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This image shows how electrons will circle around a magnetic field line one way and a proton will circle in the opposite direction.
Movie adapted from original clips provided courtesy of Professor Patricia Reiff, Rice University, Connections Program.
Basic Facts About Spiral Motion
Protons and electrons can't move across magnetic field lines very easily. So sometimes these particles are are forced to spiral around them. Electrons cirle the field line in one direction, protons in the opposite direction.
Shop Windows to the Universe Science Store!
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for you and your students. Issues of NESTA's quarterly journal, The Earth Scientist
are also full of classroom activities on different topics in Earth and space science!
You might also be interested in:
Magnetic fields can cause particles to move in these three ways: Spiral Motion Bounce Motion Drift Motion ...more
AU stands for Astronomical Units. It is a useful way to measure the distances in interplanetary space. It is the distance between the Earth and the Sun, which is about 93 million miles. For reference,...more
The solar wind is formed as the Sun's top layer blows off into space, carrying magnetic fields still attached to the Sun. Gusts form in the solar wind associated with violent events on the Sun. Particles...more
For a planet to be affected by a blob of material being ejected by the sun, the planet must be in the path of the blob, as shown in this picture. The Earth and its magnetosphere are shown in the bottom...more
The aurora we are most familiar with is the polar aurora. This is what people are talking about when they say the northern or southern lights. But there are other less-known aurora, such as SAR arcs....more
This figure shows the effect of the aurora on the atmosphere. When FAC's enter the atmosphere and create the aurora, they heat the atmosphere suddenly and abruptly. This creates an impulse which travels...more
This picture shows the flowing of particles into and out of the auroral zone, as Field-Aligned currents (FAC's) take at short-cut through the atmosphere. Some of the particles entering the auroral zone...more<|endoftext|>
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# 2. Linear Systems¶
There’s definitely, definitely, definitely no logic
To human behaviour
—“Human Behaviour” by Björk
I’m as excited as you are about making a meteor fly, so let’s take the complexity of our systems up a notch. We’ll set multiple objects in motion in this section and see what we can learn about their behavior.
# 2.1 Systems of Equations¶
## Testing Solutions¶
In the last chapter, you applied linear equations as a mathematical model for the path of a meteor. As the meteor’s $$x$$-coordinate increased, its $$y$$-coordinate increased, and we related the two using an equation like the one below.
$y = 0.5x + 50$
If you substituted $$10$$ for $$x$$, you could evaluate the right-hand side to find, or solve for, $$y$$.
$y = 0.5 \cdot 10 + 50$
$y = 55$
You could use the same approach to find every coordinate pair along the meteor’s path. Substituting the coordinate pair $$(10,55)$$ into the equation $$y=0.5x+50$$ makes both sides equal, so we call it a solution to the equation.
How about the coordinate pair $$(10,60)$$?
$60=0.5 \cdot 10 + 50$
$60 \ne 55$
No luck. The symbol $$\ne$$ means “does not equal.” $$(10,60)$$ is not a solution to this equation.
OK, so we can map out all the points along an object’s path. But what if there were multiple objects moving along multiple paths?
## Solving Equations¶
In the film Gravity, a Space Shuttle is destroyed by a barrage of space debris, forcing the survivors to find another way home to Earth before the debris completes another orbit and threatens them again. Let’s try to help out the astronauts by constructing a simulation of the Shuttle-debris system.
A Space Shuttle and debris in Earth’s orbit.
$$\mapsto$$ Outer space is an empty, infinite, two-dimensional surface. The Space Shuttle is a point moving along the straight path described by the equation $$y=0.5x+50$$. The debris is also a point moving along a straight path, this one described by the equation $$y= -0.5x+150$$.
The coordinates of both objects are ordered pairs of floating point numbers. The Space Shuttle’s initial position is $$(0,150)$$. The debris’ initial position is $$(0,50)$$.
First, paint the sky. Next, update the position of each object. Then, select a pencil color & weight. Finally, draw a point at each object’s position.
Example 2.1 A collision
At the beginning of the film, Mission Control warns the astronauts that danger is coming. In real life, the United States Space Surveillance Network tracks objects in Earth’s orbit using a variety of sensing equipment. The team records the types of objects, when they were launched, their orbits, and their sizes. NASA uses this data to construct models that predict collisions like the one in Gravity.
While we don’t have a giant network of sensors and computers, we do have all the tools we need to figure out where the paths of our simulated Shuttle and debris cross. Let’s combine the equations describing each object’s path into a system of equations.
$$y = 0.5x + 50$$
$$y = -0.5x + 150$$
A system of equations is a set of related equations. There are infinitely many possible $$(x,y)$$ coordinate pairs, but only one solves both of these equations. For example, the point $$(50,75)$$ solves the equation for the Shuttle’s path.
$$75 = 0.5 \cdot 50 + 50$$
$$75 = 25 + 50$$
$$75 = 75$$
But does it also solve the equation for the debris’ path?
$$75 = -0.5 \cdot 50 + 150$$
$$150 = -25 + 150$$
$$150 \ne 125$$
No luck. We could try guessing and checking other coordinate pairs, but we might be at it for a while. There are many good techniques for solving systems of linear equations and we need a few additional ideas to apply them. Let’s start with a simpler case.
Math that uses letters as placeholders for numbers is known as algebra. Those placeholders, or variables, can show up anywhere we establish a relationship. Take the equation below as an example.
$2 + 3 = 5$
What if I wrote the following equation instead?
$x + 3 = 5$
We’ll use this example as a starting point for finding “unknown” values like $$x$$, also called solving an equation. Given $$x+3 =5$$, I want to come up with an algorithm that solves for $$x$$. Let’s simplify again.
We know the following equation relates $$2$$, $$3$$, and $$5$$.
$2 + 3 = 5$
Figure 2.1 Three hops to the right
You could express the same relationship another way by rearranging the equation a bit.
$2 = 5 - 3$
Figure 2.2 Three hops to the left
At this point you might think, “OK, the numbers $$2$$ and $$5$$ are $$3$$ units apart. But what does that have to do with finding unknowns?” Everything, as it turns out.
We can think of the original equation $$2+3=5$$ as running “forward” from the starting point $$2$$. In this view, the second equation $$2=5-3$$ runs “backward”. Running an operation backward is known as inverting the operation. And something interesting happens when we apply an operation and its inverse together.
$2 + 3 - 3 = 2$
Figure 2.3 Hopping back and forth
We’re right back where we started! Operations and their inverses undo each other; it’s like nothing happened at all. This fact is key to solving equations.
In math, the $$=$$ symbol is an ironclad statement of equality. The left-hand side must equal the right-hand side, now and always. If you make a change to the left, you’d better do the same to the right. Let’s take the original equation one more time and invert the $$+3$$ operation.
$2 + 3 - 3 = 5 - 3$
Or simplified:
$2 = 2$
Now, let’s apply identical reasoning to the equation with the variable $$x$$.
$x + 3 = 5$
$x + 3 - 3 = 5 - 3$
$x = 2$
OK, let’s see that process one more time with a different example.
$x - 2 = 10$
$x - 2 + 2 = 10 + 2$
$x = 12$
Exercise 2.1 Given the equation $$x-5=10$$, solve for $$x$$. Then, describe the algorithm you used to solve for $$x$$ in plain English.
Muḥammad ibn Mūsā al-Khwārizmī was a mathematician and scientist. Muḥammad’s book The Compendious Book on Calculation by Completion and Balancing laid the foundation for modern algebra. He studied a range of problems in economics and natural science.
Operator Precedence
Inverting addition and subtraction seems to work just fine, but what happens when you include other operations? Let’s take the path of our Shuttle.
$y = 0.5x + 50$
How would you determine the Shuttle’s $$x$$-coordinate when its $$y$$-coordinate is $$150$$?
$150 = 0.5x + 50$
It seems like we could invert this equation to solve for $$x$$, but I’m not certain of how to proceed now that multiplication is in the mix. Figuring this out requires a brief interlude to discuss the order, or precedence, of mathematical operations.
If you come across an expression like $$2 \cdot 3 + 4$$, the mathematical community has agreed that you should multiply before adding. You can compute the value produced in this example as follows.
$$2 \cdot 3 + 4$$
$$6 + 4$$
$$10$$
We’ve seen the computation run forward, so let’s go backward. The last operation applied was $$+4$$, so let’s invert that first.
$$10 = 2 \cdot 3 + 4$$
$$10 - 4 = 2 \cdot 3 + 4 - 4$$
$$6 = 2 \cdot 3$$
You can view the multiplication as $$2$$ multiplying $$3$$ or as $$3$$ multiplying $$2$$. I’ll go with the former and undo multiplication by $$2$$.
$$\frac{6}{2} = \frac{2 \cdot 3}{2}$$
$$3 = 3$$
How about we substitute one of the operands for a variable and solve for it?
$$10 = 2x + 4$$
$$10 - 4 = 2x + 4 - 4$$
$$6 = 2x$$
$$\frac{6}{2} = \frac{2x}{2}$$
$$3 = x$$
It’s been quite a journey, but I think we’re ready to plot a course for our Shuttle.
$$150 = 0.5x + 50$$
$$150 - 50 = 0.5x + 50 - 50$$
$$100 = 0.5x$$
$$\frac{100}{0.5} = \frac{0.5x}{0.5}$$
$$200 = x$$
Exercise 2.2 Find the $$x$$-coordinate of the Shuttle when its $$y$$-coordinate is $$180$$. Then, describe the algorithm you used to solve for $$x$$ in plain English.
The order of mathematical operations is bundled up nice and neat in the acronym PEMDAS.
PEMDAS
Parentheses
Exponents
Multiplication
Division
Subtraction
Let’s focus on MDAS for now as they are key to linear relationships. Multiplication and division have the same precedence. Consider the example below.
$3 \cdot 10 \div 5$
You could compute $$3 \cdot 10 = 30$$, then divide $$30 \div 5$$ to produce $$6$$. Or you could start by dividing $$10 \div 5 = 2$$ before multiplying $$3 \cdot 2$$ to produce $$6$$. There are multiple pathways to the correct answer!
There is a similar story for addition and subtraction.
$3 + 10 - 5$
You could compute $$3 + 10 = 13$$, then subtract $$13 - 5$$ to produce $$8$$. Or you could start by subtracting $$10 - 5 = 5$$ and add $$3 + 5 = 8$$. Once again, multiple pathways!
Exercise 2.3 Evaluate the expression $$10 \cdot 5 \div 2 + 5$$.
Exercise 2.4 Evaluate the expression $$10 ÷ 5 - 2 \cdot 5$$.
## Solving Systems¶
We’re on a tight schedule to be of any help to NASA. Let’s figure out the coordinates $$(x,y)$$ where the paths of the Shuttle and debris intersect. Recall the system of equations describing the scenario.
$$y = 0.5x + 50$$
$$y = -0.5x + 150$$
There are infinitely many points along each object’s path, and the variables $$x$$ and $$y$$ are placeholders for all of them. The Shuttle and debris were moving before they collided, and they continued moving afterward. Those variables $$x$$ and $$y$$ that solved one equation at a time must now solve both equations simultaneously.
Substitution
Algebraic techniques help us solve systems of equations because we really do mean that the $$y$$ in the first equation is the very same $$y$$ in the second equation. Consider the following simplified case for a moment.
$$a = 5$$
$$b = 5$$
$$a = b$$
The same transitive property of equality applies to systems of equations.
$$y = 0.5x + 50$$
$$y = -0.5x + 150$$
$$0.5x + 50 = -0.5x + 150$$
Now we have one equation with one unknown. Let’s solve it for $$x$$!
$$0.5x + 50 = -0.5x + 150$$
$$0.5x + 0.5x + 50 = -0.5x + 0.5x + 150$$
$$x + 50 = 150$$
$$x + 50 - 50 = 150 - 50$$
$$x = 100$$
And now that we’ve found $$x$$, we can substitute the value back into one of the original equations to find $$y$$.
$$y = 0.5 \cdot 100 + 50$$
$$y = 50 + 50$$
$$y = 100$$
We also could have gone with the other equation.
$$y = -0.5 \cdot 100 + 150$$
$$y = -50 + 150$$
$$y = 100$$
Example 2.2 Predicting collision
As you may expect, there’s more than one way to solve this problem. Let’s examine a different algorithm that combines equations to find solutions.
Elimination and Back Substitution
First, I’m going to determine the value of $$y$$ by eliminating the variable $$x$$ from an equation. Unlike the substitution algorithm, which rewrote one variable in terms of another, I’ll eliminate $$x$$ by adding the top equation to the bottom equation. As usual, let’s motivate this idea by considering a simpler case. Take the equations below.
$$20 = 4 \cdot 5$$
$$2x = x + x$$
I could add the two left-hand sides to produce $$2x + 20$$. On the right-hand side, I would have $$x + x + 4 \cdot 5$$. But do these combined expressions equal each other? We’ll follow this one step-by-step.
$$2x + 20 = x + x + 4 \cdot 5$$
$$2x + 20 = x + x + 20$$
$$2x + 20 = 2x + 20$$
Those $$=$$ symbols mean that the expressions on the left- and right-hand sides are equal. When we add the same thing to each side of an equation, we maintain equality. You may hear this called the additive property of equality when you’re out at parties. Now, let’s use this property to eliminate the variable $$x$$ and solve for $$y$$.
$$y = 0.5x + 50$$
$$y = -0.5x + 150$$
$$y + y = -0.5x + 0.5x + 150 + 50$$
$$2y = 200$$
$$\frac{2y}{2} = \frac{200}{2}$$
$$y = 100$$
The $$y$$-value we just found resulted from combining information stored in separate equations. This is the $$y$$-value both equations share. But what about $$x$$? Well, we know that $$y = 100$$, so let’s substitute that part of our solution back into the system.
$$100 = 0.5x + 50$$
$$100 - 50 = 0.5x + 50 - 50$$
$$50 = 0.5x$$
$$\frac{50}{0.5} = \frac{0.5x}{0.5}$$
$$100 = x$$
You’ve now seen two of many possible algorithms for solving systems of linear equations. Practice with them a bit before we build up the logical foundations you need to explore systems of linear inequalities.
Exercise 2.5 Solve the system of equations below using Substitution. Then, solve it using Elimination and Back Substitution. Describe how each algorithm works in plain English.
$$y = 0.25x + 100$$
$$y = -0.75x + 300$$
# 2.2 Logic¶
In the novel 1984, the main character is brainwashed into accepting that $$2 + 2 = 5$$ is true. The scene depicts the ultimate flex of power by an oppressive government. Let’s lay some logical foundations to ensure we always maintain our grasp on the truth.
## Boolean Algebra¶
You just tested a few different coordinate pairs $$(x,y)$$ to determine whether or not they solved an equation. The test could only have gone one of two ways: success or failure, yes or no, true or false. There is an entire branch of algebra called Boolean algebra dedicated to studying these two truth values, which we usually write as $$1$$ (true) and $$0$$ (false). There are not infinitely many truth values like there are numbers; there are only $$1$$ and $$0$$.
A truth value can correspond to a situation in the real world. For example, I could claim, “The sun is up”. This claim happens to be false in my neck of the woods as I write this sentence. I can express this idea using the variable $$s$$ to represent sunniness.
$s = 0$
Even though the sun has already set on this particular day, the sky above me is still momentarily deep blue. I can claim “The sky is blue” and express this blueness, $$b$$, matter-of-factly.
$b = 1$
OK, we have variables with assigned values. But what can we actually do with them?
Let’s begin by combining s and b using our first logical operation: conjunction, also known as AND. The expression s∧b means “$$s$$ is true AND $$b$$ is true”. The sky above my front porch is blue, but the sun is not up, so the combined statement is false. We could write this concisely as $$s \land b = 0$$.
There is a special set of diagrams called logic gates that depict the results of applying logical operations. Each logic gate has a distinctive shape. Below is the diagram for the AND logic gate.
Figure 2.4 The AND logic gate
One of the variables $$s$$ and $$b$$ is true, and we can test for such a condition using the disjunction operation, also known as OR. A disjunction is true if at least one of its operands is true. I could claim “The sun is up OR the sky is blue” and that would be true because $$b = 1$$. We could express idea this as $$s \lor b = 1$$.
Like AND, OR also has its own logic gate.
Figure 2.5 The OR logic gate
The following truth table organizes all of the facts we’ve established about the view of the sky from my front porch.
Table 2.1 A truth table’s view of my piece of sky
$$s$$
$$b$$
$$s \land b$$
$$s \lor b$$
$$0$$
$$1$$
$$0$$
$$1$$
The final logical operation we’ll discuss is negation, also known as NOT. The NOT operation simply flips a truth value from $$1$$ to $$0$$ or from $$0$$ to $$1$$. Let’s take the variable $$s$$ and negate it using the NOT operator, $$\lnot$$.
$$s = 0$$
$$\lnot s = 1$$
NOT is a unary operation, meaning we only apply it to a single truth value at a time. AND and OR are both binary operations, meaning we have to provide a pair of operands.
Not to be left out, NOT also has its own logic gate.
Figure 2.6 The NOT logic gate
And that’s all you need to get started with logic! You can compose logical operations just like you do arithmetic operations. For example, let’s figure out how to cross the street safely using logic. I’ll define the variables $$l$$ and $$r$$ to represent vehicle traffic from the left and traffic from the right, respectively.
If you were trying to cross a busy street, you would want to avoid vehicles. In logical terms, you would check to see that both $$l = 0$$ and $$r = 0$$. “No vehicles on the left? No vehicles on the right? OK, let’s go!” This condition is easily expressed by combining operations $$\lnot l \land \lnot r = 1$$.
Exercise 2.6 Complete the following truth table for two boolean variables $$x$$ and $$y$$.
$$x$$
$$y$$
$$x \land y$$
$$x \lor y$$
$$0$$
$$0$$
$$\,$$
$$\,$$
$$0$$
$$1$$
$$\,$$
$$\,$$
$$1$$
$$0$$
$$\,$$
$$\,$$
$$1$$
$$1$$
$$\,$$
$$\,$$
Exercise 2.7 Compute the value of the expression $$\lnot (1 \land 1)$$.
Exercise 2.8 Compute the value of the expression $$(1 \land 0) \lor (1 \lor 0)$$.
Exercise 2.9 Rewrite the following logic circuit as an equivalent logical expression.
Exercise 2.10 Complete the following truth table for two boolean variables $$x$$ and $$y$$. What do you notice?
$$x$$
$$y$$
$$\lnot (x \land y)$$
$$\lnot x \lor \lnot y$$
$$\lnot (x \lor y)$$
$$\lnot x \land \lnot y$$
$$0$$
$$0$$
$$\,$$
$$\,$$
$$\,$$
$$\,$$
$$0$$
$$1$$
$$\,$$
$$\,$$
$$\,$$
$$\,$$
$$1$$
$$0$$
$$\,$$
$$\,$$
$$\,$$
$$\,$$
$$1$$
$$1$$
$$\,$$
$$\,$$
$$\,$$
$$\,$$
Mary Everest Boole
Mary Everest Boole was a mathematician and teacher. Mary wrote multiple books on teaching mathematics and edited her husband George’s book on algebraic logic. She was an innovative teacher who advocated for a playful, cooperative approach to learning mathematics.
## Branching¶
Logic is a big deal for computation, from the way the machines are physically built to the way we program them. Conditional statements let us test conditions and make decisions while a program executes. For example, let’s revisit the collision scene from Gravity.
Example 2.3 Collision revisited
The simulation works fine, but it doesn’t really convey the full drama of the situation. Let’s revise our system a bit to account for the additional debris generated upon collision.
A Space Shuttle and debris in Earth’s orbit.
$$\mapsto$$ Outer space is an empty, infinite, two-dimensional surface. The Space Shuttle is a point moving along the straight path described by the equation $$y = 0.5x + 50$$. The debris is also a point moving along a straight path, this one described by the equation $$y = -0.5x + 150$$. After colliding, each object leaves a trail of smaller debris along its path.
The coordinates of both objects are ordered pairs of floating point numbers. The Space Shuttle’s initial position is $$(0,50)$$. The debris’ initial position is $$(0,150)$$.
First, test for collision and set the alpha value for the sky. Then, paint the sky. Next, update the position of each object. After that, select a pencil color & weight. Finally, draw a point at each object’s position.
Lucky for us, you already solved this system and know that the two objects collide at $$(100,100)$$. As the sketch continues running, the Shuttle and debris continue moving to the right. We can test for this using an if statement.
if (condition) {
// things to do if condition is true
}
If statements present a logical crossroads in a program. The first line of the if statement is called a header. The header begins with if, defines the condition to test in between a pair of parentheses (), then opens a pair of {}. If the condition is true, JavaScript will execute the set of statements between the {}, also known as the if statement’s body.
if (condition) {
thingOne() // this is part of the body
thingTwo() // this too!
}
thingThree() // this is not
In the Gravity example, we’re testing whether or not the value of the variable x is greater than 100. JavaScript has the following relational operators that work as you would expect for numbers.
Table 2.2 JavaScript’s relational operators
$$Math$$
Code
English
$$=$$
===
Equal to
$$\ne$$
!==
Not equal to
$$>$$
>
Greater than
$$<$$
<
Less than
$$\ge$$
>=
Greater than or equal to
$$\le$$
<=
Less than or equal to
Applying a relational operator produces a boolean value. For example, the expression 2 + 2 === 5 produces the boolean value false because, well, math. 2 + 2 === 4, on the other hand, produces the value true. These sorts of logical expressions are called boolean expressions.
if (x > 100) {
// things to do if x > 100
}
Example 2.4 A more impactful collision
Exercise 2.11 Duplicate your original collision sketch and add a few effects. How should the visual appearance of the Shuttle and debris change after impact?
You can also test multiple conditions together. Thinking back to the meteor sketch, how about we make the meteor glow red as it passes over the middle half of the canvas? In other words, when $$x \ge 50$$ AND $$x \le 150$$.
if (x >= 50 && x <= 150) {
stroke('tomato')
}
&& is one of JavaScript’s boolean operators along with || and !. These JavaScript operators function identically to the logical operators you just studied.
Recall our earlier $$2 + 2$$ example from the novel 1984. Notice you can test for the same condition using different boolean operators.
if (2 + 2 !== 4) {
resist()
}
or
if (!(2 + 2 === 4)) {
resist()
}
Example 2.5 A colorful meteor
Reading through this code, it isn’t immediately clear that ghostwhite is meant to be the default stroke color. You could make this more explicit by adding an else clause to your conditional statement. Here is an example of the syntax.
if (condition) {
// things to do if condition is true
} else {
// things to do if condition is false
}
The conditional statement now has two distinct pathways, or branches, that may be followed depending on the truth value of the condition.
