopping first, and then have 25 minutes of nothing to do while the potatoes cook. It took 90 minutes instead of an hour.
# What’s The Verdict?
I liked each of the three meals that I made this week, which isn’t something that normally happens. I ended up freezing half of the pulled pork and half of the sausage skillet to use in later weeks.
The sausage skillet and the pulled pork are both repeat recipes that I’ve made in the past. My favorite of the three is the sausage and veggie dish, although it is the most involved–you really are cooking for the entire 60 minutes of the recipe. I’m not much of a sausage fan, so I’d like to point out that not all sausages are made the same. Most are not my thing, but I’ve found a couple that I really do like. Experiment around a bit and find one that you like because it really does make all the difference for this dish. The pulled pork is my second favorite. It’s so easy and so flavorful. A really quick dinner to reheat, as well. The chicken was good, definitely nothing wrong with it. I just preferred the other two a bit more. I plan to eat the chicken with steamed green beans and brown rice. I think that’ll be a really tasty and healthy meal.
Total Time: 2 hours and 30 minutes (not counting the 7 hour cook time for the crock pot pulled pork, which I started before I went to sleep the night before).
Overall, I’m really happy with my meals this week. No duds! Woo!
## My No. 1 “Teacher Hack” For Interactive Notebooks
Making things for interactive notebooks can be tedious, at times. If you’re like me, you use a composition notebook so students will (hopefully) resist the urge to tear out pages for scratch paper. The issue with composition notebooks, however, is their sizing. A full sheet of paper is much too large to fit, but a half sheet makes the page feel a bit empty.
Also, unless you want to make everything from scratch to perfectly fit in your interactive notebook, you’re a bit stuck on what to do to get full-sized materials you may have used in the past to fit.
My hack: print any normal sized paper at 80-85% the size and, after cutting out the paper, it will fit PERFECTLY into a composition interactive notebook. Use this hack to make the world your oyster.
Here’s how to do it:
1. Make sure your document has been saved as a PDF.
2. When you go to print, select the following setting:
Rule of Thumb:
If the margins on the original paper are 1″, print at 85%.
If the original margins on the paper are .5″, print at 80%.
If the original margins on the paper are at .25″, print at 75% (not common).
Here’s the difference it makes:
This has saved me a TON of time making interactive notebook pages, and also allows the writing space to be much larger for students. Sometimes a half-sheet can be cramped. Hopefully this teaching hack can help save you a ton of time, like it does for me!<|endoftext|>
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# If f(x)=secx then calculate f''(pi/3)?
Sep 29, 2017
$f ' ' \left(\frac{\pi}{3}\right) = 14$
#### Explanation:
We have:
$f \left(x\right) = \sec x$
Differentiate wrt $x$:
$f ' \left(x\right) = \sec x \tan x$
Differentiate wrt $x$ applying the product rule:
$f ' ' \left(x\right) = \sec x \left(\frac{d}{\mathrm{dx}} \tan x\right) + \left(\frac{d}{\mathrm{dx}} \sec x\right) \tan x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sec x \left({\sec}^{2} x\right) + \left(\sec x \tan x\right) \tan x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\sec}^{3} x + \sec x {\tan}^{2} x$
When $x = \frac{\pi}{3} \implies \tan x = \sqrt{3}$, $\sec x = 2$, And so:
$f ' ' \left(\frac{\pi}{3}\right) = {\left(2\right)}^{3} + \left(2\right) \left(3\right) = 14$<|endoftext|>
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The Arctic Monitoring and Assessment Programme (AMAP) says unless action is taken to reduce greenhouse gas emissions, ocean levels will increase by between 0.9 and 1.6 metres in this century.
That’s two to three times greater than the range predicted in the last Intergovernmental Panel on Climate Change report, which did not include the impact of Arctic ice melting.
The Snow, Water, Ice and Permaforst in the Arctic report is based on the latest scientific research into the state of the cryosphere, the part of the Earth’s surface that is seasonally or perennially frozen. Its Key Findings include:
- The past six years (2005–2010 have been the warmest period ever recorded in the Arctic Higher surface air temperature are driving changes in the cryosphere.
- There is evidence that two components of the Arctic cryosphere – snow and sea ice are interacting with the climate system to accelerate warming.
- The extent and duration of snow cover and sea ice have decreased across the Arctic. Temperatures in the permafrost have risen by up to 2 °C. The southern limit of permafrost has moved northward in Russia and Canada.
- The largest and most permanent bodies of ice in the Arctic – multiyear sea ice, mountain glaciers, ice caps and the Greenland Ice Sheet – have all been declining faster since 2000 than they did in the previous decade.
- Model projections reported by the Intergovernmental Panel on Climate Change (IPCC) in 2007 underestimated the rates of change now observed in sea ice.
- Maximum snow depth is expected to increase over many areas by 2050, with greatest increases over Siberia. Despite this, average snow cover duration is projected to decline by up to 20% by 2050.
- The Arctic Ocean is projected to become nearly ice-free in summer within this century, likely within the next thirty to forty years.
- Changes in the cryosphere cause fundamental changes to the characteristics of Arctic ecosystems and in some cases loss of entire habitats. This has consequences for people who receive benefits from Arctic ecosystems.
- The observed and expected future changes to the Arctic cryosphere impact Arctic society on many levels. There are challenges, particularly for local communities and traditional ways of life. There are also new opportunities.
- Transport options and access to resources are radically changed by differences in the distribution and seasonal occurrence of snow, water, ice and permafrost in the Arctic. This affects both daily living and commercial activities.
- Arctic infrastructure faces increased risks of damage due to changes in the cryosphere, particularly the loss of permafrost and land-fast sea ice.
- Loss of ice and snow in the Arctic enhances climate warming by increasing absorption of the sun’s energy at the surface of the planet. It could also dramatically increase emissions of carbon dioxide and methane and change large-scale ocean currents. The combined outcome of these effects is not yet known.
- Arctic glaciers, ice caps and the Greenland Ice Sheet contributed over 40% of the global sea level rise of around 3 mm per year observed between 2003 and 2008. In the future, global sea level is projected to rise by 0.9–1.6 m by 2100 and Arctic ice loss will make a substantial contribution to this.
- Everyone who lives, works or does business in the Arctic will need to adapt to changes in the cryosphere. Adaptation also requires leadership from governments and international bodies, and increased investment in infrastructure.
- There remains a great deal of uncertainty about how fast the Arctic cryosphere will change in the future and what the ultimate impacts of the changes will be. Interactions (‘feedbacks’) between elements of the cryosphere and climate system are particularly uncertain. Concerted monitoring and research is needed to reduce this uncertainty.
Joe Romm writes in Climate Progress that this report’s predictions are conservative, because they do not include feedback effects that are likely to release methane from the Arctic permafrost when temperatures rise.<|endoftext|>
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## 5.2.2 Joint Cumulative Distribution Function (CDF)
We have already seen the joint CDF for discrete random variables. The joint CDF has the same definition for continuous random variables. It also satisfies the same properties.
The joint cumulative function of two random variables $X$ and $Y$ is defined as \begin{align}%\label{} \nonumber F_{XY}(x,y)=P(X \leq x, Y \leq y). \end{align} The joint CDF satisfies the following properties:
1. $F_X(x)=F_{XY}(x, \infty)$, for any $x$ (marginal CDF of $X$);
2. $F_Y(y)=F_{XY}(\infty,y)$, for any $y$ (marginal CDF of $Y$);
3. $F_{XY}(\infty, \infty)=1$;
4. $F_{XY}(-\infty, y)=F_{XY}(x,-\infty)=0$;
5. $P(x_1<X \leq x_2, \hspace{5pt} y_1<Y \leq y_2)=$ $\hspace{60pt} F_{XY}(x_2,y_2)-F_{XY}(x_1,y_2)-F_{XY}(x_2,y_1)+F_{XY}(x_1,y_1)$;
6. if $X$ and $Y$ are independent, then $F_{XY}(x,y)=F_X(x)F_Y(y)$.
Example
Let $X$ and $Y$ be two independent $Uniform(0,1)$ random variables. Find $F_{XY}(x,y)$.
• Solution
• Since $X,Y \sim Uniform(0,1)$, we have $$\nonumber F_X(x) = \left\{ \begin{array}{l l} 0 & \quad \textrm{for } x < 0 \\ x & \quad \textrm{for } 0 \leq x \leq 1 \\ 1 & \quad \textrm{for } x > 1 \end{array} \right.$$ $$\nonumber F_Y(y) = \left\{ \begin{array}{l l} 0 & \quad \textrm{for } y < 0 \\ y & \quad \textrm{for } 0 \leq y \leq 1 \\ 1 & \quad \textrm{for } y > 1 \end{array} \right.$$ Since $X$ and $Y$ are independent, we obtain $$\nonumber F_{XY}(x,y)=F_X(x)F_Y(y) = \left\{ \begin{array}{l l} 0 & \quad \textrm{for } y < 0 \textrm{ or } x<0 \\ & \\ xy & \quad \textrm{for } 0 \leq x \leq 1, 0 \leq y \leq 1 \\ & \\ y & \quad \textrm{for } x>1, 0 \leq y \leq 1 \\ & \\ x & \quad \textrm{for } y>1, 0 \leq x \leq 1 \\ & \\ 1 & \quad \textrm{for } x>1, y > 1 \end{array} \right.$$ Figure 5.7 shows the values of $F_{XY}(x,y)$ in the $x-y$ plane. Note that $F_{XY}(x,y)$ is a continuous function in both arguments. This is always true for jointly continuous random variables. This fact sometimes simplifies finding $F_{XY}(x,y)$. The next example (Example 5.19) shows how we can use this fact.
Figure 5.7: The joint CDF of two independent $Uniform(0,1)$ random variables $X$ and $Y$.
Remember that, for a single random variable, we have the following relationship between the PDF and CDF: \begin{align}\label{} \nonumber F_X(x) &=\int_{-\infty}^{x} f_X(u)du, \\ \nonumber f_X(x) &=\frac{dF_X(x)}{dx}. \end{align} Similar formulas hold for jointly continuous random variables. In particular, we have the following:
\begin{align}\label{} \nonumber F_{XY}(x,y) &=\int_{-\infty}^{y}\int_{-\infty}^{x} f_{XY}(u,v)dudv, \\ \nonumber \\ \nonumber f_{XY}(x,y) &=\frac{\partial^2}{\partial x \partial y} F_{XY}(x,y) \end{align}
Example
Find the joint CDF for $X$ and $Y$ in Example 5.15
• Solution
• In Example 5.15, we found $$\nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} x+\frac{3}{2}y^2 & \quad 0 \leq x,y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right.$$ First, note that since $R_{XY}=\{(x,y)|0 \leq x,y \leq 1\}$, we find that \begin{align}\label{} \nonumber &F_{XY}(x,y)=0, \textrm{ for }x<0 \textrm{ or } y<0,\\ \nonumber &F_{XY}(x,y)=1, \textrm{ for }x \geq 1 \textrm{ and } y \geq 1. \end{align} To find the joint CDF for $x>0$ and $y>0$, we need to integrate the joint PDF: \begin{align}\label{} \nonumber F_{XY}(x,y) &=\int_{-\infty}^{y}\int_{-\infty}^{x} f_{XY}(u,v)dudv \\ \nonumber &=\int_{0}^{y}\int_{0}^{x} f_{XY}(u,v)dudv \\ \nonumber &=\int_{0}^{\min(y,1)}\int_{0}^{\min (x,1)} \left(u+\frac{3}{2}v^2\right) dudv. \end{align} For $0 \leq x,y \leq 1$, we obtain \begin{align}\label{} \nonumber F_{XY}(x,y) &=\int_{0}^{y}\int_{0}^{x} \left(u+\frac{3}{2}v^2\right) dudv\\ \nonumber &=\int_{0}^{y} \bigg[\frac{1}{2}u^2+\frac{3}{2}v^2u \bigg]_{0}^{x} dv\\ \nonumber &=\int_{0}^{y} \left(\frac{1}{2}x^2+\frac{3}{2}xv^2\right) dv\\ \nonumber &=\frac{1}{2}x^2y+\frac{1}{2}xy^3. \end{align} For $0 \leq x \leq 1$ and $y \geq 1$, we use the fact that $F_{XY}$ is continuous to obtain \begin{align}\label{} \nonumber F_{XY}(x,y) &=F_{XY}(x,1)\\ \nonumber &=\frac{1}{2}x^2+\frac{1}{2}x. \end{align} Similarly, for $0 \leq y \leq 1$ and $x \geq 1$, we obtain \begin{align}\label{} \nonumber F_{XY}(x,y) &=F_{XY}(1,y)\\ \nonumber &=\frac{1}{2}y+\frac{1}{2}y^3. \end{align}
The print version of the book is available through Amazon here.<|endoftext|>
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OSIRIS-REx arrived at Bennu for a close encounter of the asteroid kind.
The asteroid Bennu is one of 780,000 asteroids in our solar system. Out of all those, it was chosen as the destination for NASA's Origins, Spectra Interpretation, Resource Identification, Security, Regolith Explorer (OSIRIS-REx) spacecraft on a mission to understand our early solar system and return a sample from the asteroid's surface to Earth. After a two-year journey, OSIRIS-REx arrived at Bennu in December 2018 to begin its science mission. But what made Bennu a good source for answers? First, Bennu’s proximity to Earth meant a quicker and easier flight for OSIRIS-REx. The spacecraft used Earth’s gravity to boost itself toward the asteroid. Bennu’s size is another bonus: it allows Bennu to spin on its axis at a rate that makes it easily approachable by a spacecraft, and rich in soil that would’ve flung off of smaller, fast-spinning asteroids. Its ancient age and the fact that it is well-preserved in the vacuum of space means it contains dust from the formation of our solar system. Bennu also is rich in carbon, which could offer clues to the possible role asteroids played in life on Earth. Plus, scientists have studied Bennu from Earth, and now will see whether their predictions about it were right. This will help us understand other asteroids, particularly their trajectory, and help us deflect ones that come too close. Watch the videos to learn more.<|endoftext|>
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Search
# Household Arithmetic Relating to Profit, Interest, Discount and Commission
Welcome to Class !!
In today’s Mathematics class, We will be discussing “Simple Interest. We will so be looking at Profit and Loss. We will ten round up with Discount and Commision. We hope you enjoy the class!
### SIMPLE INTEREST
Interest is the money paid for saving a particular amount of money.
Simple interest can be calculated using the formula
Where P= principal, R=rate & T= time
Also total amount =principal +interest
Example 1
Find the simple interest on N60, 000 for 5 years at 9% per annum
Solution
Applying
where P= N 60,000 R=9% T=5 years
Example 2
A man borrows N 1,600,000 to buy a house. He is charged interest at the rate of 11% per annum. In the first year, he paid the interest on the loan. He also paid back N 100,000 of the money borrowed. How much did he pay back altogether? If he paid this money by monthly instalment, how much did he pay per month?
Solution
P= N 1,600,000 R=11% T=1 year
Interest on N 1,600,000
The total amount paid in the first year= N 100,000 + N 176,000 = N 276,000
Monthly payment =
$\frac{276,000}{12}$
= N 23,000
Classwork
1. Find the simple interest on the following:
• N 10,000 for 31/2 years at 4% per annum
• N 20,000 for 4 years at 4% per annum
2. A man got N 1,800,000 loan to buy a house. He paid interest at a rate of 9% per annum. In the first year, he paid the interest on the loan. He also paid back N 140,000 of the money he borrowed. (a) How much did he pay in the first year altogether?
#### Profit and loss percent
Profit means to gain, while loss is the inverse of profit.
% Profit is the percentage of the gain made from a particular product or item
To find the selling price of an article
At loss, Selling price = cost price – loss
At gain, Selling = cost price + profit
Example
1. A trader buys a kettle for N 800 and sells at a profit of 15%. Find the actual profit and the selling price
Solution
Profit =15 of N 800
=15 x800
= N 120
Selling price = C. + profit
= 800+120
= N 920
1. A hat is bought for N 250 and sold for N 220, what is the loss percentage
Solution
Cost price C.P = N 250
Selling price = N 220
Lost= N 250- N 220 = N 30
% Loss =
Classwork
1. Find the actual profit and the selling price of a material which cost #1000 sold at a profit of 15%
2. A farmer buys a piece of land for #40,000 and sells it for N 33,000 what is the percentage loss?
3. A car that cost N 336,000 was sold at a loss of 17½%. What is the selling price
#### Discount and Commission
A discount is a reduction in the price of goods or items. Discounts are often given for paying in cash.
Commission: this is a payment or reward for selling an item.
Examples
• Find the discount price, if a discount of 25% is given on a market price of #9,200
Solution
Discount =25% of N 9,200
• A radio cost N 5,400. A 12½% discount is given for cash. What is the cash price
Solution
Discount =12 ½% of N 5,400
• A bank charges 2½ % commission for issuing a bank draft to its customers, if a customer obtained a bank draft for N 84,000 from the bank, calculates the total cost of the bank draft.
Solution
Commission =22% of N 84,000
5/2/100 x 84,000
=5/2×100 x84, 000
= N 2,100
Total cost of bank draft = 84,000+ N 2,000 = N 86,100
Classwork
1. Find the discount price if a discount of 20% is given on a market of N 2,915
2. The selling price of a table is N 14,000, the trader gives a 25% discount for cash, what is the cash price
3. An insurance agent sells N 284,000 worth of insurance, his commission is 20%. How much money does he get?
New General Mathematics, UBE Edition, Chapter 1, pages 78-79
Essential Mathematics by A J S Oluwasanmi, Chapter 1, pages 61-64
WEEKEND ASSIGNMENT
1. Calculate the simple interest on N 200 in 2 years at 4% per annum (a) N 160n (b) N 240 (c) N 16 (d) N 260
2. Calculate the simple interest on N 20,000 for 21/2 years at 2% per annum (a) N 10 (b) N10, 000 (c) N 1000 (d) N 2000
3. Find the simple interest on N 40,000 for 1 year at 5% per annum (a) N 100 (b) N 250 (c) N 2,590 (d) N 50
4. What is the simple interest on #70,000 for 1 year at 4% per annum (a) N 2,800 (b) N 2,000 (c) N 2,400 (d) N 2,300
5. Find the simple interest on N 10,000 for 3 years at 6% per annum (a) N 1,800 (b) N 1000 (c) N1,850 (d) N 1,200
THEORY
1. Find the simple interest on the following
• N 55,000 for 4 years at 60% per annum
• N 25,000 for 3 years at 5% per annum
We have come to the end of this class. We do hope you enjoyed the class?
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
In our next class, we will be talking about Approximation of Numbers. We are very much eager to meet you there.
Get more class notes, videos, homework help, exam practice on Android [DOWNLOAD]
Get more class notes, videos, homework help, exam practice on iPhone [DOWNLOAD]
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# Examples on Highest Common Factor (H.C.F)
Highest common factor is the highest or greatest number that divides the two or three numbers without leaving any remainder. So that highest number is called the H.C.F of that group of two or three numbers.
This topic will provide examples on H.C.F (highest common factor) which will help the learners to improve their knowledge, understanding and will help in getting a clear understanding of the concepts on H.C.F better.
Here are few illustrated on Highest Common Factor (H. C.F)
1. Find the H.C.F of 75, 50 and 125
Solution:
The factors of 75 are = 5 × 5 × 3 = 5× 3
The factors of 50 are = 5 × 5 × 2 = 5× 2
The factors of 125 are = 5 × 5 × 5 = 53
Therefore, the common prime factor of 75, 50 and 125 is 5
The lowest power of 5 is 52
Therefore, H.C.F = 5= 5 × 5 = 25
2. Find the H.C.F of 64, 32, and 256
Solution:
The factors of 64 are = 2 × 2 × 2 × 2 × 2 × 2 = 26
The factors of 32 are = 2 × 2 × 2 × 2 × 2 = 25
The factors of 256 are = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 28
Therefore, the common prime factor of 64, 32 and 256 is 2
The lowest power of 2 is 25
Therefore, H.C.F = 2 × 2 × 2 × 2 × 2 = 32
3. Find the H.C.F of 11, 121, and 1331
Solution:
The factor of 11 is = 11
The factors of 121 are = 11 × 11 =112
The factors of 1331 are = 11 × 11 × 11 = 113
Therefore, the common prime factor of 11, 121, and 1331 is 11
The lowest power of 11 is 11
Therefore, H.C.F = 11
4. Find the H.C.F of 48, 96, and 12
Solution:
The factors of 48 are = 2 × 2 × 2 × 2 × 3 = 24 × 3
The factors of 96 are = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3
The factors of 12 are = 2 × 2 × 3 = 22 × 3
Therefore, the common prime factors of 48, 96 and 12 are 2 and 3
The lowest power of 2 is 22
The lowest power of 3 is 3
Therefore, H.C.F = 2 × 2 × 3 = 12
5. Find the H.C.F of 81, 27 and 33
Solution:
The factors of 81 are = 3 × 3 × 3 × 3 = 34
The factors of 27 are = 3 × 3 × 3 = 33
The factors of 33 are = 3 × 11
Therefore, the common prime factors of 81, 27 and 33 is 3
The lowest power of 3 is 3
Therefore, H.C.F = 3
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Here are few problems on the divisibility rules of 2, 3, 4, 5, 6, 7, 8, 9, and 10 which will help the learners in revising their concepts on the divisibility rules. 1. Check whether 3456 is divisible by 2? Solution: The last digit is an even number (i.e. 6) hence 3456 is
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High School Physics
# Biot-Savart Law Numericals
Last updated on August 23rd, 2023 at 10:45 am
Physics Numerical Problems Based on Biot Savart Law – The Biot-Savart law is a fundamental relationship in magnetostatics. It describes the magnetic field generated by a steady current. In this post, we will look at a set of numerical problems based on the Biot-Savart law and their step-by-step solutions.
Formulas Used
The Biot-Savart law allows us to calculate the magnetic field B at some point P due to a current-carrying conductor. It states that:
dB = (μ0/4π) (Idl x r)/r^3
Where:
dB is the incremental magnetic field at point P
μ0 is the permeability of free space
I is the current in the conductor
dl is an incremental length of the conductor
r is the displacement vector from dl to P
To find the total magnetic field B, we integrate dB over the entire length of the conductor.
## Biot-Savart Law Numericals (class 12)
Below are 10 numerical problems based on the Biot Savart law.
Problem 1: Calculate the magnetic field at point P located on the axis at a distance of 10 cm from the center of a circular wire of radius 5 cm carrying a current of 5 A.
Solution: [Biot-Savart numerical set1 Q1 Solution]
Problem 2: Find the magnetic field at a distance of 20 cm from a long, straight wire carrying a current of 10A.
Solution: [Biot-Savart-numerical set1 Q2 solution]
Problem 3: Calculate the magnetic field at the center of a square loop of side 10 cm carrying a current of 5A.
Solution: [Q3 Solution]
Problem 4: Find the magnetic field at a point on the axis of a circular coil of 100 turns and a radius of 10 cm carrying a current of 1.5A at a distance of 15cm from the center.
Problem 5: Calculate the magnetic field at the center of a square of side 30 cm carrying a current of 15A along each arm.
Problem 6: Find the magnetic field at a distance of 5cm from an infinitely long straight conductor carrying a current of 8A.
Problem 7: Calculate the magnetic field at a point on the axis at a distance of 12cm from the center of a circular loop of radius 8cm carrying a current of 3A.
Problem 8: A square conducting loop of side length L carries a current I. The magnetic field at the center of the loop is:
A) independent of L
B) proportional to L
C) inversely proportional to L
D) linearly proportional to L
Hint: B∝1/L
problem 9: A current-carrying wire is in the form of a square loop of side length 25 cm. The current in the wire is 2A. Find out the magnetic field at the center of the loop.
problem 10: The magnetic induction at the center of a current-carrying circular coil of radius 10 cm is
5.51/2 times the magnetic induction at a point on its axis. The distance of the point from the center of the coil in cm is:___________________
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error: physicsTeacher.in<|endoftext|>
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“If placed within its international context, the Sv. Nikolai’s 1808 voyage has significance for Russian expansion in North America that might be compared, for example, to the 1540 expedition of Francisco Vásquez de Coronado on the northern borderlands frontier of New Spain.” (Preface, Wreck of the Sv Nikolai)
The Sv. Nikolai (a 45-50-foot schooner,) owned by the Russian American Company, set sail from New Arkhangel (modern-day Sitka, Alaska) to explore and identify a site for a permanent Russian fur trading post on the mainland south of Vancouver Island in the Oregon Country.
Heavy seas drove the ship aground on the Washington coast just north of the mouth of the Quileute River, forcing twenty-two crew members ashore.
Over the next several months the shipwrecked crew clashed with Hohs, Quileutes and Makahs; they lived in hand-built shelters roughly 9-miles up the Hoh River.
The tribes captured and enslaved several of the crew members. In 1810, an American captain sailing for the Russian American Company ransomed the survivors. (Owens)
OK, but what about Hawaiʻi? … Let’s look back.
Throughout the years of late-prehistory, AD 1400s – 1700s, and through much of the 1800s, the canoe was a principal means of travel in ancient Hawaiʻi. Canoes were used for interisland and inter-village coastal travel.
Most permanent villages initially were near the ocean and sheltered beaches, which provided access to good fishing grounds, as well as facilitating canoe travel between villages.
With “contact” (arrival of Captain James Cook in 1778,) a new style of boat was in the islands and Kamehameha started to acquire and build them. The first Western-style vessel built in the Islands was the Beretane (1793.)
Through the aid of Captain George Vancouver’s mechanics, after launching, it was used in the naval combat with Kahekili’s war canoes off the Kohala coast. (Thrum)
Encouraged by the success of this new type of vessel, others were built. The second ship built in the Islands, a schooner called Tamana (named after Kamehameha’s favorite wife, Kaʻahumanu,) was used to carry his cargo of trade to the missions along the coast of California. (Couper & Thrum, 1886)
Then, on June 21, 1803, the Lelia Byrd, an American ship under Captain William Shaler, arrived at Kealakekua Bay with two mares and a stallion on board – they were gifts for King Kamehameha.
The captain left one of the mares with John Young (a trusted advisor of the King, who begged for one of the animals) then left for Lāhainā, Maui to give the mare and stallion to Kamehameha.
During his stay, Shaler asked Kamehameha for one of the chief’s small schooners. Wanting bigger and better, in 1805, Kamehameha traded the 45-ton Tamana and a cargo of sandalwood for the Lelia Byrd,) a “fast, Virginia-built brig of 175-tons.” It became the flagship of Kamehameha’s Navy.
Kamehameha kept his shipbuilders busy; by 1810 he had more than thirty small sloops and schooners hauled up on the shore at Waikīkī and about a dozen more in Honolulu harbor, besides the Lelia Byrd. (Kuykendall)
That, then, takes us to the Tamana and her fate.
Shaler’s agent, John Hudson, sailed the Tamana east to Baja California. Within a year, Hudson sold the Tamana to Russian Captain Pavl Slobodchikov for 150 sea otter skins.
Slobodchikov renamed the Tamana to Sv. Nikolai.
With a makeshift crew of three Hawaiians and three Americans, Slobodchikov sailed the newly-named Sv Nikolai back to Hawaiʻi, and later returned to New Arkhangel (Sitka, Alaska) in August 1807 where the boat served the Russian fur traders along the Northwest Coast of North America.
At the time, the Northwest was unsettled territory. To bypass hostile Native Americans in the Northwest, the Russian American Company contracted with American ships to carry Russian fur traders to California.
Then, the Sv. Nikolai took the fateful trip in 1808 (as noted in the introductory paragraphs, above.)
Under Nikolai Isaakovich Bulygin, the Sv. Nikolai sailed to explore the coast of Vancouver Island and select a site for a settlement on what is today the Oregon coast.
The expedition did not succeed. Near Destruction Island the ship was becalmed and they aimlessly drifted. Then, on November 1, 1808, Sv. Nikolai was pushed onto a rocky reef by a heavy squall.
The ship did not sink immediately, and everyone on board reached shore safely. At low tide the crew returned to the vessel to salvage sail canvas, food, munitions and other supplies. (NOAA)
The survivors (including Anna Petrovana Bulygin (Captain Bulygin’s wife) – reportedly the first western woman to set foot in Washington state (Cook & Black) were crossing the Hoh River and three of the group, including the captain’s wife, were captured.
The rest of the crew then followed the Hoh River inland. They spent the winter in the valley, foraging for food and constructing a boat which they hoped would take them down the river and out to the freedom of the ocean.
In February 1809, they attempted to leave in their new boat, but at the mouth of the river it capsized. All the rest of the crew was taken captive. They lived in captivity for about 18 months.
In May 1810, an American vessel arriving in Neah Bay learned of their plight and attempted to arrange their release. All but seven members of the expedition were eventually freed. However both the captain and his wife died in captivity. (NOAA)
A monument was constructed on Upper Hoh Road to commemorate the 1808 shipwreck of a Russian sailing vessel near Rialto Beach. It was created to remember the lives lost when the Russian brig Sv. Nikolai (formerly owned by King Kamehameha and known as the schooner ‘Tamana’) beached in heavy squalls along the Pacific coast of the North Olympic Peninsula.
The image shows a drawing of the Sv Nikolai aground (AssocOfWashingtonGenerals.) In addition, I have added other images in a folder of like name in the Photos section of my Facebook and Google+ pages.<|endoftext|>
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## Intermediate Algebra (12th Edition)
$2p+7$
The distributive property holds that $a(b+c)=ab+ac$ (for real numbers $a$, $b$, and $c$). We are given the expression $-(2p+5)+3(2p+4)-2p$. We can use the distributive property to simplify the terms in parentheses and combine like terms. $-(2p+5)+3(2p+4)-2p=(-1\times2p)+(-1\times5)+(3\times2p)+(3\times4)-2p=(-2p)+(-5)+(6p)+(12)-2p=(-2+6-2)p+(-5+12)=(2)p+(7)=2p+7$<|endoftext|>
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NIPBL, cohesin loading factor
The NIPBL gene provides instructions for making a protein called delangin, which plays an important role in human development. Before birth, delangin is found in the developing arms and legs, the bones of the skull and face, the spinal column, the heart, and other parts of the body.
Delangin helps control the activity of chromosomes during cell division. Before cells divide, they must copy all of their chromosomes. The copied DNA from each chromosome is arranged into two identical structures, called sister chromatids. The sister chromatids are attached to one another during the early stages of cell division by a group of proteins known as the cohesin complex. Delangin plays a critical role in the regulation of this complex. Specifically, it controls the interaction between the cohesion complex and the DNA that makes up the sister chromatids.
Researchers believe that delangin, as a regulator of the cohesin complex, also plays important roles in stabilizing cells' genetic information, repairing damaged DNA, and controlling the activity of certain genes that are essential for normal development.
More than 300 mutations in the NIPBL gene have been identified in people with Cornelia de Lange syndrome, a developmental disorder that affects many parts of the body. Mutations in this gene are the most common known cause of Cornelia de Lange syndrome, accounting for more than half of all cases.
Many different kinds of NIPBL gene mutations have been reported; most lead to the production of an abnormally short (truncated), nonfunctional version of the delangin protein from one copy of the gene in each cell. These mutations reduce the overall amount of delangin produced in cells, which likely alters the activity of the cohesin complex and impairs its ability to regulate genes that are critical for normal development. Although researchers do not fully understand how these changes cause Cornelia de Lange syndrome, they suspect that altered gene regulation probably underlies many of the developmental problems characteristic of the condition. Studies suggest that mutations leading to a nonfunctional version of delangin tend to cause more severe signs and symptoms than mutations that result in a partially functional version of the protein.
- Nipped-B homolog (Drosophila)<|endoftext|>
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The Discovery of Gold
James Marshall (Left) and John Sutter (right)
In 1847, a man named John Sutter was in desperate need of a sawmill to finish building his flouring mill and other buildings for a small village of Yerba Buena, which is now San francisco.He decided to hire James Marshall and his crew to build a sawmill near the American River in Coloma, California. This location was just forty-five miles away from Sutter's Fort in Sacramento.
Marshall and his crew were set to work.In the morning of January 24,1848, Marshall went to check his crew's progress and found what he thought was gold. He approached John Sutter, in private, requesting to close the door and lock it. Marshall took out a rag from his pocket and unfolded it. What he had presented to Mr. Sutter, was yellow metal, claiming it was gold.. Sutter didn't believe him at first, but then tested it with aqua fortis ( chemical solution to help dissolve gold). Both men had discovered they had 23 carat gold.<|endoftext|>
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NCERT Solutions for Class 9 Math Chapter 1 Number Systems are provided here with simple step-by-step explanations. These solutions for Number Systems are extremely popular among class 9 students for Math Number Systems Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 9 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 9 Math are prepared by experts and are 100% accurate.
#### Question 1:
Is zero a rational number? Can you write it in the form, where p and q are integers and q ≠ 0?
Yes. Zero is a rational number as it can be represented asetc.
##### Video Solution for Number Systems (Page: 5 , Q.No.: 1)
NCERT Solution for Class 9 math - Number Systems 5 , Question 1
#### Question 2:
Find six rational numbers between 3 and 4.
There are infinite rational numbers in between 3 and 4.
3 and 4 can be represented asrespectively.
Therefore, rational numbers between 3 and 4 are
##### Video Solution for Number Systems (Page: 5 , Q.No.: 2)
NCERT Solution for Class 9 math - Number Systems 5 , Question 2
#### Question 3:
Find five rational numbers between.
There are infinite rational numbers between.
Therefore, rational numbers betweenare
##### Video Solution for Number Systems (Page: 5 , Q.No.: 3)
NCERT Solution for Class 9 math - Number Systems 5 , Question 3
#### Question 4:
State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
(i) True; since the collection of whole numbers contains all natural numbers.
(ii) False; as integers may be negative but whole numbers are positive. For example: −3 is an integer but not a whole number.
(iii) False; as rational numbers may be fractional but whole numbers may not be. For example: is a rational number but not a whole number.
##### Video Solution for Number Systems (Page: 5 , Q.No.: 4)
NCERT Solution for Class 9 math - Number Systems 5 , Question 4
#### Question 1:
State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form, where m is a natural number.
(iii) Every real number is an irrational number.
(i) True; since the collection of real numbers is made up of rational and irrational numbers.
(ii) False; as negative numbers cannot be expressed as the square root of any other number.
(iii) False; as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.
##### Video Solution for Number Systems (Page: 8 , Q.No.: 1)
NCERT Solution for Class 9 math - Number Systems 8 , Question 1
#### Question 2:
Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
If numbers such asare considered,
Then here, 2 and 3 are rational numbers. Thus, the square roots of all positive integers are not irrational.
##### Video Solution for Number Systems (Page: 8 , Q.No.: 2)
NCERT Solution for Class 9 math - Number Systems 8 , Question 2
#### Question 3:
Show howcan be represented on the number line.
We know that,
And,
Mark a point ‘A’ representing 2 on number line. Now, construct AB of unit length perpendicular to OA. Then, taking O as centre and OB as radius, draw
an arc intersecting number line at C.
C is representing.
##### Video Solution for Number Systems (Page: 8 , Q.No.: 3)
NCERT Solution for Class 9 math - Number Systems 8 , Question 3
#### Question 1:
Write the following in decimal form and say what kind of decimal expansion each has:
(i) (ii) (iii)
(iv) (v) (vi)
(i)
Terminating
(ii)
Non-terminating repeating
(iii)
Terminating
(iv)
Non-terminating repeating
(v)
Non-terminating repeating
(vi)
Terminating
##### Video Solution for Number Systems (Page: 14 , Q.No.: 1)
NCERT Solution for Class 9 math - Number Systems 14 , Question 1
#### Question 2:
You know that. Can you predict what the decimal expansion of are, without actually doing the long division? If so, how?
[Hint: Study the remainders while finding the value of carefully.]
Yes. It can be done as follows.
##### Video Solution for Number Systems (Page: 14 , Q.No.: 2)
NCERT Solution for Class 9 math - Number Systems 14 , Question 2
#### Question 3:
Express the following in the form, where p and q are integers and q ≠ 0.
(i) (ii) (iii)
(i)
Let x = 0.666…
10x = 6.666…
10x = 6 + x
9x = 6
(ii)
Let x = 0.777…
10x = 7.777…
10x = 7 + x
(iii)
Let x = 0.001001…
1000x = 1.001001…
1000x = 1 + x
999x = 1
##### Video Solution for Number Systems (Page: 14 , Q.No.: 3)
NCERT Solution for Class 9 math - Number Systems 14 , Question 3
#### Question 4:
Let x = 0.9999…
10x = 9.9999…
10x = 9 + x
9x = 9
x = 1
##### Video Solution for Number Systems (Page: 14 , Q.No.: 4)
NCERT Solution for Class 9 math - Number Systems 14 , Question 4
#### Question 5:
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of? Perform the division to check your answer.
It can be observed that,
There are 16 digits in the repeating block of the decimal expansion of.
##### Video Solution for Number Systems (Page: 14 , Q.No.: 5)
NCERT Solution for Class 9 math - Number Systems 14 , Question 5
#### Question 6:
Look at several examples of rational numbers in the form (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Terminating decimal expansion will occur when denominator q of rational number is either of 2, 4, 5, 8, 10, and so on…
It can be observed that terminating decimal may be obtained in the situation where prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both.
##### Video Solution for Number Systems (Page: 14 , Q.No.: 6)
NCERT Solution for Class 9 math - Number Systems 14 , Question 6
#### Question 7:
Write three numbers whose decimal expansions are non-terminating non-recurring.
3 numbers whose decimal expansions are non-terminating non-recurring are as follows.
0.505005000500005000005…
0.7207200720007200007200000…
0.080080008000080000080000008…
##### Video Solution for Number Systems (Page: 14 , Q.No.: 7)
NCERT Solution for Class 9 math - Number Systems 14 , Question 7
#### Question 8:
Find three different irrational numbers between the rational numbers and.
3 irrational numbers are as follows.
0.73073007300073000073…
0.75075007500075000075…
0.79079007900079000079…
##### Video Solution for Number Systems (Page: 14 , Q.No.: 8)
NCERT Solution for Class 9 math - Number Systems 14 , Question 8
#### Question 9:
Classify the following numbers as rational or irrational:
(i) (ii) (iii) 0.3796
(iv) 7.478478 (v) 1.101001000100001…
(i)
As the decimal expansion of this number is non-terminating non-recurring, therefore, it is an irrational number.
(ii)
It is a rational number as it can be represented in form.
(iii) 0.3796
As the decimal expansion of this number is terminating, therefore, it is a rational number.
(iv) 7.478478 …
As the decimal expansion of this number is non-terminating recurring, therefore, it is a rational number.
(v) 1.10100100010000 …
As the decimal expansion of this number is non-terminating non-repeating, therefore, it is an irrational number.
##### Video Solution for Number Systems (Page: 14 , Q.No.: 9)
NCERT Solution for Class 9 math - Number Systems 14 , Question 9
#### Question 1:
Visualise 3.765 on the number line using successive magnification.
3.765 can be visualised as in the following steps.
##### Video Solution for Number Systems (Page: 18 , Q.No.: 1)
NCERT Solution for Class 9 math - Number Systems 18 , Question 1
#### Question 2:
Visualise on the number line, up to 4 decimal places.
= 4.2626…
4.2626 can be visualised as in the following steps.
##### Video Solution for Number Systems (Page: 18 , Q.No.: 2)
NCERT Solution for Class 9 math - Number Systems 18 , Question 2
#### Question 1:
Classify the following numbers as rational or irrational:
(i) (ii) (iii)
(iv) (v) 2π
(i) = 2 − 2.2360679…
= − 0.2360679…
As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number.
(ii)
As it can be represented in form, therefore, it is a rational number.
(iii)
As it can be represented in form, therefore, it is a rational number.
(iv)
As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number.
(v) 2π = 2(3.1415 …)
= 6.2830 …
As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number.
##### Video Solution for Number Systems (Page: 24 , Q.No.: 1)
NCERT Solution for Class 9 math - Number Systems 24 , Question 1
#### Question 2:
Simplify each of the following expressions:
(i) (ii)
(iii) (iv)
(i)
(ii)
= 9 − 3 = 6
(iii)
(iv)
= 5 − 2 = 3
##### Video Solution for Number Systems (Page: 24 , Q.No.: 2)
NCERT Solution for Class 9 math - Number Systems 24 , Question 2
#### Question 3:
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, . This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
There is no contradiction. When we measure a length with scale or any other instrument, we only obtain an approximate rational value. We never obtain an exact value. For this reason, we may not realise that either c or d is irrational. Therefore, the fraction is irrational. Hence, π is irrational.
##### Video Solution for Number Systems (Page: 24 , Q.No.: 3)
NCERT Solution for Class 9 math - Number Systems 24 , Question 3
#### Question 4:
Represent on the number line.
Mark a line segment OB = 9.3 on number line. Further, take BC of 1 unit. Find the mid-point D of OC and draw a semi-circle on OC while taking D as its centre. Draw a perpendicular to line OC passing through point B. Let it intersect the semi-circle at E. Taking B as centre and BE as radius, draw an arc intersecting number line at F. BF is.
##### Video Solution for Number Systems (Page: 24 , Q.No.: 4)
NCERT Solution for Class 9 math - Number Systems 24 , Question 4
#### Question 5:
Rationalise the denominators of the following:
(i) (ii)
(iii) (iv)
(i)
(ii)
(iii)
(iv)
##### Video Solution for Number Systems (Page: 24 , Q.No.: 5)
NCERT Solution for Class 9 math - Number Systems 24 , Question 5
#### Question 1:
Find:
(i) (ii) (iii)
(i)
(ii)
(iii)
##### Video Solution for Number Systems (Page: 26 , Q.No.: 1)
NCERT Solution for Class 9 math - Number Systems 26 , Question 1
#### Question 2:
Q2. Find:
(i) (ii) (iii)
(iv)
(i)
(ii)
(iii)
(iv)
##### Video Solution for Number Systems (Page: 26 , Q.No.: 2)
NCERT Solution for Class 9 math - Number Systems 26 , Question 2
#### Question 3:
Simplify:
(i) (ii) (iii)
(iv)
(i)
(ii)
(iii)
(iv)
##### Video Solution for Number Systems (Page: 26 , Q.No.: 3)
NCERT Solution for Class 9 math - Number Systems 26 , Question 3
View NCERT Solutions for all chapters of Class 9<|endoftext|>
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# Thread: Area of Parallelogram
1. ## Area of Parallelogram
A Parallelogram has area 85 $\displaystyle cm^2$. The side lengths are 10 cm and 9 cm. What are the measures of the interior angles
2. Given the area is 85cm^2. Use the base as 10 cm, so $\displaystyle 10 \cdot h = 85 \, \Rightarrow h = 8.5$
This gives one angle $\displaystyle \alpha = arcsin(\frac{8.5}{9})$
So the other angle: $\displaystyle \beta = 180 - arcsin(\frac{8.5}{9})$
3. Originally Posted by supersaiyan
A Parallelogram has area 85 $\displaystyle cm^2$. The side lengths are 10 cm and 9 cm. What are the measures of the interior angles
Let's stand the parallelogram on the table so that the long edge--10cm--is flat (horizontal). You are given the area of the parallelogram (85 sq. cm). You recall how that area was found? It's the base times the vertical height of the parallelogram. The vertical height is different from the other side (which is why the area isn't 10*9, Twig--if we were playing with squares or rectangles that would be the case though). Use the long edge and the area to find the height (Area=base*height, and you know two of these already). Then you have a right triangle and you can use what you know about right triangles to find the measure of the acute angle (there are two of them, but they're of equal measure). And you know that all four angles in any 4-sided polygon add to 360 degrees. That should get you through.
Incidentally, you can think of a square or a rectangle as special kinds of parallelograms since they do fit the definition of parallelogram. If you do this then, like I mentioned above, the height turns out to be the length of the vertical side. You may have done this already in class (I remember doing it in my geometry class), but if you didn't know the formula for the area of a parallelogram, just cut one out of a piece of paper. Then draw a vertical line from a top corner to the base (through the inside of the parallelogram). Cut along this line to get that right triangle. Lastly, slide that right triangle over to the other side of the parallelogram...SCORE! You just made a square and you can see why the area of a parallelogram must come from the base times the VERTICAL HEIGHT, and not just the product of two sides. Happy mathing!
4. Thanks, is this correct?
5. Looks good from my house!<|endoftext|>
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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Grade 7 - Mathematics4.36 Multiplication of Two Binomials - FOIL Method
The FOIL method is another way to multiply binomials. "FOIL" is an acronym to remember a set of rules to perform this multiplication. To FOIL you multiply together all of the following: F: Firsts O: Outers I: Inners L: Lasts and then you add each of these products. Example: Find the product of (x+2) and (x+3). Solution: Multiply Firsts: x.x = x2 Outers: x.3 = 3x Inners: 2.x = 2x Lasts: 2.3 = 6 Adding: x2 + 3x + 2x + 6 Simplify: x2 + 5x + 6 Example: Find the product of (3x3-5) and (7x-1). Solution: (3x3-5).(7x-1) Multiply Firsts: 3x3 . 7x = 21x4 Outers: 3x3 . -1 = -3x3 Inners: -5.7x = -35x Lasts: -5 . -1 = 5 Adding: 21x4 -3x3 -35x + 5 Directions: Solve the following problems. Also write at least ten examples of your own multiplying two binomials using FOIL method.
Q 1: (5x2-3).(x-9) = _____?5x3+45x2-3x+275x3-45x2-3x+275x3-45x2+3x+27 Q 2: (x2+9).(5x+4) = _____?5x3-4x245x+365x3+4x245x-365x3+4x245x+36 Q 3: (5x2-7).(2x+5) = _____?10x3-25x2-14x-3510x3+25x2+14x-3510x3+25x2-14x-35 Q 4: (4x2+1).(3x+4) = _____?12x3-16x2+3x+412x3+16x2-3x+412x3+16x2+3x+4 Q 5: (6x2+5).(6x-5) = _____?36x3-30x2+30x-2532x3-30x2+30x-2536x2-30x+30 Q 6: (4x2-1).(3x-4) = _____?12x3-16x2+3x+412x3+16x2-3x+412x3-16x2-3x+4 Q 7: (5x2+3).(7x-2) = _____?35x3-10x2-21x+635x3+10x2+21x-635x3-10x2+21x-6 Q 8: (5x2+4).(6x+7) = _____?30x3+35x2+24x+2832x3+35x2+24x+2830x3-35x2+24x+28 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!<|endoftext|>
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# 2008 AMC 10B Problems/Problem 23
## Problem
A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers and $b > a$. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half the area of the whole floor. How many possibilities are there for the ordered pair $(a,b)$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
## Solution
Because the unpainted part of the floor covers half the area, then the painted rectangle covers half the area as well. Since the border width is 1 foot, the dimensions of the rectangle are $a-2$ by $b-2$. With this information we can make the equation:
$\begin{eqnarray*} ab &=& 2\left((a-2)(b-2)\right) \\ ab &=& 2ab - 4a - 4b + 8 \\ ab - 4a - 4b + 8 &=& 0 \end{eqnarray*}$ Applying Simon's Favorite Factoring Trick, we get
$\begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \\ (a-4)(b-4) &=& 8 \end{eqnarray*}$
Since $b > a$, then we have the possibilities $(a-4) = 1$ and $(b-4) = 8$, or $(a-4) = 2$ and $(b-4) = 4$. This allows for 2 possibilities: $(5,12)$ or $(6,8)$ which gives us $\boxed{\textbf{(B)} \: 2}$<|endoftext|>
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# a standard deck of cards contain 52 cards. one cards is selected from the deck A) compute the probobility of randomly selecting a nine or ten B) compute the probobility of randomly selecting a nine...
a standard deck of cards contain 52 cards. one cards is selected from the deck
A) compute the probobility of randomly selecting a nine or ten
B) compute the probobility of randomly selecting a nine or ten or ace
C) compute the probobility of randomly selecting a four or spade
p (nine or ten )= type a interger or a siplified fraction
mariloucortez | Certified Educator
There are four 9 cards out of 52 cards, so the probability of getting 9 in a deck of 52 cards is P(9) = 4/52. The same goes with the probability of getting 10 in a deck of 52 cards, P(10) = 4/52. So, solving for
A) P (9 or 10) = P(9) + P(10) = 4/52 + 4/52 = 8/52 = 2/13
For B), you just have to get the probability of getting an ace. There are four aces in a standard deck of cards, so P(ace) = 4/52. Do the same formula as the first.
B) P(9 or 10 or ace) = P(9) + P(10) + P(ace) = 4/52 + 4/52 + 4/52
= 12/52 = 3/13
For C), there are four 4 cards in a deck, so the probability is P(4) = 4/52. Likewise, there are 13 spades in a deck of 52, so P(spade) = 13/52. There is a card that is 4 and spade, so P(4 and spade) = 1/52.<|endoftext|>
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What is anaphylaxis?
Anaphylaxis is a sudden, severe, potentially fatal, systemic allergic reaction that can involve various areas of the body (such as the skin, respiratory tract, gastrointestinal tract, and cardiovascular system). Symptoms occur within minutes to two hours after contact with the allergy-causing substance but, in rare instances, may occur up to four hours later. Anaphylactic reactions can be mild to life threatening. The annual incidence of anaphylactic reactions is about 30 per 100,000 persons, and individuals with asthma, eczema, or hay fever are at greater relative risk of experiencing anaphylaxis.
Anyone with a previous history of anaphylactic reactions is at risk for another severe reaction. Individuals with food allergies (particularly allergies to shellfish, peanuts, and tree nuts) and asthma may be at increased risk for having a life-threatening anaphylactic reaction. A recent study showed that teens with food allergy and asthma appear to be at highest risk for a reaction because they are more likely to dine away from home, they are less likely to carry medications, and they may ignore or not recognize symptoms.
Although an individual could be allergic to any food, such as fruits, vegetables, and meats, there are eight foods that account for 90% of all food-allergic reactions. These are: milk, egg, peanut, tree nut (walnut, cashew, etc.), fish, shellfish, soy, and wheat.
Some people have the misconception that nuts are the only cause of anaphylaxis and that “healthy” food like milk or eggs couldn’t possibly cause death, but an individual may be anaphylactic to any food.<|endoftext|>
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Cities Face Up to the Climate Challenge
Millions of people living in cities around the world already feel the impacts of climate change: heat waves, flooded streets, landslides and storms. All of these affect important infrastructure such as transportation and water supplies, ports and commerce, public health and people’s daily lives. And it is cities that are at the forefront of the response.
Experts from the Earth Institute attending the Paris climate summit are presenting a fresh report today on what’s at stake for the world’s growing urban population, and what many cities are doing to adapt. “Climate Change and Cities” is the Second Assessment Report of the Urban Climate Change Research Network, a consortium of 600 researchers from around the world based at the Center for Climate Systems Research, part of the Earth Institute at Columbia.
“Cities and their citizens already have begun to experience the effects of climate change. Understanding and anticipating these changes will help cities prepare for a more sustainable future,” the report says. “This means making cities more resilient to climate-related disasters and managing long-term climate risks in ways that protect people and encourage prosperity. It also means improving cities’ abilities to reduce greenhouse gas emissions.”
The task is daunting: Each city has its own resources, needs and political dynamics. And the challenges are different for rich and poor nations. For instance, the report notes, “Urban transport emissions are growing at 2 to 3 percent annually. The majority of emissions from urban transport is from higher-income countries. In contrast, 90 percent of the growth in emissions is from transport systems in lower-income countries.”
Tackling the problems involves work on many fronts, from urban planning to infrastructure, housing and hospitals to transit and waste removal. The problems are especially acute for coastal cities: The report projects that more than half of the global urban population will live in coastal zones by the middle of this century. Storm surges, erosion and salt water intrusion are already a problem in many places. “[S]ea level rise and climate change will likely exacerbate these hazards,” the report says. It estimates annual losses from flooding along coastlines could amount to $71 billion by 2100.
But while national leaders debate what to do about climate change, city officials around the world cannot afford to wait, and are already taking action. The report includes more than 100 case studies of what cities are doing to mitigate and adapt to climate change. The online “Case Study Docking Station” is meant to spread information about how cities are coping and offer models other cities can emulate. The report emphasizes the importance of integrating mitigation and adaptation strategies.
For instance, New York is well on the way to reaching a goal of planting a million trees by 2017 (900,000 as of August 2014, the report says). The project serves to both mitigate and adapt to climate changes. Among other benefits, the trees absorb CO2, helping to curb greenhouse gases; and by helping to lower air temperature in summer, they reduce the amount of energy used for cooling. They also improve air quality and reduce stormwater runoff.
The tree planting is one of more than 100 initiatives that are part of PlaNYC 2030, a broad strategy to support the long-term sustainability of the city. Following the devastation of Superstorm Sandy in 2012, New York also has adopted an aggressive strategy to build a more resilient shoreline, by upgrading building codes, protecting important infrastructure such as subways and power systems, raising bulkheads and building seawalls, and restoring wetlands and beach dunes.
A more dramatic case study comes from South Korea, where a whole new city is being built with sustainability in mind. New Songdo City, a $35 billion development eventually projected to have 65,000 residents and a workforce of 300,000, incorporates the highest concentration of LEED-certified buildings in the world. Forty percent of the city will be green space. It will incorporate extensive public transit, pedestrian- and bicycle-friendly design and a cutting-edge waste collection that sends garbage out through a pneumatic system (in other words: no garbage trucks).
The city “aims to generate efficient energy use through ‘ubiquitous’ technology that uses the internet to link hardware and software to monitoring systems to generate efficient resource consumption. Consequently, Songdo consumes 40 percent less energy per capita than cities of similar scale,” says the Songdo case study.
The “Climate Change and Cities” report being released Friday is an executive summary: The full report is still being prepared. But it offers some key findings regarding disaster preparation; urban planning and design; public health, water and waste systems; transportation and energy systems; financing solutions and urban governance; protecting urban ecology; and insuring equitable approaches that encompass the needs of poor and low-income residents and neighborhoods.
The report outlines five “pathways to urban transformation”:
- Disaster risk reduction and climate change adaptation are the cornerstones of resilient cities.
- Actions that reduce greenhouse gas emissions while increasing resilience are a win-win.
- Risk assessments and climate action plans co-generated with the full range of stakeholders and scientists are most effective.
- Needs of the most disadvantaged and vulnerable citizens should be addressed in climate change planning and action.
- Advancing city creditworthiness, developing robust city institutions, and participating in city networks enable climate action.
Cynthia Rosenzweig, an adjunct senior research scientist at the Center for Climate Systems Research and the NASA-Goddard Institute for Space Studies, is the report’s lead author. To see the report and find out more, visit the Urban Climate Change Research Network.<|endoftext|>
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# Understanding Limits in Calculus: Definition, Rules, Calculations, and Applications
Limits are a fundamental concept in mathematics, integral to the field of calculus. They describe the behavior of functions as they approach specific values or points. By examining how functions change, as variables get closer to a given value, limits allow us to explore and understand complex mathematical and real-world scenarios.
Whether analyzing instantaneous rates of change, optimizing designs, or modeling population growth, limits are a powerful technique in mathematics with far-reaching applications.
Here in this article, we will discuss the concept of limit with the help of definitions, rules and properties, different formulas, and real-life applications and also, discuss the topic with the help of detailed examples.
## Definition of a Limit
Mathematically definitions of the limit are:
limx→a f(x) = L
We must make precise our native notation that f(x) gets arbitrarily close to L as X gets close to a.
## Limit Calculus: Rules
If our question is more complex the rules and properties of limits are essentially used to make them easily solveable. These rules allow us to manipulate limits and solve various mathematical equations efficiently. Some of the fundamental rules include:
### • Limit of a Constant
Constant function limit is also itself a function used for constant.
limx→a c = c
### • Limit of a Sum/Difference
In the rule for difference and sum limit, we put the limit on individual functions.
limx→a [f(x) ± g(x)] = limx→a g(x) ± limx→a f(x)
### • Limit of a Product
The formula used for the product of limit is.
limx→a [f(x). g(x)] = limx→a g(x). limx→a f(x)
### • Limit of a Quotient
The formula used for the limit of the Quotient function is.
If limx→a g(x) ≠ 0 then formula of quotient is
limx→a L(x)/M(x) = limx→a L(x)/ limx→a M(x)
These rules are instrumental in simplifying complex limits and solving problems in calculus.
## Formulas of limit
Limit formulas offer specific techniques to compute limits for various functions. Some common limit formulas include:
### • The Limit of a Polynomial Function
You may get the limit of a polynomial function when “x” approaches a value “a” by changing “x” in the polynomial to “a”.
f(x) = k xn,
limx→a (x) = k. an= f(a)
### • The Limit of an Exponential Function
The limit of an exponential function, such as limx→a f(x) ex, is ea.
### • The Limit of a Trigonometric Function
By using trigonometric identities, we find the limit of trigonometric functions like sine and cosine functions.
### • The Limit of a Rational Function
To find the limit of a rational function, factor and cancel out common terms in the numerator and denominator.
### • The Limit of a Root Function
The limit of a root function like limx→a f(x) n √ f(x) can be computed by taking the nth root of the limit of f(x).
## How to find limit?
We’ll solve a few examples to understand how to find the limit.
### Example number 1
Suppose a function f(x)= x3-5×2+3x-1.
limx→-9 f(x) =?
#### Solution
Given data
f(x)= x3-5×2+3x-1
we can find the limit of a given function step by step.
Step 1:
Apply the limit
limx→-9 (x3-5×2+3x-1) = (-9)3-5(-9)2+3(-9)-1
Step 2:
Simplification
limx→-9 f(x) = (-9)3-5(-9)2+3(-9)-1
limx→-9 (x3-5×2+3x-1) = – 729 – 5(81) -27 -1
limx→-9 (x3-5×2+3x-1) = – 729 – 405 -27 -1
limx→-9 (x3-5×2+3x-1) = – 729 –433
limx→-9 (x3-5×2+3x-1) = – 1162
### Example number 2
Suppose we want to find the limit of the function f(x) = x2 – 4/ x – 2 as x approaches 2.
#### Solution
Given data
limx→2 x2 – 4/ x – 2
Step 1:
To solve this limit question we can apply the limit formula for rational functions:
limx→2 x2 – 4/ x – 2 = limx→2 (x2 – 4)/ limx→2 (x – 2) (1)
Step 2:
Now in this step, we find that both expressions limit individual like,
limx→2 (x2 – 4) = (2)2 – 4 = 4-4 = 0
limx→2 (x – 2) = 2-2 = 0
Step 3:
Now put the step 2 answer value in equation (1), and we get.
limx→2 x2 – 4/ x – 2 = 0/ 0
To find the limit in this case, we can use L’Hopital’s Rule, which allows us to evaluate limits of indeterminate forms by taking derivatives. Applying L’Hopital’s Rule, we get:
limx→2 x2 – 4/ x – 2 = limx→2 d/dx (x2 – 4)/d/dx (x – 2)
limx→2 x2 – 4/ x – 2 = limx→2 2x /1
Now put the given limit in the function, and we get
limx→2 x2 – 4/ x – 2 = 2(2) = 4
## Real-Life Applications of Limits
Understanding limits is not only essential for academic success but also for solving practical problems in fields like physics, engineering, and economics. Let’s explore some real-life applications of limits:
### Population Growth
Demographers and economists use limits to model population growth. The limit of a population function as time goes to infinity provides insights into the long-term growth or decline of a population.
### Finding Acceleration and Speed
In physics, limits are used to determine instantaneous speed and acceleration. By taking the limit of a displacement function as time approaches zero, you can find the instantaneous velocity and acceleration of an object.
### Calculating Interest Rates
In finance, limits help determine interest rates on loans and investments. By taking the limit as the compounding frequency approaches infinity, you can calculate continuous compounding interest.
### Engineering Design
Engineers use limits to optimize designs and ensure safety. Limits can help analyze the stress and strain on materials, ensuring that structures can withstand various loads.
### Computer Graphics:
In computer graphics, limits are used to render smooth curves and surfaces. Bezier curves and B-splines, for instance, rely on limits to create aesthetically pleasing graphics.
## Conclusion
Here in this article, we have discussed the concept of limit with the help of definitions, rules and properties, different formulas, and real-life applications and also, discussed the topic with the help of detailed examples. Anyone can defend this topic anywhere after studying this article in detail.<|endoftext|>
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Titration is a common laboratory method of quantitative/chemical analysis which can be used to determine the concentration of a known reactant. Because volume measurements play a key role in titration, it is also known as volumetric analysis. In 1855, the German chemist, Friedrich Mohrn, defined titration as the "weighing without scale" method, because this process allows determination of the concentration of a sample without using complex instrumentation. The process of determining the quantity of a substance A by adding measured increments of substance B, with provision for some means of recognizing (indicating) the endpoint at which essentially all of A has reacted. If the endpoint coincides with the addition of the exact chemical equivalence, it is called the equivalence point or stoichiometric or theoretical endpoint, thus allowing the amount of A to be found from known amounts of B added up to this point, the reacting weight ratio of A to B being known from stoichiometry or otherwise. Terms for varieties of titration can reflect the nature of the reaction between A and B. Thus, there are acid–base, complexometric, chelatometric, oxidation–reduction, and precipitation titrations. Additionally, the term can reflect the nature of the titrant, such as acidimetric, alkalimetric, and iodometric titrations as well as coulometric titrations, in which the titrant is generated electrolytically rather than being added as a standard solution.<|endoftext|>
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Listen to this article
“Gallina=chicken, puerta=door, ventana=window y pluma=pen”
Those are words of a traditional rhyme back in my country. Even though I didn’t know English when I learned that song, I was probably around 3 years old, I knew the letters of that song to perfection.