Figure 2.7 Flow of execution with two branches
You could reorganize your sketch to reflect this structure like so.
At this point, you might say, “Branches seem useful, but what if I want more than two in my program?” Say no more! You can chain conditionals together using an else if statement (a combination of “else” and “if”). In a chained conditional, conditions are tested in the order they are written, and only the first branch whose condition is true will execute.
if (condition1) {
thingOne()
} else if (condition2) {
thingTwo()
} else {
thingThree()
}
Example 2.6 A multicolor meteor
Exercise 2.12 Change the previous example so that the following conditions are tested in this order. Can you explain what happened?
if (x >= 50) {
stroke('tomato')
} else if (x >= 150) {
stroke('crimson')
} else {
stroke('ghostwhite')
}
Exercise 2.13 Duplicate one of your previous sketches and modify its behavior using conditional statements. Use at least one relational and one other boolean operator.
# 2.3 Iteration¶
The logical building blocks you assembled in the last two sections let you create branches and make decisions in your programs. In this section, we’ll use many of the same building blocks to form another type of control flow: repetition.
## while statements¶
In Exercise 1.18, you drew a constellation by calling the point() function repeatedly with different arguments. Let’s revisit this exercise using the constellation Orion as a starting point. We’ll focus on Orion’s Belt, which consists of the stars Alnitak, Alnilam, and Mintaka.
Orion’s Belt
$$\mapsto$$ Outer space is an empty, infinite, two-dimensional surface. Stars are points of light within this plane.
Star coordinates are ordered pairs of floating point numbers.
Table 2.3 The “Orion’s Belt” data set
Star Name
$$x$$
$$y$$
Alnitak
$$25$$
$$100$$
Alnilam
$$100$$
$$100$$
Mintaka
$$175$$
$$100$$
First, paint the night sky. Then, select a pencil color & weight. Finally, draw a point at each star’s position.
Example 2.7 Orion’s Belt
The stars Alnitak and Mintaka are actually star systems; each system is made up of multiple stars orbiting one another. I’ll use this fact as my creative license to adjust Orion’s Belt a little. For starters, how about we draw Orion’s Belt with all of the major stars in each system? Let’s keep things simple by assuming all of the stars are the same size and are aligned horizontally in the sky.
Table 2.4 Expanded “Orion’s Belt” data set
Star Name
$$x$$
$$y$$
Alnitak Aa
25
100
Alnitak Ab
50
100
Alnitak B
75
100
Alnilam
100
100
δ Ori Aa1
125
100
δ Ori Aa2
150
100
δ Ori Ab
175
100
The visual result looks good, but the code worries me a little. Notice that I wrote seven nearly identical copies of the same statement to draw the stars. This approach works fine to get started but imagine writing a program that needs to repeat an instruction dozens of times. Or millions of times. Writing each variant by hand would be tedious and error prone. JavaScript’s while statement makes repetition, or iteration, simple.
while (condition) {
// this is the loop body
// statements in here repeat while condition is true
thingOne()
thingTwo()
}
while statements are structured similarly to if statements; they have a header with a condition and a body with code that might be executed. Each statement in the body executes in order, from top to bottom, repeatedly, until the condition in the header is false. Iterative control structures like this are commonly known as loops.
Figure 2.8 Flow of execution in a while loop
Let’s use a while loop to simplify our sketch of Orion’s Belt. Reviewing the previous example, the only difference between the stars is their $$x$$-coordinates. We know where our $$x$$-coordinates start, where they stop, and how much space is between them. This is all the information we need to simplify our work by using a loop.
Example 2.9 Orion’s Belt with iteration
Not too shabby! Given an initial value for x, the while statement draws a point, then increments x by 25, and repeats this process until x is greater than 175.
We could change one line of code from the previous example to pack twice as many stars in the same region of sky.
Example 2.10 Bedazzling Orion’s Belt
Exercise 2.14 Modify the sketch above to draw a row of stars in a different arrangement. What is the initial value of your loop variable x? What condition do you test to end the loop’s execution? How much do you increment x by during each iteration?
You might say, “OK, but what if I turned my head a little? Could I draw the stars along a vertical line instead?” Sure! In this case, you could keep the value of x constant while varying y.
Example 2.11 Switching axes from $$x$$ to $$y$$
## Infinite Loops¶
while statements are powerful tools that should be used with care. Consider the example below.
while (true) {
thingOne()
thingTwo()
}
true is a keyword in JavaScript that corresponds to a boolean value of, you guessed it, true. If a while statement’s condition is always true, then it will continue looping forever, thus creating an infinite loop. Unintended infinite loops are bad news. Let’s examine a slightly modified version of the loop from the previous sketch.
let y = 25
while (y <= 175) {
point(100, y)
}
y += 25 // oops
Notice that I incremented y outside of the loop body. This means that y isn’t incremented after each iteration. Instead, the value of y will always be 25, which is always less than or equal to 175, so the condition 25 <= 175 is always true. The loop never stops executing! If you ran this code, your web browser might give you a friendly notification to stop the sketch before it grinds your computer to a halt.
Exercise 2.15 The code snippet below is meant to draw a horizontal row of points across the canvas. Instead, it creates an infinite loop. Identify the error and fix it. Explain the problem and your solution in plain English.
let x = 0
let y = 100
while (x < 200) {
point(x, y)
x += 10
}
Exercise 2.16 Create a sketch that uses a while statement to draw points along a diagonal line. Review the code snippet below as a hint.
let x = 0
while (x < 200) {
// >> compute y here
point(x, y)
x += 10
}
Exercise 2.17 In Chapter 1, we defined an algorithm for multiplying two integers $$a \cdot b$$ as repeated addition $$b + b + ... + b$$. Fill in the missing code below to express the same algorithm in JavaScript using a while loop.
let a = 5
let b = 3
let product = 0
let i = 0
while (i < a) {
// >> compute product here
i += 1
}
Exercise 2.18 Review the algorithms for integer division and exponentiation, then implement them in JavaScript using a while loop. The subtraction assignment -= and multiplication assignment *= operators might be helpful.
# 2.4 Systems of Inequalities¶
We began this chapter by analyzing a single linear equation in slope-intercept form: $$y = mx + b$$. You could substitute any real number for $$x$$ and compute the corresponding value of $$y$$, making the ordered pair $$(x,y)$$ a solution to the equation. Let’s review a concrete example.
Given the following linear equation
$y = 0.5x + 100$
compute the value of $$y$$ when $$x = 50$$.
$$y = 0.5 \cdot 50 + 100$$
$$y = 25 + 100$$
$$y = 125$$
One solution to this equation is located at $$(50,125)$$. What if we tried $$(50,126)$$ instead?
$$126 = 0.5 \cdot 50 + 100$$
$$126 = 25 + 100$$
$$126 \ne 125$$
It turns out $$(50,126)$$ is not a solution to this particular equation, but there are infinitely many other solutions. For example, we could find solutions to the left and right of $$(50,125)$$ at $$x = 49$$ and $$x = 51$$.
$$y = 0.5 \cdot 49 + 100$$
$$y = 24.5 + 100$$
$$y = 124.5$$
$$y = 0.5 \cdot 51 + 100$$
$$y = 25.5 + 100$$
$$y = 125.5$$
You could use a while loop to quickly compute and visualize all of the solutions in the interval $$0 \le x \le 200$$.
Example 2.12 Visualizing solutions to a linear equation
Mathematicians often express a change in the value of a variable with the Greek letter $$\Delta$$, pronounced “delta”. Using this notation, we can express a change in the variable $$x$$ as $$\Delta x$$. I declare the variable dx on line 11 and use it to increment x on line 15 as a nod to our standard mathematical notation.
## Testing Solutions¶
OK, the solutions to a linear equation generate a line. I wonder what shapes other linear relationships make. Let’s take the previous example and swap out the $$=$$ symbol for a $$>$$.
$y > 0.5x + 100$
You can test solutions to this linear inequality just as you did with linear equations. For example, let’s see if the coordinate pair $$(50,125)$$ produces a truth value of 1 when we substitute the values into our inequality.
$$125 > 0.5 \cdot 50 + 100$$
$$125 > 25 + 100$$
$$125 > 125$$
Uh oh. We evaluated the right-hand side of the inequality and produced a value of $$125$$. But that resulted in a false statement; $$125$$ is not greater than itself. We can conclude that $$(50,125)$$ isn’t a solution to this inequality. How about we move along the $$y-axis$$ a little to $$(50,126)$$?
$$126 > 0.5 \cdot 50 + 100$$
$$126 > 25 + 200$$
$$126 > 125$$
Success! Let’s go a little further along the $$y$$-axis to $$(50,127)$$.
$$127 > 0.5 \cdot 50 + 100$$
$$127 > 25 + 200$$
$$127 > 125$$
Interesting. Let’s try one more coordinate pair $$(50,128)$$ to see if this pattern holds.
$$128 > 0.5 \cdot 50 + 100$$
$$128 > 25 + 200$$
$$128 > 125$$
When $$x = 50$$, we can pair it with any $$y > 125$$ to solve the inequality $$y > 0.5x + 100$$. You could automate this sort of test with a while loop.
Example 2.13 Testing solutions to a linear inequality
The sketch above fixes the value of x at 50 and tests solutions to the inequality for all y values in the interval $$0 \le y < 200$$. Solutions along this column are colored black while other points are colored ghostwhite.
This is the first example we’ve seen of a while loop that includes an if statement in its body. You can put (almost) whatever code you want in the body of a while loop: function calls, arithmetic operations, if statements, and even other while loops. This last option opens up many interesting possibilities.
## Nested Loops¶
You just tested hundreds of possible solutions to the inequality $$y > 0.5x + 100$$ when $$x$$ was fixed at $$50$$. Let’s fully automate the process of testing solutions by iterating over the canvas’ $$x$$-axis just like we’re doing with its $$y$$-axis.
A quick note about the algorithm we are about to run: it is very inefficient and would normally grind your computer to a halt. By default, p5 executes each statement you place in the body of the draw() function in order, from top to bottom, repeatedly, about $$60$$ times per second. Behind the scenes, you can imagine that the code you write in draw() is executing in the body of a while statement.
setup() // all of your code bundled up
while (true) {
draw() // all of your code bundled up
}
Testing individual solutions to a linear inequality requires computing once on each $$(x,y)$$ coordinate pair before drawing a point on the canvas–that’s roughly $$200 \cdot 200 = 40000$$ operations! The algorithm is slow and produces the same results every time, so there is no need to repeat it.
In the next example and those that follow, I will call the noLoop() function once in setup() like so.
function setup() {
createCanvas(400, 400)
noLoop()
}
By calling noLoop(), you change p5’s behavior so that draw() only executes a single time. You can imagine p5 running the following code instead.
setup() // all of your code bundled up
draw() // all of your code bundled up
Now, we can compute once on each $$(x,y)$$ coordinate pair, draw the corresponding point, and produce a single image. This adjustment should keep your computer happy, but it may still take a moment for the result to appear.
Example 2.14 Visualizing a linear inequality
The control structure you just created is called a nested loop. The outer loop increments the variable x while the inner loop increments the variable y and tests for solutions along each column of your canvas. And the result of all that computation? It turns out the set of $$(x,y)$$ coordinate pairs that solve our inequality, known as the solution set, forms a triangle-shaped region with the edges of the canvas. Neat! Can we make a rectangle?
Example 2.15 The rectangle inequality
Exercise 2.19 Modify the previous example to draw a new five-sided shape with your solution set. Closed shapes made by connecting three or more straight lines are known as polygons, and a five-sided polygon is known as a pentagon.
Exercise 2.20Ellsworth Kelly’s “Austin” is a serene little chapel located on the campus of the University of Texas at Austin. Its walls feature a series of fourteen black and white marble panels that look suspiciously like linear inequalities. Use Kelly’s panels as inspiration for your own series of abstract images. How many images will you include in your series? What colors will you use? What shapes will you create?
Drawing with linear inequalities opens up a dizzying number of creative possibilities. But what if you just wanted to draw a rectangle in the middle of your canvas? Meeting this challenge requires expanding our modeling toolkit yet again.
## Solving Systems¶
When you solved your first system of linear equations, you found the point $$(x,y)$$ where two lines intersected. In other words, you found the only ordered pair $$(x,y)$$ that solved both equations simultaneously. We’ll follow a very similar train of thought to solve systems of linear inequalities.
For starters, let’s try consider the system $$y < x + 150$$ and $$y > 75$$. We can test possible solutions $$(x,y)$$ against both inequalities using the && operator.
Example 2.16 A system of linear inequalities
OK, but how would we draw that rectangle in the middle of the canvas? You can think of a rectangle as the set of points between a pair of $$x$$-values and a pair of $$y$$-values. For example, we could take all of the points where $$x > 75$$ AND $$x < 125$$ AND $$y > 100$$ AND $$y < 160$$.
Example 2.17 A more constrained system
At this point you might say, “This is great! But how do I draw multiple shapes?” Simple: define multiple systems of inequalities. I’d like to frame this part of the discussion by studying the work of another abstract painter, Piet Mondrian.
Figure 2.9 “Composition II in Red, Blue, and Yellow” Courtesy Wikimedia Foundation
The painting “Composition II”.
$$\mapsto$$ Each colored region is the solution set to a system of linear inequalities.
The boundaries for each system of inequalities are either vertical or horizontal lines.
Table 2.5 The “Composition II” data set
Name
Left $$x$$
Right $$x$$
Bottom $$y$$
Top $$y$$
Blue corner
0
35
155
200
Yellow corner
188
200
183
200
Red corner
40
200
0
150
Stripe 1
35
40
0
200
Stripe 2
0
35
60
72
Stripe 3
0
200
150
155
Stripe 4
183
188
155
200
Stripe 5
188
200
173
183
First, prime the canvas by painting it ghostwhite. Next, select the paint color by testing whether a point solves a system of inequalities. Finally, paint the point. Repeat for every point on the canvas.
Example 2.18 “Composition II”
Exercise 2.21 Spend a few minutes exploring WikiArt and find an abstract painting or painter who inspires you. Hilma af Klint and Mark Rothko are personal favorites of mine. Then, create a new sketch that uses systems of inequalities to draw your own abstraction. Is your sketch completely abstract or is it based on an object, place, emotion, etc.? What do you like most about your sketch? What was challenging about creating it?
Exercise 2.22 Visualize integer multiplication by building on your solution to Exercise 2.17. Use the starter code below to snap (very tiny) blocks together.
The noStroke() function removes edges p5 draws around text; doing so can make it easier to read some fonts. The fill() function sets the interior color of text. Note that the text() function takes three arguments: the text to be displayed, and the $$x$$- and $$y$$-coordinates where the text should appear.
The control structures you just studied make it possible to construct many useful computations. In the next chapter, we’ll raise the level of abstraction by bundling computations into functions you define. We’ll also have fun drawing with the many built-in shapes that p5 provides.<|endoftext|>
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What's cool about the universe is that if you stare at nothing long enough, you'll see something big. That's what scientists have done with the Hubble Space Telescope a few times, creating the enchanting Hubble Deep Field images with swarms of galaxies that have opened our eyes to the immensity of the cosmos.
But sometimes the universe comes to you, as it did last April 23 when a gamma ray burst now named GRB 090423 was detected by NASA's orbiting Swift satellite. Gamma ray bursts are the most violent explosions known, and occur when a star perhaps 20 times the size of our sun exhausts its fuel. Its core collapses under the force of the star's gravity and becomes a black hole while outer layers fly off. Then matter falling into that hole produces a powerful jet of gamma rays that flare outward in opposite directions. If one of those jets is pointed at us, it is so energetic that (we now know) we can detect it from a distance of 13.1 billion light years, which is how far away this burst originated. That's more than 95 percent of the distance across the known universe, and 190 million light years farther away than the previous record holder.
Astronomers estimate the explosion occurred 630 million years after the Big Bang, when the first stars were forming. Astronomers also think the first stars formed no earlier than 150 million years after the Big Bang. So GRB 090423 would have been only a few hundred million years old when it blew. That sounds like a big number until you compare it to the age of stable stars such as our sun that burn for 10 billion years before they swell into a red giant, then smolder into a white dwarf for another five billion or more.
Here's a good video on the Nature web site that explains it.<|endoftext|>
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Calculating the Sounds of Music
31.05.2017Calculating the Sounds of Music
How did machines come to make music? A short history of computer music, in the words of the curator of the “Good Vibrations” exhibition.
At a concert they attended together in the late 1950s, electrical engineer Max Mathews and his boss John Pierce had a groundbreaking idea: why not write a computer program that could generate musical sounds? They had heard works by Arnold Schönberg and Artur Schnabel that evening and, while impressed by the former, they were rather disappointed by the latter. Pierce is said to have told Mathews that a computer could do a better job and instructed him to write a program that could do just that. At the time, they were both working at Bell Laboratories in Murray Hill, New Jersey.
About a year later, Mathews presented a musical composition. It lasted a mere seventeen seconds, but it had been generated by the program that he had written: Music I. A huge IBM 704 mainframe computer was needed to perform the piece. That event, in 1957, marked the dawn of the age of computer music. Pierce's vision of a convincing musical performance wouldn’t be realized for a long time to come, as the program was initially limited to experimental applications and the sounds that it generated were very simple. What’s more, the computation times were long – several times longer than the sound sequence itself – and the program was run using low-level instructions, so it remained well out of reach for any musician not trained in programming.
The next several decades saw the development of a wide range of systems for programming music, due in no small measure to research work done at institutions such as Princeton, Stanford, MIT (Massachusetts Institute of Technology), the University of Illinois, and the Institute of Research and Coordination in Acoustics/Music (IRCAM) in Paris. As the hardware achieved faster processing times and became easier to operate, these systems gradually became more popular among musicians.
Computers learned to play music – and compose it
From the very beginning, a distinction was made between two conceptually different types of use: that of composing with the aid of a computer and that of generating sounds with the aid of a computer. In the early stages of computer music, while Bell Labs was experimenting with the generation of sound using Music I, others were exploring the use of computers as a purely compositional tool. In 1957, the same year that Music I was presented, the ILLIAC I computer at the University of Illinois, programmed by Lejaren Hiller and Leonard Isaacson, produced the “Illiac Suite” (later renamed “String Quartet No. 4”). This was the first score to be composed by a computer. For its acoustic rendition, however, the decision was made to use a traditional string quartet – a practice adopted somewhat later by Iannis Xenakis for the performance of his own computer-generated compositions.
Computer processing speeds continued to increase until, around the turn of the millennium, it became possible to synthesize sound in real time. This advancement, alongside the practice of algorithmic composition, increasingly blurred the boundaries between the two types of use to the extent that, today, it is sometimes difficult to distinguish between the process of composition and that of real-time improvisation (using programs that are themselves essentially compositions).
In the late 1960s, attention turned to interface issues. One question was how to improve computer control methods by incorporating additional sensory channels – for example, cathode ray tube monitors, a focus of research at MIT. And, conversely, how could computers be used to operate external sound generators – for example, analog synthesizers, which Max Mathews integrated into his GROOVE system from 1970.
On the technical side, these questions were essentially answered by 1983 at the latest, with the introduction of the MIDI standard – a communications protocol for the transmission of musical data and control signals. After that, it was relatively easy to use computers as controllers for electronic musical instruments and, conversely, as sound generators for external controllers, which were being developed in great numbers.
Computers became standard equipment in recording studios
At the same time, the computer – mostly in the form of personal computers – was also gaining hold in commercial music studios, where its efficient emulation of traditional analog media infrastructure soon made it indispensable. Some of the most popular emulation programs – called digital audio workstations (DAW) – were Cubase, Logic, and ProTools. Soon afterwards, the German company Steinberg established the VST (Virtual Studio Technology) standard, and the integration of software instruments in production processes became a regular feature in DAW systems.
The use of computers to generate sound is now commonplace, and has been for quite some time. Programs for generating, editing, recording, and playing sounds have long been part of the virtual toolbox used by every producer – and many musicians. Thanks to computer technology, it is now increasingly common for a single individual to master both these occupations, which were once completely independent.
The degree to which computers are used in the creative process varies greatly. They are probably employed most commonly as compact and versatile studio tools, for the most part in recording and production processes. But it is not unusual for them to be used as musical instruments, or at least as part of instrumental arrangements in which they are tasked with the instruments’ sound synthesis.
In live coding, for example (a form of live performance art that has been around for about fifteen years), sounds are generated and controlled in real time by writing source code in specialized languages such as Max/MSP, SuperCollider, ChucK or Sonic Pi. The same programming environments also are used in experimental digital interface development as sound engines – that is, as sound-generating components that make it easy to explore a wide variety of sounds. In the context of popular music, standardized configurations of hardware and software, such as Ableton’s “Live” (in conjunction with the dedicated hardware controller “Push”) or Native Instruments’ “Maschine”, are performance-ready, versatile instrumental setups that are marketed by their makers as contemporary types of musical instrument.
Lastly, following in the tradition of Hiller und Isaacson, people are experimenting with many different processes of algorithmic composition. They want to see how computers calculate compositions by themselves on the basis of different sets of rules and, of course, they want to hear the results in acoustic form. Unlike in the 1950s, however, these compositions are usually played or printed by the computer itself and output either as a sound file or as a score.
What taste in music do machines have?
The release of several computer-generated pop songs made headlines last year. The songs were the product of a Sony AI (artificial intelligence) program called Flow Machines, which works with a music database containing a huge number of scores from different genres. The self-learning program analyzes and identifies the characteristic parameters of a particular style of music, say that of the Beatles, and synthesizes the results in a song – in the case of the Beatles-inspired release, that song was Daddy’s Car. The experiment was not entirely free of human influence, however: French song writer Benoit Carré added the finishing touches in the form of lyrics, arrangement, and production. He had the following to say about the shared creative process: “The machine does not pass any judgment on what it produces. This forces me to question my own decisions. Why this melody and not the previous one? With every step I make, I see my shadow taking shape – mine or Flow Machines’, I don’t know which.” In light of such intensely cooperative compositional practice, Tom Jenkinson’s vision of music making as a collaborative process in which machines and their users are equal partners – as laid out in his 2004 manifesto Collaborating with Machines – would appear to have come to fruition.<|endoftext|>
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When highly energetic charged particles from the Sun enter the Earth’s upper atmosphere, they collide with oxygen and nitrogen atoms. This stimulates the atoms into a temporary excited state, and on returning to their normal resting state, the atoms emit visible light. Oxygen emissions are green or brownish-red, whereas nitrogen emissions are blue or red.