Music is recognized as a universal feature of human cognition: every healthy human is born with the ability to appreciate it.
Spoken language is introduced to the child as a vocal performance, and children attend to its musical features first. Without the ability to hear musically, it would be impossible to learn to speak.
Songs are a natural way to learn about language. They develop listening skills and slow down language so children can hear the different sounds in words, a key decoding skill.
Songs have repetitions and repletion is key when it comes to language development. It helps them to learn new words and information, strengthens their memory and attention.
How does singing with children help them get ready to read? (Extracted from Every Child Ready to Read)
Children love singing. A great option to encourage not only reading, but also singing, is to read books that can be sung. These can include nursery rhymes, books that promote singing, or books that can be sung to a specific tune.
A foundational early literacy skill is understanding that print has meaning. To help children make this connection, print out lyrics to favorite songs or read books that can be sung.
Letter knowledge is, at its base, a shape recognition skill, so any rhyme or song that talks about how things are the same and different can help build skills children will need to identify letters. Of course, the ABC Song helps them learn letter names and alphabetical order!
Just like books, songs have great vocabulary words, such as “tuffet” or “In a cavern, in a canyon.” Hearing new words in context helps children build their vocabularies. In addition, songs have a long tradition of being used as memory boosters! I’m sure many of you can still recite all 50 states because of a song you learned.
Listening to and singing songs is one of the best ways for children to build their phonological awareness because often each syllable of a word connects to a note. In addition, many songs and rhymes have rhyming words. Both pieces help children hear the individual parts of each word.
Many Mother Goose and other childhood songs are little stories, and listening to them helps children learn about story structure and sequencing. Even silly songs like “Little Bunny Foo Foo” have a beginning, a problem in the middle, and a resolution at the end. When children sing these songs, they become storytellers.
Singing activities to do with your child (Extracted from Earlier is Easier)
- Sing while changing your baby’s diapers.
- Sing in the car! Hearing songs and stories will help baby learn how to communicate and soon they’ll respond!
- Move, gently bounce, or hold your baby’s hand as you dance together to music.
- Rhyming and bouncing songs help babies hear and feel words and sounds so they can begin to repeat them.
- Put your baby on your lap or on a blanket on the floor and look into their eyes as you sing. Tap their hands together to the beat.
- Sing a quiet, calming song before your baby goes to sleep. How about “Twinkle Twinkle Little Star” or “Hush Little Baby?”
- Young children love to sing, so sing everywhere – in the car, in the bathtub, at the store and at the table.
- Make music with things you have in the house – pots, pans, spoons, boxes, cups. Crawl around on the floor with your child to the beat of the music.
- Songs have a note for each part of a word, so when you sing you’re helping your child hear that words have smaller parts. Clap or tap along to better hear these smaller parts.
- Sing the same quiet song at bedtime. Repetition and routine is good for young children and they will know it’s time for sleep.
- Sing a song you remember learning as a child. A song that was special to you can become special to your child too!
- The tune of “Here We Go Round the Mulberry Bush” is great for singing about your daily activities. “This is the way we brush our teeth, brush our teeth, brush our teeth. This is the way we brush our teeth so early in the morning.” (Don’t know the tune? Listen here.)
- Sing the ABC song!
- Take a song you know and change the words to something silly. Include your child’s name.
- Sing a familiar song faster…and then faster…and then slower…and slower.
- Visit your local library and borrow some kids’ music cds to sing along with in the car.
Latest posts by Pamela Mejia de Rodriguez (see all)
- 5 Early Literacy Practices for Babies - June 6, 2018
- Summer Reading 2018: Libraries Rock!The Importance of Music in Early Child Development - May 8, 2018
- Creating Full Brain Activities - May 4, 2018<|endoftext|>
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study published Monday in the journal “Proceedings of the National Academy of Sciences,” suggests rising temperatures have been responsible for significant declines in the fish that populate Africa’s Lake Tanganyika and feed the surrounding communities — and that these declines may only worsen as global warming continues to progress.
The research suggests that sustained warming throughout the decades has hindered a vital “mixing” process, common to lakes all over the world, that helps to spread nutrients and oxygen throughout the water. Typically, this process occurs when oxygen-rich water near the surface of a lake sinks down to deeper areas (a churn often aided by the wind), and nutrient-rich water rises up from the bottom to take its place.
The problem is that when the air temperature rises, it can cause water near the surface to heat up — and because warm water is less dense than cooler water, the differences in temperature between the surface and the bottom of the lake make it much more difficult for two to mix together. When this happens, some parts of the lake don’t receive enough oxygen while other parts don’t receive enough nutrients, and marine organisms in both areas suffer as a result.
This is the effect the research team believes has been occurring in Lake Tanganyika — which is a serious problem, given how important the lake is to nearby human communities.
Lake Tanganyika is one of Africa’s Great Lakes and the second largest freshwater lake in the world by volume. It’s also one of the oldest lakes in the world at about 10 million years in age, and it’s known for having some of the highest levels of endemic species, or organisms found nowhere else in the world.
Bordered by Tanzania, Burundi, Zambia and the Democratic Republic of the Congo, it’s a vital source of food in the region. The lake has been known to produce more than 200,000 tons of fish in a year, and it’s estimated that its yields may account for up to 60 percent of the animal protein consumed by humans in the region.
In recent decades, though, it’s become apparent that fish populations in the lake are not what they used to be. The question that’s been occupying scientists’ minds is what’s behind the declines. While experts have hypothesized that climate change has played a role, many researchers have also pointed to the pressure of commercial fishing — which took off in Lake Tanganyika in the mid-20th century — as a major culprit.
“There has been an ongoing debate between various scientific communities as to the relative importance of climate change and fishing pressure,” said Andrew Cohen, the new study’s lead author and a distinguished professor and expert in paleolimnology — the study of lake history using ancient sediments — at the University of Arizona.
The new study suggests that declines in the lake’s fish predate the onset of commercial fishing there, he said, meaning climate change was having a negative impact before overfishing ever became a concern.
The researchers demonstrated this effect by drawing sediment cores from various sites in the lake and analyzing them in the lab. Certain chemical components of ancient sediments can reveal a lot about a region’s climate history — it’s a method that’s commonly used by climate scientists to determine how an area has changed over thousands of years.
The researchers were also able to use certain aspects of the sediment samples to determine how algae production — the foundation of the lake’s food chain — has changed over time. And they also examined the lake’s fossil record to see how fish and mollusk populations have grown or declined throughout the ages.
After constructing a history going back about 1,500 years, the researchers found that as the lake temperatures have risen, the fish and mollusk fossils have declined, suggesting warmer temperatures have been bad for the lake’s biodiversity. The results indicate this pattern has been particularly striking since the late 19th century, when the lake began to experience sustained warming.
In fact, the study suggests that the area of suitable habitat on the lake floor has shrunk by about 38 percent since the mid-20th century.
A reduction in the mixing of the water can be identified as “the fundamental problem that this lake is facing,” Cohen said. “And the decline in fishery is in large part a consequence of this climate change phenomenon.”
This is not to say that commercial fishing has played no role in recent declines, Cohen added. Since the middle of the 20th century, the two factors have likely been working together in concert. Rather, the new study confirms that climate change also is a major culprit — and has been for a much longer period of time.
Furthermore, the results serve as a kind of warning bell for the future, given that temperatures in the region will likely continue to warm in the coming decades. Future management of the lakes fisheries will need to account for the fact that fishing is not the only force placing pressure on fish populations, and that current levels of commercial fishing may already be unsustainable there.
Cohen cautioned that the effects observed in Lake Tanganyika don’t necessarily apply to lakes all over the world. In the Arctic, for instance, there’s some evidence to suggest that rising temperatures may actually increase productivity in some lakes.
And even in other tropical lakes, there are additional factors that need to be considered when making predictions about the future. It’s possible that an increase in a region’s windiness could help counteract a rise in temperature, since wind helps to mix up the water.
“We don’t have any evidence of that in the Tanganyika region,” Cohen added. “But the importance is we need to look at these big, deep lakes in the tropics on a case-by-case basis and think about all the types of climate processes.”
Chelsea Harvey is a freelance journalist covering science. She specializes in environmental health and policy.
Read more at Energy & Environment:
original story HERE
Sign up for the Global Warming Blog for free by clicking here. In your email you will receive critical news, research and the warning signs for the next global warming disaster.
Click here to learn how global warming has become irreversible and what you can do to protect your family and assets.
To share this blog post: Go to the original shorter version of this post. Look to lower right for the large green Share button.
To view our current agreement or disagreement with this blog article, click here.<|endoftext|>
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What is Subsidence?
The Friant-Kern Canal was designed as a gravity fed facility and does not rely on pumps to move water. Subsidence (which is the gradual sinking of an area of land) has caused parts of the canal to sink in relationship to others parts. This negatively affects the canal’s ability to convey water. When the land elevation lowers, the canal must be operated at a lower flow-stage to ensure that water doesn’t overflow the banks.
Subsidence has reduced the ability to deliver water to many Friant Division Contractor’s by nearly 60%.
A water supply in peril
Read documents from the research and effort being done by the Friant Division to create a sustainable path forward for the Friant-Kern Canal.<|endoftext|>
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The development of red, painful blisters inside the mouth of a young child can be a frightening scenario and parents could be forgiven for panicking, especially if they've been supervising a twice daily, healthy oral hygiene routine. In the case of hand, foot and mouth disease (HFMD) however, oral hygiene has little to do with it developing. Here's more about HFMD, including its symptoms, how it is contracted and how parents can prevent it.
What Is Hand, Foot and Mouth Disease (HFMD) ?
HFMD is a common viral infection caused by the enterovirus, according to the Centers for Disease Control. This infection is from the coxsackievirus A16 which belong in the family of non-polio enterviruses. Children have the highest risk of contagion, although older children, teens and adults occasionally contract the virus, too.
Symptoms of the condition include:
- A fever lasting a day or two.
- Small sores in and around the mouth.
- Skin rash with blisters on the hands, feet, which might also appear on the buttocks region too.
- Poor appetite and difficulty swallowing food and water, which can cause dehydration if patients are unable to drink enough liquids due to the pain of the mouth sores.
- General sense of being unwell.
Symptoms vary between patients and adults often show no symptoms of the virus at all, although they are still able to carry and transmit the illness.
The Coxsackie virus is usually transmitted from person to person through the saliva and nasal and throat passages. The most common methods of contracting the disease is saliva, fluid from blisters, droplets from a sneeze or cough and from the stool, according to the Mayo Clinic. Because these substances occur frequently in child care settings where diapers are changed and young children often put their hands in their mouths while playing, hand, foot and mouth disease is common in these environments.
In most instances, patients begin improving in three to five days and should be fully recovered in seven to 19 days. Carriers can continue to infect others for weeks after the symptoms disappear and the virus appears to be gone.
Treatment of HFMD
Medical and dental practitioners usually recommend supportive treatments for cases of HFMD. Over-the-counter topical analgesics help relieve pain, both orally and in other areas. Acetaminophen and ibuprofen are helpful to relieve patients' headaches and sore throats. Antibiotics are prescribed if the medical practitioner has a concern about infection developing. It's essential to keep patients hydrated, and IV fluids are sometimes necessary to achieve this.
Home care includes giving the patient chips of ice, ice cream or ice pops to suck, and cold beverages to drink. Continue twice daily brushing of your young child's teeth with a toothbrush like Colgate My First toothbrush, which has extra soft bristles for gentle, yet effective cleaning.
Prevention of HFMD
Handwashing is key to reduce the risk of spreading hand, foot and mouth disease. This is especially important after using the toilet or changing a diaper, and before preparing or eating food. In child care environments, disinfect frequently used surfaces (including toys and other shared items) after regular washing. Keep any known infected children away from others to prevent spreading the virus.
Various medications are being tested for use with enterovirus infections, so a faster-acting treatment may be available in the future, notes Medscape. For now, though, your best bet for keeping your family safe from hand, foot and mouth disease is to practice good overall hygiene and to take immediate action if symptoms develop.<|endoftext|>
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# Decimal Representation of Rational Numbers
Rational numbers can be represented in decimal forms rather than representing in fractions. They can easily be represented as decimals by just dividing numerator ‘p’ by denominator ‘q’ (as rational numbers is in the form of p/q).
A rational number can be expressed as a terminating or nonterminating, recurring decimal.
For example:
(i) 5/2 = 2.5,
2/8 = 0.25,
7 = 7.0, etc., are rational numbers which are terminating decimals.
(ii) 5/9 = 0.555555555……. = 0.5 ̇,
4/3 = 1.33333….. = 1.3 ̇,
1/6 = 0.166666 ….. = 0.16 ̇
9/11 = 0.818181…… = 0.8 ̇1 ̇ etc., are rational numbers which are nonterminating, recurring decimals.
Representation of rational numbers in decimal fractions makes calculations more easier as compared to that in case of improper rational fractions.
Some of the examples below will show how rational numbers can be represented as decimal fractions:
(i) 2/3 is a rational number which can be written as 0.667 as decimal fraction.
(ii) 4/5 is rational number which can be written as 0.8 as decimal fraction.
(iii) 2/1 is a rational number which can be written as 2.0 as decimal fraction.
So, with the help of the above examples we can see that how easy is to convert rational numbers into decimal fractions.
Also we conclude that these decimal fractions that are converted can be of any type of example (i) shows that decimal fraction is non-terminating. In case of non-terminating decimal fraction we use rules of rounding off decimal fractions so as take final answer more simpler. While examples (ii) and (iii) have terminating decimal fractions, so they are to be written as such only and no use of rounding off decimals is used.
Rational Numbers
Rational Numbers
Decimal Representation of Rational Numbers
Rational Numbers in Terminating and Non-Terminating Decimals
Recurring Decimals as Rational Numbers
Laws of Algebra for Rational Numbers
Comparison between Two Rational Numbers
Rational Numbers Between Two Unequal Rational Numbers
Representation of Rational Numbers on Number Line
Problems on Rational numbers as Decimal Numbers
Problems Based On Recurring Decimals as Rational Numbers
Problems on Comparison Between Rational Numbers
Problems on Representation of Rational Numbers on Number Line
Worksheet on Comparison between Rational Numbers
Worksheet on Representation of Rational Numbers on the Number Line<|endoftext|>
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## Saturday, 20 October 2012
### Calculating Percentage Solutions
For some reason percentage solution calculation cause students some problems. However, hopefully, once you have been through this blog post you should have a good understanding of percentage solutions, and how to do the calculations.
If you want to practice your 'science maths' than have a look at: Maths4Biosciences.com
If you are having problems with percentage solution calculations then you might like to check out the Percentage Solutions course over at the Maths4Biosciences School on Fedora
What are percentage solutions?
A percentage solution is an amount or volume of something per 100 ml or 100 g of a solution. It is as simple as that. It is a percentage.
Why are they used?
Percentage solutions are just a convenient and easy way to record solution concentrations. One advantage of percentage solutions is that you don’t need to know anything about the compound in terms of molecular weight as all you need is the percentage of the required solution.
Why are their three types of percentage solutions?
This is slightly difficult to explain. However, there are three types of percentage solution:
• Percentage weight by volume (w/v)
• Percentage volume by volume (v/v)
• Percentage weight by weight (w/w)
The percentage weight by volume (w/v) is the number of grams of compound per 100 ml of solution. This type of percentage solution tends to be used when describing the amount of some powder made up in a solution. So, for example, 5 g of a powder made up to a final volume of 100 ml would be a 5% (w/v) solution. Likewise, 2.5 g of powder made up to 50 ml would also be a 5% (w/v) solution as you would have 5 g in 100 ml.
The percentage volume by volume (v/v) is the number of ml of some liquid per 100 ml of the solution. This type of percentage solution is usually used to describe a solution that is made by mixing two liquids. So, for example, 5 ml of a liquid made up to a final volume of 100 ml would be a 5% (v/v) solution. Likewise, 2.5 ml of a liquid made up to 50 ml would also be a 5% (v/v) solution as you would have 5 ml in 100 ml.
Finally, the percentage weight by weight (w/w). This one is a little more difficult to understand, but the principles as explained above are still true. A percentage weight by weight (w/w) solution can be a weight of a powder or a liquid made up in a solution to a final weight of the solution. So, for example, 5 g of a powder (or a liquid) made up in a solution that has a final weight of 100 g would be a 5% (w/w) solution. Likewise, 2.5 g of a liquid or powder made up to give a solution that weighed 50 g would also be a 5% (w/w) solution as you would have 5 g of the powder or liquid in 100 g of a solution.
If you are having problems with percentage solution calculations then you might like to check out the Percentage Solutions course over at the Maths4Biosciences School on Fedora
## Tuesday, 16 October 2012
### BGM3001:BMS3011 - lecture movies - 3d Structure of the GDP G-protein alpha subunit
The movie below shows the 3d structure of a guanosine diphosphate (GDP) bound G protein alpha-subunit.
The small green molecule in the middle is the bound GDP.
http://www.rcsb.org/pdb/cgi/explore.cgi?pdbId=1TAG
### BGM3001:BMS3011 - lecture movies - Bovine Rhodopsin
The structure of a G-protein coupled receptor is often shown as:
However, the crystal structure of bovine rhodopsin shows that the seven transmembrane spanning domains actually form a tight core
J. Mol. Biol. (2004) 342, 571–583 http://www.rcsb.org/pdb/cgi/explore.cgi?pdbId=1U19
### BGM3001:BMS3011 - lecture movies - 3d Structure - β2AR-T4L
The structure of a G-protein coupled receptor is often shown as:
Which shows a receptor with the N-terminal outside the cell, seven transmembrane spanning domains, 3 exoloops, 4 cytoloops and a C terminal inside the cell.
However, it is actually more compact with the seven transmembrane spanning domains forming a tight core.
Lower part = T4-lysozyme in place of cyto 3 Small green molecules = cholesterol; can just make out bound ligand in pocket
Science. 2007 318(5854): 1258–1265. http://www.rcsb.org/pdb/cgi/explore.cgi?pdbId=2RH1
### BGM3001:BMS3011 - lecture movies - αGs and Adenylyl Cyclase
The following animation shows αGs interacting with adenylyl cyclase.
The animation is also available as a slightly larger video....
### BGM3001:BMS3011 - lecture movies - GPCR movement of TM3 and 6
I have been asked for access to a number of the movies I used in the lectures.
The following animation shows the proposed movements of transmembrane spanning domains 3 and 6 for rhodopsin and β2 adrenergic receptor.
It should be noted that other GPCR may use different methods.
## Monday, 15 October 2012
### How do I get my marks and feedback on Blackboard
A number of the class have asked me how they can get their marks and feedback successfully on Blackboard.
The following short video explains how you can get your marks and feedback on modules that I teach on, and which have assessment (set by me) on Blackboard.
### How do I know if I have submitted successfully on Blackboard?
A number of the class have asked me how they can check that they have successfully submitted work on Blackboard.
The following video explains how you can check on modules that I teach on, and which have assessment (set by me) on Blackboard.<|endoftext|>
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The portion 64/7 is equal to 9.1428571428571 when converted come a decimal. See below detalis on exactly how to convert the portion 64/7 to a decimal value.
You are watching: 64/7 as a mixed number
### Fraction come Decimal Converter
Enter a fraction value:Ex.: 1/2, 2 1/2, 5/3, etc. Note that 2 1/2 method two and fifty percent = 2 + 1/2 = 2.5
Fraction come decimal explained:
## How to transform from fraction to decimal?
To easily convert a portion to a decimal, divide the numerator (top number) by the denominator (bottom number).
### Example 1: just how to transform 4/8 come a decimal?
Step 1:Divide 4 by 8: 4 ÷ 8 = 1 ÷ 2 = 0.5Step 2:Multiply the an outcome by 100 and add the decimal sign: 0.5 × 100%Answer: 4/8 = 50%
### Example 2: just how to convert 1 1/3 come a decimal?
Step 1:Divide 1 by 3: 1 ÷ 3 = 0.3333Step 2:Add this worth to the the essence part: 1 + 0.3333 = 1.3333Step 3:Multiply the result by 100 and include the decimal sign: 1.3333 × 100%Answer: 1 1/3 = 133.33%
Note: the result was rounded come 2 decimal places.
fractioninchesmm
1/640.01560.3969
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3/640.04691.1906
1/160.06251.5875
5/640.07811.9844
3/320.09382.3813
7/640.10942.7781
1/80.12503.1750
9/640.14063.5719
5/320.15633.9688
11/640.17194.3656
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1/40.25006.3500
fractioninchesmm
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9/320.28137.1438
19/640.29697.5406
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23/640.35949.1281
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25/640.39069.9219
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27/640.421910.7156
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31/640.484412.3031
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See more: 305/45/22 Silverado - Tire Question 305/40/22 Vs 285/45/22
fractioninchesmm
49/640.765619.4469
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11.000025.4000<|endoftext|>
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### Home > CC3 > Chapter 9 > Lesson 9.2.5 > Problem9-132
9-132.
Simplify each of the following expressions.
1. $3 \frac { 1 } { 5 } \cdot \frac { 7 } { 4 }$
Convert the mixed number into a fraction greater than one.
$\frac{16}{5}\left(\frac{7}{4}\right)$
$\frac{28}{5} = 5\frac{3}{5}$
1. $5^3\cdot(-\frac{4}{5})$
$\frac{4}{5}=(-1)(4)\left(\frac{1}{5}\right)$
Notice that the $5$ in the denominator would have an exponent of $-1$.
$5^{(3-1)}(-4)$
$5^2(-4)=-100$
1. $2 ^ { 4 } \cdot \frac { 5 } { 8 }$
This can be rewritten as:
$2^{4}\left(\frac{5}{2^{3}}\right)$
See part (b).
$10$
1. $-\frac{1}{2}\cdot3^2$
Use the Order of Operations.
1. $-\frac{5}{6}+(\frac{1}{2})^2$
Use the Order of Operations.
$-\frac{5}{6}+\frac{1}{4}$
$-\frac{20}{24}+\frac{6}{24}=-\frac{14}{24}$
1. $(-\frac{4}{5})^2-\frac{3}{50}$
Notice that squaring a negative number has the same result as multiplying two negative numbers: the product is always positive.
1. $(\frac{3}{10})^2-(-\frac{2}{5})^2$
1. $8^2(-\frac{7}{8})-\frac{1}{2}$
$8^{(2-1)}(-7)-\frac{1}{2}$
$8(-7)-\frac{1}{2}$
$(-56)-\frac{1}{2}$
$-56\frac{1}{2}$<|endoftext|>
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Tuesday, September 15, 2015
Measuring time...
Today we started by splitting everyone into permanent pairs and assigning each pair a particular robot. Our goal was to figure out how far the robot moves per rotation of each wheel so that we can predict where the robot will end up as we program it to go places. We used three different methods to find out this information. First, we used the formula for the perimeter of a circle (P):
P = π * D
where:
π = 3.14159
and D is the wheel’s diameter (diameter is a straight imaginary line that passes through the center of the circle and cuts it in half). So each pair got a tape measure and proceeded to measure the diameter of the wheel. Almost everyone arrived at 1 5/8 inches for the diameter of the robot wheels. Which created the following issue. How do we convert 1 5/8 inches to a decimal value that we can plug into a calculator? So we explained that to do this, we did the following:
5 ÷ 8 + 1 = 1.625 inches
and when we plug this value into the perimeter equation we get:
P = 3.14159 * 1.625 = 5.1 inches
Next, we asked everyone to program their robots to turn both wheels 1 rotation and asked them to measure how far the robot moved. Almost everyone measured 5 1/8 inches. Finally, we asked everyone to program their robots to turn both wheels 5 rotations and measure again. This time some kids got measurements of 25 inches, 25.5 inches and others 26 inches. So now we know how far the robot moves per each wheel rotation.
Next week, we’ll put all of what we have learned so far into practice. We will give the kids a “Rescue” mission, where the robots start from base and should be programmed to fetch a stranded Lego figure/pet from a given location and bring it back to base. It should be lots of fun. Also, if you can, go over some of these concepts with your kids at home. We went over them very quickly during the meeting, but this is where they get to apply the theoretical knowledge that they learn in school to real life situations. Try asking them to figure out how far your car moves for each rotation of your car’s wheels.<|endoftext|>
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Enveloping our planet and defending us from dangerous photo voltaic wind is an invisible magnetic discipline. Solar wind is a stream of plasma or charged particles like protons and electrons unleashed from the solar’s outer layers. Without magnetic discipline, photo voltaic wind may destroy our planet’s ambiance and make it inhabitable.
But precisely how the method works? For the primary time, researchers clarify how Earth’s magnetic discipline shoves apart the highly effective photo voltaic wind with none harm. As Earth strikes across the solar, it creates a bow-shaped wave or bow shock forward of itself, very like that in entrance of a transferring motorboat. When the electrons within the photo voltaic wind encounter the bow shock, their velocity is momentarily accelerated to such an extent that the electron stream turns into unstable and break down. This interplay slows down electrons and converts the power to warmth.
“If you were to stand on a mountaintop, you might get knocked over by a fast wind,” explained lead creator Li-Jen Chen from University of Maryland. “Fortunately, as the solar wind crashes into Earth’s magnetic field, the bow shock protects us by slowing down this wind and changing it to a nice, warm breeze. We now have a better idea how this happens.”
The discovering relies on knowledge collected by NASA’s Magnetospheric Multiscale (MMS) mission. The MMS mission consists of 4 equivalent satellites that research Earth’s magnetic discipline because it interacts with the photo voltaic wind and takes three-dimensional measurements each 30 milliseconds. The knowledge assist researchers higher perceive the position magnetic discipline performs round Earth.
“These extremely fast measurements from MMS allowed us finally to see the electron heating process in the thin shock layer,” mentioned co-author Thomas Moore from NASA’s Goddard Space Flight Center. “This is groundbreaking because now we have the ability to identify the mechanism at work, instead of just observing its consequences.”
During the mission, researchers noticed photo voltaic wind electron earlier than, throughout and after the encounter with bow shock and located that it took simply 90 milliseconds for bow shock to destabilize and break down electrons.
“The new observations of electron acceleration at the bow shock rewrite the current understanding of electron heating. For example, researchers didn’t expect that the bow shock could accelerate the solar wind electron stream to the speeds that we observed,” said Chen. “The study of electron heating is important not just for understanding how the bow shock protects Earth, but potentially for satellites, space travel and maybe exploring other planets in the future.”<|endoftext|>
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This material may consist of step-by-step explanations on how to solve a problem or examples of proper writing, including the use of citations, references, bibliographies, and formatting. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden.
Plato’s Theory of the Human Person
Plato’s metaphysical and epistemological ideas connect with his philosophical anthropology. Plato paints a picture of the human by formalizing a theory of reality cut into two. For Plato, human beings are caught in a world where truth is doubtful, sensory experiences delude us and opinion masquerades as truth. Plato devises a theory that depicts the human as a body with a soul, strung between desires rooted in this world, and a longing for the struggle that will lead him to truth in another, transcendent world. Plato envisions a society that can help people come to know greater truths, and he worries that a society ruled only by opinion will keep humankind forever trapped in a cave of our making. Plato creates a vision of an ideal society to help map out an ethical world where truth can be known, and justice can be realized....
This is only a preview of the solution. Please use the purchase button to see the entire solution<|endoftext|>
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HomeLogical ReasoningLearn to Solve Logical Series For CLAT
# Learn to Solve Logical Series For CLAT
AÂ number of events, objects, or people of a similar or related kind coming one after another constitute a series.
There is a pattern or relationship between any two successive terms of a series that determines the terms that follow. Your task will be to ascertain the logical relation between the terms of a given series in the question and calculate the answer accordingly.
Over the years, CLAT exams have featured three broad categories of series, viz., numerical series, alphabetical series and word series.
➤ NUMERICAL SERIES
Before we begin with the illustrations, let us brush up on some basic formulae and series that can prove helpful in solving number series.
Series you must know:
• Odd/ Even
(i) Series of odd numbers:1,3,5,7,….
(ii) Series of even numbers: 2,4,6,8,…
• Increasing/Decreasing series
(i)Â Series of increasing numbers:
1. A series can increase when a number is added to every term. For example: 17, 22,32,47,…In this series, each term increases by a multiple of 5. 17+5 = 22, 22+10 = 32, 32+ 15 = 47 and so on.
2. A series can also increase when a number is multiplied to every term. For example: 2, 4, 8, 16…etc. In this series, each term is being multiplied by the number 2 in order to obtain the following term.
(ii)Â Series of decreasing numbers:
1. A series can decrease when a number is being subtracted from successive terms. For example: 100, 97, 92, 85,..etc. in this series, odd numbers starting from 3 are being subtracted in order to obtain the successive term. 100-3 = 97, 97-5 = 92 and so on.
2. A series can decrease when each term is being divided in order to obtain the successive term. For example: 500, 250, 125, 52.5,…etc. In this series, each term is being successively divided by 2.
• Arithmetic progression (AP)
A sequence of the form → a, a + d, a + 2d, a + 3d, a + 4d, … a, a + d, a + 2d, a + 3d, … upto n terms refers to an arithmetic progression. The n-th term is [a + (n − 1)d].
Where, a: first term ; d: common difference; n: number of terms.
For example: 1, 5, 9, 13, 17… forms an AP with a = 1, d = 4.
• Geometric progression (GP)
A geometric progression is a series where each term is multiplied successively by a number in order to obtain the next term. The quotient of two consequtive terms is always the same and it is known as the common ratio (r).
For example: 2, 6, 18, 54,… . In this series, a = 2, r = 3.
The series is of the form: a, ar, ar², ar³,… and so on.
• Prime number series
Prime numbers are those numbers which have only two divisors, 1 and the number itself.
You are advised to go through the list of prime numbers from 1 to 100.
1, 2, 3, 5, 7, 9, 11, 13, 17, 19, 23 and so on.