As a result, a natural light display – an aurora – can be seen during the hours of darkness in the sky at high latitudes. During daylight hours, the aurora cannot be readily seen.
Normally, these spectacular light displays are restricted to Arctic and Antarctic regions, but during a geomagnetic storm, the auroral zone reaches lower latitudes.
In the southern hemisphere, auroral displays are referred to as the Aurora Australis, whereas in the northern hemisphere, it is called the Aurora Borealis.
The word ‘aurora’ is named after the Roman goddess of dawn Aurora.
Photo courtesy of Fraser Gunn<|endoftext|>
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## Monday, January 8, 2018
Using Mathematics many different types of puzzle can be created. In some of these Maths Brain Teasers one has to find the missing number. In some of the Math Puzzles one has to find the next number in the series. In some of the Maths Picture Puzzles one has to solve the Mathematical equations where variables are represented by different symbols. Today's Maths IQ Questions are based on Mathematical equations.
## How to Solve Maths IQ Questions
In each of these Maths Brain Teasers, some Mathematical equations are given. Using these equations you have to find the value of different variables in these equations and then solve the last equation.
1. Can you solve this New Year Maths equation?
2. Can you solve this Maths Equations?
3. Can you solve this Maths Picture Puzzle?
4. Can you solve this shapes Maths Puzzle?
5. Can you solve this Maths Picture Equation?
You need good Mathematical knowledge to solve these Maths Picture Equation Puzzles. Do checkout below Mentioned Maths and Logic Puzzles to shake your brain.
## Answers of Maths IQ Questions
A1.
Let us use the following symbols:
Clock...........L
Snowman.......S
Calendar.......C
L-S= 18...(1)
C- S = 6......(2)
C+ L= 28.....(3)
Subtracting eq (1) from (2)
C-L = -12......(4)
2C= 16
C= 8, From eq (2) S=2
from (1), L=20
C+L+S= 30.
A2.
Let us use the following symbols:
S..........Star
N..........Snowman
P......... Penguin
S+N = 15...(1)
N+P=20....(2)
S-N=5....(3)
2S = 20, S= 10
substituting value of S in (1).
N=5
Substituting value of N in (2)
P=15
Therefore S+N+P = 30
A3.
Let us use the following symbols:
C.........Cap
L.........Clock
N ........New year
Given:
CL+LN+NC=26..........(1)
CL+LN =18................(2)
C + N = 6.................(3)
Substituting value from (2) in (1).
NC= 8............(4)
From equation (3)
N= 6-C,
Substituting value of N in (4)
(6 -C)x C=8,
C² - 6C+ 8 = 0
C²-4C-2C +8=0
C(C-4)-2(C-4)=0
C=2, C=4
If C=2 , N=4 and if C=4, N=2 and
from eq (2)
CL+LN =18., 2L+4L=18, L= 3
CxLxN= 2x3x4 = 24.
A4.
14
Let me use the following notations for the figures
C........circle
P..........pentagon
T..........Triangle
Given
C+C+C=12..... 3C = 12.......(1)
C+P =19........ 4+P = 19....(2)
T-P = 10......... T- P = 10....(3)
From (1).... C = 4,
From (2).....P = 15
From (3)....T = 25
Therefore
C+T-P = 14
A5.
150
Let me use the following notations for the figures
S........star
C........circle
F.........flower
Given:
S x S x S = 27, S= 3...............(1)
S x S+ C= 14, 3x3 +C= 14....(2)
C x C +F= 35, ........................(3)
From (1), S=3
From (2) , C=5
From (3) F=35-25= 10
Therefore, SxFxC= 3x10x 5= 150<|endoftext|>
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by Janice Florent
Computers can read text on a screen but images, graphs, and charts are meaningless to visually impaired users. Alternative text (alt text) is an alternate method for supplying information about images, graphs, and charts to users who are visually impaired.
Adding alt text to images, graphs, and charts is an important part of making them accessible. Most images, graphs, and charts can be made accessible using alt text descriptions. However, complex images, graphs, and charts require more detailed description than the limited one or two brief sentences that are used in the alt text.
What should you do when you have a complex image, graph, or chart whose meaning cannot be conveyed with alt text alone? There are several ways to handle complex images where a short description is not possible. The best solution is to include a thorough description of the complex image in the content of the page, immediately before or after the image. If you don’t want to add more content to your page, another preferred alternative is to create another web page with the thorough description of the complex image and link to it near the image.
You may be wondering what exactly should you include in the thorough description. The Diagram Center (Digital Image And Graphic Resources for Accessible Materials) is an excellent resource that provides comprehensive guidelines to make it easier for you to make complex images accessible to all learners.
Here are a few additional resources to help you with describing complex images:
- Effective practices for description of science content
- How do we access meaning in art? (Describing art images)
- Open University's guidelines for describing visual teaching materials
It is extremely important for students with disabilities to have access to accessible course content. Describing complex images utilizing these tips is good course design. Even though you may not have a student with a disability currently enrolled in your course, you will find students without disabilities will take advantage of accessible content as well.<|endoftext|>
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Upper and Lower Bounds
Key Questions
Best way is to find it from the graph of the function.
Explanation:
The range of a function can be found in many ways , best way and a concrete way is by plotting its graph and determining.
How you draw the graph depends on you, a generic way is by differentiating the function to get the critical points or the points of maxima or minima . Then you double differentiate the function to see if the critical point is a maxima or minima by putting the critical point in the double differentiated function. If the output is positive then is a minima , if its negative then its a maxima and if its zero then the test fails,we have to go for high derivative tests. The global maxima points give you the upper bound.
Also we can determine the point of inflection for the given curve to determine the convexity or concavity changes of the given function to make the graph more precise .
Thus, having plotted the graph we can tell the upper bound of the function by merely looking at the maximum value of y for all values of x.
• A function $f$ is bounded in a subset $U$ of its domain if there exist constants $M , m \in \mathbb{R}$ such that
$m \le f \left(x\right) \le M ,$ for all $x \in U .$
For example,
1. $f \left(x\right) = \sin \left(x\right)$ is bounded in $\mathbb{R}$ because
$- 1 \le \sin \left(x\right) \le 1 ,$ for all $x \in \mathbb{R}$.
2. $f \left(x\right) = {x}^{2}$ is bounded in $\left[0 , 1\right]$ because
$0 \le {x}^{2} \le 1 ,$ for all $x \in \left[0 , 1\right] .$
See below.
Explanation:
First of all, let's get rid of infinities: a function can tend to $\setminus \pm \setminus \infty$ either at the extreme points of its domain or because of some vertical asymptote.
So, you should first of all check
${\lim}_{x \setminus \to {x}_{0}} f \left(x\right)$
for every point ${x}_{0}$ at the boundary of the domain. For example, if the domain is $\left(- \setminus \infty , \setminus \infty\right)$, you should check
${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right)$
If the domain is like $\setminus m a t h \boldsymbol{R} \setminus \setminus \setminus \left\{2 \setminus\right\}$ you should check
${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right) , \setminus q \quad {\lim}_{x \setminus \to {2}^{\setminus} \pm} f \left(x\right)$
and so on. If any of these limits is $- \setminus \infty$, the function has no finite lower bound.
Else, you can check the derivative: when you set $f ' \left(x\right) = 0$, you will find points of maximum of minimum. For every $x$ which solves $f ' \left(x\right) = 0$, you should compute $f ' ' \left(x\right)$. If $f ' ' \left(x\right) > 0$, the point is indeed a minumum.
Now, in the most general case, you have a collection of points ${x}_{1} , \ldots , {x}_{n}$ such that
$f ' \left({x}_{i}\right) = 0 , \setminus q \quad f ' ' \left({x}_{i}\right) > 0$ for every $i = 1 , . . , n$
Which means that they are all local minima of your function. The lower bound of the function, i.e. the global minimum, will be the smallest image of those points: you just need to compare
$f \left({x}_{1}\right) , \ldots , f \left({x}_{n}\right)$, and choose the smallest one.
• If real numbers $L$ and $U$ satisfy the inequalities
$L \le f \left(x\right) \le U$,
for all $x$ in the domain of $f$, then $L$ is a lower bound of $f$ and $U$ is an upper bound of $f$.
I hope that this was helpful.<|endoftext|>
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# Eigenvalues of a Hermitian Matrix are Real Numbers
## Problem 202
Show that eigenvalues of a Hermitian matrix $A$ are real numbers.
(The Ohio State University Linear Algebra Exam Problem)
We give two proofs. These two proofs are essentially the same.
The second proof is a bit simpler and concise compared to the first one.
## Proof 1.
Let $\lambda$ be an arbitrary eigenvalue of a Hermitian matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Then we have
$A\mathbf{x}= \lambda \mathbf{x}. \tag{*}$ Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we obtain
\begin{align*}
\bar{\mathbf{x}}^{\trans}(A\mathbf{x}) &= \bar{\mathbf{x}}^{\trans}(\lambda \mathbf{x})\\
&=\lambda \bar{\mathbf{x}}^{\trans}\mathbf{x}\\&
=\lambda ||\mathbf{x}||.
\end{align*}
We also have
\begin{align*}
\bar{\mathbf{x}}^{\trans}(A\mathbf{x})=(A\mathbf{x})^{\trans} \bar{\mathbf{x}}=\mathbf{x}^{\trans}A^{\trans}\bar{\mathbf{x}}.
\end{align*}
The first equality follows because the dot product $\mathbf{u}\cdot \mathbf{v}$ of vectors $\mathbf{u}, \mathbf{v}$ is commutative.
That is, we have
$\mathbf{u}\cdot \mathbf{v}=\mathbf{u}^{\trans}\mathbf{v}=\mathbf{v}^{\trans}\mathbf{u}=\mathbf{v}\cdot\mathbf{u}.$ We applied this fact with $\mathbf{u}=\bar{\mathbf{x}}$ and $\mathbf{v}=A\mathbf{x}$.
Thus we obtain
$\mathbf{x}^{\trans}A^{\trans}\bar{\mathbf{x}}=\lambda ||\mathbf{x}||.$ Taking the complex conjugate of this equality, we have
$\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}=\bar{\lambda} ||\mathbf{x}||. \tag{**}$ (Note that $\bar{\bar{\mathbf{x}}}=\mathbf{x}$. Also $\overline{||\mathbf{x}||}=||\mathbf{x}||$ because $||\mathbf{x}||$ is a real number.)
Since the matrix $A$ is Hermitian, we have $\bar{A}^{\trans}=A$.
This yields that
\begin{align*}
\bar{\lambda} ||\mathbf{x}|| &\stackrel{(**)}{=} \bar{\mathbf{x}}^{\trans}A\mathbf{x}\\
& \stackrel{(*)}{=}\bar{\mathbf{x}}^{\trans} \lambda \mathbf{x}\\
&=\lambda ||\mathbf{x}||.
\end{align*}
Recall that $\mathbf{x}$ is an eigenvector, hence $\mathbf{x}$ is not the zero vector and the length $||\mathbf{x}||\neq 0$.
Therefore, we divide by the length $||\mathbf{x}||$ and get
$\lambda=\bar{\lambda}.$
It follows from this that the eigenvalue $\lambda$ is a real number.
Since $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that all the eigenvalues of the Hermitian matrix $A$ are real numbers.
## Proof 2.
Let $\lambda$ be an arbitrary eigenvalue of a Hermitian matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Then we have
$A\mathbf{x}= \lambda \mathbf{x}. \tag{*}$
Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we obtain
\begin{align*}
\bar{\mathbf{x}}^{\trans}(A\mathbf{x}) &= \bar{\mathbf{x}}^{\trans}(\lambda \mathbf{x})\\
&=\lambda \bar{\mathbf{x}}^{\trans}\mathbf{x}\\&
=\lambda ||\mathbf{x}||.
\end{align*}
Now we take the conjugate transpose of both sides and get
$\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}=\bar{\lambda}||\mathbf{x}||. \tag{***}$ Since $A$ is a Hermitian matrix, we have $\bar{A}^{\trans}=A$.
Then the left hand side becomes
\begin{align*}
\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}&=\bar{\mathbf{x}}^{\trans}A\mathbf{x}\\
& \stackrel{(*)}{=} \bar{\mathbf{x}}^{\trans}\lambda\mathbf{x}\\
&=\lambda ||\mathbf{x}||. \tag{****}
\end{align*}
Therefore comparing (***) and (****) we obtain
$\lambda ||\mathbf{x}||=\bar{\lambda}||\mathbf{x}||.$ Since $\mathbf{x}$ is an eigenvector, it is not the zero vector and the length $||\mathbf{x}||\neq 0$.
Dividing by the length $||\mathbf{x}||$, we obtain $\lambda=\bar{\lambda}$ and this implies that $\lambda$ is a real number.
Since $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that every eigenvalue of the Hermitian matrix $A$ is a real number.
## Corollary
Every real symmetric matrix is Hermitian. Thus, as a corollary of the problem we obtain the following fact:
Eigenvalues of a real symmetric matrix are real.
## Related Question.
Problem. Let $A$ be an $n\times n$ real symmetric matrix.
Prove that there exists an eigenvalue $\lambda$ of $A$ such that for any vector $\mathbf{v}\in \R^n$, we have the inequality
$\mathbf{v}\cdot A\mathbf{v} \leq \lambda \|\mathbf{v}\|^2.$
Note that the inequality makes sense because eigenvalues of $A$ are real by Corollary.
For a proof of this problem, see the post “Inequality about Eigenvalue of a Real Symmetric Matrix“.
## The Ohio State University Linear Algebra Exam Problems and Solutions
Check out more problems from the Ohio State University Linear Algebra Exams ↴
The Ohio State University Linear Algebra.
### 4 Responses
1. 04/30/2017
[…] that the eigenvalues of a real symmetric matrix are real. (See the corollary in the post “Eigenvalues of a Hermitian matrix are real numbers“.) Let $lambda$ be a (real) eigenvalue of $A$ and let $mathbf{x}$ be a corresponding real […]
2. 06/15/2017
[…] that the eigenvalues of a real symmetric matrices are all real numbers and it is diagonalizable by an orthogonal […]
3. 07/19/2017
[…] The proof is given in the post Eigenvalues of a Hermitian Matrix are Real Numbers […]
4. 07/28/2017
[…] that all the eigenvalues of a symmetric matrices are real numbers. Let $lambda_1, dots, lambda_n$ be eigenvalues of […]
##### Cosine and Sine Functions are Linearly Independent
Let $C[-\pi, \pi]$ be the vector space of all continuous functions defined on the interval $[-\pi, \pi]$. Show that the...
Close<|endoftext|>
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# (solved)Question 14 SSC-CGL 2020 March 3 Shift 1
If $\displaystyle x^{2a} = y^{2b} = z^{2c}$ and x² = yz, then (ab + bc + ca)/bc = ?
(Rev. 20-Jan-2024)
## Categories | About Hoven's Blog
,
### Question 14SSC-CGL 2020 Mar 3 Shift 1
If $\displaystyle x^{2a} = y^{2b} = z^{2c}$ and x² = yz, then (ab + bc + ca)/bc = ?
### Solution in Brief
Observe that if a =b = c = x = y = z = 1, all the given conditions are satisfied. So with a = b = c = 1, the required expression (ab + bc + ca)/bc becomes 3 ans!
### Solution in Detail
Let $\displaystyle x^{2a} = y^{2b} = z^{2c} = k$
Take $\displaystyle x^{2a} = k$
$\displaystyle \implies x = k^{1/2a}$
Likewise $\displaystyle y = k^{1/2b}$ and $\displaystyle z = k^{1/2c}$
Given $\displaystyle x^2 = yz$
$\displaystyle \implies (k^{1/2a})^2 = k^{1/2b} \cdot k^{1/2c}$
$\displaystyle \implies k^{2/2a} = k^{(1/2b + 1/2c)}$
Bases are same, so powers must also be same
$\displaystyle \therefore \frac {2}{2a} = \frac{1}{2b} + \frac{1}{2c}$
Add $\displaystyle \frac{1}{2a}$ to both sides
$\displaystyle \frac{3}{2a} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c}$
$\displaystyle \therefore \frac{3}{2a} = \frac{bc + ca + ab}{2abc}$
Canceling $\displaystyle 2a$,
$\displaystyle 3 = \frac{bc + ca + ab}{bc}\:\underline{Ans}$<|endoftext|>
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# Formulas for the Algebra Questions on the GRE Quantitative Reasoning Test
## What’s on the Algebra section of the GRE?
If you are preparing for the algebra questions on the GRE Quantitative Reasoning Test, chances are that you are overwhelmed by the huge amount of content you need to cover. But you don’t need to feel that way!
The algebra questions cover topics such as linear and quadratic equations, factoring, simplifying algebraic expressions, exponents, and setting up equations to solve word problems.
## Math Formulas for the GRE
Algebra can be easily summarized in a few rules that you need to follow and at Union Test Prep, we have the essential formulas you’ll need for Algebra. You can see them all at a glance, and be prepared to succeed at the GRE Quantitative Reasoning Test.
You can also check out our free practice tests, flashcards, and study guides if you want to dig deeper.
Remember that this chart only shows formulas for one part of the test: Algebra. Access our other three formula charts for the GRE Quantitative Reasoning section, here:
Arithmetic
Geometry
Data Analysis
## Algebra Formula Chart for the GRE Quantitative Reasoning Test
Category Formula Symbols Comment
Algebra $$x+ a = b \rightarrow x= b - a$$
$$x-a = b \rightarrow x = b+a$$
$$x\cdot a = b \rightarrow x = b \div a$$
$$x \div a = b \rightarrow x = b \cdot a$$
$$x^a = b \rightarrow x = \sqrt[a]{b}$$
$$\sqrt[a]{x} = b \rightarrow x = b^a$$
$$a^x = b \rightarrow \dfrac{\log{b}}{\log{a}}$$
a,b = constants
x = variable
Algebra $$x^a \cdot x^b = x^{a+b}$$ a,b,x = any real number
Algebra $$\dfrac{x^a}{x^b} = x^{a-b}$$ a,b,x = any real number
Algebra $$(x^a)^b = x^{a\cdot b}$$ a,b,x = any real number
Algebra $$(x \cdot y)^a = x^a \cdot y^a$$ a,b,y = any real number
Algebra $$x^1 = x$$ x = any real number
Algebra $$x^0 = 1$$ x = any real number
Algebra $$x^{-a} = \dfrac{1}{x^a}$$ a,x = any real number
Algebra $$x^{\frac{a}{b}} = \sqrt[b]{x^a} = (\sqrt[b]{x})^a$$ a,b,x = any real number
Algebra $$\dfrac{x}{\sqrt{y}} \cdot \dfrac{\sqrt{y}}{\sqrt{y}} = \dfrac{x\sqrt{y}}{y}$$ x, y = any real number
Linear
Equations
$$Ax+By=C$$ A, B, C = any real number
y = dependent variable
x = independent variable
Standard form
Linear
Equations
$$y = mx+b$$ y = dependent variable
m = slope
x = independent variable
b = y axis intercept
Slope-Intercept
Form. Try to convert
any given linear
equation to this
format.
Linear
Equations
$$m = \dfrac{y_2-y_1}{x_2-x_1}$$ m = slope
$$y_n$$ = independent variable (point n)
$$x_n$$ = dependent variable (point n)
This is a rearranged
version of the
point-slope form.
Linear
Equations
$$y - y_1 = m(x-x_1)$$ $$(x_1, y_1)$$ = point on the line
y = dependent variable
x = independent variable
m = slope
Point slope form
Equations
$$x = \frac{-b\pm\sqrt{b^2-4 \cdot a \cdot c}}{2 \cdot a}$$ a, b = constants
c = constant (y-axis intercept)
x = variable
equation in form
$$ax^2+bx+c=0$$
Equations
$$(a\pm b)^2= a^2 \pm 2ab + b^2$$ a,b = constants or variables Square of a sum
or difference
Equations
$$a^2 - b^2 = (a+b)\cdot (a-b)$$ a,b = constants or variables Difference
of squares
Computing
Interest
$$SI = P \cdot IR \cdot t$$ SI = Simple Interest
P = Principal (Amount borrowed)
IR = Interest Rate
t = time (same units as in IR)
Computing
Interest
$$A_{SI} = P + SI = P \cdot (1+IR\cdot t)$$ $$A_{SI}$$ = Future value to be paid (for SI)
P = Principal (Amount borrowed)
SI = Simple Interest
IR = Interest Rate
t = time (same units as in IR)
Computing
Interest
$$A_{CI} = P \cdot \big{(}1+\dfrac{IR}{n}\big{)}^{n\cdot t}$$ $$A_{CI}$$ = Future value to be paid (for CI)
P = Principal (Amount borrowed)
IR = Interest Rate
n = # of times interest is compounded per unit,t
t = time (same units as in IR)<|endoftext|>
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# Completing the square – getting rid of the square (difficult)
When it isn’t obvious how to get rid of the square
Complete the following square:
Remember the above picture
NOTE:
1. The coefficient (or number) in front of the x must be a 1
2. The next square must be ½ of the next term
3. Multiply out into all of the boxes
This method uses a visual model to solve a quadratic equation
For example:
x^2+2x-8=0
Visually x^2 can look like an area of a table:
The 2x can be represented by:
And this will therefore complete the last square as:
NOTE:
This is the same as (x+1)^2
At the moment if you add each area together you get:
x^2+1x+1x+1times1
=x^2+2x+1
Originally we had x^2+2x-8=0
We now have x^2+2x+1=0
Always plot this on a number line
Original number - New number
-8-1=-9 (we need to -9 )
So x^2+2x-8=0
Is the same as (x+1)^2-9=0
Which can now be solved
(x+1)^2-9=0
(x+1)^2=9
x+1=+-sqrt9
(Don't forget the root of anything can be + or -)
x+1=+-3
x=-1+-3
x=-1+3\ \ \ or\ \ \ x=-1-3
x=2\ \ \ or\ \ \ x=-4
## Now check
x^2+2x-8=0
If x=2 2^2+2times2-8=0 Which is correct
If x=-4 (-4)^2+2times(-4)-8=0
16-8-8=0 Which is correct
(If they don't add up to zero you can be assured that it is wrong)
The roots of x^2+2x-8=0 are x=2 and x=-4<|endoftext|>
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Rest in space…
Rest in space…
Prior to 2015, scientists knew little about Pluto, mainly because it is dim and small from Earth’s perspective, not to mention 4.67 billion miles away.
When NASA’s New Horizons space probe flew by the far-away dwarf planet, imaging it in unprecedented detail, the historic mission raised more questions than answers. For one, the probe’s findings raised suspicions that some of Pluto’s mountains were formed on a bedrock of water ice.
According to a new study, computer simulations provide have provided compelling evidence that an insulating layer near the surface is keeping a subsurface ocean from freezing beneath Pluto’s ice. In other words, there could be a liquid ocean on the planet.
A team of Japanese scientists published the study, proposing that such an otherworldly idea is possible because a thin layer of ice containing trapped gas molecules, known as gas hydrates, at the bottom of the ice shell could be insulating the ocean. By calculating Pluto’s temperature and the thickness of the ice shell, the scientists concluded that the gas hydrates would be enough to maintain a subsurface ocean.
Understanding how a subsurface ocean can exist on Pluto will provide scientists with invaluable information to better understand how similar bodies of water can exist on other planets, too. Liquid water oceans are thought to exist inside icy satellites of gas giants such as Europa and Enceladus. Understanding the survival of subsurface oceans is of fundamental importance not only to planetary science but also to astrobiology.
Scientists have been repeatedly surprised and bewildered by the data New Horizons collected and processed from its flyby in 2015. Even the initial photos showed unexpected complexity of the dwarf planet.
The study is published in in the journal Nature Geoscience.
In 2010, an analysis of images from NASA’s Lunar Reconnaissance Orbiter (LRO) found that the Moon shriveled like a raisin as its interior cooled, leaving behind thousands of cliffs called thrust faults on the lunar surface. A new analysis of archival data from seismometers deployed during the Apollo missions gives the first evidence that these thrust faults are still active and likely producing moonquakes today as the Moon continues to gradually cool and shrink.
Researchers found that a number of the quakes recorded in the Apollo data happened very close to the faults seen in the LRO imagery. The LRO images also show physical evidence of geologically recent fault movement, such as landslides and tumbled boulders.
Astronauts placed five seismometers on the Moon’s surface during the Apollo 11, 12, 14, 15, and 16 missions. The Apollo 11 seismometer operated only for three weeks, but the four remaining instruments recorded 28 shallow moonquakes from 1969 to 1977. On Earth, the quakes would have ranged in magnitude from about 2 to 5.
Using the revised location estimates from their new algorithm, scientists found that the epicenters of eight of the 28 shallow quakes were within 19 miles of faults visible in the LRO images. This was close enough for the team to conclude that the faults likely caused the quakes.
The researchers also found that six of the eight quakes happened when the Moon was at or near its apogee, the point in the Moon’s orbit when it is farthest from Earth. This is where additional tidal stress from Earth’s gravity causes a peak in the total stress on the Moon’s crust, making slippage along the thrust faults more likely.
The LRO imaged more than 3,500 fault scarps on the Moon. Some of these images show landslides or boulders at the bottom of relatively bright patches on the slopes of fault scarps or nearby terrain. Brighter areas indicate regions that are freshly exposed by an event such as a moonquake.
Other LRO fault images show fresh tracks from boulder falls, suggesting that quakes sent these boulders rolling down their cliff slopes. Such tracks would be erased relatively quickly, in terms of geologic time, by the constant rain of micrometeoroid impacts on the Moon.
The study appears in the journal Nature Geoscience.
Last month, NASA’s Mars Odyssey orbiter captured a new thermal image of Phobos, the larger of Mars’ two moons. Each color in the full-moon image represents a temperature range detected by Odyssey’s Thermal Emission Imaging System (THEMIS) camera. Each observation is done from a slightly different angle or time of day, providing new kinds of data.
The new, full-moon view is better for studying material composition, whereas earlier half-moon views are better for looking at surface textures. With the half-moon views, scientists could see how rough or smooth the surface is and how it’s layered. With the new full-moon views, scientists can gather data on what minerals are in it, including metals.
Iron and nickel are two such metals. Depending on how abundant the metals are, and how they’re mixed with other minerals, researchers might be able to determine whether Phobos is a captured asteroid or a pile of Mars fragments blasted into space by a giant impact long ago.
Human exploration of Phobos has been discussed in the space community as a distant, future possibility, and a Japanese sample-return mission to the tiny moon is scheduled for launch in the 2020s. These and future observations could help future missions identify hazards and find safe areas to land of the surface.
New data from the Spitzer and Hubble space telescopes show that in particular wavelengths of infrared light, some of the first galaxies to form in the Universe (less than 1 billion years after the Big Bang) were considerably brighter than astronomers anticipated.
No one yet knows for sure when the first stars in our Universe burst to life. Evidence suggests that between 100 million and 200 million years after the Big Bang, the Universe was filled mostly with neutral hydrogen gas that had perhaps just begun to coalesce into stars, which then began to form the first galaxies.
By about 1 billion years after the Big Bang, the Universe had become a sparkling firmament. Something else had changed, too: electrons of the omnipresent neutral hydrogen gas had been stripped away in a process known as ionization.
The Epoch of Reionization, the changeover from the Universe full of neutral hydrogen to one filled with ionized hydrogen, is well documented.
Before this Universe-wide transformation, long-wavelength forms of light, such as radio waves and visible light, traversed the Universe more or less unencumbered. But shorter wavelengths, including ultraviolet light, X-rays and gamma rays, were stopped short by neutral hydrogen atoms. These collisions would strip the neutral hydrogen atoms of their electrons, ionizing them.
But what could have produced enough ionizing radiation to affect all the hydrogen in the Universe? Was it individual stars? Giant galaxies?
If either were the culprit, those early cosmic colonizers would have been different than most modern stars and galaxies, which typically don’t release high amounts of ionizing radiation. Then again, perhaps something else entirely caused the event, such as quasars.
Researchers found that early galaxies were particularly bright in two specific wavelengths of infrared light produced by ionizing radiation interacting with hydrogen and oxygen gases within the galaxies. This implies that the galaxies were dominated by young, massive stars composed mostly of hydrogen and helium. They contain very small amounts of heavy elements, like nitrogen, carbon and oxygen, compared to stars found in modern galaxies.
These stars were not the first stars to form in the Universe (those would have been composed of hydrogen and helium only) but were still members of very early generations of stars.
The Epoch of Reionization wasn’t an instantaneous event, so while the new results are not enough to close the book on this cosmic event, they do provide new details about how the Universe evolved during this time and how the transition played out.
The findings are published in the Monthly Notices of the Royal Astronomical Society.<|endoftext|>
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# Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.3
You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.3
Question 1.
How many Triangles are there in each of the following figures?
Solution:
(i) The single portions A, B, C, D, E, F, G and H are 8 triangles.
Taking the combination of 2 E & F, F & G & H and E & 11 are 4 triangles.
There are 8 + 4 = 12 triangles.
(ii) The smaller portions A, B, C, D, E, F, G and H are 8 triangles.
Combining 2 at a time B & C, D & F, F & G, A & H are 4 triangles.
Taking 4 at a time
Combination of B, C, D and E; D, E, F and G; F, G, H and A; H, A, B and C are 4 triangles.
Total triangles = 8 + 4 + 4 = 16.
(iii) Every part A, B, C, D, E, F, G, H, I, J, K, L, M, N, O and P are 16 triangles.
Combining two at a time. A & B, B & C, C & D, D & A, J & K, L & M, N & O, P & I are 8 triangles.
Taking the combination of 4, we have I, J, E & A; K, L, F and B, M, N, G and C, P, O, H and D are 4 triangles.
Taking 8 parts together we have QRS, RST, STQ, and TRQ are 4 triangles.
Total triangles = 16 + 8 + 4 + 4 = 32 triangles.
(iv)
Taking smaller triangles we have AFB, BFG, BGC, CGH, HCD, DHI, DIE, EIJ, EJAand AJF ⇒ 10 triangles.
Taking the combination of 2.
CGD, CID, DJE, DHE, EIA, EFA, AGB, AJB, BHC and BFC are 10 triangles.
Combining 3 at a time we have CED, DAE, EBA, ACB, BDC, CEF, DBJ, GDA, EBH, ACI ⇒ 10 triangles.
Other triangles are BCE, CAD, DBE, ADB and ACE are 5 triangles.
Total 10 + 10 + 10 + 5 = 35 triangles.
Question 2.
Find the number of dots in the tenth figure of the following pattern.
Solution:
(i) The number of dots given are 1, 3, 6, 10, 15,…
1st number = 1
2nd number = 1 + 2 = 3
3rd number = 3 + 3 = 6
4th number = 6 + 4 = 10
5th number = 10 + 5 = 15
6th number will be 15 + 6 = 21
7th number will be 21 + 7 = 28
8th number will be 28 + 8 = 36
9th number will be 36 + 9 = 45
10th number will be 45 + 10 = 55
Number of dots in the 10th figure = 55
(ii) The number of dots given are 1, 4, 9, 11, 16, 25,…
The no. of dots are in the pattern 1 × 1, 2 × 2, 3 × 3, 4 × 4, 5 × 5, …… 10 × 10
The number of dots in the 10th figure = 100.
Question 3.
(i) Draw the next pattern.
(ii) Prepare a table for the number of dots used for each pattern.
(iii) Explain the pattern.
(iv) Find the number of dots in the 25th pattern.
Solution:
First number is 2
2nd number is 2 + 3 = 5
3rd number is 5 + 4 = 9
4th number is 9 + 5 = 14
5th number is 14 + 6 = 20
(iv) Number of dots in the 25th pattern is
The number of dots in the 25th pattern = 350.
Question 4.
Count the number of squares in each of the following figures?
Solution:
Smaller squares 4 × 4 = 16 (As numbered).
As a whole bigger = 4
Total squares = 20
As we see in the figure in the middle 1, 2, 3, and 4 are 4 small squares.
Also, we have 9 and 10 = 2 big squares.
Outer squares 5, 6, 7, & 8 = 4.
Total = 4 + 4 + 1 + 1 = 10 squares.
Question 5.
How many circles are there in the following figure?
Solution:
There are 7 Circles.
Question 6.
Find the minimum number of straight lines used in forming the following figures.
Solution:
There are 5 horizontal lines and 5 vertical lines.
Total of 10 lines minimum needed.
There are 12 straight lines used which is a minimum.<|endoftext|>
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Concentrate on Key Points
As more and more students are included in the general education curriculum, teachers need to find ways of meeting the individual
learning needs of all the students. By understanding how to differentiate instruction in mixed ability classrooms, teachers can adapt their existing curriculum to meet the vast majority of needs. This can happen for students both with, and without. disabilities. Let’s take a look at some off the options.
One of the easiest things you can do is target your instruction so that you are only teaching the key points of a lesson. Some teachers refer to this as removing the “fluff" from the curriculum. We all know that there are things we like to teach, but is the information really necessary? Take a deeper look into your curriculum and see what parts can be removed. For students with learning disabilities, for example. it may not be necessary to address higher level thinking skills. Instead, focus on teaching them the basics that they will definitely need to know to demonstrate mastery of the standard.
Slow it Down or Speed it Up
Another option for students with disabilities is to slow down your curriculum. If you really want your students to master the concepts that you are teaching them, then it may be necessary to work at a slower pace. Use achievement to dictate learning and lesson planing. If you match their learning speed, you will better meet their needs.
For students needing more of a challenge, you could consider actually moving at a faster pace. This is sometimes referred to as "compacting" the curriculum. Test students at the beginning of a new unit to determine what they already might know, and what they still need to learn. By spending less time covering skills which have already been mastered, you will be able to move through the curriculum at a faster pace.
Change it Up
The way you present the curriculum is another method in which you can differentiate. Teachers can sometimes get into a routine of presenting the same material in the same way – no matter what the needs of their students. For example, a teacher may be used to reading notes and having students copy them. This, however, may not meet the needs of all students. Make sure that you routinely present the material in order to meet the needs of different types of learners, such as, visual, auditory, and kinesthetic.
Try Different Methods of Assessment
There are some additional things that you can do in the classroom to assist in meeting the individual learning needs of all students. This includes varying the types of assessments you use. Change tests so that students are only tested on major things, or require less repetition. Employ oral assessments in addition to traditional paper-and-pencil exams. Use a variety of multiple choice, fill-in-the-blank, true and false, and essay questions. Finally, don't forget to give the students adapted copies of notes to help them focus their studying as they prepare for exams.
As you can see there are several ways that you can do curriculum differentiation in your classroom. While you may be a little hesitant to make changes, knowing how to differentiate instruction in mixed-ability classrooms will allow students to learn better. For additional assistance, consider co-planning with a special education teacher. They are experts at differentiation, and will be more than willing to help you help meet the needs of the students in your classroom.<|endoftext|>
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You are on page 1of 4
LECTURE 6
Partial Derivatives and Differentiability
1. Partial Derivatives
Definition 6.1.
Let f : U
real-valued function
f
( x)
xi
Rn R be a real-valued function.
Then the ith
partial derivative
is the
f ( x + he i ) f ( x )
= hlim
0
h
f (x1 , . . . , xi1 , xi + h, xi+1 , . . . , xn ) (x1 , . . . , xi1 , xi , xi+1 , . . . , xn )
= lim
h
6.2. Note that limit used in the above definition is just the limit of a function of a single variable.
Indeed, fix x = (x1 , . . . , xi1 , xi, xi+1 , . . . , xn ) and define
F (h) = (x1 , . . . , xi1 , h, xi+1 , . . . , xn )
Then
dF
f
F ( x i + h) F ( x i )
(
x) = lim
h 0
x
h
dh
Remark
= xi
This observation in fact tells us precisely how to compute the i partial derivative of f . We regard all
the coordinates except xi to be fixed (i.e., constants) and then use ordinary calculus to differentiate the
resulting function of xi (regarded as a function of a single variable).
Example 6.3. Let f (x, y) = cos(x2 y2 ). Then
i
th
f
x
f
y
sin(x2 y2 )(2xy2 )
= sin(x2 y2 )(2x2 y)
The ordinary derivative of a function f of a single variable x tells us the rate of change of a function as
x increases. Similarly, the partial derivative is interpretable as a rate of change: the partial derivative
df
dx
f
( x)
xi
is the rate at which the function f changes as one moves away from the point x in the direction e .
i
2. Differentiability of Functions of Several Variables
Recall that in the case of a function of a single variable, a function f (x) is differentiable only if it is
continuous; but that continuity does not guarantee differentiability. Intuitively, continuity of f (x) requires
that its graph be a continuous curve; and differentiability requires also that there is always a unique tangent
vector to the graph of f (x). In other words, a function f (x) is differentiable if and only if its graph is a
smooth continuous curve with no sharp corners (a sharp corner would be a place where there would be two
possible tangent vectors).
1
2. DIFFERENTIABILITY OF FUNCTIONS OF SEVERAL VARIABLES
If we try to extend this graphical picture of differentiability to functions of two or more variables, it would be
natural to think of a differentiable function of several variables as one whose graph is a smooth continuous
surface, with no sharp peaks or folds. Because for such a surface it would always be possible to associate a
unique tangent plane at a given point.
However, differentiability in this sense turns out to be a much stronger condition than the mere existence
of partial derivatives. For the existence of a partial derivatives at a point x0 requires only a smooth approach
to the point f (x0 ) along the direction of the coordinate axes. We have seen examples of functions that
are discontinuous even though
f ,y
lim
0 f x,
y 0
(
0) = lim
(0
both exist. For example the function
f (x, y) =
has this property, and in fact, both
f
x
and
f
y
x y )2
x2 + y 2
exist and are continuous functions at the point (0 0).
With this sort of phenomenon in mind we give the following definition of differentiability.
Definition .
6.4
if
1.
R2 R
of two variables x and y is
differentiable
at (x0 , y0 )
f
f
Both x and y exist at the point (x0 , y0 ).
2.
(x,y )
Remark
(x0,y0)
lim
f (x , y ) (y y )
f (x, y) f (x0 , y0 ) f
(
x
,
y
)
(
x
x
)
0
0
0
0
0
0
x
y
|(x x0 )2 + (y y0 )2 | 12
6.5. The limit condition simply means that
F (x, y) = f (x0 , y0 ) +
=0
f
f
(
x x0 ) +
(y y0 )
x (x0 ,y0 )
y (x0 ,y0 )
is a good approximation to f (x, y) near the point (x0 , y0 ). To make contact with our graphical interpretation
of differentiability, we simply note that the graph of F (x, y) is a plane (it is linear in the variables x and y).
To make this completely obvious, recall that the solution set of any equation of form
Ax + By + Cz = D
is a plane in R3
; so taking
f
x
(x0 ,y0 )
f
y (x0 ,y0 )
f (x0 , y0 )
f
x0
x (x0 ,y0 )
f
y0
y (x0 ,y0 )
we see that the equation of the graph of F (x, y)
z = F (x, y) = Ax By + D
is a plane. Thus the limit condition is saying that the graph of f (x, y) coincides with the graph of a plane
as (x, y) approaches the point (x0 , y0 ). This observation motivates the following definition.
Definition 6.6.
plane in
Let f : R2
f (x0 , y0 ) +
f
f
( x x0 ) +
(y y0 )
x (x0 ,y0 )
y (x0 ,y0 )
is called the
at
x0 .
Rn
to
Rm
our notation.
Definition .
from Rn to
Rn
If f is any function from
Rn defined by
6.7
f ( x) =
to
R we define the
f
f
f
( x) ,
( x) , . . . ,
( x)
x1
x2
xn
of f to be the function
f (x), , . . . , f (x)
Of course, for this definition to make sense all the partial derivatives x
xn
1
Definition 6.8.
Let U be an open set in Rn and let f be
differentiable at a point x0 U if the partial derivatives
a function from U to
R.
must exist.
We say that f is
f
f (x0 + hei ) f (x0 )
(x0 ) = lim
h
0
xi
h
all exist and if
0 =
lim
x x0
f ( x ) f ( x 0 ) f ( x 0 ) ( x x0 )
x x0
Rn to Rm .
Definition 6.9.
Let U be an open set in Rn and let f be a function from U to R. We say that f is
differentiable at a point x0 U if the partial derivatives
fj
f j ( x0 + he i ) f j ( x0 )
(x0 ) = lim
h 0
xi
h
all exist and
0 =
lim
n
f (x) f (x0 ) j=0 fj (x0 ) (x x0 )ej
x x0
x x0
Example 6.10. Let f (x, y) = (xy, x + y). Then
T (x, y) =
and so for
f1
x
f2
x
f1
y
f2
y
=
to be differentiable at (1 0) we require
f
j=0 fj x0 x x0 ej
x x0
x x0
f (x) f (x0 ) (f1 (x0 ) (x x0 ),0) (0, f2 (x0 ) (x x0 ))
x x0
x x0
lim
lim
x ) f (x 0 )
2. DIFFERENTIABILITY OF FUNCTIONS OF SEVERAL VARIABLES
Now
f ( x)
f (x 0 )
= (xy, x + y)
= f(1, 0) = (0, 1)
xy ,
xy
= x
y
f1
f1 x0
f2 x0
(
f2
x x0
x x0
x + y ),
= (
x + y)
y, x)
x
y
(0, 1) (x 1, y 0) = y
,
(1 1)
x
(
f1 x0
(
= (1 1)
) =
f1
f2 x0
(
(1 0) = (0 1)
) =
f2
(1 0) = (1 1)
, y 0) = x 1 + y
0
=
=
=
(xy, x + y) (0, 1) (y, 0) (0, x 1 + y)
(x 1, y
(x,y )(1,0)
(xy 0 y 0, x + y 1 0 x + 1 y)
lim
(x,y )(1,0)
(x 1, y
(xy y, 0)
lim
(x,y )(1,0) ( x 1, y
(xy y)2
lim
(x,y )(1,0)
(x 1)2 + y2
lim
Unfortunately as we have seen in the examples we discussed in the lecture on limits and continuity, it is not
so easy to see whether or not this limit exists. Luckily we have a theorem at our disposal that makes it a
lot easier to decide questions of differentiability.
Theorem 6.11. Let f : U
neighborhood of a point
Rn Rm . Suppose that the derivatives
in U . Then f is differentiable at
x.
fj<|endoftext|>
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# MA.4.FR.2.3
Explore the addition of a fraction with denominator of 10 to a fraction with denominator of 100 using equivalent fractions.
### Examples
is equivalent to which is equivalent to .
### Clarifications
Clarification 1: Instruction includes the use of visual models.
Clarification 2: Within this benchmark, the expectation is not to simplify or use lowest terms.
General Information
Subject Area: Mathematics (B.E.S.T.)
Strand: Fractions
Status: State Board Approved
## Benchmark Instructional Guide
• NA
### Vertical Alignment
Previous Benchmarks
Next Benchmarks
### Purpose and Instructional Strategies
The purpose of this benchmark is to connect fraction addition to decimal addition through decimal fractions. This will be the first opportunity for students to create common denominators to add fractions. This benchmark continues the work of equivalent fractions (MA.3.FR.1.2) by having students rename fractions with denominators of 10 as equivalent fractions with denominators of 100 (MA.4.FR.1.1). Students who can generate equivalent fractions can adapt this new procedure to develop strategies for adding fractions with unlike denominators in grade 5 (MA.5.FR.2.1).
• Instruction may include students shading decimal grids (10 × 10 grids) to support their understanding.
• Subtraction of decimal fractions is not a requirement of grade 4.
### Common Misconceptions or Errors
• Students often will add the numerators and the denominators without finding the like denominator. Students will need visual models to understand what the like denominator means.
### Strategies to Support Tiered Instruction
• Instruction includes opportunities to explore the addition of a fraction with a denominator of 10 to a fraction with a denominator of 100 using visual models to help understand equivalent fractions. Students use visual models to make sense of equivalent fractions when finding like denominators. The teacher provides clarification that students must find the like denominator before adding.
• For example, the teacher displays the problem $\frac{\text{4}}{\text{10}}$ + $\frac{\text{23}}{\text{100}}$ and asks students to share what they notice about this expression. Students identify that the denominators are different. The teacher guides students to shade decimal grids to represent the understanding that $\frac{\text{4}}{\text{10}}$ is equivalent to $\frac{\text{40}}{\text{100}}$. This is repeated with similar addition problems and solved while supporting students as they use the visual models to problems that have denominators of 10 and 100.
• For example, the teacher displays the problem $\frac{\text{24}}{\text{100}}$ + $\frac{\text{3}}{\text{10}}$ , asking students to share what they notice about this expression. Students identify that the denominators are different. The teacher guides students to use place value blocks and shaded decimal grids to represent the problem and solve, having tens rods represent tenths, and one's cubes represent hundredths. Students are supported as they use the visual models to understand that $\frac{\text{3}}{\text{10}}$ is equivalent to $\frac{\text{30}}{\text{100}}$. This is repeated with similar addition problems that have denominators of 10 and 100.