• Fibbonachhi series
It is a series of numbers in which each number ( Fibonacci number ) is the sum of the two preceding numbers. The simplest is the series 1, 1, 2, 3, 5, 8, etc.
Now let us proceed to certain examples for better understanding:
Example 1. Examine the following numbers and identify the next number:Â [CLAT 2016]
45, 43, 40, 36, 31, 25, ….
1. 23
2. 29
3. 17
4. 18
Explanation:Â In the above series, numbers starting from 2 are being successively subtracted.
45-2 = 43, 43-3 = 40, 40-4 = 36, 36-5 =31, 31-6 = 25, 25-7= 18.
Example 2.  0,3,8,15… [CLAT 2014]
1. 24
2. 26
3. 35
4. none
Explanation:Â In this series, odd numbers are being added successively. 0+3 = 3, 3+5= 8, 8+7= 15, 15+9= 24.
Example 3. POQ, SRT, VUW,… [CLAT 2014]
1. XYZ
2. XZY
3. YZX
4. YXZ
Explanation:Â In all the terms of the series, the first two alphabets have been interchanged. The alphabet O comes before P but it has been put in second place. Similarly, X comes before Y but should be put in the second place. Hence, the answer will be YXZ.
Example 4Â : B, C, E, G, K, M, Q, S,______Â [CLAT 2011]
What is the next alphabet in this sequence?
(a) T
(b) U
(c) V
(d) W
A B C D E F G H I J K L M 1 2 3 4 5 6 7 8 9 10 11 12 13 Z Y X W V U T S R Q P O N 26 25 24 23 22 21 20 19 18 17 16 15 14
By matching the alphabets to their corresponding numerical values, it can be seen that the sequence of prime numbers is being followed. The next prime number after 19 (S) is 23. Hence, the next alphabet should be W.
Example 5. Z, X, T, N,______Â [CLAT 2011]
What is the next alphabet in this sequence?
(a) E
(b) F
(c) G
(d) H
Explanation: Even numbers starting from 2 are being subtracted successively. 26 (Z) -2= 24 (x), 24(X) – 4= 20 (T), 20(T) – 6 = 14(N), 14(N) – 8= 6 (F)
Example 6. ‘Apple, Application,_______, Approval, Apricot, April’ [CLAT 2011]
Which of the following best film in the blank?
(a) Arrogant
(b) April
(c) Appropriate
(d) Apiary.
Explanation: Since till the fourth word,  words with ‘pp’ are being used, therefore, the word here should be appropriate.
Example 7.:  ‘Gym, hymn, lynx, pygmy, rhythm’ [CLAT 2011]
Which of the following words does not belong to the above set?
(a) Myrrh
(b) Mythic
(c) Flyby
(d) Syzygy
Explanation: The words given in the series have either 1 or 2 ys’. The word ‘syzygy’ has 3 ys.
Example 8:. 2,3,4,6,7,8,9,11,12,13,14,15,17,18,19,20,21 ? [CLAT 2010]
1. 22
2. 23
3. 24
4. 25
Explanation: On observing the series we notice that the term after every 3rd, 4th, 5th… term is being skipped.
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23
The next term will be skipped after 6 terms. Hence, 22 will be next.
First published on January 9, 2021.Â
Cracking the CLAT!<|endoftext|>
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“Campylobacter” is the most common foodborne illness throughout the world and one of the most common diarrheal illnesses in the United States. An estimated 2.4 million persons become infected per year and an approximate 125 deaths per year result. Campylobacter is a general term that includes multiple strains of bacterium. The symptoms associated with Campylobacter food poisoning illness are diarrhea, which can be bloody, stomach cramps and pain, fever, nausea and vomiting. Symptoms usually arise within two to five days after a person is exposed to the bacteria and persist for about one week, though some infected persons have no symptoms at all. Thus, the range of reactions can span from no outward symptoms to death in rare cases. Illnesses generally occur in isolated instances; outbreaks in a specific geographical or community population are less common.
How Campylobacter Food Poisoning is Contracted
The main food item with which Campylobacter food poisoning is associated is poultry. Individuals generally contract the bacteria either when food preparation surfaces become cross-contaminated by raw poultry juices, or when undercooked poultry is ingested. Infection may also arise from other sources, such as contaminated water or unpasteurized milk. Water can become contaminated by infected feces. Milk may be contaminated if the cow it was drawn from was infected. Contaminated water sources are a common cause of outbreaks. The bacteria are sensitive to oxygen and drying, thus freezing foods is usually effective in reducing or eliminating their numbers. Infections occur more frequently in summer months than in the winter.
Persons Most Commonly Affected
The immunocompromised may be at a greater risk than other individuals; the bacteria can spread to the blood stream and cause a life-threatening infection that weakened immune systems cannot effectively fight. Travelers to foreign areas are more at risk, as Campylobacter food poisoning is common in developing countries. Infection in infants have been reported where children have been exposed to poultry packaging in grocery shopping carts.
Rare Severe Effects
Some individuals infected may develop arthritis over time. In few instances, a serious disease, “Guillain-Barré Syndrome,” may result. Though rare, the syndrome may affect an individual’s nerves several weeks following recovery from the main symptoms. The infected individual’s immune system may attack the body’s nerves, which can result in short-term paralysis capable of lasting for weeks.
Infection with Campylobacter is diagnosed as “Campylobacteriosis.” Medical laboratories identify the bacteria by conducting a test on an infected person’s stool. Though most infected individuals recover without treatment, antibiotics may be prescribed for those with more severe symptoms and will ultimately shorten the duration of the symptoms.
If you have contracted Campylobacter food poisoning, contact The Food Poisoning Lawyers @ the Merman Law Firm immediately. We will work tirelessly to recover for your pain, suffering, lost wages and medical bills.<|endoftext|>
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## Finding the greatest common divisor: basic terms
To learn to find the greatest common divisor of two or more numbers, you need to understand that are natural, simple and complex numbers.
Natural is any number that is used when counting whole objects.
If a natural number can be divided only by itself and one, it is called simple.
All natural numbers can be divided by itself and one, but the only even Prime number is 2, all the rest can be divided into two. Therefore, a simple can only be odd numbers.
Primes quite a lot, a complete list of them does not exist. For finding GCD you can use special table with such numbers.
Most positive integers can share not only in the unit themselves, but also on other numbers. For example, the number 15 can be divided by 3 and 5. All are called divisors of the number 15.
Thus, the divisor of any natural number a is a number by which it can be divided without a remainder. If the number has more than two positive divisors, it is called composite.
The number 30 can be identified such factors as 1, 3, 5, 6, 15, 30.
You can see that 15 and 30 have the same divisors 1, 3, 5, 15. The greatest common divisor of those two numbers is 15.
Thus, a common divisor of A and B is a number that you can divide them evenly. Most can be considered the maximum number that can be divided.
To solve problems we use such abbreviated inscription:
GCD (A; B).
For example, GCD (15; 30) = 30.
To write down all the divisors of a natural number, applies to the record:
D (15) = {1, 3, 5, 15}
D (9) = {1, 9}
GCD (9; 15) = 1
In this example, the natural numbers have only one common divisor. They are called coprime, respectively, the unit is their greatest common divisor.
## How to find the greatest common divisor of numbers
To find the GCD of multiple numbers, you need to:
find all divisors of each natural number individually, that is to decompose them into factors (primes);
- select all the same factors in these numbers.
- multiply them together.
For example, to calculate the greatest common divisor of 30 and 56, it is necessary to record the following:
30 = 2 * 3 * 5
70 = 2 * 5 * 7
Not to be confused with the decomposition, it is convenient to write the multiplier with the help of vertical bars. In the left part from the line you need to place the dividend and the right divider. Under severable, specify the resulting quotient.
So, in the right column will be all you need to solve the factors.
The same dividers (multipliers) for convenience to emphasize. They should rewrite and multiply and write the greatest common divisor.
30/2 70/2
15/5 35/5
7 3
NOD (30; 56) = 2 * 5 = 10
So easy actually to find the greatest common divisor of numbers. With a little practice you can do it virtually automatic.<|endoftext|>
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Scientists at the University of Tampere (Finland) and the Karolinska Institutet (Sweden) have demonstrated that an enterovirus vaccine can protect against virus-induced diabetes in a mouse model for Type 1 diabetes.
Type 1 diabetes is increasing worldwide and to date, the exact causes of the disease are not known. Enteroviruses (the most common virus affecting humans) are one of the environmental factors that have been touted as a potential cause of the disease. However, no firm evidence exists proving their role. In this study, Stone et al. have taken an initial step to determine the involvement of these viruses through testing the efficacy of a novel prototype vaccine in preventing experimental Type 1 diabetes after enterovirus infection. Vaccination of at risk-individuals with such a vaccine and the subsequent monitoring of disease onset could reveal a role for these viruses in Type 1 diabetes. Furthermore, if enteroviruses were involved, vaccination with an enterovirus vaccine would provide a viable preventative treatment for virus-induced Type 1 diabetes.
Type 1 diabetes is the most common, chronic, life-threatening disease in children. Finland and Sweden have the highest incidence of Type 1 diabetes in the world with more than 1 in 200 suffering from the disease in Sweden. The disease is caused by the destruction of the cells in the pancreas which produce insulin, the hormone responsible for regulating blood sugar levels. There are currently no preventative treatments and this disease requires life-long insulin injections for survival. Type 1 diabetes is also associated with severe and life-shortening complications including cardiovascular disease, loss of eyesight and a risk of lower limb amputations, and as such, it puts a heavy burden on both the affected individual and the health care system at large.
Currently, there are no commercially available vaccines for human use available which target the enteroviruses that are associated with Type 1 diabetes in humans. For the first time, this study by Stone et al. reports that the vaccine described prevents diabetes induced by enteroviruses in a clinically relevant model for Type 1 diabetes. Furthermore, it also protected against other signs of infection that were seen in control animals and it had no adverse effects on vaccinated animals.
"These exciting results showing that the vaccine completely protects against virus-induced diabetes indicate the potential that such a vaccine has for elucidating the role of enteroviruses in human Type 1 diabetes" says Prof. Malin Flodström-Tullberg at the Karolinska Institutet whose group were responsible for the pre-clinical studies.
The researchers are keen to continue pursuing these studies as confirmed by Dr Vesa Hytönen, who produced the prototype vaccine: "The model described in this paper provides an excellent platform to test further enterovirus vaccines which contain a greater number of potential diabetogenic viruses. Through these proof-of-concept studies we hope to develop and experimentally validate an enterovirus vaccine similar to the commonly used poliovirus vaccine, which has the potential to establish whether enteroviruses play a role in Type 1 diabetes."
Work is currently ongoing at the University of Tampere to develop a vaccine that targets a greater number of viruses than the single virus vaccine described in this study, and all of the proposed viruses have been implicated in Type 1 diabetes. This experimental model provides an exciting opportunity for further proof-of-concept studies before progress to a clinical set-up in humans.
"The experiments here are important steps towards the clinical use of novel enterovirus vaccines. Such a vaccine is under further development by Vactech Ltd. and its collaborator Provention Bio for testing in a clinical setting", says professor Heikki Hyöty, University of Tampere, an author in this study and one of the pioneers in this research field.
The research was performed in collaboration between University of Tampere and Karolinska Institutet and was supported by Tekes (Finnish Funding Agency for Innovation) and by Barndiabetesfonden (the Swedish Child Diabetes Foundation). The Tekes-funded consortium Therdiab includes, besides the University of Tampere and Karolinska Institutet, several Finnish Biotech companies including Vactech Ltd., which is an important partner for enterovirus vaccine development.
For more information, please contact: Professor Malin Flodström-Tullberg, Karolinska Institutet, [email protected] +46 76 947 4569; Associate professor Vesa Hytönen, University of Tampere, [email protected] +358 40 190 1517.
The original source: VM Stone et al. A Coxsackievirus B vaccine protects against virus-induced diabetes in an experimental mouse model of type 1 diabetes. Published in Diabetologia (the journal of the European Association for the Study of Diabetes [EASD]).
UNIVERSITY OF TAMPERE PRESS RELEASE 20 Nov 2017<|endoftext|>
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Task 1 (P1)
Which part of a cell controls all the activities that happen in the cell? What does it contain that enables it to carry out its function?
The nucleus is the control centre of the cell. A cell’s nucleus is able to control the other activities in a cell by expressing certain segments of its DNA, which creates proteins that perform specific activities. Proteins can vary from enzymes to structural components, but they are needed for almost all processes in a cell and the human body.
The nucleus directs all cellular activities by controlling the synthesis of proteins. The nucleus contains encoded instructions for the synthesis of proteins in a helical molecule called deoxyribonucleic acid (DNA). The cell’s DNA is packaged within the nucleus in a structural form called chromatin. Chromatin consists of DNA wound tightly around spherical proteins called histones. When the cell prepares to divide, the DNA unwinds from the histones and assumes the shape of chromosomes, the X-shaped structures visible within the nucleus prior to cell division. Chromatin packaging of DNA allows all of the cell’s DNA to fit into the combined space of the nucleus. If DNA was not packaged into chromatin, it would spill out over a space about 100 times as large as the cell itself.
The first step in protein synthesis begins in the nucleus. Within the nucleus, DNA is translated into a molecule called messenger ribonucleic acid (mRNA). mRNA then leaves the nucleus through the nuclear pores. Once in the cytoplasm, mRNA attaches to ribosomes (either bound to endoplasmic reticulum or free in the cytoplasm) and initiates protein synthesis. Proteins made for export from the cell function as enzymes that participate in all the body’s chemical reactions. Because enzymes are essential for all the body’s chemical processes-from cellular respiration to digestion-direction of the synthesis of these enzymes in essence controls all the activities of the body. Therefore, the nucleus, which contains the instructions for the synthesis of these proteins, directs all cellular activities and thus all body processes.
Draw a flow diagram below that shows the relationship between a cell, the nucleus, chromosomes, genes and DNA.
(a) Write a description of what you understand by the term “gene”.
Include any appropriate diagrammatic illustration to support your text
(maximum 100 words).
A gene is a length of DNA that codes for a specific protein. So, for example, one gene will code for the protein insulin, which is important role in helping your body to control the amount of sugar in your blood.Genes are the basic unit of genetics. Human beings have 20,000 to 25,000 genes. These genes account for only about 3 per cent of our DNA. The function of the remaining 97 per cent is still not clear, although scientists think it may have something to do with controlling the genes.
(b) Research how wheat has changed since the 1950s.
What was the name of the man who enabled today’s wheat to be grown successfully?
What did he do to enable wheat to be grown around the
world and save 100 million people from starvation?
Dr. Norman Borlaug Studies :- Borlaug studied plant biology and forestry at the University of Minnesota and earned a Ph.D. in plant pathology there in 1941. From 1944 to 1960 he served as a research scientist at the Rockefeller Foundation’s Cooperative Mexican Agricultural Program in Mexico. Borlaug’s work was founded on earlier discoveries of ways to induce genetic mutations in plants. These methods led to modern plant breeding, with momentous results that included the tailoring of crop varieties for regions of climatic extremes. At a research station at Campo Atizapan he developed strains of grain that dramatically increased crop yields. Borlaug ultimately developed short-stemmed (“dwarf”) wheat, a key element in the Green Revolution in developing countries.
How is this connected to genes?
University of Minnesota alumnus Norman Borlaug left an indelible mark on the world. The late agronomist’s work in developing new varieties of wheat starting in the 1940s spawned the “Green Revolution,” and is credited with saving at least a billion lives.
For his unparalleled contributions in using science to feed the world, Borlaug was awarded the Nobel Peace Prize in 1970, the Presidential Medal of Freedom (1977), the Public Welfare Medal (2002) by the National Academy of Sciences, the National Medal of Science (2005), and the Congressional Gold Medal (2007). In 2014 a statue of him was placed in the National Statuary Hall of the United States Congress; he is the sole scientist represented there.
Borlaug also received more than 50 honorary degrees, was inducted into the collegiate National Wrestling Hall of Fame, and helped introduce Little League baseball in Mexico. And just last month, at the “9 Billion and Counting” conference on campus, he was lauded as being “the single greatest graduate of a land-grant institution.”
Beyond all the accolades, Borlaug was a plant pathologist/breeder, a teacher of scientists, a scientific collaborator, and a team builder. His unique teaching and training methods for young scientists became part
If you complete the above correctly then you will achieve AC. P1
Task 2 (M1)
An example of Natural Selection – The case of the English peppered Moth
The English peppered moth rests on tree trunks during the day. Its speckled grey colouring provides a good camouflage on the lichen-covered trunks of the trees. In about 1850, a black English peppered moth was discovered in Manchester.
It was not an advantage for these moths to be black as they stood out against the white lichen and were therefore easily seen by predators.
With the spread of the Industrial Revolution and the expanding use of steam power, pollution became a problem in large cities. The trees once covered in lichen turned black. The number of black moths increased dramatically and by 1895 they made up about 98% of the population of English peppered moths. How did the moths get to be black?
Two different varieties of the peppered moth. The distribution of peppered moths in
They are on a tree that is covered in lichen. the 1950s. The black variety is in
Industrial areas and the light variety is
in rural farm areas
Using the information given above, and some of your own research, answer the following questions:
What changes occurred in the genes of the peppered moth, and how did the changes occur, that led to the creation of the black moth? (Max 100 words).
The evolution of the peppered moth is an evolutionary instance of directional colour change in the moth population as a consequence of air pollution during the Industrial Revolution. The frequency of dark-coloured moths increased at that time, an example of industrial melanism.
Explain why the population of black moths was so low in the 1850s. (Max 100 words).
Industrialisation and domestic coal fires had caused sooty air pollution which had killed off lichens and blackened urban tree trunks and walls. So now it was the pale form of the moth that was more obvious to predators, while the melanic form was better camouflaged and more likely to survive and produce offspring.
Explain why the population of black moths increased to 98% of the whole population by 1895. (Max 100 words).
The evolution of the peppered moth is an evolutionary instance of directional colour change in the moth population as a consequence of air pollution during theIndustrial Revolution. The frequency of dark-coloured moths increased at that time, an example of industrial melanism.
What happened to the numbers of light moths between 1850 and 1895? Look at the map that shows the distribution of moths between 1895 and 1950? Explain the pattern shown. (Max 150 words).
The melanic phenotype is due to underlying homozygous (BB) and heterozygous (Bb) dominant genotypesIn the mid 1950’s, air pollution controls were introduced in BritainWhen smoke pollution decreased in Britain, natural selection acted very quickly to favor survival of the wild type peppered morphs as bird predation eliminated melanic forms in progressively less polluted forestsThe frequency of the melanic form has declined ever since
If you complete the above correctly then you will achieve AC. M1
Task 3 (D1)
Explain how a genetic code leads to the formation (synthesis) of a protein.
You need to explain in detail the sequence of events involved in protein synthesis. Draw a diagram to represent this.
Transcription • Information transcribed from DNA into RNA – mRNA carries information for protein structure, but other RNA molecules formed in same way • RNA polymerase binds to promoter nucleotide sequence at point near gene to be expressed • DNA helix unwinds • RNA nucleotides assemble along one DNA strand (sense strand) in complementary sequence to order of bases on DNA beginning at start codon (AUG – methionine) • Transcription of DNA sense strand ends at terminator nucleotide sequence • mRNA moves to ribosome • DNA helix rewinds From: Tortora, GJ & Grabowski SR (2000) Principles of Anatomy and Physiology (9th Ed). New York: John Wiley & Sons. P88.
11. Transcriptional control • Each cell nucleus contains all genes for that organism but genes only expressed as needed • Transcription regulated by transcription factors – Proteins produced by their own genes • If transcription factors promote transcription – activators • If transcription factors inhibit transcription – repressors • General transcription factors interact with RNA polymerase to activate transcription of mRNA – Numerous transcription factors required to initiate transcription – General transcription factors set base rate of transcription – Specific transcription factors interact with general transcription factors to modulate rate of transcription • Some hormones also cause effects by modulating rate of gene transcription
12. Protein synthesis occurs in ribosomes
13. Protein synthesis occurs in ribosomes
Explain why proteins are important in the functioning of organisms? Give examples of at least three specific proteins and their functions.
Proteins are large, complex molecules that play many critical roles in the body. They do most of the work in cells and are required for the structure, function, and regulation of the body’s tissues and organs. … Antibodies bind to specific foreign particles, such as viruses and bacteria, to help protect the body.
Strengthens bones, ligaments, and tendons (a fibrous protein)
Provides stretch in skin, blood vessels, and lung tissue (a fibrous protein)
Forms structure of hair and nails and water proofs the skin (a fibrous protein)
DystrophinReinforces parts of muscle cells (a fibrous protein)
Forms blood clots (a fibrous protein)
Actin and Myosin
Are involved in contraction of muscle cells, division in all cells, and transport of substances within cells (a fibrous protein)
Explain how proteins influence the physical attributes of individuals within a species (i.e. variation).
Relationship Between DNA Bases Genes, Proteins and Traits
By John Brennan; Updated April 26, 2018
Although you might have heard people talk about a gene for red hair, green eyes or other characteristics, it’s important to remember that genes code for proteins, not traits. While your genetic makeup does indeed determine physical traits like eye color, hair color and so forth, your genes affect these traits indirectly by way of the proteins created via DNA.
Your DNA carries information in the sequence of base pairs of its nucleotides. These biological molecules, the building blocks of DNA, are often abbreviated with the first letter of their names: adenine (A), thymine (T), guanine (G) and cytosine (C).
The types and sequence of nucleotides in DNA determine the types and sequence of nucleotides in RNA. This in turn determines the types and order of amino acids included in proteins. Specific three-letter groups of RNA nucleotides code for specific amino acids. The combination TTT, for example, codes for the amino acid phenylalanine. Regulatory regions of the gene also contribute to protein synthesis by determining when the gene will be switched on or off.
SCIENCING VIDEO VAULT
In active genes, genetic information determines which proteins are synthesized and when synthesis is turned on or off. These proteins fold into complicated three-dimensional structures, somewhat like molecular origami.
Because each amino acid has specific chemical characteristics, the sequence of amino acids determine the structure and shape of a protein. For example, some amino acids attract water, and others are repelled by it. Some amino acids can form weak bonds to each other, but others cannot. Different combinations and sequences of these chemical characteristics determine the unique three-dimensional folded shape of each protein
Structure & Function
The structure of a protein determines its function. Proteins that catalyze (accelerate) chemical reactions, for example, have “pockets,” which can bind specific chemicals and make it easier for a particular reaction to occur.
Variations in the DNA code of a gene can change either the structure of a protein or when and where it is produced. If these variations change the protein structure, they could also change its function. For example, a single, specific mutation in hemoglobin — the oxygen-carrying protein abundant in your red blood cells — affects oxygen transport and is enough to cause sickle-cell anemia.
Variations in a gene can affect traits in several ways. Variations in proteins involved in growth and development, for example, can give rise to differences in physical features like height. Pigments of skin and hair color are produced by enzymes, proteins that catalyze chemical reactions. Variations in both the structure and quantity of the proteins produced give rise to different amounts of skin and hair pigment and therefore different colors of hair and skin.
If you complete the above correctly then you will achieve AC. D1
P1: Describe how the functioning or organisms relates to the genes in their cells
M1: Describe how variation within a species brings about evolutionary change
D1: Explain how genes control variation within a species using a simple coded message<|endoftext|>
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# 11.10: Calculus with Power Series
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Now we know that some functions can be expressed as power series, which look like infinite polynomials. Since calculus, that is, computation of derivatives and antiderivatives, is easy for polynomials, the obvious question is whether the same is true for infinite series. The answer is yes:
Theorem 11.9.1
Suppose the power series
$f(x)=\sum_{n=0}^\infty a_n(x-a)^n$
has radius of convergence $$R$$. Then
\eqalign{ f'(x)&=\sum_{n=0}^\infty na_n(x-a)^{n-1},\cr \int f(x)\,dx &= C+\sum_{n=0}^\infty {a_n\over n+1}(x-a)^{n+1},\cr }
and these two series have radius of convergence $$R$$ as well.
Example 11.9.2
\eqalign{ {1\over 1-x} &= \sum_{n=0}^\infty x^n\cr \int{1\over 1-x}\,dx &= -\ln|1-x| = \sum_{n=0}^\infty {1\over n+1}x^{n+1}\cr \ln|1-x| &= \sum_{n=0}^\infty -{1\over n+1}x^{n+1}\cr }
when $$|x| < 1$$. The series does not converge when $$x=1$$ but does converge when $$x=-1$$ or $$1-x=2$$. The interval of convergence is $$[-1,1)$$, or $$0 < 1-x\le2$$, so we can use the series to represent $$\ln(x)$$ when $$0 < x\le2$$.
For example
$\ln(3/2)=\ln(1--1/2)= \sum_{n=0}^\infty (-1)^n{1\over n+1}{1\over 2^{n+1}}$
and so
$\ln(3/2)\approx {1\over 2}-{1\over 8}+{1\over 24}-{1\over 64} +{1\over 160}-{1\over 384}+{1\over 896} ={909\over 2240}\approx 0.406 .$
Because this is an alternating series with decreasing terms, we know that the true value is between $$909/2240$$ and $$909/2240-1/2048=29053/71680\approx .4053$$, so correct to two decimal places the value is $$0.41$$.
What about $$\ln(9/4)$$? Since $$9/4$$ is larger than 2 we cannot use the series directly, but $$\ln(9/4)=\ln((3/2)^2)=2\ln(3/2)\approx 0.82,$$ so in fact we get a lot more from this one calculation than first meets the eye. To estimate the true value accurately we actually need to be a bit more careful. When we multiply by two we know that the true value is between $$0.8106$$ and $$0.812$$, so rounded to two decimal places the true value is $$0.81$$.
## Contributors
This page titled 11.10: Calculus with Power Series is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Guichard via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.<|endoftext|>
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First aid is the assistance given to an injured or sick person in need of urgent medical assistance. First aid applies to a broad range of medical situations and consists both of specific knowledge and skills (for example, what to do for each type of injury or illness) and the ability to assess a situation and make appropriate decisions (such as when to call for emergency medical assistance).
First Aid Kit
Preparedness is a key element of first aid. While every home, auto, and boat should be equipped with a basic emergency kit that includes first aid supplies and a first aid manual, special circumstances may necessitate more advanced or specific degrees of preparation for an emergency. For example, residents of certain geographic areas where natural disasters (such as hurricanes, earthquakes, tornados, floods, landslides, or tsunamis) may occur should prepare for emergencies by assembling disaster preparedness kits such as earthquake kits, flood kits, andevacuation kits.
Travelers should also prepare to administer first aid in the region they plan to visit. In many developed countries, this may amount to packing a standard first aid kit and manual with your belongings. Medicines to prevent motion sickness and advanced awareness about the management of traveler's diarrhea are also helpful to travelers. In the developing world or in remote regions, a travel medicine kit should include medicines and supplies that may not be available for purchase as well as any specific products (such as insect repellant to prevent mosquito and tick bites) needed for your destination. It's also important for travelers in foreign countries to learn how to access emergency services by telephone.
CPR and First Aid Certification
Regardless of your level of skill or degree of first aid training, if you find yourself in a true medical emergency, always call 911 for emergency medical assistance immediately. Likewise, if you are involved in any medical situation that is beyond your personal abilities to provide first aid, you should never hesitate to summon emergency medical assistance right away.
Health Solutions From Our Sponsors
Medically reviewed by Avrom Simon, MD; Board Certified Preventative Medicine with Subspecialty in Occupational Medicine
Kasper, D.L., et al., eds. Harrison's Principles of Internal Medicine, 19th Ed. United States: McGraw-Hill Education, 2015.<|endoftext|>
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This article includes a list of references, but its sources remain unclear because it has insufficient inline citations. (September 2012) (Learn how and when to remove this template message)
|Date||September 9, 1924|
|Goals||$2 daily wage|
8 hour day
|Parties to the civil conflict|
The Hanapēpē Massacre (also called the Battle of Hanapēpē since both sides were armed) happened on September 9, 1924. Toward the end of a long-lasting strike of Filipino sugar workers on Kauaʻi, Hawaiʻi, local police shot dead nine strikers and fatally wounded seven, strikers shot and stabbed three sheriffs to death and fatally wounded one; a total of 20 people died. The massacre brought an end to armed protests in Hawaii.
By the 1920s, the sugarcane plantation owners in Hawaiʻi had become disillusioned with both Japanese and Filipino workers. They spent the next few years trying to get the U.S. Congress to relax the Chinese Exclusion Act so that they could bring in new Chinese workers. Congress prevented the importation of Chinese labor.
But organized labor in the 1920s' U.S. mainland supported the Congress in this action, so that for a while it looked as though militant unionism on the sugarcane plantations was dead. To oppose organized labor, the Hawaiian Territorial Legislature passed the Criminal Syndicalism Law of 1919, Anarchistic Publications law of 1921, and the Anti-Picketing Law of 1923.
These laws, with penalties of up to 10 years in prison, increased the discontent of the workers. The Filipinos, who were rapidly becoming the dominant plantation labor force, had deep-seated grievances: as the latest immigrants they were treated most poorly. Although the planters had claimed there was a labor shortage and they were actively recruiting workers from the Philippines, they wanted only illiterate workers and turned back any arrivals who could read or write, as many as one in six.
Strikes and Massacre
By 1922 Filipino labor activist Pablo Manlapit had organized a new Filipino Higher Wage Movement which numbered some 13,000 members. In April 1924, it called for a strike on the island of Kauaʻi, demanding $2 a day in wages and reduction of the workday to 8 hours. As they had previously, the plantation owners used armed forces, the National Guard, and strike breakers paid a higher wage than the strikers demanded. Again workers were turned out of their homes. Propaganda was distributed to whip up racism. Spying and infiltration of the strikers' ranks was acknowledged by Jack Butler, executive head of the Hawaiian Sugar Planters' Association.
Strike leaders were arrested in attempts to disrupt workers' solidarity, and people were bribed to testify against them. On September 9, 1924, outraged strikers seized two strike breakers at Hanapēpē and prevented them from going to work. The police, armed with clubs and guns, came to union headquarters to rescue them. Filipino strikers were armed only with homemade weapons and knives.
The Associated Press flashed the story of what followed across the United States in the following words: Honolulu. - Twenty persons dead, unnumbered injured lying in hospital, officers under orders to shoot strikers as they approached, distracted widows with children tracking from jails to hospitals and morgues in search of missing strikers - this was the aftermath of a clash between cane strikers and workers on the McBryde plantation, Tuesday at Hanapepe, island of Kauai. The dead included sixteen Filipinos and four policemen.
After the massacre police rounded up all male protesters they could find, and a total of 101 Filipino men were arrested. 76 were brought to trial, and of these 60 received four-year jail sentences. Pablo Manlapit was charged with subornation of perjury and was sentenced to two to ten years in prison. The Hawaiʻi Hochi claimed that he had been railroaded into prison, a victim of framed-up evidence, perjured testimony, racial prejudice and class hatred. Shortly thereafter, he was paroled on condition that he leave Hawaiʻi. After eight months the strike disintegrated.
The Federationist, the official publication of the American Federation of Labor, reported that in 1924 the ten leading sugar companies listed on the New York Stock Exchange paid dividends averaging 17 percent. From 1913 to 1923, the eleven leading sugar companies paid cash dividends of 172.45 percent, and most of them issued large stock dividends.
After the 1924 strike, the labor movement in Hawaiʻi dwindled, but did not die, and discontent among the workers rarely surfaced again. Pablo Manlapit, who had been imprisoned and exiled, returned to the islands in 1932 and started a new labor organization, this time hoping to include other ethnic groups. But the time was not ripe in the Depression years. There were small nuisance strikes in 1933 that made no headway and involved mostly Filipinos. Protests since the massacre have discouraged carrying guns at demonstrations.
- List of massacres in the United States
- Murder of workers in labor disputes in the United States
- List of incidents of civil unrest in the United States
- Chapin, Helen Geracimos (1996). "Suppressing the News and Contributing to a Masscre". Shaping History: The Role of Newspapers in Hawai'i. University of Hawaii Press. pp. 131–138. ISBN 978-0-8248-1718-3.
- Alegado, Dean (October 1997). "Blood in the Fields: The Hanapepe Massacre and the 1924 Filipino Strike". Filipinas Magazine. Retrieved 24 February 2018 – via Positively Filipino.
- Chang, Lester (April 9, 2006). "Massacre strengthened labor unions". The Garden Island. Retrieved 26 February 2018.
- Soboleski, Hank (September 10, 2006). "Pablo Manlapit and the Hanapepe Massacre". The Garden Island. Retrieved 25 February 2018.
- Matsuoka, Fumitaka; Fernandez, Eleazar S (2003). Realizing the America of Our Hearts: Theological Voices of Asian Americans. Chalice Press. p. 91. ISBN 978-0-8272-3251-8.
- Raymundo, Rizaline R (2003). Tomorrow's Memories: A Diary, 1924-1928. University of Hawaii Press. p. 233. ISBN 978-0-8248-2688-8.
- Beechert, Edward D. Working in Hawaii: A Labor History. Honolulu: University of Hawaii Press, 1985; ISBN 978-0-8248-0890-7
- Monrayo, Angeles. Tomorrow's Memories: A Diary, 1924-1928. Rizaline R. Raymunod, ed. Honolulu: University of Hawaii Press, 2003; ISBN 0-8248-2671-X
- "Plaque Remembers Hanapepe Massacre", Associated Press, September 10, 2006.; accessed May 23, 2007.
- Reinecke, John E. The Filipino Piecemeal Sugar Strike of 1924-1925. Honolulu: University of Hawaii Press, 1997; ISBN 0-8248-1896-2
- Aquino, Belinda A. Understanding the ‘Hanapepe Massacre.’ "The Filipino Century Beyond Hawaii," International Conference On The Hawaii Filipino Centennial, Honolulu, Hawaii.<|endoftext|>
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South Carolina State History
The earliest history of South Carolina is similar to a lot of the Eastern United States; it was populated 13,000 years ago by archaic communities, which went on to follow the same big changes that affected most cultures in the area. By the time of European contact there were 29 nations of native people in South Carolina. The nations of South Carolina mostly fell under two large culture groups, the Eastern Siouan and Cusaboan peoples, though other groups lived in the area.
Following exploration of the coast in 1521 by Francisco de Gordillo, the Spanish tried unsuccessfully to establish a colony near present-day Georgetown in 1526, and the French also failed to colonize Parris Island near Fort Royal in 1562. Although there was an older charter from 1629, King Charles II of England chartered the colony in 1663 to wealthy aristocrats in exchange for their political support back in England. The colony was named Carolina after King Charles, and it acted as a buffer zone between Spanish territory and the other English colonies. The first English settlement was made in 1670 at Albemarle Point on the Ashley River, but poor conditions drove the settlers to the site of Charleston (originally called Charles Town).
South Carolina, officially separated from North Carolina in 1729. Charleston, despite its size and influence in the colonies, wasn't incorporated until after the Revolution. The city was run by a governor representing the Crown, and the state was a bastion of Loyalism to the British government. It was the scene of extensive military action during the Revolution and again during the Civil War.
In the 1800s, South Carolina played a leading role in Southern hostility to the federal government, and in the entrenchment of slavery in the Deep South. South Carolina was the largest slave state by percentage of people in slavery, and had the strictest laws toward freeing slaves; the government worked hard to ensure the dominance of the minority white population over the black majority (related, the huge Black population is credited with the preservation of African traditions in the Carolina Lowcountry). In 1832 this contributed to the Nullification Crisis. In 1861, South Carolina became the first state to secede from the Union. The Civil War began in 1861 as South Carolina troops fired on federal Fort Sumter in Charleston Harbor.
After the Confederate loss in the Civil War, South Carolina became the only state with a black majority congress. This saw a brief period of rapidly expanding rights for Black Americans that was swiftly reversed during the Jim Crow era. However, despite its long history as a bedrock slave state, South Carolina didn't see the levels of violence that plagued Mississippi and Alabama during desegregation. Likewise South Carolina was relatively swift among former Confederate states to remove the Confederate flag from government buildings after incidents of racial violence.
Historic points of interest include Fort Sumter National Monument, Fort Moultrie, Fort Johnson, and aircraft carrier USS Yorktown in Charleston Harbor; the Middleton, Magnolia, and Cypress Gardens in Charleston; and Cowpens National Battlefield. VIsitors might also be interested in the Hilton Head resorts, and the Riverbanks Zoo and Botanical Garden in Columbia.
South Carolina Culture and Interesting Facts
Myrtle Beach & the Grand Strand
Myrtle Beach is likely South Carolina's biggest tourist attraction, as the center of the Grand Strand string of beaches. The beaches collectively draw over 14 million visitors every year—that's nearly three times the population of South Carolina—and Myrtle Beach specifically hosts a lot of famous attractions. The recently opened boardwalk, the nearby Family Kingdom Amusement Park, and the abundant shopping have led Myrtle Beach to receive regular national acclaim. That's a pretty big achievement, considering the town used to be almost entirely off the map. The idea to develop the beach into a tourist area only started in the early 1900s, and the city of Myrtle Beach was only incorporated in the 1950s.
Charleston, South Carolina
The city of Charleston is one of the most historically important cities in the United States. It was one of the first major cities in the country, founded back in 1680, and was at one point the fifth largest city in the nation. For rather unsavory reasons, the city continued to have an outsize influence on the country even as the rest of the country grew and expanded; Charleston was the largest slave port in the United States, sometimes second to Savannah, Georgia. The slave trade in Charleston led to South Carolina being the state with the largest Black majority population and being the only state with a majority of legislators being slaveholders. The Civil War started in Charleston when secessionists attacked Fort Sumter. Today the city is famous for much nicer reasons, including its lovely and well-preserved architecture, its food culture, and its healthy arts scene. The city has received international attention and praise as a tourist destination.
South Carolina Plantations
The plantations of South Carolina are another popular tourist destination. The plantations were home to the aristocratic elite, who built large country mansions in which to live comfortably as they profited off the forced labor of slaves. These former slaveholding mansions embody a lot of important history in the South. They offer a glimpse into the historic culture of the area including the artistic tastes and values of the slaveholding class, the economic realities that shaped American history, and the violent injustices inflicted on America's Black population. Some plantations are just catered to tourists interested in the design and art, but many of the best plantation museums allow visitors to dive into the deep history of slavery in the United States.
Gullah Culture in the Lowcountry
Gullah culture refers to the unique culture of Black people living in the Lowcountry of South Carolina and Georgia. Gullah people are most culturally distinct from other Black Americans due to their use of their own creole language, also called Gullah, which heavily draws on different West and Central African languages. South Carolina was a Black-majority state, and more than anywhere else Gullah culture has heavily influenced the whole of South Carolina. The unique style of Gullah art and cuisine makes up some of South Carolina's most powerful cultural legacy. One particular type of art (or craft), Sweetgrass basket weaving, was recognized as a symbol of the state.
Famous South Carolina Natives and Residents
Joe Frazier prize fighter;
Althea Gibson tennis champion;
Dizzy Gillespie jazz trumpeter;
DuBose Heyward poet, playwright, and novelist;
Andrew Jackson president;
Jesse Jackson civil rights leader;
Eartha Kitt singer;<|endoftext|>
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# What is the equivalent of 1 5
Fractions equivalent to 1/5: 2/10, 3/15, 4/20, 5/25 and so on … Fractions equivalent to 2/5: 4/10, 6/15, 8/20, 10/25 and so on …
## What does 5 over 10 equal?
We do this by first finding the greatest common factor of 5 and 10, which is 5. Then, we divide both 5 and 10 by the greatest common factor to get the following simplified fraction: 1/2. Therefore, this equation is true: 5/10 = 1/2. If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction.
## How do you determine whether fractions are equivalent?
• Equivalent fractions may look different, but they have the same value.
• You can multiply or divide to find an equivalent fraction.
• Adding or subtracting does not work for finding an equivalent fraction.
• If you multiply or divide by the top of the fraction, you must do the same to the bottom.
More items…
## What are the rules of equivalent fractions?
Equivalent Fractions Rule. A rule stating that if the numerator and denominator of a fraction are multiplied by the same nonzero number, the result is a fraction that is equivalent to the original fraction. This rule can be represented as: a//b = (n * a)//(n * b).
## How do you find the equivalent expression?
equivalent expressionshave the same value but are presented in a differentformat using the properties of numbers eg, ax + bx = (a + b)x are equivalent expressions. Strictly, they are not “equal”, hence we should use 3 parallel lines in the”equal” rather than 2 as shown here.
## What is the equivalent of 3 5?
6/10Decimal and Fraction Conversion ChartFractionEquivalent Fractions3/56/1018/304/58/1024/301/62/126/365/610/1230/3623 more rows
## What is 1/5 equivalent to as a ratio?
Fractions equivalent to 1/5 are 2/10, 3/15, 4/20, 5/25, … Fractions equivalent to 2/5 are 4/10, 6/15, 8/20, 10/25, …
## What is 1/4 equivalent to as a fraction?
Answer: The fractions equivalent to 1/4 are 2/8, 3/12, 4/16, etc. Equivalent fractions have the same value in their reduced form. Explanation: Equivalent fractions can be written by multiplying or dividing both the numerator and the denominator by the same number.
## What is a equivalent to?
1 : equal in force, amount, or value also : equal in area or volume but not superposable a square equivalent to a triangle. 2a : like in signification or import. b : having logical equivalence equivalent statements.
## What is 1/5 into a decimal?
0.2Answer: 1/5 as a decimal is expressed as 0.2.
## What is the percentage for 1 5?
20%Fraction to percent conversion tableFractionPercent1/520%2/540%3/560%4/580%41 more rows
## What is 1/3 the same as?
Answer: The fractions equivalent to 1/3 are 2/6, 3/9, 4/12, etc. Equivalent fractions have the same value in the reduced form. Explanation: Equivalent fractions can be written by multiplying or dividing both the numerator and the denominator by the same number.
## What is 1/6 equal to as a fraction?
Equivalent Fractions ChartUnit FractionEquivalent Fractions1/32/6, 3/9, 4/12..1/42/8, 3/12, 4/16..1/52/10, 3/15, 4/20,..1/62/12, 3/18, 4/24,..4 more rows
## What is 3/4 equivalent to as a fraction?
Equivalent fractions of 3/4 : 6/8 , 9/12 , 12/16 , 15/ Equivalent fractions of 1/5 : 2/10 , 3/15 , 4/20 , 5/
## What is equivalent calculator?
Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds.
## How do I find equivalent fractions?
How to Find Equivalent Fractions. Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction, and you’ll create an equivalent fraction.
## What is equivalent formula?
The equivalent of any substance is given by the charge it carries on itself. Thus, the equivalent weight is obtained by the ratio of the molar mass of the substance and the number of equivalents. It can be mathematically represented as. Equivalent weight = Molar Mass number of equivalents .
## What is the meaning of 1 by 4?
Noun. 1. quarter – one of four equal parts; “a quarter of a pound” fourth part, one-fourth, one-quarter, quartern, twenty-five percent, fourth.
## What is 1 over 4 as a decimal?
0.25a) The long division Let us do the division 1 ÷ 4 . Hence the decimal form of the fraction is 0.25.
## What is a fraction equivalent to 4 6?
2/3So 2/3, which is equal to 4/6, is also equal to 6/9. All three of these are equivalent fractions, 2/3, 4/6, and 6/9.
## What is a fraction equivalent to 2 6?
1/ 3Both 1/ 3 and 2/ 6 are equivalent fractions having the same value.
## How to make a fraction equivalent?
Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction , and you’ll create an equivalent fraction.
## What are Equivalent Fractions?
Equivalent fractions are fractions with different numbers representing the same part of a whole. They have different numerators and denominators, but their fractional values are the same.
## What are Equivalent Fractions?
Two or more fractions are said to be equivalent if they are equal to the same fraction when simplified. For example, the equivalent fractions of 1/5 are 5/25, 6/30, and 4/20, which on simplification, result in the same fraction, that is, 1/5.
## How to Determine if Two Fractions are Equivalent?
We need to simplify the given fractions to find whether they are equivalent or not. Simplification to get equivalent numbers can be done to a point where both the numerator and denominator should still be whole numbers. There are various methods to identify if the given fractions are equivalent. Some of them are:
## How to write 6/8 as an equivalent fraction?
In order to write the equivalent fraction for 6/8, let us multiply the numerator and denominator by 2 and we will get (6 × 2)/ (8 × 2) = 12/16. Therefore, 6/8 and 12/16 are equivalent fractions. Now, let us get another equivalent fraction for 6/8, by dividing it by a common number, say, 2. After dividing the numerator and denominator by 2 and we will get (6 ÷ 2)/ (8 ÷ 2) = 3/4. Therefore, 6/8 and 3/4 are equivalent fractions.
## How to multiply the numerator and the denominator?
Multiply the numerator and the denominator by the same number.
## What is the LCM of 2/6 and 3/9?
The denominators of the fractions, 2/6 and 3/9 are 6 and 9. The LCM of the denominators 6 and 9 is 18. Let us make the denominators of both fractions 18, by multiplying them with suitable numbers.
## Is a decimal the same as a fraction?
The decimal values of both the fractions are the same and hence, they are equivalent.
## Do equivalent fractions get reduced to the same fraction?
All equivalent fractions get reduced to the same fraction in their simplest form as seen in the above example. Explore the given lesson to get a better idea of how to find equivalent fractions and how to check if the given fractions are equivalent.
## What is 1/5 as a percentage?
Now we can see that our fraction is 20/100, which means that 1/5 as a percentage is 20%.
## What is 50% in fraction form?
“Percent” means per hundred, and so 50% is the same as saying 50/100 or 5/10 in fraction form.
## What is the equivalent of (3+7)+2?
The expression equivalent to (3+7)+2 is 12.
## What is equivalent expression calculator?
Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds.
## What is the equivalent fraction of 2/3?
For example, if we multiply the numerator and denominator of 2/3 by 4 we get. 2/3 = 2×4 / 3×4 = 8/12 which is an equivalent fraction of 2/3.
## Is 3y+3 true?
True, because when the numbers are in the parethasis that means you multiply what is outside of the parenthasis like the three by the numbers inside of the parenthasis so 3 x y and 3 x 1, therefore 3y+3 and 3 (y+1) is true.
## Is 3y+3 a simplified expression?
The expressions 3y+3 and 3 (y+1) are equivalent expressions. Because 3 (y+1) can be simplified as 3y+3.
## How to find equivalent ratios?
As we previously mentioned, Equivalent Ratios are two ratios that express the same relationship between numbers. The Equivalent Ratio Calculator provides a table of equivalent ratios that have the same relationship between each other and directly with the ratio you enter into the calculator. We will look at how to calculate equivalent ratios shortly, first lets look at how to use the free online equivalent ratio calculator: 1 Enter a Ratio into the equivalent ratio calculator, for example, you could enter 7:25 2 Select the number of equivalent ratios that you would like to see in the table of results 3 The equivalent ratio calculator will calculate as you type and produce a lis of equivalent ratios in a table below the calculator 4 [Optional] Print or email the Table of Equivalent Ratios for later use
## What is a ratio?
A ratio is a direct comparison of one number against another. A ratio calculator looks to define the relationship that compares between those two numbers
## Is there a formula for equivalent ratios?
As equivalent ratios have the same value there is technically no equivalent ratio formula but the following equivalent ratio formula will help you with the manual math calculations.
## How many values are in a percentage?
Although the percentage formula can be written in different forms, it is essentially an algebraic equation involving three values.
## How to find the percentage difference between two numbers?
The percentage difference between two values is calculated by dividing the absolute value of the difference between two numbers by the average of those two numbers. Multiplying the result by 100 will yield the solution in percent, rather than decimal form. Refer to the equation below for clarification.
## How to calculate percentage increase and decrease?
Percentage increase and decrease are calculated by computing the difference between two values and comparing that difference to the initial value. Mathematically, this involves using the absolute value of the difference between two values, and dividing the result by the initial value, essentially calculating how much the initial value has changed.
## What is 1.5 GPA?
1.5 GPA is equivalent to 70% on percentile scale. 1.5 GPA is considered as ‘C’ grade. Grade point average famously known as GPA is the calculated average of grades you earn during a stated period of time, it could be a term, semester or session.
## Is a 1.5 GPA good?
The answer is No . The national average for a GPA is around 3.0 and a 1.5 GPA puts you below that average. A 1.5 GPA means that you’ve gotten only C-s and D+s in your high school classes so far. Since this GPA is significantly below a 2.0, it will make things very difficult for you in the college application process. Remember that the 3.0 national average represents all students, not just students applying to college, so the average GPA for students applying to colleges is naturally higher than the national average.
## What does 1.0 mean in FTE?
For example, 1.0 is typically the FTE representation of an individual’s full work/school day, while 0.5 would indicate half of the original figure, generally in reference to the workday. In reference to an individual, 0.5 usually refers to the fact that the worker renders less than a full day of work or the student attends less than a full day of classes.
## What is a full time equivalent?
What is Full Time Equivalent (FTE)? Full Time Equivalent (FTE) refers to the unit of measurement equivalent to an individual – worker or student – one unit of a work or school day , applicable in a variety of contexts. In most cases, full time equivalents measure an employee. or student and/or their workload.<|endoftext|>
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In mathematics, a triangular prism is a three-dimensional solid shape with two identical ends connected by equal parallel lines, and have 5 faces, 9 edges, and 6 vertices.
where “b” is the length of the base, “h” is the height of the triangle, “s1, s2, s3” are the respective length of each side of the triangle, and H is the height of the prism (which is also the length of the rectangle).
Given the base, the height of the triangle, height of prism and the length of each side of triangle base and the task is to calculate the surface area of the triangular prism.
Input: b = 3, h = 4, s1 = 3, s2 = 6, s3 = 6, Ht = 8
Output: The area of triangular prism is 132.000000
Input: b = 2, h = 3, s1 = 4, s2 = 5, s3 = 6, Ht = 8
Output: The area of triangular prism is 126.000000
Formula for calculating the surface area:
As stated above, the prism contains two triangles of the area (1/2)*(b)*(h) and three rectangles of the area H*s1, H*s2 and H*s3.
Now after adding all the terms we get the total surface area:
SA = b * h + (s1 + s2 + s3 ) * H
The area of triangular prism is : 132.000000
- Program to find volume and surface area of pentagonal prism
- Calculate Volume, Curved Surface Area and Total Surface Area Of Cylinder
- Surface Area and Volume of Hexagonal Prism
- Program to calculate Volume and Surface area of Hemisphere
- Program to find the Volume of a Triangular Prism
- Calculate Volume and Surface area Of Sphere
- Calculate volume and surface area of Torus
- Calculate volume and surface area of a cone
- Program for Surface Area of Octahedron
- Program for Volume and Surface Area of Cube
- Program for Volume and Surface Area of Cuboid
- Program to find the surface area of the square pyramid
- Program for Volume and Surface area of Frustum of Cone
- Program to calculate Area Of Octagon
- Program to calculate area of Enneagon
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.<|endoftext|>
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General Robert E. Lee’s Army of Northern Virginia fought the Union’s Army of the Potomac, which was commanded by Major General Joseph Hooker, at the Battle of Chancellorsville in Virginia between April 30 and May 6, 1863. The Battle of Chancellorsville is known as one of General Lee’s greatest victories of the American Civil War because, despite being massively outnumbered, Lee forced Hooker’s army to retreat. But General Lee suffered a significant loss during the battle when one of his most brilliant strategists, General Thomas Jonathan “Stonewall” Jackson, was shot and died shortly after.
General “Stonewall” Jackson was shot in his left arm and right hand at the Battle of Chancellorsville on May 2, 1863. Although the bullets were removed and his left arm was amputated immediately, he died eight days later. Investigators studied the projectiles that were removed from Jackson’s body in order to verify who fired the lethal shots. This case is significant because it is one of the earliest applications of ballistic principles in a death investigation.
Thomas Jonathan Jackson (January 21, 1824-May 10, 1863) graduated from West Point in 1846 and fought with the U.S. Army in the Mexican American War. When the American Civil War started in 1861, he trained new recruits for the Confederate Army and was given command of an infantry regiment. “Stonewall” Jackson earned his famous moniker at the First Battle of Bull Run in July of 1861 when the then brigadier general charged his troops forward to reinforce the Confederate defensive line against a Union attack “like a stone wall.” Jackson is probably most famous for his Shenandoah Valley Campaign in the spring of 1862 during which he marched his troops more than 650 miles in order to keep the Union Army from attacking the Confederate capital in Richmond, VA.
From April 30 to May 6, 1863, Jackson fought under the command of General Robert E. Lee at the Battle of Chancellorsville. As the fighting started to dwindle at sundown on May 2nd, Jackson and his aids returned to the Confederate lines. As Lee and his men approached, a North Carolina regiment mistook them for Union troops and fired on them. Jackson was hit in his left arm right below the shoulder and in his right hand.1
Dr. Hunter McGuire, the chief surgeon of the Army of Northern Virginia, amputated Stonewall Jackson’s left arm in an attempt to save his life. But Jackson died anyway eight days later on May 10, 1863 and McGuire attributed his death to complications from pneumonia.1
The bullets in Jackson’s left arm and right hand were removed during surgery and were later examined during an investigation into his death to confirm it was the result of friendly fire. It was noteworthy that the projectiles were round balls fired from a smoothbore musket because there were two categories of muskets used during the Civil War, those with smoothbore barrels and those with rifled barrels. 1,2
A smoothbore weapon does not have rifled grooves etched into the barrel. Smoothbore muskets had a shorter range and were less accurate than their rifled counterparts. Rifling, or spiral grooves etched into the walls of the inside of the barrels, was invented in the 16th century but was not part of widespread gun production until the 19th century. This was an important innovation in firearms manufacture because when a bullet passed through the barrel the rifling caused the bullet to spin, which made the shot more stable and accurate. By the time Jackson was shot, most of the muskets used by the Confederate Army were smoothbore and most rifles used by the Union army had rifled barrels.
Ballistic analysis eliminated Union troops as the source of the shot and confirmed witness accounts that Jackson was injured by and eventually killed by his own men.
Most of Stonewall Jackson’s body was sent to his family in Lexington, VA where it was buried. His left arm, however, was buried in a private cemetery at Ellwood Manor, not far from the Chancellorsville battlefield. According to Ramona Martinez in “The Curious Fate Of Stonewall Jackson’s Arm,” Union solders exhumed the limb in 1864 and reburied it in an unknown location. Then in 1903, one of Jackson’s officers erected a granite headstone at what was thought to be the appendage’s final resting place but it’s “unclear” if it marks the exact spot or if it is buried nearby.
- Smith, B.C. (1975). “The Last Illness and Death of General Thomas Jonathan (Stonewall) Jackson.” VMI Alumni Review.
- Albin, M.S. (2001). “The Wounding, Amputation and Death of Thomas Jonathan “Stonewall” Jackson Some Medical and Historical Insights.” Bulletin of Anesthesia History, Vol. 19, Number 4.
- Martinez, R. (2012). The Curious Fate Of Stonewall Jackson’s Arm. Retrieved from: http://www.npr.org/2012/06/28/155804965/the-curious-fate-of-stonewall-jacksons-arm<|endoftext|>
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Gödel numbers are constructed as follows. First we identify the elements we are to represent, and then we assign a natural number to each one. Finally, we assign a natural number to each possible sequence of the elements; these last numbers are our Gödel numbers.
### Elements
A system P is made up of two main types of elements:
##### Primitive signs
These represent all the operations that we can perform in the system; more complex operations can be constructed from them, as from the Peano postulates. (Not all these symbols show up in all browsers, so I'll explain in words after each usage.)
##### Variables
These are classed as:
### Numbering elements
To number the primitive signs, they are assigned the first 7 odd numbers:
• "0" (zero) = 1
• "succ" (successor) = 3
• "¬" (not) = 5
• "⊦" (logical or) = 7
• "∀" (for all) = 9
• "(" (left parenthesis) = 11
• ")" (right parenthesis) = 13
Each variable of type n is uniquely assigned a number of the form pn where p is a prime greater than 13; each variable of a given type will have a different value of p.
This allows us to represent every sequence of basic signs as a sequence of natural numbers.
### Numbering sequences
Finally, given the sequence (n1, n2, ... nk), which can be mapped to a given sequence of signs, we seek to represent this sequence as a single number. We do this by mapping (n1, n2, ... nk) to 2n1.3n2 ... pknk, where pk is the kth prime. This product of powers of primes is the Gödel number of the expression.
To obtain the components from a given Gödel number, we need simply to represent it as a product of prime factors and decode in the opposite sequence to the above; there is only one possible prime factorization for any natural number, and hence each Gödel number represents a unique expression.
### Example
The expression "succ(n)", where n is a variable of type 1, can be expressed by the sequence (1,17), where "succ" is represented by 1, and "n" is represented by 17 to the power of 1. We convert this to a single number by calculating 21.317 = 258280326. This rather large number 258280326 is the Gödel number of "succ(n)"; however it is not usually necessary to calculate a Gödel number, but just to know that one can be calculated.
### Reference
Kurt Gödel. "On Formally Undecidable Propositions of Principia Mathematica and Related Systems". [www.research.ibm.com/people/h/hirzel/papers/canon00-goedel.pdf]. 1931. Section 2.2.<|endoftext|>
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# Difference between revisions of "Roots of unity"
This is an AoPSWiki Word of the Week for June 20-26
The Roots of unity are a topic closely related to trigonometry. Roots of unity come up when we examine the complex roots of the polynomial $x^n=1$.
## Solving the equation
First, we note that since we have an nth degree polynomial, there will be n complex roots.
Now, we can convert everything to polar by letting $x = re^{i\theta}$, and noting that $1 = e^{2\pi ik}$ for $k\in \mathbb{Z}$, to get $r^ne^{ni\theta} = e^{2\pi ik}$. The magnitude of the RHS is 1, making $r^n=1\Rightarrow r=1$ (magnitude is always expressed as a positive real number). This leaves us with $e^{ni\theta} = e^{2\pi ik}$.
Taking the natural logarithm of both sides gives us $ni\theta = 2\pi ik$. Solving this gives $\theta=\frac{2\pi k}n$. Additionally, we note that for each of $k=0,1,2,\ldots,n-1$ we get a distinct value for $\theta$, but once we get to $k > n-1$, we start getting coterminal angles.
Thus, the solutions to $x^n=1$ are given by $x = e^{2\pi k i/n}$ for $k=0,1,2,\ldots,n-1$. We could also express this in trig form as $x=\cos\left(\frac{2\pi k}n\right) + i\sin\left(\frac{2\pi k}n\right) = \mathrm{cis }\left(\frac{2\pi k}n\right).$
## Geometry of the roots of unity
All of the roots of unity lie on the unit circle in the complex plane. This can be seen by considering the magnitudes of both sides of the equation $x^n = 1$. If we let $x = re^{i\theta}$, we see that $r^n = 1$, since the magnitude of the RHS of $x^n=1$ is 1, and for two complex numbers to be equal, both their magnitudes and arguments must be equivalent.
Additionally, we can see that when the nth roots of unity are connected in order (more technically, we would call this their convex hull), they form a regular n-sided polygon. This becomes even more evident when we look at the arguments of the roots of unity.
## Properties of roots of unity
Listed below is a quick summary of important properties of roots of unity.
• They occupy the vertices of a regular n-gon in the complex plane.
• For $n>1$, the sum of the nth roots of unity is 0. More generally, if $\zeta$ is a primitive nth root of unity (i.e. $\zeta^m\neq 1$ for $1\le m\le n-1$), then $\sum_{k=0}^{n-1} \zeta^{km}=\begin{cases} n & {n\mid m}, \\ 0 & \mathrm{otherwise.}\end{cases}$
• If $\zeta$ is a primitive nth root of unity, then the roots of unity can be expressed as $1, \zeta, \zeta^2,\ldots,\zeta^{n-1}$.
• Also, don't overlook the most obvious property of all! For each $n$th root of unity, $\zeta$, we have that $\zeta^n=1$
## Uses of roots of unity
Roots of unity show up in many surprising places. Here, we list a few:<|endoftext|>
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Support Force (Normal Force)
When standing on the ground is pulling you down, but you aren’t falling. In fact you are in so the ground must be providing a supporting that balances your weight. The ground provides that force in response to caused by your weight. When solid objects push back against forces that are deforming them we call that responsive push-back the .
Push your finger down into your palm and feel the resistance from your palm.
That’s the .
The normal force is a , meaning it only exists in response to a push from another object. When you pull your finger away from your palm, the normal force from your palm goes away.
In the diagram below, we see a person placing a bag of dog food on a table. When the bag of dog food is placed on the table, and the person lets go, how does the table exert the necessary to balance the of the bag? While you wouldn’t see it with your naked eye, the table sags slightly under the load (weight of the bag). This would be noticeable if the load were placed on a thin plywood table, but even a sturdy oak table deforms when a force is applied to it. That resistance to deformation causes a much like a deformed spring (or a trampoline or diving board). When the load is placed on the table, the table sags until the restoring force becomes as large as the weight of the load, putting the load in . The table sags quickly and the sag is slight, so we do not notice it, but it is similar to the sagging of a trampoline or a hammock when you climb on.