Determine the equivalent fraction.
• $\frac{\text{5}}{\text{10}}$ = $\frac{\text{}}{\text{100}}$
• $\frac{\text{31}}{\text{100}}$ + $\frac{\text{5}}{\text{10}}$
### Instructional Items
Instructional Item 1
An expression is shown. What is the value of the expression?
• $\frac{\text{3}}{\text{10}}$+ $\frac{\text{32}}{\text{100}}$
*The strategies, tasks and items included in the B1G-M are examples and should not be considered comprehensive.
## Related Courses
This benchmark is part of these courses.
5012060: Mathematics - Grade Four (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
7712050: Access Mathematics Grade 4 (Specifically in versions: 2014 - 2015, 2015 - 2018, 2018 - 2022, 2022 and beyond (current))
5012055: Grade 3 Accelerated Mathematics (Specifically in versions: 2019 - 2022, 2022 and beyond (current))
5012015: Foundational Skills in Mathematics 3-5 (Specifically in versions: 2019 - 2022, 2022 and beyond (current))
## Related Access Points
Alternate version of this benchmark for students with significant cognitive disabilities.
MA.4.FR.2.AP.3: Explore the addition of a fraction with denominator of 10 to a fraction with denominator of 100 using visual models to find equivalent fractions.
## Related Resources
Vetted resources educators can use to teach the concepts and skills in this benchmark.
## Formative Assessments
Tenths and Hundredths:
Students are asked if an equation involving the sum of two fractions is true or false. Then students are asked to find the sum of two fractions.
Type: Formative Assessment
Students express a fraction with a denominator of 10 as an equivalent fraction with a denominator of 100 and are then asked to add the fraction to another fraction with a denominator of 100.
Type: Formative Assessment
Hundredths and Tenths:
Students are asked if an equation is true or false. Then students are asked to find the sum of two fractions.
Type: Formative Assessment
Seven Tenths:
Students express a fraction with a denominator of 10 as an equivalent fraction with a denominator of 100 and are then asked to add the fraction to another fraction with a denominator of 100.
Type: Formative Assessment
## Original Student Tutorial
Fractions at the Fair: Equivalent Tenths and Hundredths:
Learn about equivalent 10ths and 100ths and how to calculate these equivalent fractions at the fair in this interactive tutorial.
Type: Original Student Tutorial
## Perspectives Video: Expert
B.E.S.T. Journey:
What roles do exploration, procedural reliability, automaticity, and procedural fluency play in a student's journey through the B.E.S.T. benchmarks? Dr. Lawrence Gray explains the path through the B.E.S.T. maththematics benchmarks in this Expert Perspectives video.
Type: Perspectives Video: Expert
The purpose of this task is adding fractions with a focus on tenths and hundredths.
Expanded Fractions and Decimals:
The purpose of this task is for students to show they understand the connection between fraction and decimal notation by writing the same numbers both ways. Comparing and contrasting the two solutions shown below shows why decimal notation can be confusing. The first solution shows the briefest way to represent each number, and the second solution makes all the zeros explicit.
## Tutorial
Adding Two Fractions with Denominators 10 and 100:
The Khan Academy tutorial video presents a visual fraction model for adding 3/10 + 7/100 .
Type: Tutorial
## MFAS Formative Assessments
Students express a fraction with a denominator of 10 as an equivalent fraction with a denominator of 100 and are then asked to add the fraction to another fraction with a denominator of 100.
Hundredths and Tenths:
Students are asked if an equation is true or false. Then students are asked to find the sum of two fractions.
Seven Tenths:
Students express a fraction with a denominator of 10 as an equivalent fraction with a denominator of 100 and are then asked to add the fraction to another fraction with a denominator of 100.
Tenths and Hundredths:
Students are asked if an equation involving the sum of two fractions is true or false. Then students are asked to find the sum of two fractions.
## Original Student Tutorials Mathematics - Grades K-5
Fractions at the Fair: Equivalent Tenths and Hundredths:
Learn about equivalent 10ths and 100ths and how to calculate these equivalent fractions at the fair in this interactive tutorial.
## Student Resources
Vetted resources students can use to learn the concepts and skills in this benchmark.
## Original Student Tutorial
Fractions at the Fair: Equivalent Tenths and Hundredths:
Learn about equivalent 10ths and 100ths and how to calculate these equivalent fractions at the fair in this interactive tutorial.
Type: Original Student Tutorial
The purpose of this task is adding fractions with a focus on tenths and hundredths.
Expanded Fractions and Decimals:
The purpose of this task is for students to show they understand the connection between fraction and decimal notation by writing the same numbers both ways. Comparing and contrasting the two solutions shown below shows why decimal notation can be confusing. The first solution shows the briefest way to represent each number, and the second solution makes all the zeros explicit.
## Tutorial
Adding Two Fractions with Denominators 10 and 100:
The Khan Academy tutorial video presents a visual fraction model for adding 3/10 + 7/100 .
Type: Tutorial
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this benchmark.<|endoftext|>
| 4.5 |
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2020 AMC 12B Problems/Problem 22
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Problem 22
What is the maximum value of $\frac{(2^t-3t)t}{4^t}$ for real values of $t?$
$\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}$
Solution
First, substitute $2^t = x (log_2{x} = t)$ so that $$\frac{(2^t-3t)t}{4^t} = \frac{xlog_2{x}-3(log_2{x})^2}{x^2}$$
Notice that $$\frac{xlog_2{x}-3(log_2{x})^2}{x^2} = \frac{log_2{x}}{x}-3(\frac{log_2{x}}{x})^2.$$
When seen as a function, $\frac{log_2{x}}{x}-3(\frac{log_2{x}}{x})^2$ is a synthesis function that has $\frac{log_2{x}}{x}$ as its inner function.
If we substitute $\frac{log_2{x}}{x} = p$, the given function becomes a quadratic function that has a maximum value of $\frac{1}{12}$ when $p = \frac{1}{6}$.
Now we need to check that $\frac{log_2{x}}{x}$ can have the value of $\frac{1}{6}$ in the range of real numbers.
In the range of (positive) real numbers, function $\frac{log_2{x}}{x}$ is a continuous function whose value gets infinitely smaller as $x$ gets closer to 0 (as $log_2{x}$ also diverges toward negative infinity in the same condition). When $x = 2$, $\frac{log_2{x}}{x} = \frac{1}{2}$, which is larger than $\frac{1}{6}$.
Therefore, we can assume that $\frac{log_2{x}}{x}$ equals to $\frac{1}{6}$ when $x$ is somewhere between 1 and 2 (at least), which means that the maximum value of $\frac{(2^t-3t)t}{4^t}$ is $\textbf{(C)}\ \frac{1}{12}$.<|endoftext|>
| 4.75 |
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# Convert revolution / (minute • second) to revolution / square hour
Learn how to convert 1 revolution / (minute • second) to revolution / square hour step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(\dfrac{revolution}{minute \times second}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{revolution}{square \text{ } hour}\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(\dfrac{radian}{square \text{ } second}\right)$$
$$\text{Left side: 1.0 } \left(\dfrac{revolution}{minute \times second}\right) = {\color{rgb(89,182,91)} \dfrac{π}{30.0}\left(\dfrac{radian}{square \text{ } second}\right)} = {\color{rgb(89,182,91)} \dfrac{π}{30.0}\left(\dfrac{rad}{s^{2}}\right)}$$
$$\text{Right side: 1.0 } \left(\dfrac{revolution}{square \text{ } hour}\right) = {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}\left(\dfrac{radian}{square \text{ } second}\right)} = {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}\left(\dfrac{rad}{s^{2}}\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(\dfrac{revolution}{minute \times second}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{revolution}{square \text{ } hour}\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{π}{30.0}} \times {\color{rgb(89,182,91)} \left(\dfrac{radian}{square \text{ } second}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}}} \times {\color{rgb(125,164,120)} \left(\dfrac{radian}{square \text{ } second}\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} \dfrac{π}{30.0}} \cdot {\color{rgb(89,182,91)} \left(\dfrac{rad}{s^{2}}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{rad}{s^{2}}\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{π}{30.0}} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{rad}{s^{2}}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{rad}{s^{2}}\right)}}$$
$$\text{Conversion Equation}$$
$$\dfrac{π}{30.0} = {\color{rgb(20,165,174)} x} \times \dfrac{π}{6.48 \times 10^{6}}$$
Cancel factors on both sides
$$\text{Cancel factors}$$
$$\dfrac{{\color{rgb(255,204,153)} \cancel{π}}}{30.0} = {\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{π}}}{6.48 \times 10^{6}}$$
Switch sides
$${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{6.48 \times 10^{6}} = \dfrac{1.0}{30.0}$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{6.48 \times 10^{6}}{1.0}\right)$$
$${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{6.48 \times 10^{6}} \times \dfrac{6.48 \times 10^{6}}{1.0} = \dfrac{1.0}{30.0} \times \dfrac{6.48 \times 10^{6}}{1.0}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{1.0}} \times {\color{rgb(99,194,222)} \cancel{6.48}} \times {\color{rgb(166,218,227)} \cancel{10^{6}}}}{{\color{rgb(99,194,222)} \cancel{6.48}} \times {\color{rgb(166,218,227)} \cancel{10^{6}}} \times {\color{rgb(255,204,153)} \cancel{1.0}}} = \dfrac{{\color{rgb(255,204,153)} \cancel{1.0}} \times 6.48 \times 10^{6}}{30.0 \times {\color{rgb(255,204,153)} \cancel{1.0}}}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{6.48 \times 10^{6}}{30.0}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x} = 216000 = 2.16 \times 10^{5}$$
$$\text{Conversion Equation}$$
$$1.0\left(\dfrac{revolution}{minute \times second}\right) = {\color{rgb(20,165,174)} 2.16 \times 10^{5}}\left(\dfrac{revolution}{square \text{ } hour}\right)$$<|endoftext|>
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One of the most useful features that modern technology has brought us is mapping software that uses the Global Positioning System (GPS) network. This worldwide satellite navigation system uses a network of satellites and receivers to determine location, and it’s used in many vehicles and smartphones to help people reach their destinations in the least amount of time. While anyone who has ever been lost in a sketchy neighborhood will tell you just how vital this tool is, there is also a pretty big downside: it may be hurting your brain.
A study from University College London found that relying on GPS navigation essentially turns off the parts of your brain that you would have used to simulate different routes had you not been using GPS. This conclusion was reached after studying 24 volunteers who navigated a simulation of central London’s Soho district, while undergoing brain scans.
The scientists were looking at activity in the part of the brain responsible for navigation and memory known as the hippocampus, along with the part involved in decision-making and planning, the prefrontal cortex. Scientists say that the hippocampus helps you simulate journeys on potential paths, while the prefrontal cortex helps you to choose which way to go. The researchers mapped the streets of London in an effort to learn more about how these parts of the brain would react to them. (RELATED: Read more news headlines about brain function and brain health at BRAIN.news.)
When the study’s volunteers navigated the streets manually, spikes in activity were seen, in the prefrontal cortex and hippocampus alike, when they entered new streets. When there were more options to choose from, their brain activity rose accordingly.
For example, entering a dead-end street brought about a reduction in activity, while entering a junction where several streets converge would boost activity in the hippocampus significantly. However, when the volunteers were simply following instructions given by the GPS system on the same streets, there was no additional brain activity. The researchers say this indicates the brain has “switched off its interest in the streets around us.”
They also looked at street networks in big cities around the globe to determine the difficulty of navigating them. Manhattan’s grid system requires comparatively little effort, given the fact that you can only go left, right or straight at each junction, whereas navigating London can be much more taxing on the hippocampus, given its complex layout with multiple junctions and roundabouts. This could also explain why London cab drivers have bigger hippocampi than average people, but the city’s bus drivers, who tend to follow pre-determined routes, do not. The study was published in the Nature Communications journal.
This information can be used by smart tech companies, architects and developers to design spaces that people will find easy to navigate. For example, they can give care homes and hospitals layouts that will not create confusion for people with dementia. At a time when Alzheimer’s is rising dramatically – with today’s figure of more than 5 million Americans living with the disease, which is expected to climb to 16 million by 2050 – this knowledge can be incredibly useful in improving people’s quality of life and avoiding causing undue stress.
Having a GPS system is incredibly useful, but if you want to keep your brain functioning optimally, try to work out routes yourself as much as possible and stick to using the GPS as a backup or for emergencies. Reserve your mapping apps for those times when you’re running late for an important work meeting, for example, but consider letting your brain do the work when you’re looking for that new park for your family’s picnic this weekend.<|endoftext|>
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David Bailey has created the most amazing set of Escher-type tessellations, many of which will appeal to your primary students. There is a vast range to explore from geometric designs, birds and animals. A tessellation is when you put tiles together on a flat surface so that there are no gaps or overlaps. Decorative tessellating patterns have been found in many ancient cultures, such as Ancient Rome, Egypt and throughout many monuments in the Middle East. A “tessella” is a Latin word for a small square. The Dutch graphic artist MC Escher created magnificent tessellations using irregular interlocking shapes. Escher had travelled extensively through Italy and also Spain where he was particularly inspired by the tilings he saw at The Alhambra.
Why not encourage your students to explore some of David Bailey’s work too? He has created a fantastic website packed full of information about tessellations with his own drawings plus real-life photographs. http://www.tess-elation.co.uk/new-hom<|endoftext|>
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# Thread: Help with solving these Trig Identities... Part 2
1. ## Help with solving these Trig Identities... Part 2
How do I solve:
Using Double angle identities
1. Find tan2x when secx = sq.root(65); sin x > 0
Using Pythagorean identities
1. [(1+ cot^3 x) / ( 1 + cot x)] + cot x = csc^2 x
Thank you very much for your time! =)
2. Hello, coolacu!
Using Double angle identities
$\displaystyle 1.\;\;\text{Find }\tan2x\text{ when }\sec x = \sqrt{65},\;\sin x > 0$
We have: .$\displaystyle \tan2x \:=\:\frac{\sin2x}{\cos2x} \;=\;\frac{2\sin x\cos x}{\cos^2\!x - \sin^2\!x}$ .[1]
We are given: .$\displaystyle \sec x \:=\:\frac{\sqrt{65}}{1} \:=\:\frac{hyp}{adj}$
From Pythagorus, we have: .$\displaystyle opp \:=\:8$
Hence: .$\displaystyle \sin x \:=\:\frac{8}{\sqrt{65}},\;\cos x \:=\:\frac{1}{\sqrt{65}}$
Substitute into [1]: .$\displaystyle \tan2x \;=\;\frac{2\left(\frac{8}{\sqrt{65}}\right)\left( \frac{1}{\sqrt{65}}\right)} {\left(\frac{1}{\sqrt{65}}\right)^2 - \left(\frac{8}{\sqrt{65}}\right)^2}$ $\displaystyle =\;\frac{\frac{16}{65}}{\frac{1}{65} - \frac{64}{65}} \;=\;\frac{\frac{16}{65}}{\text{-}\frac{63}{65}} \;=\;-\frac{16}{63}$
Using Pythagorean identities
$\displaystyle 1.\;\text{Prove: }\:\frac{1+ \cot^3\!x}{1 + \cot x} + \cot x \;= \;\csc^2\!x$
The numerator is the sum of two cubes . . . factor it.
$\displaystyle \frac{(1+\cot x)(1 - \cot x + \cot^2\!x)}{1 + \cot x} + \cot x \;=\; (1-\cot x + \cot^2\!x) + \cot x \;=\;$ $\displaystyle 1 + \cot^2\!x \;=\;\csc^2\!x$<|endoftext|>
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## How to Add and Subtract Fractions
Fractions are generally used to define any whole number into equal parts. While writing a fraction, there are two numbers involved. The number at the top is called the numerator, while that at the bottom is called a denominator. There are three types of fractions. They are:
Proper Fractions: In proper fractions, the denominator is greater than the numerator. For ex: $$\frac{3}{4} \ , \ \frac{1}{3}$$
Improper Fraction: As the name suggests, these fractions are “top-heavy” or the numerator is greater than the denominator. For ex: $$\frac{7}{8} \ , \ \frac{5}{2}$$
Mixed-Fraction: Mixed fractions are another type of improper fractions where there is a whole number as well as a fraction part. For ex: $$1\frac{1}{3} \ , \ 2\frac{3}{7}$$
To, add or subtract fractions, we need to follow certain criterions. First, we need to look for like and un-like fractions.
Addition & Subtraction with Like Fractions
Two fractions whose denominators are same are called like fractions. So, to add or subtract, just perform addition/subtraction between the numerator part, and then write the answer over the common denominator.
Example $$\frac{a}{b} + \frac{c}{b} = \frac{a+c}{b}$$
So, for example, $$\frac{2}{4} + \frac{3}{4} = \frac{5}{4}$$
Addition and Subtraction with Unlike Fractions
For un-like fractions or fractions with different denominators, we will perform the operation like this.
Example $$\frac{a}{b} + \frac{c}{d} = \frac{ad+cb}{bd}$$
Also, $$\frac{a}{b} - \frac{c}{d} = \frac{ad-cb}{bd}$$
So, for example, $$\frac{2}{5} - \frac{1}{6} = \frac{12-5}{30} = \frac{7}{30}$$
finally, you may have a fraction that can be reduced to a simpler fraction. it's always best to reduce simplest form when you can. Learn more
### Exercises for Add or Subtract Fractions
1) $${7 \over 4} \ + \ {5 \over 6} =$$
2)$${8 \over 10} \ + \ {2 \over 3} =$$
3)$${9 \over 8} \ + \ {8 \over 2} =$$
4)$${5 \over 3} \ - \ {3 \over 5} =$$
5)$${6 \over 3} \ - \ {8 \over 8} =$$
6)$${5 \over 1} \ - \ {5 \over 4} =$$
7)$${3 \over 2} \ - \ {7 \over 8} =$$
8)$${9 \over 3} \ - \ {3 \over 5} =$$
9)$${7 \over 11} \ + \ {16 \over 17} =$$
10)$${8 \over 9} \ + \ {15 \over 10} =$$
1)$${7 \over 4} \ + \ {5 \over 6} =$$$$\ \color{red}{{7 \times 6 \ + \ 5 \times 4 \over 4\times6} = }$$$${\color{red}{62 \over 24}}$$$$\color{red}{ = {62 \div 2 \over 24 \div 2} = {31 \over 12}}$$
GCF(62,24) = 2
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. $${7 \over 4} \ + \ {5 \over 6} =$$$$\ {7 \times 6 \ + \ 5 \times 4 \over 4\times6} =$$$${62 \over 24}$$
Then, simplify the result. $${62 \over 24} = {31 \over 12}$$
2)$${8 \over 10} \ + \ {2 \over 3} =$$$$\ \color{red}{{8 \times 3 \ + \ 2 \times 10 \over 10 \times 3} = }$$ $${\color{red}{44 \over 30}}$$$$\color{red}{ = {44 \div 2 \over 30 \div 2} = {22 \over 15}}$$
GCF(44,30) = 2
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. $${8 \over 10} \ + \ {2 \over 3} =$$$$\ {8 \times 3 \ + \ 2 \times 10 \over 10\times3} =$$ = $${44 \over 30}$$
Then, simplify the result. $${44 \over 30} = {22 \over 15}$$
3)$${9 \over 8} \ + \ {8 \over 2} =$$$$\ \color{red}{{9 \times 2 \ + \ 8 \times 8 \over 8\times2} = }$$ $${\color{red}{82 \over 16}}$$$$\color{red}{ = {82 \div 2 \over 16 \div 2} = {41 \over 8}}$$
GCF(82,16) = 2
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. $${9 \over 8} \ + \ {8 \over 2} =$$$$\ {9 \times 2 \ + \ 8 \times 8 \over 8\times2} =$$$${82 \over 16}$$
Then, simplify the result. $${82 \over 16} = {41 \over 8}$$
4)$${5 \over 3} \ - \ {3 \over 5} =$$$$\ \color{red}{{5 \times 5 \ - \ 5 \times 3 \over 3\times5} = }$$ $${\color{red}{10 \over 15}}$$$$\color{red}{ = {10 \div 5 \over 15 \div 5} = {2 \over 3}}$$
GCF(10,15) =5
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. $${5 \over 3} \ - \ {3 \over 5} =$$$$\ {5 \times 5 \ - \ 5 \times 3 \over 3\times5} =$$$${15 \over 10}$$
Then, simplify the result. $${2 \over 3} = 1$$
5)$${6 \over 3} \ - \ {8 \over 8} =$$$$\ \color{red}{{6 \times 8 \ - \ 8 \times 3 \over 3\times8} = }$$ $${\color{red}{24 \over 24}}$$$$\color{red}{ = 1}$$
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. $${6 \over 3} \ - \ {8 \over 8} =$$$$\ {6 \times 8 \ - \ 8 \times 3 \over 3\times8} =$$$${24 \over 24}$$
Then, simplify the result. $${15 \over 15} = 1$$
6)$${5 \over 1} \ - \ {5 \over 4} =$$$$\ \color{red}{{5 \times 4 \ - \ 5 \times 1 \over 1\times4} = }$$ $${\color{red}{15 \over 4}}$$
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. $${5 \over 1} \ - \ {5 \over 4} =$$$$\ {5 \times 4 \ - \ 5 \times 1 \over 1\times4} =$$$${15 \over 4}$$
7)$${3 \over 2} \ - \ {7 \over 8} =$$$$\ \color{red}{{3 \times 8 \ - \ 7 \times 2 \over 2\times8} = }$$ $${\color{red}{10 \over 16}}$$$$\color{red}{ = {10 \div 2 \over 16 \div 2} = {5 \over 8}}$$
GCF(10,16) = 2
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. $${3 \over 2} \ - \ {7 \over 8} =$$$$\ {3 \times 8 \ - \ 7 \times 2 \over 2\times8} =$$$${10 \over 16}$$
Then, simplify the result. $${10 \over 16} = { 5 \over 8}$$
8)$${9 \over 3} \ - \ {3 \over 5} =$$$$\ \color{red}{{9 \times 5 \ - \ 3 \times 3 \over 3\times5} = }$$ $${\color{red}{36 \over 15}}$$$$\color{red}{ = {36 \div 3 \over 15 \div 3} = {12 \over 5}}$$
GCF(36,15) = 3
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction.$${9 \over 3} \ - \ {3 \over 5} =$$$$\ {9 \times 5 \ - \ 3 \times 3 \over 3\times5} =$$$${36 \over 15}$$
Then, simplify the result. $${36 \over 15} = { 12 \over 5}$$
9)$${7 \over 11} \ + \ {16 \over 17} =$$$$\ \color{red}{{7 \times 17 \ + \ 16 \times 11 \over 11\times17} = }$$ $${\color{red}{295 \over 187}}$$
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. $${7 \over 11} \ + \ {16 \over 17} =$$$$\ {7 \times 17 \ + \ 16 \times 11 \over 11\times17} =$$$${295 \over 187}$$
10)$${8 \over 9} \ + \ {15 \over 10} =$$$$\ \color{red}{{8 \times 10 \ + \ 15 \times 9 \over 9\times10} = }$$ $${\color{red}{215 \over 90}}$$$$\color{red}{ = {215 \div 5 \over 90 \div 5} = {43 \over 18}}$$
GCF(215,90) = 5
Solution
For “unlike” fractions, the first step is to find the same denominator before you can add or subtract fractions with different denominators.
To find the same denominator, Multiply two denominators, and each numerator by the denominator of the other fraction. $${8 \over 9} \ + \ {15 \over 10} =$$$$\ {8 \times 10 \ + \ 15 \times 9 \over 9\times10} =$$$${215 \over 90}$$
Then, simplify the result. $${215 \over 90} = { 43 \over 18}$$
## Add and Subtract Fractions Quiz
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## Pages
### Constructing a quadratic equation from three given points
Given points (x1, y1), (x2, y2) and (x3, y3), three quadratic functions of the form ax2 + bx + c = y can be formed:
• y1 = a(x1)2 + b(x1) + c
• y2 = a(x2)2 + b(x2) + c
• y3 = a(x33)2 + b(x3) + c
These equations can be solved simultaneously for a, b and c. The resulting values of a, b and c obtained from solving the above three equations are put in the quadratic function y = ax2 + bx + c and thus the quadratic function, whose graph passes through (x1, y1), (x2, y2) and (x3, y3), is obtained. For example, given three points (1, 1), (-2, 1) and (-1, -1), the following three equations can be formed:
• 1 = a(1)2 + b(1) + c
• 1 = a(-2)2 + b(-2) + c
• -1 = a(-1)2 + b(-1) + c
on solving the above equations simultaneously, the values of a, b and c are obtained as follows:
• a = 1
• b = 1
• c = -1
Substituting the values of a, b and c in the quadratic function y = ax^2 + bx +c, the function y = x^2 + x - 1 is obtained. The following graph shows the parabola of y = x^2 + x - 1 and the points (1, 1), (-2, 1) and (-1, -1) on it.
In short, the process of obtaining a quadratic equation from its three given points involves three steps:
• Form three equations (one equation from each point) form ax2 + bx + c = 0
• Solve the three equations simultaneously to get the values of a, b and c
• Put the values of a, b and c in the equation ax2 + bx + c = 0 to obtain the quadratic equation<|endoftext|>
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# What is a ring graph?
## What is a ring graph?
A graph G is a ring graph if each block of G which is not a bridge or a vertex can be constructed inductively by starting from a single cycle and then in each step adding an H-path that meets graph H in the previous step in two adjacent vertices. Ring graphs were first introduced by Gitler et al.
## What’s a circle graph called?
A pie chart, sometimes called a circle chart, is a way of summarizing a set of nominal data or displaying the different values of a given variable (e.g. percentage distribution). This type of chart is a circle divided into a series of segments.
How do you use a circle graph?
In a pie chart or circle graph, the size of each sector will be proportional to the actual value of the data it represents as seen in the images. Percentages of the total of the sample are usually represented in the sectors. One of the more common uses for circle graphs or pie charts is poll results and surveys.
What is the join of two graphs?
The join of two graphs G and H is a graph formed from disjoint copies of G and H by connecting each vertex of G to each vertex of H.
### Why would you use a circle graph?
A circle graph, or a pie chart, is used to visualize information and data. A circle graph is usually used to easily show the results of an investigation in a proportional manner. The arcs of a circle graph are proportional to how many percent of population gave a certain answer.
### What is ring in linear algebra?
Definition. A ring is a set R equipped with two binary operations + (addition) and ⋅ (multiplication) satisfying the following three sets of axioms, called the ring axioms. R is an abelian group under addition, meaning that: (a + b) + c = a + (b + c) for all a, b, c in R (that is, + is associative).
What is the central vertex?
A vertex is central in G if its greatest distance from any other vertex is as small as possible.
How do you write an equation for a circle graph?
Explanation: The equation of a circle is \displaystyle (x – h)^2 + (y – k)^2 = r^2, in which (h, k) is the center of the circle and r is its radius. Because the graph of the circle is centered at (0, 0), h and k are both 0. Because the radius is 3, the right side of the equation is equal to 9.<|endoftext|>
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# How do you convert given units?
## How do you convert given units?
Summary: to convert units, construct a fraction that is equal to 1, multiply the original measurement by that fraction, and simplify.
How do I cancel out units?
Unit Cancellation is just a method of converting numbers to different units. Let the units tell you whether you should multiply or divide by a conversion factor. Notice that we put the first conversion factor with miles on the bottom and ft on the top so that miles would cancel out.
### What is conversion formula?
Conversion Rate = Total number of conversions / Total number of unique visitors * 100. Conversion Rate = Total number of conversions / Total number of leads * 100.
What are the 3 units of volume?
Three common units of volume are:
• cubic centimeters.
• liters.
• gallons.
## What is the formula of capacity?
You find the volume V of a rectangular container by measuring its length (l), width (w) and height (h) and multiplying these quantities. You express the result in cubic units.
How do you convert ratios?
To convert a part-to-part ratio to fractions:
1. Add the ratio terms to get the whole. Use this as the denominator. 1 : 2 => 1 + 2 = 3.
2. Convert the ratio into fractions. Each ratio term becomes a numerator in a fraction.
3. Therefore, in the part-to-part ratio 1 : 2, 1 is 1/3 of the whole and 2 is 2/3 of the whole.
### What is a unit analysis?
Unit analysis means using the rules of multiplying and reducing fractions to solve problems involving different units. Example 1: A desk is 412 feet long. Write this relationship as a fraction, and multiply it by the given length.
How do you cancel out units in an equation?
When you multiply a number or variable by 1, the value is unchanged. Step D restates the example problem. In Step E, multiply both sides of the equation by 1 and substitute the left side’s 1 with the value in step C. Step F is the unit cancellation step.
## Can you cross multiply with different units?
Changing units introduces a multiplicative factor. You convert inches to centimeters, for example, by multiplying by 2.54. But when you multiply or divide quantities with different units, the result does scale multiplicatively.
What are the formulas for temperature conversion?
Temperature Conversion Formulas
Temperature Scale Convert to by…
Fahrenheit °F = 1.8°C + 32°
Celsius °C =5/9(°F-32°)
Rankine °R = 1.8K + 0.6° °R = °F + 460°
Kelvin K =5/9(°R-0.6°) K = °C + 273°
### How to convert units of area to units of volume?
The objective of the lesson is to convert units of measure between dimensions including area and volume . The area is measured in square units (unit 2 ). Convert one square yard to square feet.
How to calculate the volume of a cubic foot?
So, 1 cubic foot is a cube with side length 12 inches. You can use the formula for the volume of a prism, V = b w h , to convert cubic units. Since 1 foot = 12 inches, replace b , w , and h with 12 . V = ( 12 ) ( 12 ) ( 12 ) = 1 , 728. So, 1 cubic foot is equal to 1,728 cubic inches.
## How to convert one unit to another unit?
The first thing to do is to list the units you have and the units to which you want to convert. In this case, we have units in meters and we want to convert to kilometers. Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio that expresses how many of one unit are equal to another unit.
How to convert units of measure between dimensions?
The objective of the lesson is to convert units of measure between dimensions including area and volume . The area is measured in square units (unit 2 ). Example 1: Convert one square yard to square feet. A square yard is a square with a side length of one yard. We know that, 1 yard = 3 feet.<|endoftext|>
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# How do you find the vertical, horizontal and slant asymptotes of: f(x)=((x-2)^2(2x+5)) /( (x-3)^2(x-2))?
Feb 24, 2017
Vertical asymptote at $x = 3$
Hole at $x = 2$
No slant asymptote
Horizontal asymptote at $y = 2$
#### Explanation:
Holes come when you can cancel a term because it is in the numerator and denominator: x - 2 = 0; x = 2
Vertical asymptotes come from setting all terms except the term that causes the hole D(x) = 0 : x-3 = 0; x = 3
Horizontal asymptotes are determined by the degrees of the numerator and denominator: $\frac{N \left(x\right)}{D \left(x\right)} = \frac{{a}_{n} {x}^{n}}{{b}_{m} {x}^{m}}$
If $n = m$ then the horizontal asymptote is at $y = {a}_{n} / {b}_{n}$
In the function: $f \left(x\right) = \frac{2 {x}^{3} + \ldots}{{x}^{3} + \ldots} , m = n = 3$,
so there is a horizontal asymptote at $y = 2$
To have a slant asymptote : $m + 1 = n$
To summarize:
Vertical asymptote at $x = 3$
Hole at $x = 2$
No slant asymptote
Horizontal asymptote at $y = 2$<|endoftext|>
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1 §12.4 The Definite Integral The student will learn about the area under a curve defining the definite integral.
Presentation on theme: "1 §12.4 The Definite Integral The student will learn about the area under a curve defining the definite integral."— Presentation transcript:
1 §12.4 The Definite Integral The student will learn about the area under a curve defining the definite integral.
2 Introduction We have been studying the indefinite integral or the antiderivative of a function. We now introduce the definite integral. We now introduce the definite integral. This integral will be the area bounded by f (x), the x axis, and the vertical lines x = a and x = b with notation:
3 Area One way we can determine the area we want is by filling it with rectangles and calculating the sum of the areas of the rectangles. The width of each rectangle is Δx = 1. If we use the left end then the height of each of the four rectangles is f (1), f (2), f (3) and f(4) respectively. The sum of the rectangles is then L 4 = f (1) Δx + f (2) Δx + f (3) Δx + f (4) Δx f (x) =.5 x 2 + 2 1 2 3 4 5 Let L 4 denote that we are using 4 rectangles and we are determining the height on the left side of the rectangle L 4 = 2.5 + 4 + 6.5 + 10 = 23
4 Area We can repeat this using the right side of the rectangle to determine the height The width of each rectangle is again Δx = 1. The height of each of the four rectangles is now f (2), f (3), f (4) and f(5) respectively. The sum of the rectangles is then R 4 = 4 + 6.5 + 10 + 14.5 = 35 f (x) =.5 x 2 + 2 1 2 3 4 5 The average of these two areas would be an even better approximation. Area (23 + 35)/2 = 29.
5 More on Area The previous average of 29 is very close to the actual area of 28.666…. Our accuracy could be improved if we let the number or rectangles increase and Δx would then become smaller. The error in our process can be calculated if the function is monotone. That is if the function is only increasing or only decreasing. f (x) =.5 x 2 + 2 1 2 3 4 5
6 Error of the Area If we calculate the left and right areas as before using n rectangles and if a is the left side of the area and b the right side of the area then: Error =|Area – L n | |f (b) – f (a)| Δx = f (x) =.5 x 2 + 2 1 2 3 4 5 For our previous example this is computed as: Error 12 Theorem 1 |R n – L n |
7 Theorem 2 If f (x) is either increasing or decreasing on [a,b], then its left and right sums approach the same real number as n → ∞. The number approached as n → ∞ by the left and right sums in this theorem is the area between the graph of f and the x axis from x = a to x = b.
8 Approximating Area From the previous area example f (x) =.5 x 2 +2 with a = 1 and b = 5 and Δx = 1. R 4 = 4 + 6.5 + 10 + 14.5 = 35 f (x) =.5 x 2 + 2 1 2 3 4 5 L 4 = 2.5 + 4 + 6.5 + 10 = 23 Error 12 With error bound
9 Approximating Area continued f (x) =.5 x 2 + 2 1 2 3 4 5 If we wanted a particular accuracy, say 0.05 we could use this last formula to calculate, n, the number of rectangles needed. And solving for n yields n = 960. Meaning that 960 rectangles would be need to guarantee an error that does not exceed 0.05 !!!!
10 Definite Integral as Limit of Sums Let f be a function on interval [a, b]. Partition [a, b] into n subintervals of equal length ∆ x = (b – a)/n with endpoints a = x 0 < x 1 < x 2, … < x n – 1 < x n = b, Then This last sum is called a Riemann sum and each c k is required to be in the subinterval [x k – 1, x k ].
11 A Visual Presentation of the Above a = x 0 x n = bx 1 x 2 x n -1... c 1 c 2 c n f (c 1 ) f (c 2 ) Δx The area under the curve is approximated by the Riemann sum
12 Area Let’s revisit our original area problem and calculate the Riemann sum using the midpoints for c k. The width of each rectangle is again Δx = 1. The height of each of the four rectangles is now f (1.5), f (2.5), f (3.5) and f(4.5) respectively. The sum of the rectangles is then S 4 = 3.125 + 5.125 + 8.125 + 12.125 = 28.5 f (x) =.5 x 2 + 2 1 2 3 4 5 This is quite close to the actual area of 28.666....
13 Definite Integral Theorem 3. Let f be a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as n → ∞. Definition. Let f be a continuous function on [a, b]. The limit I of the Riemann sums for f on [a, b] is called the definite integral of f from a to b, denoted The integrand is f (x), the lower limit of integration is a, and the upper limit of integration is b.
14 Negative Values If f (x) is positive for some values of x on [a,b] and negative for others, then the definite integral symbol Represents the cumulative sum of the signed areas between the graph of f (x) and the x axis where areas above are positive and areas below negative. y = f (x) a bA B
15 Examples Calculate the definite integrals by referring to the figure with the indicated areas. Area A = 3.5 Area B = 12 y = f (x) a bA B c - 3.5 12 - 3.5 + 12 = 8.5
16 Definite Integral Properties
17 Examples If A)B)C)D) F)
18 Summary. We summed rectangles under a curve using both the left and right ends and the centers and found that as the number of rectangles increased accuracy of the area under the curve increased. We found error bounds for these sums and average. We defined the definite integral as the limit of these sums and found that it represented the area between the function and the x axis.. We learned how to deal with areas under the x axis.
19 Practice Problems §12.4; 1, 3, 7, 9, 13, 17, 21, 25, 29, 33, 47, 51, 55.
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# Eric plans to compete in a triathlon. He can average 2 miles per hour
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Eric plans to compete in a triathlon. He can average 2 miles per hour [#permalink]
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27 Mar 2019, 23:16
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Eric plans to compete in a triathlon. He can average 2 miles per hour in the 1/4-mile swim and 6 miles per hour in the 3-mile run. His goal is to finish the triathlon in 2 hours. To accomplish his goal what must his average speed in miles per hour, be for the 15-mile bicycle ride?
(A) 120/11
(B) 11
(C) 56/5
(D) 45/4
(E) 12
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Posts: 450
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Eric plans to compete in a triathlon. He can average 2 miles per hour [#permalink]
### Show Tags
27 Mar 2019, 23:54
Main Concept:
If D=Distance; r=rate; t=time
r*t=D
1) Swimming part:
D=$$\frac{1}{4}$$ miles r=2 m/h t=$$\frac{1}{4}$$ /2 =$$\frac{1}{8}$$ hour
2) Running part:
D=3 miles r=6 m/h t=$$\frac{3}{6}$$=$$\frac{1}{2}$$ hour
3) Bicycle part:
D=15 miles t=2-($$\frac{1}{2}$$+$$\frac{1}{8}$$)=$$\frac{11}{8}$$ hour r=15/ $$\frac{11}{8}$$=$$\frac{120}{11}$$
IMO
Ans: A
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Re: Eric plans to compete in a triathlon. He can average 2 miles per hour [#permalink]
### Show Tags
28 Mar 2019, 03:29
Bunuel wrote:
Eric plans to compete in a triathlon. He can average 2 miles per hour in the 1/4-mile swim and 6 miles per hour in the 3-mile run. His goal is to finish the triathlon in 2 hours. To accomplish his goal what must his average speed in miles per hour, be for the 15-mile bicycle ride?
(A) 120/11
(B) 11
(C) 56/5
(D) 45/4
(E) 12
solve for indvidual events and determine time lapse of each event
event 1 ; time ; 1/4 * 1/2 hrs' 1/8 hrs
event 2; time ; 3/6 ; 1/2 hrs
total time left ; 2-(1/8+1/2); 11/8 hrs
so for 15 mile even he has to have speed of
15 * 8/11 ; 120 /11
IMO A
Re: Eric plans to compete in a triathlon. He can average 2 miles per hour [#permalink] 28 Mar 2019, 03:29
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# Eric plans to compete in a triathlon. He can average 2 miles per hour
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# Partial fractions integration
Re-express $\dfrac{6x^5 + x^2 + x + 2}{(x^2 + 2x + 1)(2x^2 - x + 4)(x+1)}$ in terms of partial fractions and compute the indefinite integral $\dfrac{1}5{}\int f(x)dx$ using the result from the first part of the question.
-
## Hint
Use $$\dfrac{6x^5 + x^2 + x + 2}{(x^2 + 2x + 1)(2x^2 - x + 4)(x+1)}=\frac{A}{(x+1)^3}+\frac{B}{(x+1)^2}+\frac{C}{x+1}+\frac{Dx+E}{2x^2-x+4}+F$$ and solve for $A,B,C,D,E$ and $F$.
## Explanation
Note that this partial fraction decomposition is a case where the degree of the numerator and denominator are the same. (Just notice that the sum of the highest degree terms in the denominator is $2+2+1=5.$) This means that the typical use of partial fraction decomposition does not apply.
Furthermore, notice that $(x^2+2x+1)$ factorizes as $(x+1)^2$, which means the denominator is really $(x+1)^3(2x^2-x+4)$. This means the most basic decomposition would involve denominators of $(x+1)^3$ and $(2x^2-x+4)$. However, we can escape that by using a more complicated decomposition involving the denominators $(x+1)^3$, $(x+1)^2$, $(x+1)$, and $(2x^2-x+4)$. The $F$ term is necessary for the $6x^5$ term to arise in the multiplication, intuitively speaking.
More thoroughly stated, the $F$ term is needed because of the following equivalency between $6x^5+x^2+x+2$ and the following expression: \begin{align} &A(2x^2-x+4)\\ +&B(2x^2-x+4)(x+1)\\ +&C(2x^2-x+4)(x+1)^2\\ +&[Dx+E](x+1)^3\\ +&F(2x^2-x+4)(x+1)^3. \end{align}
Notice how the term $F(2x^2-x+4)(x+1)^3$ is the only possible term that can give rise to $6x^5$? That is precisely why it is there.
## Hint 2
### Part 1
In the integration of $\frac{1}{5}\int f(x)dx$, first separate the integral, $\int f(x)dx$ into many small integrals with the constants removed from the integrals. That is, $\int f(x) dx$ is $$A\int \frac{1}{(x+1)^3}dx+B\int \frac{1}{(x+1)^2}dx+C\int \frac{1}{x+1}dx+D\int \frac{x}{2x^2-x+4}dx\quad+E\int \frac{1}{2x^2-x+4}dx+F\int 1dx.$$
### Part 2
Next, use the substitution $u=x+1$ with $du=dx$ on the first three small integrals:
$$A\int \frac{1}{u^3}du+B\int \frac{1}{u^2}du+C\int \frac{1}{u}du+D\int \frac{x}{2x^2-x+4}dx+E\int \frac{1}{2x^2-x+4}dx\quad +F\int 1dx.$$
### Part 3
To deal with the integral $\int \frac{1}{2x^2-x+4}dx$, you must complete the square. This is done as follows:
\begin{align} \int \frac{1}{2x^2-x+4}dx&=\int \frac{\frac{1}{2}}{x^2-\frac{1}{2}x+2}dx\\ &=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2+\frac{31}{16}}dx. \end{align}
To conclude, you make the trig substitution $x-\frac{1}{4}=\frac{\sqrt{31}}{4}\tan \theta$ with $dx=\left(\frac{\sqrt{31}}{4}\tan \theta+\frac{1}{4}\right)'d\theta=\frac{\sqrt{31}}{4}\sec^2\theta d\theta$. This gives you:
$$\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2+\frac{31}{16}}dx=\frac{1}{2} \int \frac{\frac{\sqrt{31}}{4}\sec^2 \theta}{\frac{31}{16}\tan^2\theta+\frac{31}{16}}d\theta.$$
-
The degree of the numerator and denominator is the same; so you have to do the division first. Also note $x^2+2x+1=(x+1)^2$; so your factorization is incorrect. – David Mitra Dec 1 '12 at 20:10
There is a little problem with what you wrote. The remark of @DavidMitra regards the initial fraction - there the degree of the numerator and denominator is the same, which implies that there should be some constant factor, i.e: $$\frac{A}{(x+1)^3}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)}+\frac{Dx+E}{2x^2-x+4}+F$$ – Dennis Gulko Dec 1 '12 at 20:26
@Limitless Note that if you simplify the right hand side of the equation in your last comment, the degree of the numerator will be $4$ (and not $5$ as it should be). Generally, for the Partial Fraction decomposition method (as given here for example) to succeed, the degree of the numerator must be smaller than the degree of the denominator. – David Mitra Dec 1 '12 at 20:28
@Limitless: look at my edited remark. If you expand what you wrote, you get a polynomial of degree $4$ in the numerator. – Dennis Gulko Dec 1 '12 at 20:30
@Limitless The usual algorithm taught would give the wrong decomposition. However, you can correct it as Dennis writes in his comment above (in this case by adding a constant term $F$). Things will be ok then. But note the $F$ is added for this particular case. For other rational functions, a different correction term might be needed. It would be easier, in my opinion, just to do the division (polynomial division) in these cases where the degree of the numerator is at least the degree of the denominator. – David Mitra Dec 1 '12 at 20:35<|endoftext|>
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GCSE Maths Algebra Equations Solving Equations Quadratic Equations
# Solving Quadratic Equations By Factorising
Here we will learn about solving quadratic equations by factorising including how to solve quadratic equations by factorising when a = 1 and when a > 1.
There are also solving quadratic equations worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What is solving quadratic equations by factorising?
Solving quadratic equations by factorising allows us to calculate values of the unknown variable in a quadratic equation using factorisation.
To do this we make sure the equation is equal to 0, factorise it into brackets and then solve the resulting linear equations.