If you place an object on a table the from the table supports the of the object. For this reason normal force is sometimes called support force. However, normal is another word for , so we will stick with normal force because it reminds us of the important fact that the normal force always acts at an angle of 90° to the surface. That does not mean the normal force always point vertically, nor is it always equal to an object’s weight. If you push horizontally on the wall, the wall pushes back (keeping your hand from moving through the wall). The force from the wall is a normal force, but it acts horizontally and is not equal to your weight.
In each situation pictured above the is not equal to body . In the left image the normal force is less than body weight, and acting horizontally. In the middle image the normal force is less than body weight and acting at an angle. In the right image the normal force on the drill is more than it’s own weight because Master Sgt. Sangster is also pushing down on the drill. The normal force on Master Sgt. Sangster’s feet is less than her weight because she is also receiving an upward normal force from the drill handle.
Often (N) is used as a symbol for normal force, but we are using N to abbreviate for the force unit , so instead we will use . The normal force comes up so often students often accidentally begin to refer to normal force as “natural force” instead, so watch out for that possible source of confusion.
- OpenStax University Physics, University Physics Volume 1. OpenStax CNX. Jul 11, 2018 http://[email protected] ↵
- "Garscon Plancher" by Obiwancho , Wikimedia Commons is licensed under CC BY-SA 3.0 ↵
- "U.S. Air Force Chief Master Sgt. Suzan Sangster", Wikimedia Commons is in the Public Domain, ↵
- "Trek on the Viedma Glacier" by Liam Quinn , Wikimedia Commons is licensed under CC BY-SA 2.0 ↵
attraction between two objects due to their mass as described by Newton's Universal Law of Gravitation
the state being in equilibrium (no unbalanced forces or torques) and also having no motion
any interaction that causes objects with mass to change speed and/or direction of motion, except when balanced by other forces. We experience forces as pushes and pulls.
reduction in size caused by application of compressive forces (opposing forces applied inward to the object).
the outward force supplied by an object in response to being compressed from opposite directions, typically in reference to solid objects.
a type of force supplied by an object in response to application of a different force on the object. Friction is a reactive force
the force of gravity on on object, typically in reference to the force of gravity caused by Earth or another celestial body
a force that tends to move a system back toward the equilibrium position
a state of having no unbalanced forces or torques
at an angle of 90° to a given line, plane, or surface
a system of physical units ( SI units ) based on the meter, kilogram, second, ampere, kelvin, candela, and mole<|endoftext|>
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Scientists found three new species of fish in one of the deepest parts of the ocean, and these animals are so soft and squishy that they disintegrate if brought to the surface. Researchers captured remarkable footage that shows the fish in their alien home environment.
The new species, all snailfish, are adapted for life in ultradeep water, where temperatures are intensely cold and pressures far higher than any human could survive. Scientists filmed the fish in their natural environment as part of an international expedition to remotely explore the Atacama Trench, off the coast of Peru, and the discovery will be presented at the ongoing Challenger Conference at Newcastle University in the United Kingdom.
"As the footage clearly shows, there are lots of invertebrate prey down there, and the snailfish are the top predator. They seem to be quite active and look very well-fed," Thomas Linley, a researcher at Newcastle University who was involved in the expedition, said in a statement. [In Photos: Spooky Deep-Sea Creatures]
In the conditions present about 4.7 miles (7.5 kilometers) below the ocean surface, a squishy body is helpful in withstanding cold and extreme pressures, Linley said. So, the hardest objects in the snailfishes' bodies are their teeth and the bones in their inner ears, and the creatures have only minimal structural body parts.
"Without the extreme pressure and cold to support their bodies, they are extremely fragile and melt rapidly when brought to the surface," Linley said.
The researchers termed the three species the pink, purple and blue Atacama snailfish. The scientists did manage to remotely trap one specimen using one of the team's deep-sea probes after it followed some prey into its chamber. That specimen didn't survive the trip to the surface, but researchers have preserved its remains and, according the statement, it's in "very good condition" for study.
The researchers also filmed a rare "munnopsid" using their underwater probe. A sort of crustacean about the size of a human hand, a munnopsid swims upside down and somersaults to walk on its long legs.
Originally published on Live Science.<|endoftext|>
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# Binomial Probability Probability Simple Theoretical Probability Experimental Probability
• Slides: 15
Binomial Probability
Probability • • • Simple Theoretical Probability Experimental Probability Independent Probability Dependent Probability Conditional Probability
Binomial Probability • Two mutually exclusive outcomes sucesses (p) failures (1 -p) • Sometimes called a Bernoulli trial • Trials must be independent
Binomial Probability • Must calculate and multiply 3 separate factors • • • The number of ways to select success The probability of failure
Formula There are several different notations for Binomial Probability. Each one will give the correct answer. It depends upon which source you use as to which notation is used.
Formula Notation #1 • The probability of achieving exactly k successes in n trials is shown as P(k successes in n trials)= n = number of trials k = number of successes n – k = number of failures p = probability of success in one trial q = 1 – p = probability of failure in one trial
Formula Notation #2 • The probability of an event, p, occurring exactly r times: n = number of trials r = number of specific events you wish to obtain p = probability that the event will occur q = probability that the event will not occur (q = 1 - p, the complement of the event)
Example If you flip a coin 11 times, what is the probability of getting exactly 2 heads. 3 factors to calculate = ways to choose success Sucess = heads (2) and Probability of heads = ½ Failure = tails (9) and Probability of tails= ½
• Calculate Notice that the exponents add to 11. Notice that out of 11 trials we are looking for 2 successes- therefore 9 failures.
Binomial Example 2 • A large lot of manufactured items contains 10% defectives. In a random sample of 6 items, what is the probability that exactly 2 are defective? • What are the 3 factors to calculate? Number of ways for defective = success- 2 and probability of success=. 1 Failure- 4 and probability of failure =. 9
Binomial Example 3 Sixty percent of the workers in a plant belong to a union. A random sample of 12 is chosen. Find the probability that exactly 4 belong to a union. Name the 3 factors to calculate Ways to choose success (belonging to the union) How many successes? Probability of success? How many failures? Probability of failure?
Helpful Information Decimals are easier than fractions with these large calculations Pascal’s Triangle may be used in place of the combination formula Make sure that you are defining your successes and failures by the information in the problem- not by your own definition of success and failure
More to Come We will be working on binomial distributions next week when we learn about making dotplots and discussing sample distribution.<|endoftext|>
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Dive deeper into higher grammar concepts and composing exercises featured in this work-text. Introduce your 8th grader to noun clauses, gerunds, and infinitives, while reviewing the 8 parts of speech plus diagramming. With multiply exercises, including 67 original sentence-writing exercises, help your child develop an excellent grasp of grammar. In addition, 22 composition practices are included to prepare your 8th grader with his research paper on a great American. With a condensed handbook provided in this book, your students will have all the tools needed to improve their overall oral and verbal communication skills.<|endoftext|>
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COMMON FACTORS
1 / 21
# COMMON FACTORS - PowerPoint PPT Presentation
COMMON FACTORS. ax + ay = a(x + y). Think Distributive property backwards. Work down, and show all steps!. Problem 1. 3x + 3y. = 3(x + y). Problem 2. 5x - 25. =5(x – 5). Problem 3. ax - ay. = a(x – y). Problem 4. 2x 2 - 10x. =2x(x- 5). Problem 5. x 2 y - 3x 2.
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## PowerPoint Slideshow about ' COMMON FACTORS' - agostino-kearns
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Presentation Transcript
COMMON FACTORS
ax + ay = a(x + y)
Think Distributive property backwards
Work down,
and show all steps!
Problem 1
3x + 3y
= 3(x + y)
Problem 2
5x - 25
=5(x – 5)
Problem 3
ax - ay
= a(x – y)
Problem 4
2x2 - 10x
=2x(x- 5)
Problem 5
x2y - 3x2
=x2(y – 3)
Problem 6
x2z + y2z2
=z(x2 + y2z)
Problem 7
8x - 16y
= 8(x - 2y)
Problem 8
4a + 20b
= 4(a + 5b)
Problem 9
3x - 6y + 12
=3(x - 2y + 4)
Problem 10
5 + 15n + 45m
= 5(1 + 3n + 9m)
Problem 11
13a2 - 169a
=13a(a – 13)
Problem 12
8x - 56x3
= 8x(1 - 7x2)
Problem 13
14x2 + 35x4
=7x2(2 + 5x2)
Problem 14
y3 - 3y2 + 17y4
= y2(y - 3 + 17y2)
Problem 15
3x3 + 3x2 + 6x
= 3x(x2 + x + 2)
Problem 16
x3 + 3x2y + 3xy
= x( x2 + 3xy + 3y)
Problem 17
4a4b - 16a2b2 + 4ab4
= 4ab(a3 - 4ab + b3)
Problem 18
15x2y2+ 225x3y3 + 15x4y4
=15x2y2(1 + 15xy + x2y2)
Problem 19
15x3 +24x2 + 36x
=3x( 5x2 + 8x + 12)
Problem 20
a3y3 + a2y2 + ay
= ay(a2y2 + ay +1)<|endoftext|>
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|Other names||Carbon dioxide solution|
|Molar mass||62.03 g/mol|
|Solubility in water||exists only in solution|
|Acidity (pKa)||6.36 (see text)
|Except where noted otherwise, data are given for
materials in their standard state
(at 25 °C, 100 kPa)
Carbonic acid (ancient name acid of air or aerial acid) is a weak acid with the formula H2CO3. It is formed in small amounts when carbon dioxide is dissolved in water, and it is usually found only in solution. The salts of carbonic acids are called bicarbonates (or hydrogen carbonates, when the anion is HCO3−) and carbonates (when the anion is CO32−).
Carbon dioxide dissolved in water is in equilibrium with carbonic acid:
The equilibrium constant at 25°C is Kh= 1.70×10−3, which indicates that most of the carbon dioxide is not converted into carbonic acid and stays as CO2 molecules. In the absence of a catalyst, the equilibrium is reached quite slowly.
The rate constants are 0.039 s−1 for the forward reaction (CO2 + H2O → H2CO3) and 23 s−1 for the reverse reaction (H2CO3 → CO2 + H2O).
Carbonic acid has two acidic hydrogens and so two dissociation constants:
Care must be taken when quoting and using the first dissociation constant of carbonic acid. The value quoted above is correct for the H2CO3 molecule, and shows that it is a stronger acid than acetic acid or formic acid: this might be expected from the influence of the electronegative oxygen substituent. However, carbonic acid only ever exists in solution in equilibrium with carbon dioxide, and so the concentration of H2CO3 is much lower than the concentration of CO2, reducing the measured acidity. The equation may be rewritten as follows (c.f. sulfurous acid):
This figure is quoted as the dissociation constant of carbonic acid, although this is ambiguous: it might better be referred to as the acidity constant of carbon dioxide, as it is particularly useful for calculating the pH of CO2 solutions.
Carbonic acid plays a very important role in mammalian blood. It is an intermediate during the transfer of carbon dioxide from the lungs to the blood and vice versa. The conversion of carbon dioxide into carbonic acid is catalyzed by an enzyme (carbonic anhydrase), which enhances the reaction rate by a factor of nearly a billion. Carbonic acid dissociates in the blood (as in other solutions), to produce mainly H+ and HCO3- (bicarbonate) ions. This dissociation is an equilibrium reaction and it helps control the pH level of blood.
For a long time, researchers found it impossible to obtain pure hydrogen bicarbonate (H2CO3) at room temperature (about 20 °C, or about 70 °F). However, in 1991, scientists at NASA's Goddard Space Flight Center (USA) succeeded in making the first pure H2CO3 samples. They did so by exposing a frozen mixture of water and carbon dioxide to high-energy radiation, and then warming to remove the excess water. The carbonic acid that remained was characterized by infrared spectroscopy.
The ability to produce carbonic acid by irradiating a solid mixture of H2O and CO2 has led to suggestions that H2CO3 might be found in outer space, where frozen ices of H2O and CO2 are common, as are cosmic rays and ultraviolet light, to help them react.
The same carbonic acid polymorph (denoted beta-carbonic acid) was prepared by a cryotechnique at the University of Innsbruck: alternating layers of glassy aqueous solutions of bicarbonate and acid were heated in vacuo, which causes protonation of bicarbonate, and the solvent was subsequently removed. A second polymorph (denoted alpha-carbonic acid) was prepared by the same technique at the University of Innsbruck using methanol rather than water as a solvent.
The researchers at Innsbruck reported that pure, water-free carbonic acid is highly stable in the gas phase, with a calculated half-life of 180,000 years. Yet, according to their calculations, the presence of a single molecule of water causes a molecule of carbonic acid to quickly decompose to carbon dioxide and water.
At a given temperature, the composition of a pure carbonic acid solution (or of a pure CO2 solution) is completely determined by the partial pressure of carbon dioxide above the solution. To calculate this composition, account must be taken of the above equilibria between the three different carbonate forms (H2CO3, HCO3− and CO32−) as well as of the equilibrium between dissolved CO2 and H2CO3 with constant (see above) and of the following equilibrium between the dissolved CO2 and the gaseous CO2 above the solution:
The corresponding equilibrium equations together with the relation and the neutrality condition result in six equations for the six unknowns [CO2], [H2CO3], [H+], [OH−], [HCO3−] and [CO32−], showing that the composition of the solution is fully determined by . The equation obtained for [H+] is a cubic whose numerical solution yields the following values for the pH and the different species concentrations:
|(atm)||pH||[CO2] (mol/L)||[H2CO3] (mol/L)||[HCO3−] (mol/L)||[CO32−] (mol/L)|
|10−8||7.00||3.36 × 10-10||5.71 × 10−13||1.42 × 10−9||7.90 × 10−13|
|10−6||6.81||3.36 × 10−8||5.71 × 10−11||9.16 × 10−8||3.30 × 10−11|
|10−4||5.92||3.36 × 10−6||5.71 × 10−9||1.19 × 10−6||5.57 × 10−11|
|3.5 × 10−4||5.65||1.18 × 10−5||2.00 × 10−8||2.23 × 10−6||5.60 × 10−11|
|10−3||5.42||3.36 × 10−5||5.71 × 10−8||3.78 × 10−6||5.61 × 10−11|
|10−2||4.92||3.36 × 10−4||5.71 × 10−7||1.19 × 10−5||5.61 × 10−11|
|10−1||4.42||3.36 × 10−3||5.71 × 10−6||3.78 × 10−5||5.61 × 10−11|
|1||3.92||3.36 × 10−2||5.71 × 10−5||1.20 × 10−4||5.61 × 10−11|
|2.5||3.72||8.40 × 10−2||1.43 × 10−4||1.89 × 10−4||5.61 × 10−11|
|10||3.42||0.336||5.71 × 10−4||3.78 × 10−4||5.61 × 10−11|
As noted above, [CO32−] may be neglected for this specific problem, resulting in the following very precise analytical expression for [H+]:
All links retrieved January 10, 2017.
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History of Astronomy From the Roman Empire to the Present, Part 4
Before passing on to the more important part of this work, it is only just to record the fact that the first practical work in triangulation since the time of Hipparchus was performed by Jean Picard and J. and D. Cassini, between Paris and Dunkirk toward the end of the 17th century; when Newton was working out his theories.
At this time the Copernican theory of astronomy was well established, and was accepted by all the scientific world, though it is probable that the public in general found it difficult to reconcile the idea of an earth careering through space at prodigious speed with common sense and reason. Even the most ardent followers of Copernicus and Galileo recognised this difficulty, and some strove to find a satisfactory explanation.
Nearly a hundred years ago Kepler had suggested that some kind of unknown force must hold the earth and the heavenly bodies in their places, and now Sir Isaac Newton, the greatest mathematician of his age, took up the idea and built the Law of Gravitation.
The name is derived from the Latin word “ gravis,” which means “ heavy,” “having weight,” while the Law of Gravitation is defined as “That mutual action between masses of matter by virtue of which every such mass tends toward every other with a force varying directly as the product of the masses, and inversely as the square of their distances apart.”
Reduced to simplicity, gravitation is said to be “That which attracts every thing toward every other thing.” That does not tell us much; and yet the little it does tell us is not true ; for a thoughtful observer knows very well that every thing is not attracted towards every other thing. . . The definition implies that it is a force ; but it does not say so, for that phrase “ mutual action ” is ambiguous, and not at all convincing.
The Encyclopaedia Britannica tells us that “The Law of Gravitation is unique among the laws of nature, not only for its wide generality, taking the whole universe into its scope, but in the fact that, so far as is yet known, it is absolutely modified by any condition or cause whatever.”
Here again we observe that the nature of gravitation is not really defined at a 11; we’re are told that masses of matter tend toward each other, but no reason is given why they do so, or should do so; while to say that “it is absolutely unmodified by any condition or cause whatever” is one of the most unscientific statements it is possible to make. There is not any thing or force in the universe that is absolute! no thing that goes its own way and does what it will without regard to other forces or things.
The thing is impossible; and it is not true; wherefore it has fallen to me to show where the inconsistency in it lies.
The name given to this mutual action means “weight,” and weight is one of the attributes of all matter. Merely to say that anything is matter or material implies that it has weight, while to speak of weight implies matter. Matter and weight are inseparable, they are not laws, but elemental facts.
But it has been suggested that gravitation is a force, indeed we often hear it referred to as the force of gravitation; but force is quite a different thing than weight, it is active energy expressed by certain conditions and combinations of matter. It acts. All experience and observation goes to prove that material things fall to earth because they possess the attribute of weight, and that an object remains suspended in air or space only so long as its weight is overcome by a force, which is contrary. And when we realize these simple facts we see that gravitation is in reality conditioned and modified by every other active force, both great and small.
Again, gravitation is spoken of as a pull, an agent of attraction that robs weight of its meaning, something that brings all terrestrial things down to earth while at the same time it keeps the heavenly bodies in their places and prevents them falling toward each other or apart. The thing is altogether too wonderful, it is not natural; and the theory is scientifically unsound…
Every man, however great his genius, must be limited by the conditions that surround him; and science in general was not sufficiently advanced two hundred years ago to be much help to Newton, so that— for lack of information which is ordinary knowledge to us having in the 20th century— he fell into the error of attributing the effects of “weight” and “force” to a common cause, which— for want of a better term— he called gravitation; but I have not the slightest doubt that if he were living now he would have arrived at the following more reasonable conclusions: That terrestrial things fall to earth by “gravis,” weight; because they are matter; while the heavenly bodies (which also are matter) do not fall because they are maintained in their courses by magnetic or electric force.
Another figure of great prominence in the early part of the eighteenth century was Dr. Hailey, who survived Sir Isaac Newton by some fifteen years, and it is to him that we owe nearly all the methods of measuring distance which are used in astronomy at the present day. So far no one had seriously considered the possibility of measuring the distance to the sun planets or stars since Hipparchus had failed— away back in the second century B.C.— but now, since the science had made great strides, it occurred to Dr. Hailey that it might be possible at least to find the distance from the earth to the sun, or to the nearest planet.
Remembering the time-honoured dogma that the stars are infinitely distant, inspired by the magnificence of the Copernican conception of the universe, and influenced— no doubt— by the colossal suggestions of Ole Roemer, he tried to invent some means of making a triangulation on a gigantic scale, with a base-line of hitherto unknown dimensions.
Long years ago Kepler had worked out a theory of the distances of the planets with relation to each other, the principle of which— when expressed in simple language and in round figures— is as follows : “ If we knew the distance to any one of the planets we could use that measurement as a basis from which to estimate the others. Thus Venus is apparently about twice as far from the sun as Mercury, while the earth is about three times and Mars four times as far from the sun as Mercury, so that should the distance of the smallest planet be— let us say— 50 million miles, then Venus would be 100, the Earth 150, and Mars 200 millions of miles.”
This seems to be the simplest kind of arithmetic, but the whole of the theory of relative distance goes to pieces because Kepler had not the slightest idea of the linear distance from the earth to anything in the firmament, and based all his calculations on time, and on the apparent movements of the planets in azimuth, that is— to right or left of the observer, and to the right or left of the sun.
Necessity compels me to state these facts in this plain and almost brutal fashion, but it is my sincere hope that no reader will suppose that I under-estimate the genius or the worth of such men as Newton and Kepler; for it is probable that I appreciate and honour them more than do most of those who blindly worship them with less understanding. I only regret that they were too ready to accept Copernican astronomy as though it were an axiom, and did not
put it to the proof; and that, as a consequence, their fine intelligence and industry should have been devoted to the glorification of a blunder.
Kepler’s work was of that high order which only one man in a million could do, but nevertheless, his calculations of the relative distances of the planets depends entirely upon the question whether they revolve round the sun or not ; and that we shall discover in due course.
However, Dr. Hailey had these theories in mind when he proposed to measure the distance to Mars at a time when the planet reached its nearest point to earth (in opposition to the sun), and then to multiply that distance by three (approximate), and in that manner estimate the distance of the sun.
He proceeded then to invent what is now known as the “ Diurnal Method of Measurement by Parallax,” which he described in detail in the form of a lecture to contemporary astronomers, introducing it by remarking that he would probably not be living when next Mars came into the required position, but others might at that time put the method into practice.
He began by saying that “If it were possible to place two observers at points diametrically opposite to each other on the surface of the earth (as A and B in dia^am 5), both observers— looking along their respective horizons— would see Mars at the same time, the planet being between them, to the east of one observer and to the westward of the other. In these circumstances the diameter of the earth might be used as a base-line, the observers at A and B might take simultaneous observations, and the two angles obtained, on being referred to the base-line, would give the distance of the planet.”
But this was in the reign of George II, long before the invention of steamships, cables or telegraphs, and Dr. Hailey knew that it was practically impossible to have B taking observations in the middle of the Pacific Ocean, so he proposed to overcome the difficulty by the following expedient: He suggested that both the observations could be taken by a single observer, using the same observatory, thus— “ Let an observer at A take the first observation in the evening, when Mars will be to his east: let him then wait twelve hours, during which time the rotation of the earth will have carried him round to B. He may then take his second observation. Mars being at this time to his west, and the two angles thus obtained— on being referred to the base-line— will give the distance of the planet.”
This proposition is so plausible that it has apparently deceived every astronomer from that day to this, and it might even now deceive the reader himself were it not that he knows I have some good reason for describing it here. It is marvellously specious; it does not seem to call for our examination; and yet it is all wrong! and Dr. Hailey has a world of facts against him. He is at fault in his premises, for if the planet was visible to one of the observers it must be above his horizon, and, therefore, could not be seen at the same time by the other; since it could not be above his horizon also. (See diagram 5.)
Again, his premises are in conflict with Euclid, because he supposes Mars to be midway between A and B, that is between their two horizons, which are parallel lines 8,000 miles apart throughout their entire length, and so it is obvious that if the planet— much smaller than the earth— was really in that position it could not be seen by either of the observers.
The alternative which Dr. Hailey proposes is as fallacious as his premises, for he overlooks the fact that— according to Copernican astronomy— during the twelve hours while the earth has been rotating on its axis it has also travelled an immense distance in its orbit round the sun. The results are:
That an observer starting from arrive at B, but must arrive in time at a point somewhere about three-quarters of a million miles beyond it, as shown in diagram 6.
The observer loses his original base-line, which was the diameter of the earth, and does not know the length of his new one, A, G, because the distance of the sun and the dimensions of the orbit had never previously been measured. 3. The angle of view from G is entirely different from the one intended from B.
Mars itself has moved along its orbit during the twelve hours, to a new position which is very uncertain.
The triangulation which was intended is utterly lost, and the combined movements of the earth and Mars, plus the two lines of sight, make up a quadrilateral figure, which of course contains angles of 360 degrees, and by means of which no measurement whatever is possible.
In conclusion. Dr. Hailey was mistaken when he supposed that two observations made from a single station with an interval of twelve hours between them, were equivalent to two observations taken simultaneously by A and B…
The actual attempt to measure the distance to Mars by the use of this Diurnal Method will be dealt with in the proper order of events, but for the present— what more need I say concerning such ingenious expedients?
A curious example of theorising to no useful purpose is the “ Theory of the Aberration of Light,” which is regarded by some as one of the pillars of astronomy. It aims to show that if the velocity of the earth were known the velocity of light could be found, while at the same time it implies the reverse that if the velocity of light were known we could find at what speed the earth is travelling round the sun. If Bradley intended to prove anything by this theory it was that the apparent movement of the stars proves that the earth is in motion ; which surely is egging the question.
The fact that the theory of the Aberration of Light has no scientific value whatever is very well shown by
the following quotation from its author: “ If the observer be stationary at B (see dia. 7) the star will appear to be in the direction B, S; if, however, he traverses the line B A in the same time as light passes from the star to his eye the star will appear in the direction A. S.”
That is true, but it would be no less true if the star itself had moved to the right while the observer remained at B, but why did he say “if he moves from B to A in the same time as it takes light to pass from the star to his eye”? It is a needless qualification, for if the observer moves to A he will see the star at the same angle whether he walks there at three miles an hour or goes there by aeroplane at a mile a minute. It has nothing to do with the speed of light, and the velocity of light has nothing to do with the direction of the star, it is merely posing, using words to no purpose.<|endoftext|>
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# Elliptic vs. Hyperbolic Paraboloids: Definitions & Equations
Paraboloids are three-dimensional objects that are used in many science, engineering and architectural applications. In this lesson, we explore the elliptic paraboloid and the hyperbolic paraboloid.
## The ‘Oid’ in a Paraboloid
In the word asteroid ‘oid’ means like , put together with ‘aster’, or star, and you get star-like. Similarly a paraboloid is an object resembling a parabola, which will be explained in the next section. In this lesson we explore the two types of paraboloids: the elliptic paraboloid and the hyperbolic paraboloid.
## Looking at Elliptic Paraboloids
We have two words here: ellipse and parabola. The ellipse is a circle that has been stretched in one direction. The parabola is the curve that looks like the particular orientation of the capital letter U.
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Writing the equation for an ellipse, you see
Unlike a circle which has one radius, an ellipse has two: the ‘a’ and the ‘b’ in the equation. First, we will examine how to go from the equation of the ellipse, to the equation of an elliptic paraboloid. Check the following:
It’s the same equation as the ellipse except the 1 on the right-hand side is now a z. We are in three dimensions. In addition to x and y we have a z. For a moment, look at the left-hand side of the equation. Do you see only a positive result even for negative x and/or y? An option is to allow for the curve to open in the negative z direction. We do this by writing –z on the right-hand side. Here’s the plot of the elliptic paraboloid with the +z. Please note the figure has been cut off at some positive value of z. In reality, the elliptic paraboloid continues to infinity in the z direction.
‘Paraboloid’ means like a parabola. Let’s look at the elliptic paraboloid very carefully. Imagine you are at the arrow point on the x-axis and you are looking towards the origin along the x-axis. The origin is where the three axes cross. Do you see a parabola? Another strategy is to visualize a ‘slice’ of the elliptic paraboloid. The slice is parallel to the yz plane. The slice cuts the x-axis at some point. Is the following parabola what you see?
Sitting at the arrowhead of the y-axis and looking towards the origin gives us another view. The slice parallel to the xz plane will pass through the y-axis at some point. From this viewpoint, the x-axis is increasing from right to left, but there’s still another parabola from this view:
Did you notice this second parabola is wider? Does this make sense from the 3D plot of the elliptic paraboloid?
Imagine being in deep space and looking down at the orbit of an asteroid. We would see an ellipse. The same thing happens as we move to the top of the z-axis and look down from overhead. We will see an ellipse in the xy plane. Positive y is to the right, and positive x is down.
Just like the asteroid which burns as bright as a star when it enters the atmosphere, you are becoming the ‘rock star’ of paraboloids! There is one more paraboloid type to look at, the hyperbolic paraboloid.
## Looking at Hyperbolic Paraboloids
The equation for the hyperbola is a lot like that of an ellipse, except there’s a minus sign instead of the plus.
This type of hyperbola will have a right and left opening. To get the hyperbola to open up and down, we reverse the order of x and y. Just as before, to get the equation for the three-dimensional paraboloid, we replace the 1 on the right-hand side with a z. Replacing z with –z is an option although it has the same effect as reversing the order of the x and y terms.
Here’s a plot for the hyperbolic paraboloid, keeping in mind that the plot actually goes to infinity.
Once again we search for parabolas. First, we sight down the x-axis into the yz plane. A parabola appears:
The other parabola is in the xz plane when we take a ‘slice’ parallel to this plane and cut the y-axis:
Did you notice how this parabola is in the opposite direction? Does this make sense when you look at the 3D plot of the hyperbolic paraboloid?
The last view is from above. In a plane parallel to the xy plane we see a hyperbola:
And just like an asteroid, the view of these paraboloids can be pretty spectacular!
## Lesson Summary
Paraboloids are three-dimensional figures resembling parabolas. In this lesson we have explored the two types of paraboloids: the elliptic paraboloid and the hyperbolic paraboloid.
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Right at this second it’s September 2nd, with 2 days until the college football season starts, once again with too many teams in the Big Ten. In honor of all of these twos and toos, we present you a GMAT problem that features too many twos:
What is the value of 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8?
(A) 2^9
(B) 2^10
(C) 2^11
(D) 3(2^10)
(E) 3(2^11)
Afternoon Update:
Great solutions, everyone. This question brings up an important point about exponents – we only have a few “core competencies” when it comes to performing with algebra, and those are:
-Multiplying/dividing exponents with common bases
-Finding patterns (units digits, relationships between adding/subtracting common terms, etc.)
-Setting common bases equal to equate exponents
Outside of that, there’s very little that we can do without the use of a calculator. So, in order to take advantage of what we do well, we should find ways when we see exponents to:
-Find common bases
-Multiply (using factorization to turn addition/subtraction into multiplication)
Here, we’re asked to add several terms together…that’s not something that we do well with exponents. However, by blending our abilities to factor terms (to get to multiplication) and to see patterns, we can attack this question relatively efficiently:
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
Combine the 2s to be 4, or 2^2, and you have:
2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
Now we can add them together, and we have 2(2^2), or 2^3, simplifying the entire statement to:
2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
Notice that we’ll be able to combine two more terms, the two 2^3 terms, to be 2(2^3) or 2^4, leaving:
2^4 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
By now hopefully you’ve seen a pattern (patterns come up frequently in exponent questions) – the first two terms will add to the third, and then adding those will add to the fourth:
2^4 + 2^4 (the first two) = 2^5 (so now we have two of the third term):
2^5 + 2^5 + 2^6 + 2^7 + 2^8
Do that again and we’ll have:
2^6 + 2^6 + 2^7 + 2^8
If we repeat the pattern, we’ll end up with:
2^8 + 2^8 = 2(2^8) = 2^9. Therefore, the correct answer is A.