E.g.
Solve the following quadratic equation by factorising:
$x^{2}-2x-15 =0$
\begin{align} x^{2} – 2x – 15 &= 0 \\ (x-5)(x + 3) &= 0 \end{align}
So,
\begin{align} x – 5&=0 \\ x &= 5 \end{align}
and
\begin{align} x + 3 &= 0 \\ x &= -3 \end{align}
E.g.
Solve the following quadratic equation by factorising:
$2 x^2+7 x – 4 = 0$
\begin{align} 2x^{2} + 7x – 4 &= 0 \\ (2x – 1)(x + 4) &= 0 \end{align}
So,
\begin{align} 2 x – 1&=0 \\ x &= \frac{1}{2} \end{align}
and
\begin{align} x + 4 &= 0 \\ x &= -4 \end{align}
## How to solve quadratic equations
In order to factor a quadratic algebraic equation we need to make sure it is in the form of the general quadratic equation:
$ax^{2}+bx+c=0$
We must ensure the quadratic equation is equal to 0, rearranging it if necessary.
NOTE: Quadratic equations are a type of polynomial equation because they consist of two or more algebraic terms.
## How to solve a quadratic equation by factorising
First make sure that the equation is equal to 0.
Step 1: Fully factorise the quadratic.
Step 2: Set each bracket equal to 0.
Step 3: Solve each linear equation.
Step-by-step guide: Solving linear equations
### Related lessons on quadratic equations
Solving quadratic equations by factorising is part of our series of lessons to support revision on quadratic equations and solving equations. You may find it helpful to start with the main solving equations lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
## Examples of solving quadratic equations
### Example 1: solve a quadratic equation by factorising when a = 1
Solve
$x^{2}-8x+15=0$
1. Fully factorise the quadratic equation.
$(x-3)(x-5)=0$
2We know that if two values multiply together to get 0, at least one of them must be 0. So set each bracket equal to 0.
$x-3=0\qquad \qquad x-5=0$
3Solve each equation to find x.
The opposite of -3 is +3, so +3 to both sides of the equation.
The opposite of -5 is +5, so +5 on left hand side, and on the right hand side.
When we solve a quadratic equation we normally have two solutions.
We call these the solutions, or roots, of the quadratic equation.
When we plot these values on an x, y grid we get a special ‘U’ shaped curve called a parabola.
We can see the real roots of the quadratic equation are where the quadratic graph crosses the x-axis.
We can check that our solution is correct by substituting it into the original equation.
### Example 2: solve a quadratic equation by factorising when a > 1
Solve
$2x^{2}+5x+3=0$
We know that if two values multiply together to get 0, at least one of them must be 0. So put each bracket equal to 0.
Solve each equation to find x.
### Common misconceptions
• Solve by factorising into two brackets
E.g.
\begin{aligned} x^{2}-2x-24&=0\\ (x-6)(x+4)&=0 \end{aligned}
Do not try and square root the quadratic otherwise you will not get all the solutions!
• The order of the brackets doesn’t matter
When we multiply two values the order doesn’t matter.
E.g.
$2\times 3=3\times 2$
It is exactly the same here:
$(x-6)(x+4)$
means
$(x-6)\times (x+4)$
So,
$(x-6)(x+4)=0$
is the same as
$(x+4)(x-6)=0$
• Forgetting to solve after factorising
Don’t forget to set the factorised expression equal to zero and solve it.
Always check you have answered the question.
• Factorising or factoring?
The term factorising can sometimes be written as ‘factoring’ or ‘factorization’.
• Difference of two squares
If the quadratic equation involves two square terms (perfect squares) that are subtracted from each other, you will need to use the ‘difference of two squares’ to factorise it.
$4 x^{2}-16=0 \\ (2 x+4)(2 x-4)=0 \\ 2 x+4=0, \quad 2 x-4=0$
• Equation cannot be factorised
If a quadratic equation cannot be factorised, we can still solve it by using the quadratic formula.
To work out the number of real solutions a quadratic equation has we can use the discriminant. The derivation of the quadratic formula is fascinating, we will explore it more when we learn about ‘completing the square’.
• Equation should be in the below form before trying to factorise
$a x^{2}+b x+c=0$
It is possible to factorise a quadratic equation without ensuring that it is in the standard form of a quadratic equation, however this can be challenging.
• When we solve a quadratic equation by factorisation at GCSE we will always get real numbers that are rational numbers
### Practice solving quadratic equations questions
1. Solve: {x}^2+5x+6=0
x=3 x=-2
x=3 x=2
x=-3 x=-2
x=-3 x=2
{x}^2+5x+6=0 can be factorised as (x+2)(x+3)=0 . Setting each bracket equal to zero and solving leads to the required solutions.
2. Solve: {x}^2-x-20=0
x=-5 x=-4
x=5 x=4
x=-5 x=4
x=5 x=-4
{x}^2-x-20=0 can be factorised as (x-5)(x+4)=0 . Setting each bracket equal to zero and solving leads to the required solutions.
3. Solve: 2{x}^2+3x-9=0
2{x}^2+3x-9=0 can be factorised as (2x-3)(x+3)=0 . Setting each bracket equal to zero and solving leads to the required solutions.
4. Solve: 3{x}^2-9x+6=0
x=2 x=1
x=-2 x=-1
x=2 x=-1
x=-2 x=1
3{x}^2-9x+6=0 can be factorised as 3(x-1)(x-2)=0 . Setting each bracket equal to zero and solving leads to the required solutions.
### Solving quadratic equations GCSE questions
1. Factorise x^{2}-x-30
(2 marks)
(x+5) or (x-6)
(1)
(x+5) (x-6)
(1)
2. Hence, or otherwise, solve x^{2}-x-30=0
(1 mark)
x=-5 and x=6
(1)
3. Solve 2 x^{2}-5 x-3
(3 marks)
(2x+1) (x-3)=0
(1)
2x+1=0 and x-3=0
(1)
x=-\frac{1}{2} \text { and } x=3
(1)
### Did you know?
• Did you know that Al-Khwarizmi (Abu Ja’far Muhammad ibn Musa al-Khwarizmi) was one of the first people in history to write about algebra? He lived in Baghdad in around 780 to 850 AD and he wrote a book called “Hisab Al-jabr w’al-muqabala”, in which we get the word ‘algebra’ (meaning ‘restoration of broken parts’) from.
Copyright (c) 2021 by khaled (https://codepen.io/ksawalme/pen/xLvGOv)
• Did you know the ancient babylonians could solve quadratic equations using a method equivalent to the quadratic formula, despite not using algebraic notation!
• Did you also know that the ancient Greek mathematician Euclid used geometric methods to solve quadratic equations way back in 300BC! His book, “The Elements”, is one of the most studied books in human history.
The history of mathematics is amazing!
## Learning checklist
You have now learned how to:
• Solve quadratic equations algebraically by factorising
• Solve quadratic equations by finding approximate solutions using a graph
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme.<|endoftext|>
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# Area of parametric equations
### Area of parametric equations
In this section, we will learn find the area under the curve of parametric equations. This still involves integration, but the integrand looks changed. The integrand is now the product between the second function and the derivative of the first function. We will examine the different types of parametric equations with a given range, and learn how to find the area of each one.
#### Lessons
Normally we know that the area under the curve from $a$ to $b$ is $\int_{a}^{b} f(x)dx$. However, what about parametric equations?
Let the curve be defined by the parametric equations $x=f(t)$, $y=g(t)$ and let the value of $t$ be increasing from $\alpha$ to $\beta$. Then we say that the area under the parametric curve is:
$A = \int_{a}^{b} y \; dx=\int_{\alpha}^{\beta} g(t)f'(t)dt$
However, if the value of $t$ is increasing from $\beta$ to $\alpha$ instead, then the area under the parametric curve will be:
$A = \int_{a}^{b} y \; dx=\int_{\beta}^{\alpha} g(t)f'(t)dt$
Be careful when determining which one to use!
• 1.
Area of Parametric Functions Overview
• 2.
Finding the Area Given the Range of the Parameter
Find the area under the curve of the parametric curve $x=t^2+1$
$y=t^3+t^2+4$, where $1 \leq t \leq 3$.
Assume that the curve traces perfectly from left to right for the range of the parameter $t$.
• 3.
Find the area enclosed of the given parametric curve $x=a \cos (\theta)$, $y= b \sin (\theta)$, where $0 \leq \theta \leq 2 \pi$ and $a, b$ are constants.
• 4.
Find the area under the curve of the parametric equations $x=t-\frac{1}{t}$, $y=t+\frac{1}{t}$, where $\frac{1}{2} \leq t \leq 2$.<|endoftext|>
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Test: Ratio and Proportion- 4 - CLAT MCQ
# Test: Ratio and Proportion- 4 - CLAT MCQ
Test Description
## 20 Questions MCQ Test Quantitative Techniques for CLAT - Test: Ratio and Proportion- 4
Test: Ratio and Proportion- 4 for CLAT 2024 is part of Quantitative Techniques for CLAT preparation. The Test: Ratio and Proportion- 4 questions and answers have been prepared according to the CLAT exam syllabus.The Test: Ratio and Proportion- 4 MCQs are made for CLAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Ratio and Proportion- 4 below.
Solutions of Test: Ratio and Proportion- 4 questions in English are available as part of our Quantitative Techniques for CLAT for CLAT & Test: Ratio and Proportion- 4 solutions in Hindi for Quantitative Techniques for CLAT course. Download more important topics, notes, lectures and mock test series for CLAT Exam by signing up for free. Attempt Test: Ratio and Proportion- 4 | 20 questions in 15 minutes | Mock test for CLAT preparation | Free important questions MCQ to study Quantitative Techniques for CLAT for CLAT Exam | Download free PDF with solutions
Test: Ratio and Proportion- 4 - Question 1
### Divide Rs.2340 into three parts, such that first part be double that of second part and second part be 1/3 of the third part.Find the Third part amount?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 1
Answer – B.Rs.1170 Explanation : First: Second: Third = 2:1:3 Third part = 3*2340/6 = 1170
Test: Ratio and Proportion- 4 - Question 2
### The ratio of income of A and B is 2:3. The sum of their expenditure is Rs.8000 and the amount of savings of A is equal to the amount of expenditure of B.What is the their ratio of sum of income to their sum of savings?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 2
Answer -A.5:3 Explanation : 2I-E + E = 8000
I = 4000
Sum of their Income = 5*I = 5*4000 = 20,000 Sum of their Savings = 20000-8000 = 12,000 20000:12000 = 5:3
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Test: Ratio and Proportion- 4 - Question 3
### There are 2 containers of equal capacity. The ratio of milk to water in the first container is 4:5 and in the second container is 3:7.If they are mixed up then the ratio of milk to water in the mixture will be
Detailed Solution for Test: Ratio and Proportion- 4 - Question 3
Answer – D.67:113 Explanation : 4+5 = 9=> 40:50
3+7 = 10=> 27:63
40+27 : 50:63 = 67:113
Test: Ratio and Proportion- 4 - Question 4
There are two numbers. When 25% of the first number is added to the second number, the resultant number is 1.5times th first number.What is the ratio of 1st number to the 2nd number ?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 4
Answer – C.4:5 Explanation : A+25/100 + B = 1.5A
A/4 + B = 15A/10
10A+40B/40 =60A/40
10A+40B = 60A
50A = 40B
A/B = 4/5
Test: Ratio and Proportion- 4 - Question 5
A bag contains 10p,25p and Rs50p coins in the ratio of 5:2:1 respectively. If the total money in the bag is Rs.120.Find the number of 25p coins in that bag?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 5
Answer – A.160 Explanation : 10*5 : 25*2 : 50*1 = 50:50:50 = 1:1:1
120/3 = Rs.40 Rs. 1 = 4 Rs.40 = 4*40 = 160 coins
Test: Ratio and Proportion- 4 - Question 6
The ratio of Ganesh’s age and his mother’s age is 5:12.The difference of their ages is 21.The ratio of their ages after 4 years will be
Detailed Solution for Test: Ratio and Proportion- 4 - Question 6
Answer – D.19:40 Explanation : 12x – 5x = 21 7x = 21 X = 3
5:12 = 15:36
After 4 years = 19:40
Test: Ratio and Proportion- 4 - Question 7
The ratio of students of three classes is 2:3:4. If 15 students are increased in each classes then their ratio turns into 13:18:23. What was the total number of students in all the three classes originally ?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 7
Answer – C.225 Explanation : 50:75:100
15 students increased 65:90:115 => 13:18 :23
Total no of students = 50+75+100 = 225
Test: Ratio and Proportion- 4 - Question 8
Ravi and Govind have money in the ratio 5 : 12 and Govind and Kiran also have money in the same ratio 5 : 12. If Ravi has Rs. 500, Kiran has
Detailed Solution for Test: Ratio and Proportion- 4 - Question 8
Answer – B.Rs.2880 Explanation : Ravi : Kiran = 5/12* 5/12 = 25/144 Kiran = 144*500/25 = 2880
Test: Ratio and Proportion- 4 - Question 9
A town with a population of 1000 has provision for 30days, after 10 days 600 more men added, how long will the food last at the same rate ?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 9
Answer – C.12 ½ days Explanation : 1000*20/1600 = 12 1/2 days
Test: Ratio and Proportion- 4 - Question 10
A man spends Rs.2480 to buy lunch box Rs.120 each and bottles at Rs.80 each,What will be the ratio of maximum number of bottles to lunch box are bought ?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 10
Answer – A.13:12 Explanation : Check the ans using option 13*80+ 12*120 = 1040+1440 = 2480
Test: Ratio and Proportion- 4 - Question 11
Three cars travel same distance with speeds in the ratio 2 : 4 : 7. What is the ratio of the times taken by them to cover the distance?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 11
B) 14 : 7 : 4
Explanation: s = d/t Since distance is same, so ratio of times: 1/2 : 1/4 : 1/7 = 14 : 7 : 4
Test: Ratio and Proportion- 4 - Question 12
Section A and section B of 7th class in a school contains total 285 students.Which of the following can be a ratio of the ratio of the number of boys and number of girls in the class?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 12
B) 10 : 9
Explanation: The number of boys and girls cannot be in decimal values, so the denominator should completely divide number of students (285).
Check each option: 6+5 = 11, and 11 does not divide 285 completely. 10+9 = 19, and only 19 divides 285 completely among all.
Test: Ratio and Proportion- 4 - Question 13
180 sweets are divided among friends A, B, C and D in which B and C are brothers also such that sweets divided between A and B are in the ratio 2 : 3, between B and C in the ratio 2 : 5 and between C and D in ratio 3 : 4. What is the number of sweets received by the brothers together?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 13
B) 84
Explanation: A/B = N1/D1 B/C = N2/D2 C/D = N3/D3
A : B : C : D = N1*N2*N3 : D1*N2*N3 : D1*D2*N3 : D1*D2*D3
A/B = 2/3 B/C = 2/5 C/D = 3/4
A : B : C : D
2*2*3 : 3*2*3 : 3*5*3 : 3*5*4
4 : 6 : 15 : 20
B and C together = [(6+15)/(4+6+15+20)] * 180
Test: Ratio and Proportion- 4 - Question 14
Number of students in 4th and 5th class is in the ratio 6 : 11. 40% in class 4 are girls and 48% in class 5 are girls. What percentage of students in both the classes are boys?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 14
B) 54.8%
Explanation: Total students in both = 6x+11x = 17x Boys in class 4 = (60/100)*6x = 360x/100 Boys in class 5 = (52/100)*11x = 572x/100 So total boys = 360x/100 + 572x/100 = 932x/100 = 9.32x % of boys = [9.32x/17x] * 100
Test: Ratio and Proportion- 4 - Question 15
Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?
Detailed Solution for Test: Ratio and Proportion- 4 - Question 15
E) 5 : 7
Explanation: Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg Total brass = 30+40 = 70 kg So copper in mixture is (50+70) – 70 = 50 kg So copper to brass = 50 : 70
Test: Ratio and Proportion- 4 - Question 16
Ratio of A and B is in the ratio 5 : 8. After 6 years, the ratio of ages of A and B will be in the ratio 17 : 26. Find the present age of B.
Detailed Solution for Test: Ratio and Proportion- 4 - Question 16
A) 72
Explanation: A/B = 5/8 , A+6/B+6 = 17/26
Solve both, B = 72
Test: Ratio and Proportion- 4 - Question 17
A bag contains 25p, 50p and 1Re coins in the ratio of 2 : 4 : 5 respectively. If the total money in the bag is Rs 75, find the number of 50p coins in the bag.
Detailed Solution for Test: Ratio and Proportion- 4 - Question 17
D) 40
Explanation: 2x, 4x, 5x (25/100)*2x + (50/100)*4x + 1*5x = 75 x = 10, so 50 p coins = 4x = 40
Test: Ratio and Proportion- 4 - Question 18
A is directly proportional to B and also directly proportional to C. When B = 6 and C = 2, A = 24. Find the value of A when B = 8 and C = 3.
Detailed Solution for Test: Ratio and Proportion- 4 - Question 18
D) 48
Explanation: A directly proportional B, A directly proportional to C: A = kB, A = kC Or A = kBC When B = 6 and C = 2, A = 24: 24 = k*6*2 k = 2 Now when B = 8 and C = 3: A = 2*8*3
Test: Ratio and Proportion- 4 - Question 19
A is directly proportional to B and also inversely proportional to the square of C.When B = 16 and C = 2, A = 36. Find the value of A when B = 32 and C = 4.
Detailed Solution for Test: Ratio and Proportion- 4 - Question 19
C) 18
Explanation: A = kB, A = k/C2 Or A = kB/ C2
When B = 16 and C = 2, A = 36: 36 = k*16/ 22 k = 9 Now when B = 32 and C = 4: A = 9*32/ 42
Test: Ratio and Proportion- 4 - Question 20
A is directly proportional to the inverse of B and also inversely proportional to C. When B = 36 and C = 9, A = 42. Find the value of A when B = 64 and C = 21.
Detailed Solution for Test: Ratio and Proportion- 4 - Question 20
A) 24
Explanation: A = k√B, A = k/C Or A = k√B/C When B = 36 and C = 9, A = 42: 42 = k√36/9 k = 63 Now when B = 64 and C = 21: A = 63*√64/21
## Quantitative Techniques for CLAT
56 videos|104 docs|95 tests
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## Quantitative Techniques for CLAT
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We've updated our
TEXT
Z-Scores
Objectives
1. Here you will learn how z-scores can be used to evaluate how extreme a given value is in a particular set or population.
2. Here you will learn to evaluate z-scores as they relate to probability.
3. Here you will learn to calculate the probability of a z-score between two others.
Part I
Using the Empirical Rule can give you a good idea of the probability of occurrence of a value that happens to be exactly one, two or three to either side of the mean, but how do you compare the probabilities of values that are in between standard deviations?
Watch This: Maths Tutorial: Z scores
The British video below is very clear and easy to follow. It is worth noting, particularly for US students, that the instructor uses the notation x rather than µ for mean, and pronounces z as “zed.” https://youtu.be/2JjaWQZChqs
Z-scores are related to the Empirical Rule from the standpoint of being a method of evaluating how extreme a particular value is in a given set. You can think of a z-score as the number of standard deviations there are between a given value and the mean of the set. While the Empirical Rule allows you to associate the first three standard deviations with the percentage of data that each SD includes, the z-score allows you to state (as accurately as you like), just how many SDs a given value is above or below the mean. Conceptually, the z-score calculation is just what you might expect, given that you are calculating the number of SDs between a value and the mean. You calculate the z-score by first calculating the difference between your value and the mean, and then dividing that amount by the standard deviation of the set. The formula looks like this: [latex-display]\displaystyle\text{z-score}=\frac{(\text{value}-\text{mean})}{\text{standard deviation}}=\frac{(x=\mu)}{\sigma}\\[/latex-display] In this lesson, we will practice calculating the z-score for various values. In the next lesson, we will learn how to associate the z-score of a value with the probability that the value will occur.
Example 1
What is the z-score of a value of 27, given a set mean of 24, and a standard deviation of 2?
Solution
To find the z-score we need to divide the difference between the value, 27, and the mean, 24, by the standard deviation of the set, 2. [latex-display]\displaystyle\text{z-score}=\frac{27-\mu}{\sigma}=\frac{27-24}{2}=\frac{3}{2}\\[/latex-display] [latex-display]\displaystyle\text{z-score of }27=+1.5\\[/latex-display] This indicates that 27 is 1.5 standard deviations above the mean.
Example 2
What is the z-score of a value of 104.5, in a set with µ = 125 and σ = 6.2?
Solution
Find the difference between the given value and the mean, then divide it by the standard deviation. [latex-display]\displaystyle\text{z-score}=\frac{104.5-\mu}{\sigma}=\frac{104.5-125}{6.2}=\frac{-20.5}{6.2}\\[/latex-display] [latex-display]\displaystyle\text{z-score of }104005=-3.306\\[/latex-display] Note that the z-score is negative, since the measured value, 104.5, is less than (below) the mean, 125.
Example 3
Find the value represented by a z-score of 2.403, given µ = 63 and σ = 4.25.
Solution
This one requires that we solve for a missing value rather than for a missing z-score, so we just need to fill in our formula with what we know and solve for the missing value: [latex-display]\displaystyle\text{z-score}=\frac{x-\mu}{\sigma}\\[/latex-display] [latex-display]\displaystyle2.403=\frac{x-63}{4.25}\\[/latex-display] [latex-display]\displaystyle10.213=x-63\\[/latex-display] [latex-display]73.213=x\\[/latex-display] 73.213 has a z-score of 2.403
Intro Problem Revisited
Using the Empirical Rule can give you a good idea of the probability of occurrence of a value that happens to be right on one of the first three standard deviations to either side of the mean, but how do you compare the probabilities of values that are in between standard deviations? The z-score of a value is the count of the number of standard deviations between the value and the mean of the set. You can find it by subtracting the value from the mean, and dividing the result by the standard deviation.
Vocabulary
The z-score of a value is the number of standard deviations between the value and the mean of the set.
Guided Practice
1. What is the z-score of the price of a pair of skis that cost $247, if the mean ski price is$279, with a standard deviation of \$16?
2. What is the z-score of a 5-scoop ice cream cone if the mean number of scoops is 3, with a standard deviation of 1 scoop?
3. What is the z-score of the weight of a cow that tips the scales at 825 lbs, if the mean weight for cows of her type is 1150 lbs, with a standard deviation of 77 lbs?