When approaching exponent problems, keep your core competencies in mind: factor, multiply, find common bases, and look for patterns. These strategies will help you turn complicated problems into efficient solutions.
Plan on taking the GMAT soon? We have GMAT prep courses starting around the world next week!. And, as always, be sure to find us on Facebook and follow us on Twitter!
### 2 Responses
1. Tom says:
2^n are pretty easy numbers to work with add up. To be honest, I looked at the numbers for a few seconds, couldn’t think of a slick method, and just added them up. Took about 30 seconds.
My paper looked like this:
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
Writing what each term is underneath (just multiply previous by 2), then adding up each term:
4 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
4 8 16 32 64 128 256
8 16 32 64 128 256 512
512 = 2^9
2. Vaishno says:
Or you can simply use Geometric progression. GP starts from the second term in our equation. S = 2+(2^8-1)/2-1 —–> 2^9-2.
Add the first term and you get – 2+2^9-2 and that gives you 2^9.
The assumption here is that you already know the GP formula.<|endoftext|>
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# In how many ways can 4 girls and 3 boys be seated in a row so that no two boys are together?
Question:
In how many ways can 4 girls and 3 boys be seated in a row so that no two boys are together?
Solution:
The seating arrangement would be like this: B G B G B G B G B So, 4 girls
can seat among the four places. Number of ways they can seat is $={ }^{4} \mathrm{P}_{4}=24$ Boys have to seat among the ' $B$ ' areas. So, there are 5 seats available for 3 boys. The number of ways the 3 boys can seat among the 5 places is $={ }^{5} \mathrm{P}_{3}=60$ Therefore, the total number of ways they can seat in this manner is $=(24 \times 60)=1440$.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Surface Area and Volume of Pyramids
## Surface area and volume of solids with a base and lateral faces that meet at a common vertex.
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Pyramids
What if you wanted to know the volume of an Egyptian Pyramid? The Khafre Pyramid is the second largest pyramid of the Ancient Egyptian Pyramids in Giza. It is a square pyramid with a base edge of 706 feet and an original height of 407.5 feet. What was the original volume of the Khafre Pyramid? After completing this Concept, you'll be able to answer this question.
### Watch This
CK-12 Foundation: Chapter11PyramidsA
### Guidance
A pyramid has one base and all the lateral faces meet at a common vertex. The edges between the lateral faces are lateral edges. The edges between the base and the lateral faces are called base edges. If we were to draw the height of the pyramid to the right, it would be off to the left side.
When a pyramid has a height that is directly in the center of the base, the pyramid is said to be regular. These pyramids have a regular polygon as the base. All regular pyramids also have a slant height that is the height of a lateral face. Because of the nature of regular pyramids, all slant heights are congruent. A non-regular pyramid does not have a slant height.
##### Surface Area
Using the slant height, which is usually labeled \begin{align*}l\end{align*}, the area of each triangular face is \begin{align*}A=\frac{1}{2} bl\end{align*}.
Surface Area of a Regular Pyramid: If \begin{align*}B\end{align*} is the area of the base and \begin{align*}P\end{align*} is the perimeter of the base and \begin{align*}l\end{align*} is the slant height, then \begin{align*}SA=B+\frac{1}{2} Pl\end{align*}.
If you ever forget this formula, use the net. Each triangular face is congruent, plus the area of the base. This way, you do not have to remember a formula, just a process, which is the same as finding the area of a prism.
##### Volume
Recall that the volume of a prism is \begin{align*}Bh\end{align*}, where \begin{align*}B\end{align*} is the area of the base. The volume of a pyramid is closely related to the volume of a prism with the same sized base.
###### Investigation: Finding the Volume of a Pyramid
Tools needed: pencil, paper, scissors, tape, ruler, dry rice or sand.
1. Make an open net (omit one base) of a cube, with 2 inch sides.
2. Cut out the net and tape up the sides to form an open cube.
3. Make an open net (no base) of a square pyramid, with lateral edges of 2.45 inches and base edges of 2 inches. This will make the overall height 2 inches.
4. Cut out the net and tape up the sides to form an open pyramid.
5. Fill the pyramid with dry rice. Then, dump the rice into the open cube. How many times do you have to repeat this to fill the cube?
Volume of a Pyramid: If \begin{align*}B\end{align*} is the area of the base and \begin{align*}h\end{align*} is the height, then the volume of a pyramid is \begin{align*}V=\frac{1}{3} Bh\end{align*}.
The investigation showed us that you would need to repeat this process three times to fill the cube. This means that the pyramid is one-third the volume of a prism with the same base.
#### Example A
Find the slant height of the square pyramid.
Notice that the slant height is the hypotenuse of a right triangle formed by the height and half the base length. Use the Pythagorean Theorem.
\begin{align*}8^2+24^2&=l^2\\ 64+576&=l^2\\ 640&=l^2\\ l&= \sqrt{640} = 8 \sqrt{10}\end{align*}
#### Example B
Find the surface area of the pyramid from Example A.
The surface area of the four triangular faces are \begin{align*}4\left ( \frac{1}{2} bl \right ) =2(16)\left ( 8 \sqrt{10} \right )=256 \sqrt{10}\end{align*}. To find the total surface area, we also need the area of the base, which is \begin{align*}16^2 = 256\end{align*}. The total surface area is \begin{align*}256 \sqrt{10}+256 \approx 1065.54\end{align*}.
#### Example C
Find the volume of the pyramid.
\begin{align*}V=\frac{1}{3} (12^2)12=576 \ units^3\end{align*}
Watch this video for help with the Examples above.
CK-12 Foundation: Chapter11PyramidsB
#### Concept Problem Revisited
The original volume of the pyramid is \begin{align*}\frac{1}{3} (706^2)(407.5)\approx 67,704,223.33 \ ft^3\end{align*}.
### Guided Practice
1. Find the area of the regular triangular pyramid.
2. If the lateral surface area of a square pyramid is \begin{align*}72 \ ft^2\end{align*} and the base edge is equal to the slant height, what is the length of the base edge?
3. Find the area of the regular hexagonal pyramid below.
4. Find the volume of the pyramid.
5. Find the volume of the pyramid.
6. A rectangular pyramid has a base area of \begin{align*}56 \ cm^2\end{align*} and a volume of \begin{align*}224 \ cm^3\end{align*}. What is the height of the pyramid?
1. The area of the base is \begin{align*}A=\frac{1}{4} s^2 \sqrt{3}\end{align*} because it is an equilateral triangle.
\begin{align*}B & =\frac{1}{4} 8^2 \sqrt{3}=16 \sqrt{3}\\ SA& = 16 \sqrt{3}+\frac{1}{2} (24)(18)=16 \sqrt{3}+216 \approx 243.71\end{align*}
2. In the formula for surface area, the lateral surface area is \begin{align*}\frac{1}{2} Pl\end{align*} or \begin{align*}\frac{1}{2} nbl\end{align*}. We know that \begin{align*}n = 4\end{align*} and \begin{align*}b = l\end{align*}. Let’s solve for \begin{align*}b\end{align*}.
\begin{align*}\frac{1}{2} nbl & = 72 \ ft^2\\ \frac{1}{2} (4) b^2 & = 72\\ 2b^2 & = 72\\ b^2 & =36\\ b & = 6\end{align*}
Therefore, the base edges are all 6 units and the slant height is also 6 units.
3. To find the area of the base, we need to find the apothem. If the base edges are 10 units, then the apothem is \begin{align*}5 \sqrt{3}\end{align*} for a regular hexagon. The area of the base is \begin{align*}\frac{1}{2} asn = \frac{1}{2} \left ( 5\sqrt{3} \right )(10)(6)=150 \sqrt{3}\end{align*}. The total surface area is:
\begin{align*}SA& = 150 \sqrt{3}+\frac{1}{2}(6)(10)(22)\\ & = 150 \sqrt{3}+660 \approx 919.81 \ units^2\end{align*}
4. In this example, we are given the slant height. For volume, we need the height, so we need to use the Pythagorean Theorem to find it.
\begin{align*}7^2+h^2&=25^2\\ h^2&=576\\ h&=24\end{align*}
Using the height, the volume is \begin{align*}\frac{1}{3} (14^2 )(24)=1568 \ units^3\end{align*}.
5. The base of this pyramid is a right triangle. So, the area of the base is \begin{align*}\frac{1}{2} (14)(8)=56 \ units^2\end{align*}.
\begin{align*}V=\frac{1}{3} (56)(17) \approx 317.33 \ units^3\end{align*}
6. The formula for the volume of a pyramid works for any pyramid, as long as you can find the area of the base.
\begin{align*}224&=56h\\ 4& = h\end{align*}
### Explore More
Fill in the blanks about the diagram below.
1. \begin{align*}x\end{align*} is the ___________.
2. The slant height is ________.
3. \begin{align*}y\end{align*} is the ___________.
4. The height is ________.
5. The base is _______.
6. The base edge is ________.
Find the area of a lateral face and the volume of the regular pyramid. Leave your answer in simplest radical form.
Find the surface area and volume of the regular pyramids. Round your answers to 2 decimal places.
1. A regular tetrahedron has four equilateral triangles as its faces. Find the surface area of a regular tetrahedron with edge length of 6 units.
2. Using the formula for the area of an equilateral triangle, what is the surface area of a regular tetrahedron, with edge length \begin{align*}s\end{align*}?
For questions 13-15 consider a square with diagonal length \begin{align*}10\sqrt{2} \ in\end{align*}.
1. What is the length of a side of the square?
2. If this square is the base of a right pyramid with height 12, what is the slant height of the pyramid?
3. What is the surface area of the pyramid?
A regular tetrahedron has four equilateral triangles as its faces. Use the diagram to answer questions 16-19.
1. What is the area of the base of this regular tetrahedron?
2. What is the height of this figure? Be careful!
4. Challenge If the sides are length \begin{align*}s\end{align*}, what is the volume?
A regular octahedron has eight equilateral triangles as its faces. Use the diagram to answer questions 20-22.
1. Describe how you would find the volume of this figure.
3. Challenge If the sides are length \begin{align*}s\end{align*}, what is the volume?
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 11.5.
### My Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
lateral edges
Edges between the lateral faces of a prism.
lateral faces
The non-base faces of a prism.
Cone
A cone is a solid three-dimensional figure with a circular base and one vertex.
Pyramid
A pyramid is a three-dimensional object with a base that is a polygon and triangular faces that meet at one vertex.
Vertex
A vertex is a point of intersection of the lines or rays that form an angle.
Volume
Volume is the amount of space inside the bounds of a three-dimensional object.
Cavalieri's Principle
States that if two solids have the same height and the same cross-sectional area at every level, then they will have the same volume.
Base Edge
The base edge is the edge between the base and the lateral faces of a prism.
Slant Height
The slant height is the height of a lateral face of a pyramid.
Apothem
The apothem of a regular polygon is a perpendicular segment from the center point of the polygon to the midpoint of one of its sides.<|endoftext|>
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# 44540 as a roman numeral
Here you will see step by step solution to convert 44540 number to roman numeral. How to write 44540 as a roman numeral? 44540 as a roman numeral written as MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMDXL, please check the explanation that how to convert 44540 in roman number.
## Answer: 44540 in roman numeral
= MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMDXL
### How to convert 44540 in roman number?
To convert the 44540 to roman number simply expand the 44540 from hindu-arabic number to roman numerals, then replace the all numbers of expanded form with respective roman numerals. That's how simple it is to convert in roman numeral.
#### Solution for 44540 to roman numeral
Given number is => 44540
1. After expanding 44540, here we got -
= 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 500 + 40
2. Replace all numbers in expanded form with their roman numeral.
= M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + M + D + XL
3. After simplify the roman numeral.
= MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMDXL
Final conclusion is -
44540 = MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMDXL or MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMDXL = 44540
Hence, the 44540 is in roman numeral form written as MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMDXL or MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMDXL roman number written as 44540.<|endoftext|>
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# Prominent Steps of How to Solve Ratios With Useful Examples
A ratio is one of the parts of a mathematical word practiced to match the number of amounts to the amount of other numbers. This is usually practiced in both math and expert conditions. This post on how to solve ratios has illustrated a few of the daily instances when one can practice ratios:
• If one measures the winnings on a play
• While one distributes a pack of sweets honestly among their friends
• When a person changes its Dollars to Pounds or vice-versa while going on vacation
• If one works how many glasses of beer they require for a party
• During the calculation of how much cost they need to must pay on their income
Ratios can be normally utilized to connect two quantities, though individuals might also be utilized to analyze multiple measures. Besides this, ratios are often involved in numeric values’ reasoning tests, where people can be performed in several methods. That is why one is capable of recognizing and planning the ratios; however, individuals are manifested.
## Numerous methods to understand how to solve ratios
Ratios can normally be given as two or more numeric terms classified with a colon, for instance, 9:2 or 1:5 or 5:3:1. Though these might also be presented in many other methods, three examples are expressed differently.
### Scaling a ratio
Ratios are very useful in a number of ways, and the basic reason for this is that it allows us to range the quantity. It indicates rising or reducing the quantity of anything. This is unusually beneficial for something such as scale maps or models, where very high amounts can be changed to enough fewer illustrations, which are yet perfect.
Scaling is additionally essential for raising or reducing the number of components into a chemical reaction or recipe. Ratios might be estimated higher or lower by multiplying each section of the ratio with the equivalent product. This is the most useful point for how to solve ratios. Let’s take an instance:
George requires to cook pancakes for nine mates, but his recipe only produces sufficient pancakes for three mates. What amount of the elements will he require to utilize?
Pancake Ingredients (works 3)
• 300ml milk
• 100g flour
• 2 large eggs
To check how to solve ratios, one needs first to recognize the ratio. It has a three-part degree, whereby:
• 300ml milk
• 100g flour
• 2 large eggs
= 300:100:2
Besides this, one requires to work how much they require to balance the ingredients with.
As the recipe is for three individuals, but George wants a recipe for severe nine persons.
Because 9/3 = 3, the required ratio must be estimated by three (it is seldom represented with 3). Now, he requires to multiply every ingredient of the ratio with 3:
• 300 x 3 = 900
• 100 x 3 = 300
• 2 x 3 = 6
Hence, to get adequate pancakes for nine persons, George will require 900ml milk, 300g flour, and 6 eggs. This is one method for how to solve ratios; now, let’s move to the other two methods that are listed below.
### Reducing the ratios
Seldom ratio can not be shown in its most manageable structure that addresses it more difficult to manage. For instance, if a person has 6 hens, and all together lay 42 eggs each day. It can be interpreted as the ratio 6:42 (or given as a portion that will show: 6/42).
Decreasing a ratio implies changing the ratio into a standard form, making it more accessible to practice. This is executed by dividing each quantity of numbers into a ratio with the highest number that it can divide by. Let’s take an example of it:
Stella has 17 birds, and all eat 68kg of seed per week. Sam has 11 birds, and all eat 55kg seed per week. Find out who has the greediest birds?
To answer how to solve ratios, one should first recognize and analyze these two ratios:
• Stella’s ratio = 17:68, explain it by dividing each number with 17, which provides a ratio as 1:4
• Sam’s ratio = 11:55, analyze it by dividing each number with 11, which provides a ratio as 1:5
It implies that Stella’s birds eat 4kg seed per week, while Sam’s birds eat 5kg seed per week. Hence, Sam’s birds are greedier.
### Analyzing unknown values from existing ratios
This is another method that ratios are individually beneficial because this allows the learners to work for unknown and new measures depending on a known (existing) ratio. There are several methods for determining these kinds of problems. Initiate with using the cross-multiplication.
Mandeep and Gabriel are going to get married. Both have estimated that all require 40 glasses of wine for the 80 guests. At the moment, both get to know that another 10 guests are coming to attend their marriage. Find out how much wine do both require in total?
Initially, one requires to work on the ratio of the glass of wine with guests. They practiced = 40 wine:80 guests.
Then analyze it as 1 wine:2 guest (also we can say that 0.5 glass wine/guest).
Both have 90 guests who are going to attend (80 + the extra 10 = 90). Therefore, one requires multiplying 90 with 0.5 = 45 glasses of wine. See for the contents in the sort of problem that can seldom demand the total order and the extra ordered. This is how to solve ratios effectively.
See also Math Vs Statistics: Top 9 Important Points One Should Know
### Things that you should remember while solving ratios
• Remember, one is studying the ratio of the best method. For instance, the ratio of colors can be represented as 3 reds to 9 blue can be represented as 3:9, not 9:3. The initial article in the statement arrives initially.
• Be accurate with understanding the contents. For instance, individuals often make errors with topics like “Sam has 10 animals and 5 birds. Determine the ratio of animals to birds.” It is fascinating to tell the ratio is 10:5, but it will be wrong as the problem demands the ratio of animals to birds. One requires determining the total number of pets (10 + 5 = 15). Therefore the right ratio will be 10:15 (or 2:3).
• Avoid putting off with decimals or units. The sources will be the equivalent, whether all connect to complete fractions, numbers, m2, or £. Assure one should take note of the units in the views and change them to similar units. For instance, if one requires a ratio of 100g to 0.50kg, then change each unit to either kilos or grams.
## Conclusion
To sum up the post on how to solve ratios, we can say that three different methods can be used to solve them. Besides these methods, some common mistakes can be done by learners. Therefore, try to remember these and avoid them while solving ratios. Ratios have significant uses in day-to-day lives that are beneficial to solve various daily problems. So, learn the methods to solve ratio problems and get the benefits of these to overcome daily numeric problems. If you think that you need help then you can tell us that I need help with my math homework. And, Get the best and affordable help with my homework math with the help of us.
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By Lea Burger at October 29 2018 10:38:33
It's helpful having printable worksheets for something like this, because parents often go through quite a few of these before the child masters writing the numbers or letters correctly. What is my goal? It must be specific, challenging and attainable. Many people set but don't achieve their goals because the goals are simply too vague, too small or too big. For instance, "I'm going to get in shape this year" is a very poor goal.
But are you also aware that math can be fun if you put some thrill and excitement to it? It can be achieved if you incorporate math in fun activity like a game. Summarizing - Summarizing is essential in processing and categorizing all of the information obtained. Students must be able to identify main ideas, discriminate essential and nonessential information, and build this new information into their current schema. So which reading for comprehension worksheets are best? Any activity or worksheet that reinforces one or more of these six general reading for comprehension strategies would be an appropriate use of instructional time in any classroom or homework assignment.
At the grassroots level, teachers in schools are given a packed curriculum for the year. Schools try to teach the students a number of procedures without delving much into its finer details. Hence, the student is left in a confounding position as to when a particular procedure must be used. The key ingredient to understanding math is constant practice and math assignment help. Unfortunately, this is not a common scenario among the popular math classes.Connect The Letter To The Correct Sound/Word: These are activities where you draw a line between a letter and the picture items that start with that letter. For instance, you'd draw a line from the letter A to the word "Apple" and the letter L to the word "Lemon". This activity is good, but takes a lot of monitoring to make sure that students are correctly connecting the letters. It's best as a homework activity, where parents can help to make sure their children are correctly connecting the letters to the words.
What are the Parts of a Worksheet? Worksheets consists of four primary parts. A cell is the most commonly used part within an Excel workbook. Cells are where users can enter data to be used within formulas and charts later on. Worksheets are the individual "pages" of an Excel file. A Worksheet is basically just a computer representation of a very large piece of paper. It is organized into columns and rows, with the columns denoted by alphabetical letters (A, B...AB, AC, AD,...etc) and rows denoted by numbers.
lego math worksheets
5th grade math test prep worksheets
5th grade math common core worksheets<|endoftext|>
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Students break down each nutrient and develop a comprehensive lesson plan and activities to teach their classmates about their assigned nutrient. Students use various methods and technologies to cover their nutrient as well as developing their own test, lesson plans, lab plan, and food lab recipe.
Time Frame: Varies by group, at least 5 weeks (Periods)
Class Size: 25
- To learn and teach the selected nutrient taking the teacher role and adapting to student-led teaching peers.
- Computer Lab
- Book resources
- Foods lab
- Go over the project guidelines, student guided research and evaluation, student direct-instruction, assessments, cooking/foods lab.
- Assign or let students choose groups, assign presentation days.
- Allow for class time to work on the project. The project is broken into steps in the packet.
- Nutrient Project (Word)<|endoftext|>
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Discuss why George is described as 'curious'. Tell the students a little about the author and illustrator. Ask what they think the H stands for in H.A.Rey's name (Hans). Although only his name appears on the early books, Margret (note the spelling), was responsible for most of the story writing.
For one of the Curious George Crafts, provide the students with a paper with a picture of Curious George on it and the sentences. "This is George. George is very curious. He is curious about you. What is your name?" Encourage students to tell about themselves and draw a picture.
Did the student understand the concept of curious?
Did the student understand the duties of an author and an illustrator?
What do Monkeys Eat?
Ask the students what monkeys like to eat – bananas. Provide paper banana shapes of different sizes and ask students to measure them using a variety of non standard units e.g. How many small blocks to measure the length of each banana shape ? How many paper clips to measure the length of each banana shape?
Did student use vocabulary pertaining to measurement : shorter, longer, the same length?
Did student compare lengths?
Yellow is George’s Favorite Color
Ask students why they think that George's favorite color is probably yellow (his friend the man in the yellow hat, the color of bananas his favorite food)
Compile a list of things that are yellow. Provide students with large sheets of paper and invite them to make up a picture showing some of these things. Encourage them to print the word beside each of their illustrations. Print the word YELLOW as a title for the picture.
Provide materials for paper headbands and ask students to make up their own yellow hats.
Did the student correctly spell the words from the list?
Did the student understand the concept of labeling?
This Monkey is WANTED!
Discuss with students what information a 'Wanted' poster would contain. For fun ask students to make up a 'Wanted' poster for George titled 'Wanted for Being Curious'. Details that may be included – color of eyes, age, height, weight, hair color, what crime he has committed, description, what to do with him when found.
Did the student describe George the monkey?
A Miniature Puppet Theater
Demonstrate how to make a miniature puppet theater. Take an empty detergent bottle and cut off the top. Cut a rectangle out of the front of the bottle (where the label is placed). Draw small pictures of the characters e.g. George and the man in the yellow hat. Color them and cut them out. Attach drinking straws or pipecleaners to the backs of the drawings and then lower the 'puppets' into the detergent bottle. The front opening is the stage and you can manipulate the puppets with the straws or pipecleaners. Encourage the students to make up their own dialogue and for the more able students they could write their own scripts. Share with the class.
Did the student use his or her imagination?
Did the student give a 'voice' to the puppets?
Did the student share with others?
What is a celebration? Ask for ideas to celebrate George being a hero.
Provide bananas and ice cream, paper dishes, various toppings and enjoy banana splits.
Provide light card, crayons, markers, wool. Invite students to make a medal to hang around George's neck.
Sing the song '5 Little Monkeys Jumping on the Bed' – make up different words for the song e.g.' 5 Little Monkeys Swinging in a Tree.'
Make a graph showing the favorite Curious George book.
Curious George books are still favorites of young children after more than 60 years. They provide the basis for many Curious George craft ideas and fun activities. These activities can be linked to many areas of the curriculum in interesting ways.
Did the students show evidence of their ENJOYMENT of the unit about Curious George?
Did the unit encourage students to read more Curious George books?<|endoftext|>
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He understands the psychology of war matters, and he is intent on trying to get the war over as quickly as possible.
We continue our series on Great Captains with a look at William Tecumseh Sherman, one of the more controversial figures in the Civil War due to his actions in the South during the latter stages. A brilliant leader who understood well the impact that war has on soldiers and societies, Sherman was credited by Liddell-Hart as being the first “modern” general. But as the architect of a brutal campaign that severly weakened the Confederacy, Sherman also invoked fear and anger from enemies and friends alike. War College professors Jacqueline E. Whitt and Andrew A. Hill take a close look at Sherman and his legacy and one of histories Great Captains.
Jacqueline E. Whitt is Professor of Strategy at the U.S. Army War College and Andrew A. Hill is the Chair of Strategic Leadership at the U.S. Army War College. The views expressed in this presentation are those of the speakers and do not necessarily reflect those of the U.S. Army War College, U.S. Army, or Department of Defense.
Photo: Cropped portrait of William T. Sherman by Mathew Brady, listed as between 1865 and 1880. Digitally enhanced from original negative.<|endoftext|>
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BRUNO VELLUTINISponges have been generally considered our most ancient animal relatives. But that title may actually belong to members of the anatomically more complex phylum Ctenophora, or comb jellies, according to a report published today (December 12) in Science. This family tree reshuffle has come about thanks to the whole-genome sequencing of the comb jelly Mnemiopsis leidyi.
“It’s very exciting because there’s been a lot of debate about early animal evolution,” said John Finnerty, a professor of biology at Boston University, who was not involved in the study. Part of the reason for the debate was the lack of a whole genome sequence from a comb jelly. “The major early animal lineages had all been represented by at least one species with a sequenced genome with the exception of the Ctenophores,” Finnerty said, “so this really was a key piece of missing evidence.”
As their name implies, comb jellies are gelatinous animals with combs—bands of cilia running along their bodies that enable them to swim about in the sea. Based on their morphology, comb jellies had traditionally been considered close relatives of the Cnidaria—the phylum that contains jellyfish and sea anemones. Such morphological analyses had also classed the sponges, or Porifera, which are non-motile and lack muscles and nerves, as the most primitive members of the animal family.
But a recent comparison of multiple genes from members of the five major animal phlya—Ctenophora, Cnidaria, Porifera, Placozoa (tiny flat animals that live on the seabed), and Bilateria (which includes humans, insects, flatworms, and more)—indicated that comb jellies might oust sponges for the rank of most primitive.
“If the split between Ctenophores and all other animals was the earliest split in animal evolution, it suggests some unintuitive facts about evolution,” said Finnerty. “For example, that sponges, which are very simple animals that lack a nervous system and lack muscle cells, actually came from an ancestor that had those features.” Ctenophores, he explained, “tend to be really active animals—they are swimmers, they have smooth muscle, they have a nervous system, they prey on zooplankton. . . . They would seem to be representative of a much more complex condition, anatomically and behaviorally.”
Consequently, the suggestion that the Ctenophora might be the earliest branch of the animal family tree “was and has remained controversial,” said Andy Baxevanis, who is head of the Computational Genomics Unit at the National Human Genome Research Institute and led the new study. “It became obvious to us that having a whole genome sequence for the Ctenophores would go along way to resolving the issue.”
The team extracted DNA from M. leidyi embryos, sequenced the entire genome, and compared it with whole genome sequences of 12 other member species of other animal phyla. They applied two methods of comparative statistical analyses—one called maximum likelihood, the other called Bayesian. From the maximum likelihood analysis, “Ctenophores always came out at the base,” said Baxevanis. Using the Bayesian approach, however, “what you end up seeing is a new clade that is composed of both the Ctenophores and the sponges [at the base], so you can’t determine which came first.”
So the team performed a gene content analysis, whereby they determined the presence or absence of a set of orthologous genes across species. Taking this approach, the team found “unequivocally that the tree supports having Ctenophores at the base.” With two out of the three approaches giving the same result, Baxevanis said, “we feel pretty confident that Ctenophores are the sister group to the rest of the extant animals we studied.”
The placing of Ctenophores at the base of the animal family tree not only suggests, counterintuitively, that sponges and placozoans have lost their ancestral nervous systems, it also suggests that muscle development may have evolved twice within the animal kingdom. In the only other animal phyla besides Ctenophores to contain muscle cells, Bilateria, muscle develops from mesoderm tissue during embryogenesis. Surprisingly, however, the genome sequence of M. leidyi lacked the genes for mesoderm development almost entirely.
“Like any good genome that has been sequenced, the results of this study generate more questions than answers,” said Antonis Rokas, a professor of biological sciences at Vanderbilt University in Nashville, Tennessee. “If the placement of Ctenophores is correct,” he added, “it really changes quite a bit of the text book interpretation that we have about the evolution of morphological complexity.”
J.F. Ryan et al., “The genome of the Ctenophore Mnemiopsis leidyi and its implications for cell type evolution,” Science, doi: 10.1126/science.1242592, 2013.<|endoftext|>
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# 1975 AHSME Problems/Problem 30
## Problem 30
Let $x=\cos 36^{\circ} - \cos 72^{\circ}$. Then $x$ equals
$\textbf{(A)}\ \frac{1}{3}\qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 3-\sqrt{6} \qquad \textbf{(D)}\ 2\sqrt{3}-3\qquad \textbf{(E)}\ \text{none of these}$
## Solution
Using the difference to product identity, we find that $x=\cos 36^{\circ} - \cos 72^{\circ}$ is equivalent to $$x=\text{-}2\sin{\frac{(36^{\circ}+72^{\circ})}{2}}\sin{\frac{(36^{\circ}-72^{\circ})}{2}} \implies$$ $$x=\text{-}2\sin54^{\circ}\sin(\text{-}18^{\circ}).$$ Since sine is an odd function, we find that $\sin{(\text{-}18^{\circ})}= \text{-} \sin{18^{\circ}}$, and thus $\text{-}2\sin54^{\circ}\sin(\text{-}18^{\circ})=2\sin54^{\circ}\sin18^{\circ}$. Using the property $\sin{(90^{\circ}-a)}=\cos{a}$, we find $$x=2\cos(90^{\circ}-54^{\circ})\cos(90^{\circ}-18^{\circ}) \implies$$ $$x=2\cos36^{\circ}\cos72^{\circ}.$$ We multiply the entire expression by $\sin36^{\circ}$ and use the double angle identity of sine twice to find $$x\sin36^{\circ}=2\sin36^{\circ}\cos36^{\circ}\cos72^{\circ} \implies$$ $$x\sin36^{\circ}=\sin72^{\circ}\cos72^{\circ} \implies$$ $$x\sin36^{\circ}=\frac{1}{2}\sin144^{\circ}.$$ Using the property $\sin(180^{\circ}-a)=\sin{a}$, we find $\sin144^{\circ}=\sin36^{\circ}.$ Substituting this back into the equation, we have $$x\sin36^{\circ}=\frac{1}{2}\sin36^{\circ}.$$ Dividing both sides by $\sin36^{\circ}$, we have $$x=\boxed{\textbf{(B)}\ \frac{1}{2}}$$
1975 AHSME (Problems • Answer Key • Resources) Preceded byProblem 28 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions<|endoftext|>
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Englisc, which the term English is derived from, means 'pertaining to the Angles'. In Old English, this word was derived from Angles (one of the Germanic tribes who conquered parts of Great Britain in the 5th century). During the 9th century, all invading Germanic tribes were referred to as Englisc. It has been hypothesised that the Angles acquired their name because their land on the coast of Jutland (now mainland Denmark) resembled a fishhook. Proto-Germanic*anguz also had the meaning of 'narrow', referring to the shallow waters near the coast. That word ultimately goes back to Proto-Indo-European *h₂enǵʰ-, also meaning 'narrow'.
Another theory is that the derivation of 'narrow' is the more likely connection to angling (as in fishing), which itself stems from a Proto-Indo-European (PIE) root meaning bend, angle. The semantic link is the fishing hook, which is curved or bent at an angle. In any case, the Angles may have been called such because they were a fishing people or were originally descended from such, and therefore England would mean 'land of the fishermen', and English would be 'the fishermen's language'.<|endoftext|>
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Mathematics
Easy
Question
# Identify the equation which matches the following graph
Hint:
## The correct answer is: Y = –2x – 1
### GIVEN-The given line passes through the points (-1,1) & (0,-1) as shown in the given graph.TO FIND-Equation of the given line.SOLUTION-From the given graph, we observe that-(-1,1) & (0,-1) are 2 points on the line.We know that when a line passes through a given point, the coordinates of that point satisfies the equation of the said line.Hence, we substitute the coordiantes of both the points in the given options to find the correct equation, for which LHS = RHS.a. y = 2x - 1We substitute (-1,1) in the given equation- y = 2x - 1∴ 1 = 2 (-1) - 1∴ 1 = -2 - 1∴ 1 ≠ -3∴ LHS ≠ RHSSince (-1,1) does not satisfy the given equation, y = 2x - 1 is not the equation of the given line.b. x2 + y2 = 1We substitute (-1,1) in the given equation- x2 + y2 = 1∴ (-1)2 + (1)2 = 1∴ 1 + 1 = 1∴ 2 ≠ 1∴ LHS ≠ RHSSince, (-1,1) does not satisfy the given equation, x2 + y2 = 1 is not the equation of the given line.c. y = -2x - 1We substitute (-1,1) in the given equation- y = -2x - 1∴ 1 = -2 (-1) - 1∴ 1 = 2 - 1∴ 1 = 1∴ LHS = RHSWe substitute (0,-1) in the given equation- y = -2x - 1∴ -1 = -2 (0) - 1∴ -1 = 0 - 1∴ -1 = -1∴ LHS = RHSSince both (-1,1) and (0,-1) satisfy the given equation, y = -2 x - 1 is the equation of the given line.FINAl ANSWER-Option 'c' i.e. 'y = -2x - 1' is the correct answer to the given question.
Alternatively, we can use the 2 point formula to find the equation of the given line-
When we have 2 points that lie on a given line then we can find the equation of the said line by using the 2-point formula-
(y-y1) = (y2-y1) * (x-x1)
(x2-x1)
In the given question, x1 = -1, y1 = 1, x2 = 0 & y2 = -1
∴ (y-y1) = (y2-y1) * (x-x1)
(x2-x1)
∴ (y - 1) = (-1 - 1) * [x - (-1)]
[0 - (-1)]
∴ (y - 1) = -2 * (x + 1)
0 + 1
∴ (y - 1) = -2/1 * (x + 1)
∴ y - 1 = -2x - 2
∴ y = -2x - 2 + 1
∴ y = -2x - 1<|endoftext|>
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1. ## Mathematical Induction Proofs
I have absolutely no idea how to do this problem. PLEASE SHOW ME HOW TO DO IT!!!!!
Verify by mathematical induction.
(1^3)+(3^3)+(5^3)+...+(2n-1)^3 = n^2[2(n^2)-1]
2. Originally Posted by zuic
Verify by mathematical induction.
(1^3)+(3^3)+(5^3)+...+(2n-1)^3 = n^2[2(n^2)-1]
Most mathematical induction problems are done the same way. Some may require a little ingenuity on the second step than others, but the overall format is still the same.
start by verifying that P(1) is true.
then assume P(n) is true.
show that P(n) implies P(n + 1) is true.
Then the conclusion will follow by Mathematical induction.
care to take a shot at it?
3. Hello,
Firstly, prove that it's true for n=0, then, prove that : (1^3)+(3^3)+(5^3)+...+(2n-1)^3+(2(n+1)-1)^3=(n+1)^2[2(n+1)^2-1], supposing that (1^3)+(3^3)+(5^3)+...+(2n-1)^3 = n^2[2(n^2)-1] (induction hypothesis)
This is a proof by induction.
So, you have $\displaystyle (1^3)+(3^3)+(5^3)+...+(2n-1)^3+(2(n+1)-1)^3$
$\displaystyle =\underbrace{(1^3)+(3^3)+(5^3)+...+(2n-1)^3}_{n^2[2(n^2)-1]} +(2n+1)^3$
$\displaystyle =n^2[2n^2-1]+(2n+1)^3$
Develop and retrieve $\displaystyle (n+1)^2(2(n+1)^2-1)$
4. First, test your formula for a trivial value of n
[Math] \sum_{i=1}^{1}{(2i-1)^3} = 1 = 1^2(2 \times 1^2-1) [/tex]
Assume it work for all integer value of n up to a a particular value say k.
so we assume that [Math] \sum_{i=1}^{k}{(2i-1)^3} = k^2(2 k^2-1) [/tex] is true.
then attempt to prove that the formula is also true for k+1.
[Math] \sum_{i=1}^{k+1}{(2i-1)^3} = \sum_{i=1}^{k}{(2i-1)^3} + (2(k+1) -1)^3 [/tex]
[Math] \Rightarrow k^2(2 k^2-1) +(2k +1)^3 [/tex]
..... (omitting some algebra) .....
[Math] 2k^4 + 8k^3 + 11k^2 + 6k +1 [/tex]
now when you factor this remember your expecting k+1 to be a factor, (it's not cheating to use your expected result to help factor an expression).
[Math]\Rightarrow (k+1)(2k^3+6k^2+5k+1) [/tex]
[Math]\Rightarrow (k+1)^2(2k^2+4k+1) [/tex]
[Math]\Rightarrow (k+1)^2(2(k+1)^2 -1 ) [/tex]
If true of k , it is also true for k+1, we have shown that it is true for k=1 so it is therefore. also true for k=2 and hence all integers.
Bobak
5. $\displaystyle 1^{3}+3^{3}+5^{3}+.....+(2n-1)^{3}=n^{2}(2n^{2}-1)$
Check for n=1: $\displaystyle 1^{3}=(1)^{2}(2(1)^{2}-1)$, TRUE.
Assume $\displaystyle P_{k}$ is true, then the induction hypothesis is:
$\displaystyle 1^{3}+3^{3}+5^{3}+....+(2k-1)^{3}=k^{2}(2k^{2}-1)$
Now we have to show that $\displaystyle P_{k+1}$ is true. Add $\displaystyle (2k+1)^{3}$ to both sides, which is the next odd cube.
$\displaystyle 1^{3}+3^{3}+5^{3}+.....+(2k-1)^{3}+(2k+1)^{3}=k^{2}(2k^{2}-1)+(2k+1)^{3}$
$\displaystyle k^{2}(2k^{2}-1)+(2k+1)^{3}=2k^{4}+8k^{3}+11k^{2}+6k+1$$\displaystyle =(k+1)^{2}(2k^{2}+4k+1)=(k+1)^{2}(2(k+1)^{2}-1)$
and the induction is complete.<|endoftext|>
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Summary problem for rational functions | mathXplain
Összes egyetemi tantárgy
Legnépszerűbb tantárgyak:
Contents of this Calculus 1 episode:
Integrating rational functions, Partial fractions, First type partial fractions, Second type partial fractions, Decompose it into partial fractions, Factorization, Linear denominator, Non-factorable quadratic denominator, f’/f and arctan forms, Polynomial division.
Text of slideshow
Here is a summary problem where we can see all important steps.
Bad news. First, we need polynomial division, because the degree of the numerator must be less than the degree of the denominator.
This polynomial division is just like long division we learned in elementary school.
Like 25:7=3 and the remainder is 4.
In other words
Polynomial division is just like this.
quotient remainder
Here comes the polynomial division:
Everything is OK so far.
But this is not the end, yet.
We multiply the quotient with the divisor,
and subtract that from the dividend.
And then we divide again, and repeat this until the degree of the dividend becomes less than that of the divisor.
Hey, it is less now, so we are done.
We integrate the first two members, and then we proceed to the fraction where the degree of the numerator is less than that of the denominator.
Again, we need to factorize the denominator into linear and irreducible quadratic factors. First, we factor out x2.
Then we check whether the remaining quadratic part could be factorized. The answer is no. It is because there is no real solution to the equation.
On the other hand, x2 can be factorized.
Now come the decomposing into partial fractions.
If the denominator contains a square of some ax+b linear polynomial, then we decompose into partial fractions, such that
the denominator of one of the partial fractions is ax+b,
and the other is (ax+b)2.
Now we have to figure out the numerators. The denominator of the first fraction seems to be linear, so the numerator is some A.
The denominator of the second fraction is a square of a linear expression, so the numerator here is also some A, but since A is already used, let's call it B.
Finally, the denominator of the third fraction is a quadratic polynomial, so its numerator should be in the form of Ax+B, but since A and B are already used, Cx+D will do.
Now we calculate the values of A, B, C and D.
We multiply by the denominators.
Next, expand the parentheses.
And then we check how many x3 terms, x2 terms, x terms and constants are on the right side.
Because there are exactly the same on the left side as well.
The first two terms are very easy to integrate.
The third term becomes this:
The first term, as we wanted, f’/f, while the second term leads to arctangent.
And then it is done.
Finally, one more example:
First of all, we need to factorize the denominator into linear or irreducible quadratic factors. The factorization isn't trivial at all, because the denominator does not have a real root. The product form:
Then:
The factors in the denominator will be the denominators of the partial fractions. Since both factors are irreducible quadratic expressions, it seems we will get a sum of two Type II partial fractions:
Next, we move on to finding A, B, C, and D.
Multiplication:
and then conversion
finally the usual system of equations:
The solutions: , thus
We will integrate the two fractions separately.
The first fraction:
This will turn into a linear substitution of arctgx:
The second fraction, due to symmetry reasons:
The solution of the problem is the sum of the expressions found before:
Let's not forget though, that the method of integrating rational functions should be regarded as the last resort, and used only when all other methods failed to work. This latter problem was solved earlier, somewhat faster, using S4!
# Summary problem for rational functions
17
Let's see this
Calculus 1 episode
Enter the world of simple math.
• It is very well priced and better and more understandable than many private tutors.
Mark, 22
• It is the best clear, interpretable, usable learning opportunity for the lowest price.
Ellen, 23
• My seventh-grader brother learned to derive, which is quite an evidence that it is explained clearly.
George, 18
• It was recommended by senior students with the title 'mandatory'.
Richard, 19<|endoftext|>
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# If both roots of the Quadratic Equation are similar then prove that
If both roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$.
Things should be known:
• Roots of a Quadratic Equations can be identified by:
The roots can be figured out by: $$\frac{-b \pm \sqrt{d}}{2a},$$ where $$d=b^2-4ac.$$
• When the equation has equal roots, then $d=b^2-4ac=0$.
• That means $d=(b-c)^2-4(a-b)(c-a)=0$
• What is $(b-c)^2 + 4bc$? – Alex B. Jan 4 '13 at 15:22
• So have you tried working with your expression for $d$? – Mark Bennet Jan 4 '13 at 15:22
• Yes, I tried to evaluate the $d$, It was (not completely evaluated) $$d= (b-c)^2-4(a-b)(c-a)=0$$ Is it right? And after this I am not able to understand that how to proceed now. Can you help? – Saharsh Jan 4 '13 at 15:27
• Expand $(-2a+b+c)^2$ . Do you see some similarity? – Inquisitive Jan 4 '13 at 15:48
As the two roots are equal the discriminant must be equal to $0$.
$$(b-c)^2-4(a-b)(c-a)=(a-b+c-a)^2-4(a-b)(c-a)=\{a-b-(c-a)\}^2=(2a-b-c)^2=0 \iff 2a-b-c=0$$
Alternatively, solving for $x,$ we get $$x=\frac{-(b-c)\pm\sqrt{(b-c)^2-4(a-b)(c-a)}}{2(a-b)}=\frac{c-b\pm(2a-b-c)}{2(a-b)}=\frac{c-a}{a-b}, 1$$ as $a-b\ne 0$ as $a=b$ would make the equation linear.
So, $$\frac{c-a}{a-b}=1\implies c-a=a-b\implies 2a=b+c$$
The expression for $d$ factors as $(2a-b-c)^2$, so we must have $2a-b-c=0$.
The easiest way to see this is to let $x=a-b,y=c-a$ and note that $b-c=-x-y$: $d=(-x-y)^2-4xy=x^2+y^2+2xy-4xy=(x-y)^2$.
Alternative solution: Note that $(a-b)+(b-c)+(c-a)=0$, i.e. $x=1$ is a root of the equation. If both roots are equal, it means that the other root is also $1$, showing that $(a-b)x^2+(b-c)x+(c-a)=(a-b)(x-1)^2$.
Equating the constant term, we find: $c-a=a-b$, which means $2a=b+c$. (This can be done with the coefficient of $x$ too).
Edit: $(b-c)^2=4(a-b)(c-a)$, so $a-b$ and $c-a$ are either both nonnegative or both nonpositive. In the former case, put $x=a-b$ and $y=c-a$; in the latter case, put $x=-(a-b)$ and $y=-(c-a)$. Now the A.M. $\ge$ G.M. inequality says that $$\frac{x+y}2\ge\sqrt{xy},\ \text{ i.e. } (x+y)^2\ge4xy$$ for $x,y\ge0$, and equality holds if and only if $x=y$.
• Sorry, but I didn't get the answer. Can you clarify what's A.M. and G.M. – Saharsh Jan 4 '13 at 15:36<|endoftext|>
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# Adding 1 – Digit Number | Addition of One Digit Numbers | How to Quickly Add One Digit Numbers?
The addition is one of the arithmetic operations in numbers. The symbol of the addition operation is ‘ + ‘. We can also call addition sometimes as a plus, add, the sum of the numbers, sum, and put together with. If we add the numbers with the 1 – digit number, the resultant value will be shown with the sign of equivalent. That is ‘=”. It is easy to add single-digit numbers compared to other operations. Adding 1 – Digit Numbers in different methods are explained on this page to help students to learn easily.
Do Read:
## Addition of One-Digit Numbers
The mathematical expression for adding 1- digit number is like one digit + one digit = resultant. For example, if you take the digits 4 and 4, then the mathematical expression appears as below.
We can write it as 4 plus 4 is equal to 8.
In the early stages of the children only, we have to develop the mathematical skills very clearly. If we explain the addition process by telling stories, games, by calculating the objects like counting toys, images, stars, and etc.. Most of the students can easily understand the concept and they are ready to add the numbers easily by imagining it in their mind.
We are providing more activities that are related to adding 1 – digit number. By following our concept of Adding 1 – digit number and practicing the problems, a student can perform addition very quickly.
Example:
1. There are two girls are playing with the balls. Girl A is playing with the 3 balls and Girl B is playing with the 2 balls. Count the total number of balls.
Solution:
Girl A is playing with the 3 balls. So,
Girl B is playing with the 2 balls
Now, the total number of balls with the two girls are
Therefore, the total number of balls is 5.
Theoretical Explanation:
By using the numbers, Girl A is playing with 3 balls.
Girl B is playing with the 2 balls.
So, the total number of balls with Girl A and Girl B is
3 + 2 = 5.
### Adding 1 – Digit Number Activities
Add the 1 digit numbers by playing with the below images. We can add the numbers with the help of our hand fingers also. Like
Example 1.
Add 5 + 2?
If you want to add 5 and 2, then open your 5 fingers on one hand and open two fingers on another hand first. Then, start counting the numbers of opened fingers from 1. Count the opened fingers and note down the final answer.
5 + 2 = 7.
Example 2.
Add 2 + 2?
If you want to add 2 and 2, then open your 2 fingers first and open the other two fingers next. Then, start counting the numbers of opened fingers from number 1. Count the opened fingers and note down the resultant answer.
2 + 2 = 4.
Example 3.
Add 4 + 0?
If you want to add 4 and 0, then open your 4 fingers on one hand. You need not open any finger if the addend is 0. Then, start counting the numbers of opened fingers from number 1. Count the opened fingers and note down the last answer.
4 + 0 = 4.
### Adding 1- Digit number on a Number Line
The addition of one-digit numbers is easy with the number line. First, take a number line with numbers up to required numbers. Then, recognize the first number on the number line. After that note down the second number. Count the numbers as per the given second number from the first number. Point out the number where the counting stopped. Draw a curve from the first number to the last number on the number line. The last number is the resultant answer of your addition.
By using the above number line also, we can add the 1 –digit numbers. From the above number line,
4 + 8 = 12.
### Adding 1 – Digit Number with Images
In this section, we need to add the number of triangles and find the equivalent value. Check out all the below images and find the addition of numbers.
1.
In the given image, two triangles are present. So, add 1 triangle with other triangles. 1 + 1 = 2.
2.
In the given image, there is one blue triangle and two yellow triangles. So, add the number of blue triangles with the number of yellow triangles. 1 + 2 = 3.
3.
From the given image, there are two blue triangles and two yellow triangles. So, add the number of blue triangles with the number of yellow triangles. 2 + 2 = 4.
4.
In the given image, there is two blue triangle and three yellow triangles. So, add the number of blue triangles with the number of yellow triangles. 2 + 3 = 5.
### Addition of 1 – Digit Numbers in Columns
While adding the 1 – digit number with the two numbers in Columns, follow the below procedure and add the numbers.
(i) first, we have to add the 1-digit number with the 1’s place value of 2 – digit number.
(ii) If we get any carry value, then add that carry to the 10’s place number.
(iii) Otherwise, note down the same value of 10’s place number in the resultant with the addition of 1’s place with 1 – digit number.
Examples:
1. 1 + 2
1
+ 2 (no carry)
———
3
2. 5 + 9
5
+9 (carry generated)
———
14
3. 2 + 3
2
+3 (no carry)
———
5
4. 5 + 5
5
+5 (carry generated)
———
10
5. 2 + 7
2
+7 (no carry)
———
9
### Addition of 1 – Digit Numbers Horizontally
You can also add one-digit numbers horizontally by placing one number next to the other number. Place the addition symbol or plus symbol in between them. The below examples can help you to add one-digit numbers Horizontally.
1. 6 + 1 = 7.
2. 9 + 4 = 13
3. 8 + 2 = 10
4. 2 + 4 = 6
5. 7 + 6 = 13
### Adding 1- Digit Number Table
Students can easily remember and quickly answer the addition of 1 – digit number by following the below table. In the below table, we are adding the 1 – digit numbers to other numbers. And, you can observe the sum of the numbers also in that same table.
Scroll to Top
Scroll to Top<|endoftext|>
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# Finding Area
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Students will be able to add and multiply square units in order to find the area of a rectangle.
(10 minutes)
• Pick up six square tiles. Using a projector, display three of them in a connected row, and display the other three in a scattered group. Ask students to identify which group shows an area, or square unit total, of three. (You'll be defining the term "area" more clearly later on in the lesson.)
• After hearing their answers, explain that both groups have a total area of three. Each tile is a Square unit, and regardless of how they're positioned, three tiles will always equal three square units.
• On your display, write "3 sq. units" above the row of tiles and "1 sq. unit" above each tile in the scattered group.
(15 minutes)
• Display a 4 square unit rectangle (e.g. four rows of one block, 2 rows of 2 blocks, or 1 row of 4 blocks). Ask students what they think the area of the shape is. Have those who volunteer answers explain their reasoning.
• After hearing their responses, display a different 4 square unit rectangle. Ask students to compare this rectangle to the previous one. Ask them what they think the area of this rectangle is.
• After hearing their responses, display the final 4 square unit model. Remind students that the area of a figure is the sum of its unit squares. Ask students to count the number of unit squares in this model as you point at each of its tiles.
• Make sure that students understand that all three of the models have an area of 4 square units.
• Ask the class for suggestions on more efficient ways to find area.
• Once a student suggests multiplication, begin modeling an example of it on the board. As you're making the model, explain that rows and columns can be used as part of multiplication equations. These equations, in turn, can be used to find the areas of shapes.
• Draw an arrow going across each row of your model, then number each row. For example, if you're using a model with two rows of three blocks, you'd draw two arrows and number them "1" and "2." Above the model, you'd write "2 rows of ___."
• Circle one of the rows to show how many units are in each row. Continuing the previous example, you'd fill in the blank with "3." This results in the statement "2 rows of 3."
• Replace the words "rows of" with a multiplication symbol, and you'll end up with "2 x 3." The answer to this equation, six, is the area of the shape.
• Explain that the AreaOf a shape is equal to its length multiplied by its width. Write "Area = Length x Width" on the board.
(10 minutes)
• Have students partner up. Let them know that for this portion of the lesson they'll be working with a classmate to create various rectangular shapes using square tiles, recreate the shapes on graph paper, and write matching multiplication equations.
• Distribute one-inch square tiles and a sheet of one-inch graph paper to each pair.
• Ask students to use their tiles and construct a rectangle with an area of 6 square units.
• Once students have finished, allow volunteers to share their model with the class by drawing it on the board.
• Have students share until all four possible models have been represented on the board.
• Ask partners to copy and shade in their rectangles on their graph paper.
• Have each pair write a multiplication equation to go with their shape. Then, have them find and record its area. For example, a pair with a model that has one row of six would write "6 x 1 = 6" and "Area = 6 sq. units."
(15 minutes)
• Have students repeat the activity on their own, with rectangles of varying sizes and areas.
• Remind students to save space on their graph paper by drawing their shapes close to one another.
• Enrichment:Give advanced students an extra challenge by having them work out different multiplication equations on another sheet of graph paper. Some problems you can assign are: 9 x 6 = ___, 3 x 7 = ___, 5 x 12 = ___, and 8 x 4 = ___. Ask them to find and record the area of each answer by decomposing a factor. For example, 9 x 6 = (5 x 6) + (4 x 6) = 30 + 24 = 54. Area = 54 sq. units. Students should also draw and label a visual representation for each problem. (A 9 x 6 array would be divided into two sections: one labeled "5 x 6" and another labeled "4 x 6.")
• Support:If there are only a few struggling students, give them one-on-one support. If there are multiple students in need of assistance, pull them aside for small group instruction. Briefly review the lesson with them and, if needed, reduce their workload.
• An interactive whiteboard could be used to present visual aids in a more streamlined manner. Its square shape tool could be used to represent square inch tiles within the lesson. Visual representations can be made in advance and hidden behind the shade tool. The clone tool would allow for the infinite creation of square tiles when putting together visual models.
(10 minutes)
• To assess their understanding, sit near pairs during Independent Working Time and make visual observations.
• Prior to closing the lesson, ask each student to create a model with an area of 16 square units. Allow volunteers to share multiplication equations that match their models.
(5 minutes)
• Ask students to share what methods they thought were most useful for finding area.
• Collect student work, and review it later to identify any students who need additional support.
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# A rectangular garden is to be twice as long as it is…
## Solution:
We first draw a diagram of our garden. The length of the rectangular garden is twice as long is it is wide. Thus, we know:
length = 2(width)
Thus, we can label length as 2w, where w = width. Let’s plug these values into the diagram.
We are given that 360 yards of fencing will completely enclose the garden; thus, the perimeter of the garden is 360 yards. The formula for the perimeter of a rectangle is:
perimeter = 2(length) + 2(width)
Knowing that length = 2w, width = w, and perimeter = 360 we can plug all this info into our perimeter equation:
360 = 2(2w) + 2w
360 = 4w + 2w
360 = 6w
60 = w
Since the width of the garden is 60 yards, the length is 2 x 60 = 120 yards.<|endoftext|>
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12. Learning, Memorizing, and Mental Play¶
There is no faster way of memorizing than to memorize when you are first learning a piece and, for a difficult piece, there is no faster way of learning than memorizing it. Start memorizing by learning how the music should sound: melody, rhythm, etc. Then use the sheet music to find and memorize each key on the piano for each note on the sheet music; this is called keyboard memory – you memorize how you play this piece on the piano, complete with the fingering, hand motions, etc. Some pianists use photographic memory, in which they photographically memorize the sheet music. If one were to take a sheet of music and try to memorize it note for note, this task would be impossibly difficult even for concert pianists. However, once you know the music (melody, chord structure, etc.), it becomes easy for everyone! This is explained in 6. Memorizing, where you will find more detailed discussions on how to memorize. I prefer keyboard memory to photographic memory because it helps you to find the notes on the piano without having to “read” the music in your head. Memorize each section that you are practicing for technique while you are repeating them so many times in small segments, HS. The procedures for memorizing are basically the same as those for technique acquisition. For example, memorization should be started HS, difficult sections first, etc. If you memorize later, you will have to repeat the same procedure again. It might appear that going through the same procedure a second time would be simpler. It is not. Memorizing is a complex task (even after you can play the piece well); therefore, students who try to memorize after learning a piece will either give up or never memorize it completely. This is understandable; the effort required to memorize can quickly reach the point of diminishing returns if you can already play the piece.
Two important items to memorize are the time signature (see b. What is Rhythm? (Beethoven’s Tempest, Op. 31, #2, Appassionata, Op. 57)) and
key signature (see d. Scales: Origin, Nomenclature and Fingerings). The time signature is easy to understand
and will help you to play with the correct rhythm. The key signature (how many
sharps or flats) is more complex because it does not tell you the precise key
(scale) that it is in (
C major, etc.). If you know that the composition is
in a major or minor scale, the key signature tells you the key; for example if
the key signature has no sharps or flats (as in Für Elise), it is in either
C major or
A minor (see d. Scales: Origin, Nomenclature and Fingerings). Most students know the major
scales; you will need to know more theory to figure out the minor keys;
therefore, only those with enough theory knowledge should memorize the key. If
you are not sure, memorize only the key signature. This key is the basic
tonality of the music around which the composer uses chord progressions to
change keys. Most compositions start and end with the base tonality and the
chords generally progress along the circle of fifths (see Ch. Two, 2.b). So
far, we know that Für Elise is either in
C major or
A minor. Since it
is somewhat melancholy, we suspect a minor. The first 2 bars are like a fanfare
that introduces the first theme, so the main body of the theme begins on bar 3,
which starts with
A, the tonic of
A minor! Moreover, the final chord is
also on the tonic of
A minor. So it is probably in
A minor. The only
A minor is
G# (see Table 1.III.5.b Ascending Harmonic Minor Scales), which we find in
bar 4; therefore we conclude that it is in
A minor. When you understand
these details, you can really memorize well.
Let’s revisit the time signature, which is
3/8; three beats per measure (bar),
an eighth per beat. Thus it is in the format of a waltz but musically, it
should not be played like a dance but much more smoothly because it is
melancholy and hauntingly romantic. The time signature tells us that bars like
bar 3 must not be played as two triplets because there are 3 beats. However,
there is no need to overly accent the first beat of every bar like a Viennese
Waltz. The time signature is clearly useful for playing musically and
correctly. Without the time signature, you can easily form incorrect rhythmic
habits that will make your playing sound amateurish.
Once students develop memorizing-learning routines that are comfortable for them, most of them will find that learning and memorizing together takes less time than learning alone, for difficult passages. This happens because you eliminate the process of looking at the music, interpreting it, and passing the instructions from the eyes to the brain and then to the hands. Material memorized when young (before about age 20) is almost never forgotten. This is why it is so critical to learn fast methods of technique acquisition and to memorize as many pieces as possible before reaching the later teen years. It is easier to memorize something if you can play it fast; therefore, if you have difficulty memorizing it initially at slow speed, don’t worry; it will become easier as you speed it up.
The only way to memorize well is to learn Mental Play (MP). In fact, MP is the logical and ultimate goal of all these practice methods that we are discussing because technique alone will not enable you to perform flawlessly, musically, and without getting nervous. Read III.6.j for more details on MP. With MP, you learn to play the piano in your mind, away from the piano, complete with accurate fingering and your concept of how you want the music to sound. You can use keyboard memory or photographic memory for MP, but I recommend keyboard memory for beginners because it is more efficient; for advanced players, keyboard memory and photographic memory are the same, since if you can do one, the other follows naturally. Whenever you memorize a small section, close your eyes and see if you can play it in your mind without playing it on the piano. Once you have memorized an entire piece HS, you should also be able to play the complete piece HS in your head. This is the time to analyze the structure of the music, how it is organized and how the themes develop as the music progresses. With practice, you will find that it requires only a small investment of time to acquire MP. Best of all, you will also discover that once solid MP is established, your memory is as good as it can get; you will have confidence that you will be able to play without mistakes, blackouts, etc., and will be able to concentrate on music. MP also helps technique; for example, it is much easier to play at a fast speed after you can mentally play it at that speed; very often, the inability to play fast originates in the brain. One benefit of MP is that you can practice it at any time, anywhere, and can greatly increase your effective practice time.
Memory is an associative process. Super memorizers (including some savants) and all concert pianists who can memorize hours of music depend on algorithms with which to associate their memory (whether they know it or not). Musicians are especially fortunate in this regard because music is precisely such an algorithm. Nonetheless, this “memory trick” of using music as an algorithm to memorize is seldom formally taught to music students; instead, they are often advised to keep repeating “until the music is in the hands”, which is one of the worst methods of memory because, as we shall see in d. Hand Memory, repetition results in “hand memory” which is a false type of memory that can lead to many problems, such as blackouts. With MP, you associate the music in your mind with how you produce it at the piano. It is important to practice MP without playing the piano because you can acquire “sound memory” (just as you can acquire “hand memory”) and use the sound of the piano as a crutch for recall, and sound memory can cause the same problems associated with hand memory.
Why are memory and MP so important? They not only solve the practical problems of technique and performance but also advance your musicianship and increase intelligence. You can speed up a computer by adding memory; similarly, you can increase your effective intelligence by improving your memory. In fact, one of the first signs of mental deterioration, such as Alzheimer’s, is loss of memory. It is now clear that many of those “amazing feats” of great musicians such as Mozart were simple byproducts of strong MP, and that such skills can be learned. More on MP in j. Establishing Permanent Memory, Mental Play.<|endoftext|>
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