4. What is the z-score of a measured value of 0.0034, given µ = 0.0041 and σ = 0.0008?
Solutions
1. First find the difference between the measured value and the mean, then divide that difference by the standard deviation: [latex-display]\frac{247-279}{16}=\frac{-32}{16}\\[/latex-display] z-score = −2
2. This one is easy: The difference between 5 scoops and 3 scoops is +2, and we divide that by the standard deviation of 1, so the z-score is +2.
3. First find the difference between the measured value and the mean, then divide that difference by the standard deviation: [latex-display]\frac{825\text{ lbs}-1150\text{ lbs}}{771\text{ lbs}}=\frac{-325}{77}\\[/latex-display] z-score = −4.2407
4. First find the difference between the measured value and the mean, then divide that difference by the standard deviation: [latex-display]\frac{0.0034-0.0041}{0.0008}=\frac{-0.0007}{0.0008}\\[/latex-display] z-score = −0.875
Part II
Knowing the z-score of a given value is great, but what can you do with it? How does a z-score relate to probability? What is the probability of occurrence of a z-score less than +2.47?
Watch This: The Normal Distribution
The video below provides a demonstration of how to use a z-score probability reference table, as we do in this lesson. The table he uses in the video is slightly different, but the concept is the same. https://youtu.be/rEmNUkKSpbU
Since z-scores are a measure of the number of SDs between a value and the mean, they can be used to calculate probability by comparing the location of the z-score to the area under a normal curve either to the left or right. The area can be calculated using calculus, but we will just use a table to look up the area. I believe that the concept of comparing z-scores to probability is most easily understood with a graphic like the one we used in the lesson on the Empirical Rule, so I included one below. Be sure to review the examples to see how the scores work. Like the graphic we viewed in the Empirical Rule lesson, this one only provides probability percentages for integer values of z-scores (standard deviations). In order to find the values for z-scores that aren’t integers, you can use a table like the one below. To find the value associated with a given z-score, you find the first decimal of your z-score on the left or right side and then the 2nd decimal of your z-score across the top or bottom of the table. Where they intersect you will find the decimal expression of the percentage of values that are less than your sample (see example 4).
Table 1 (scroll to see all values)
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 Z
0.0 0.5 0.504 0.508 0.512 0.516 0.5199 0.5239 0.5279 0.5319 0.5359 0.0
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.1
0.2 0.5793 0.5832 0.5871 0.591 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.2
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.648 0.6517 0.3
0.4 0.6554 0.6591 0.6628 0.6664 0.67 0.6736 0.6772 0.6808 0.6844 0.6879 0.4
0.5 0.6915 0.695 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.719 0.7224 0.5
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.6
0.7 0.758 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.7
0.8 0.7881 0.791 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.8
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.834 0.8365 0.8389 0.9
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.0
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.877 0.879 0.881 0.883 1.1
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.898 0.8997 0.9015 1.2
1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.3
1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.4
1.5 0.9332 0.9345 0.9357 0.937 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.5
1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.6
1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.7
1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.8
1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.975 0.9756 0.9761 0.9767 1.9
2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.0
2.1 0.9821 0.9826 0.983 0.9834 0.9838 0.9842 0.9846 0.985 0.9854 0.9857 2.1
2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.989 2.2
2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 2.3
2.4 0.9918 0.992 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 2.4
2.5 0.9938 0.994 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.5
2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.996 0.9961 0.9962 0.9963 0.9964 2.6
2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.997 0.9971 0.9972 0.9973 0.9974 2.7
2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.998 0.9981 2.8
2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986 2.9
3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.999 0.999 3.0
3.1 0.999 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993 3.1
3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995 3.2
3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997 3.3
3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998 3.4
3.5 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 3.5
3.6 0.9998 0.9998 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 3.6
3.7 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 3.7
3.8 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 3.8
3.9 1 1 1 1 1 1 1 1 1 1 3.9
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 Z
Z-score tables like the one above describe the probability that a given value, or any value less than it, will occur in a given set. This particular table assumes you are looking to find the probability associated with a positive z-score. You may have additional work to do if the z-score is negative.
• To find the percentage of values greater than a negative Z score, just look up the matching positive Z score value.
• To find the percentage of values less than a negative z-score, subtract the chart value from 1.
• To find the percentage of values greater than a positive z-score, subtract the chart value from 1.
Example 4
What is the probability that a value with a z-score less than 2.47 will occur in a normal distribution?
Solution
Scroll up to the table above and find “2.4” on the left or right side. Now move across the table to “0.07” on the top or bottom, and record the value in the cell: 0.9932. That tells us that 99.32% of values in the set are at or below a z-score of 2.47.
Example 5
What is the probability that a value with a z-score greater than 1.53 will occur in a normal distribution?
Solution
Scroll up to the table of z-score probabilities again and find the intersection between 1.5 on the left or right and 3 on the top or bottom, record the value in the cell: 0.937. That decimal lets us know that 93.7% of values in the set are below the z-score of 1.53. To find the percentage that is above that value, we subtract 0.937 from 1.0 (or 93.7% from 100%), to get 0.063 or 6.3%.
Example 6
What is the probability of a random selection being less than 3.65, given a normal distribution with µ = 5 and σ = 2.2?
Solution
This question requires us to first find the z-score for the value 3.65, then calculate the percentage of values below that z-score from a reference.
1. Find the z-score for 3.65, using the z-score formula: $\frac{(x-\mu)}{\sigma}=\frac{3.65-5}{2.2}=\frac{-1.35}{2.2}\approx{-0.61}\\$
2. Now we can scroll up to our z-score reference above and find the intersection of 0.6 and 0.01, which should be .7291.
3. Since this is a negative z-score, and we want the percentage of values below it, we subtract that decimal from 1.0 (reference the three steps highlighted by bullet points below the chart if you didn’t recall this), to get 1 − .7291 = .2709
There is approximately a 27.09% probability that a value less than 3.65 would occur from a random selection of a normal distribution with mean 5 and standard deviation 2. 2.
Concept Problem Revisited
Knowing the z-score of a given value is great, but what can you do with it? How does a z-score relate to probability? What is the probability of occurrence of a z-score less than 2.47? A z-score lets you calculate the probability that a randomly selected value will be greater or less than a particular value in a set. To find the probability of a z-score below +2.47, using a reference such as the table in the lesson above:
1. Find 2.4 on the left or right side
2. Move across to 0.07 on the top or bottom.
3. The cell you arrive at says: 0.9932, which means that apx 99.32% of the values in a normal distribution will occur below a z-score of 2.47.
Vocabulary
Z-score table: a table that associates the various common z-scores between 0 and 3.99 with the decimal probability of being less than or equal to that z-score.
Guided Practice
1. What is the probability of occurrence of a value with z-score greater than 1.24?
2. What is the probability of z < −.23?
3. What is P(Z < 2.13)?
Solutions
1. Since this is a positive z-score, we can use the value for z = 1.24 directly from the table, and just express it as a percentage: 0.8925 or 89.25%
2. This is a negative z-score, and we want the percentage of values greater than it, so we need to subtract the value for z = +0.23 from 1: 1 − 0.591 = .409 or 40.9%
3. This is a positive z-score, and we need the percentage of values below it, so we can use the percentage associated with z = +2.13 directly from the table: 0.9834 or 98.34%
Part III
Do z-score probabilities always need to be calculated as the chance of a value either above or below a given score? How would you calculate the probability of a z-score between −0.08 and +1.92?
Watch This: Reading Probabilities from the Z-table
https://youtu.be/tnVr1Qbk4YU
To calculate the probability of getting a value with a z-score between two other z-scores, you can either use a reference table to look up the value for both scores and subtract them to find the difference, or you can use technology. In this lesson, which is an extension of Z-scores and Z-scores II, we will practice both methods. Historically, it has been very common to use a z-score probability table like the one below to look up the probability associated with a given z-score:
Table 2 (scroll to see all values)
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 Z
0.0 0.5 0.504 0.508 0.512 0.516 0.5199 0.5239 0.5279 0.5319 0.5359 0.0
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.1
0.2 0.5793 0.5832 0.5871 0.591 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.2
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.648 0.6517 0.3
0.4 0.6554 0.6591 0.6628 0.6664 0.67 0.6736 0.6772 0.6808 0.6844 0.6879 0.4
0.5 0.6915 0.695 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.719 0.7224 0.5
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.6
0.7 0.758 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.7
0.8 0.7881 0.791 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.8
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.834 0.8365 0.8389 0.9
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.0
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.877 0.879 0.881 0.883 1.1
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.898 0.8997 0.9015 1.2
1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.3
1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.4
1.5 0.9332 0.9345 0.9357 0.937 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.5
1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.6
1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.7
1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.8
1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.975 0.9756 0.9761 0.9767 1.9
2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.0
2.1 0.9821 0.9826 0.983 0.9834 0.9838 0.9842 0.9846 0.985 0.9854 0.9857 2.1
2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.989 2.2
2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 2.3
2.4 0.9918 0.992 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 2.4
2.5 0.9938 0.994 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.5
2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.996 0.9961 0.9962 0.9963 0.9964 2.6
2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.997 0.9971 0.9972 0.9973 0.9974 2.7
2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.998 0.9981 2.8
2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986 2.9
3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.999 0.999 3.0
3.1 0.999 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993 3.1
3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995 3.2
3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997 3.3
3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998 3.4
3.5 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 3.5
3.6 0.9998 0.9998 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 3.6
3.7 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 3.7
3.8 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 3.8
3.9 1 1 1 1 1 1 1 1 1 1 3.9
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 Z
Since the proliferation of the Internet, however, you can also use a free online calculator such as one of these three:
Example 7
What is the probability associated with a z-score between 1.2 and 2.31?
Solution
To evaluate the probability of a value occurring within a given range, you need to find the probability of both the upper and lower values in the range, and subtract to find the difference.
• First find z = 1.2 on the z-score probability reference above: .8849. Remember that value represents the percentage of values below 1.2.
• Next, find and record the value associated with z = 2.31: .9896
• Since approximately 88.49% of all values are below z = 1.2 and approximately 98.96% of all values are below z = 2.31, there are 98.96% −88.49% = 10.47% of values between.
Example 8
What is the probability that a value with a z-score between -1.32 and +1.49 will occur in a normal distribution?
Solution
Let’s use the online calculator at mathportal.org for this one. When you open the page, you should see a window like this: All you need to do is select the radio button to the left of the first type of probability, input “−1.32” into the first box, and 1.49 into the second. When you click Compute, you should get the result P(−1.32 < Z < 1.49) = 0.8385 Which tells us that there is approximately and 83.85% probability that a value with a z-score between 1.32 and 1.49 will occur in a normal distribution. Notice that the calculator also details the steps involved with finding the answer:
1. Estimate the probability using a graph, so you have an idea of what your answer should be.
2. Find the probability of z < 1.49, using a reference. (0.9319)
3. Find the probability of z < −1.32, again, using a reference. (0.0934)
4. Subtract the values: 0.9319 − 0.0934 = 0.8385 or 83.85%
Example 9
What is the probability that a random selection will be between 8.45 and 10.25, if it is from a normal distribution with µ = 10 and σ = 2?
Solution
This question requires us to first find the z-scores for the value 8.45 and 10.25, then calculate the percentage of value between them by using values from a z-score reference and finding the difference.
1. Find the z-score for 8.45, using the z-score formula: $\frac{(x-\mu)}{\sigma}\\$ $\frac{8.45-10}{2}=\frac{-1.55}{2}\approx{-0.78}\\$
2. Find the z-score for 10.25 the same way: $\frac{10.25-10}{2}=\frac{0.25}{2}\approx{.13}\\$
3. Now find the percentages for each, using a reference (don’t forget we want the probability of values less than our negative score and less than our positive score, so we can find the values between): P(Z < −0.78) = .2177 or 21.77% P(Z < .13) = .5517 or 55.17%
4. At this point, let’s sketch the graph to get an idea what we are looking for:
5. Finally, subtract the values to find the difference: .5517 − .2177 = .3340 or about 33.4% There is approximately a 33.4% probability that a value between 8.45 and 10.25 would result from a random selection of a normal distribution with mean 10 and standard deviation 2.
Concept Problem Revisited
Do z-score probabilities always need to be calculated as the chance of a value either above or below a given score? How would you calculate the probability of a z-score between −0.08 and +1.92? After this lesson, you should know without question that z-score probabilities do not need to assume only probabilities above or below a given value, the probability between values can also be calculated. The probability of a z-score below −0.08 is 46.81%, and the probability of a z-score below 1.92 is 97.26%, so the probability between them is 97.26%−46.81% = 50.45%.
Vocabulary
z-score: a measure of how many standard deviations there are between a data value and the mean. z-score probability table: a table that associates z-scores to area under the normal curve. The table may be used to associate a Z-score with a percent probability.
Guided Practice
1. What is the probability of a z-score between −0.93 and 2.11?
2. What is P(1.39 < Z < 2.03)?
3. What is P(−2.11 < Z < 2.11)?
Solutions
1. Using the z-score probability table above, we can see that the probability of a value below −0.93 is .1762, and the probability of a value below 2.11 is .9826. Therefore, the probability of a value between them is .9826−.1762 = .8064 or 80.64%
2. Using the z-score probability table, we see that the probability of a value below z = 1.39 is .9177, and a value below z = 2.03 is .9788. That means that the probability of a value between them is .9788 − .9177 = .0611 or 6.11%
3. Using the online calculator at mathportal.org, we select the top calculation with the associated radio button to the left of it, enter “−2.11” in the first box, and “2.11” in the second box. Click Compute to get “ .9652,” and convert to a percentage. The probability of a z-score between −2.11 and +2.11 is about 96.52%.
Practice Questions
1. Given a distribution with a mean of 70 and standard deviation of 62, find a value with a z-score of −1.82.
2. What does a z-score of 3.4 mean?
3. Given a distribution with a mean of 60 and standard deviation of 98, find the z-score of 120.76.
4. Given a distribution with a mean of 60 and standard deviation of 21, find a value with a z-score of 2.19.
5. Find the z-score of 187.37, given a distribution with a mean of 185 and standard deviation of 1.
6. What does a z-score of −3.8 mean?
7. Find the z-score of 125.18, given a distribution with a mean of 101 and standard deviation of 62.
8. Given a distribution with a mean of 117 and standard deviation of 42, find a value with a z-score of −0.94.
9. Given a distribution with a mean of 126 and standard deviation of 100, find a value with a z-score of −0.75.
10. Find the z-score of 264.16, given µ = 188 and σ = 64.
11. Find a value with a z-score of −0.2, given µ = 145 and σ = 56.
12. Find the z-score of 89.79 given µ = 10 and σ = 79.
Find the probabilities, use the table from the lesson above.
1. What is the probability of a z-score less than +2.02?
2. What is the probability of a z-score greater than +2.02?
3. What is the probability of a z-score less than −1.97?
4. What is the probability of a z-score greater than −1.97?
5. What is the probability of a z-score less than +0.09?
6. What is the probability of a z-score less than −0.02?
7. What is P(Z < 1.71)?
8. What is P(Z > 2.22)?
9. What is P(Z < −1.19)?
10. What is P(Z > −2.71)?
11. What is P(Z < 3.71)?
12. What is the probability of the random occurrence of a value greater than 56 from a normally distributed population with mean 62 and standard deviation 4.5?
13. What is the probability of a value of 329 or greater, assuming a normally distributed set with mean 290 and standard deviation 32?
14. What is the probability of getting a value below 1.2 from the random output of a normally distributed set with µ = 2.6 and σ = .9?
Find the probabilities, use the table from the lesson or an online resource.
1. What is the probability of a z-score between +1.99 and +2.02?
2. What is the probability of a z-score between −1.99 and +2.02?
3. What is the probability of a z-score between −1.20 and −1.97?
4. What is the probability of a z-score between +2.33 and −0.97?
5. What is the probability of a z-score greater than +0.09?
6. What is the probability of a z-score greater than −0.02?
7. What is P(1.42 < Z < 2.01)?
8. What is P(1.77 < Z < 2.22)?
9. What is P(−2.33 < Z < −1.19)?
10. What is P(−3.01 < Z < −0.71)?
11. What is P(2.66 < Z < 3.71)?
12. What is the probability of the random occurrence of a value between 56 and 61 from a normally distributed population with mean 62 and standard deviation 4.5?
13. What is the probability of a value between 301 and 329, assuming a normally distributed set with mean 290 and standard deviation 32?
14. What is the probability of getting<|endoftext|>
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# Angle Relationships
54 %
46 %
Information about Angle Relationships
Education
Published on January 9, 2009
Author: sandio55
Source: authorstream.com
Slide 1: Algebra and Geometry Using Angle Relationships Slide 2: What do we know about 3x complementary angles? x + 10 You could write this equation: (3x) + (x+ 10) = 90 Complementary Angles Their sum is 90º. Slide 3: Let’s solve for x: 3x + x + 10 = 90 4x + 10 = 90 -10 -10 4x = 80 4 4 x = 20 If x = 20, then 3x = 3(20) = 60º and x + 10 = 20 + 10 = 30º Check: 60 + 30 = 90 Yes, you need to find each angle and yes, you should double check that your answer makes sense. Slide 4: What do we know about supplementary angles? Supplementary angles add up to 180º Write an equation you could use to solve for x in this problem. 4x + x + 30 = 180 Supplementary Angles Slide 5: 4x + x + 30 = 180 5x + 30 = 180 -30 -30 5x = 150 x = 30 Now find the measure for each angle: 4x = 4(30) = 120º x + 30 = 30 + 30 = 60º 120 + 60 = 180, so that works! 4x x + 30 Slide 6: Vertical angles are congruent. How will that help us set up and equation? Congruent means the same so the angles are equal. 4x + 40 = 9x - 10 4x + 40 9x - 10 Vertical Angles Slide 7: 4x + 40 = 9x – 10 -4x -4x___ 40 = 5x - 10 +10 + 10 50 = 5x 10 = x Find each angle. They should be the same. 4x + 40 4(10) + 40 40 + 40 80º 9x - 10 9(10) – 10 90 - 10 80º 80 80 Slide 8: Alternate Interior Angles are congruent, too. 2x + 20 4x - 10 2x + 20 = 4x – 10 -2x -2x 20 = 2x – 10 +10 +10 30 = 2x 15 = x 2x + 20 4x – 10 2(15) + 20 4(15) – 10 30 + 20 60 – 10 50º 50º Alternate Angles Slide 9: Alternate Exterior Angles are also congruent. 6x 4x + 40 6x = 4x + 40 -4x -4x 2x = 40 x = 20 6x 4x + 40 6(20) 4(20) + 40 º 80 + 40 120º Slide 10: Corresponding Angles are congruent, too! 2x + 10 x + 35 2x + 10 = 1x + 35 -1x -1x 1x + 10 = 35 - 10 -10 x = 25 2x + 10 x + 35 2(25) + 10 25 + 35 50 + 10 60º 60º Corresponding Angles Slide 11: In summary….. Complementary angles add up to 90 (corner) Supplementary angles add up to 180 (straight) In all the others the angles are congruent, including Vertical angles X Alternate interior angles Z Corresponding angles F Alternate exterior angles
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# How do you solve 1/ sqrt 8 = 4^(m – 2)?
Feb 14, 2016
The answer is $m = 1 \frac{1}{4}$
#### Explanation:
When solving exponential equations (or inequalities) first you have to find a suitable common base. In this case it would be $2$ because $8 = {2}^{3}$ and $4 = {2}^{2}$.
Now you have to write the equation using calculated base:
$\frac{1}{\sqrt{{2}^{3}}} = {2}^{2 \cdot \left(m - 2\right)}$
Now you can use the property of powers which says that $\sqrt[n]{a} = {a}^{\frac{1}{n}}$
$\frac{1}{2} ^ \left(\frac{3}{2}\right) = {2}^{2 m - 4}$
Next property to use is: $\frac{1}{{a}^{x}} = {a}^{- x}$
${2}^{- \frac{3}{2}} = {2}^{2 m - 4}$
Now since we have the equality of 2 powers with equal base we can write it as the equality of exponents:
$- \frac{3}{2} = 2 m - 4$
$2 m = 4 - \frac{3}{2}$
$2 m = 2 \frac{1}{2}$
$2 m = \frac{5}{2}$
$m = \frac{5}{4} = 1 \frac{1}{4}$<|endoftext|>
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Maria Huey
2022-01-05
You have two solid steel spheres. Sphere 2 has twice the radius of sphere 1. By what factor does the moment of inertia ${I}_{2}$ of sphere 2 exceed the moment of inertia ${I}_{1}$ of sphere 1?
Robert Pina
Firstly, $I\propto m{r}^{2}$
$m=\frac{4}{3}\pi {r}^{3}{\rho }_{steel}$
As sphere 2 has twice the radius, same destiny< its mass is greater by a factor of ${2}^{3}=8$
The extra mass is also distributed farther from the center $\left({r}_{2}=2{r}_{1}\right)$, thus,
$I\propto m{r}^{2}⇒{I}_{2}\propto \left(8{m}_{1}\right){\left(2{r}_{1}\right)}^{2}=32{I}_{1}$
Eliza Beth13
To solve the problem, we need to find the factor by which the moment of inertia of sphere 2 exceeds the moment of inertia of sphere 1. Let's denote the radius of sphere 1 as ${r}_{1}$ and the radius of sphere 2 as ${r}_{2}$.
Given that sphere 2 has twice the radius of sphere 1, we have ${r}_{2}=2{r}_{1}$.
The moment of inertia of a solid sphere is given by the formula:
$I=\frac{2}{5}m{r}^{2}$,
where $m$ is the mass of the sphere and $r$ is its radius.
Since both spheres are made of solid steel, the masses of the spheres will be proportional to their volumes. The volume of a sphere is given by:
$V=\frac{4}{3}\pi {r}^{3}$.
So, the mass $m$ of a sphere will be proportional to ${r}^{3}$.
Let's assume the mass of sphere 1 is ${m}_{1}$, and the mass of sphere 2 is ${m}_{2}$. We can write the following equation:
$\frac{{m}_{2}}{{m}_{1}}={\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{3}$.
Substituting ${r}_{2}=2{r}_{1}$ into the equation, we have:
$\frac{{m}_{2}}{{m}_{1}}={\left(\frac{2{r}_{1}}{{r}_{1}}\right)}^{3}={2}^{3}=8$.
So, the mass of sphere 2 is eight times the mass of sphere 1.
Now, let's find the ratio of the moments of inertia:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}{m}_{2}{r}_{2}^{2}}{\frac{2}{5}{m}_{1}{r}_{1}^{2}}=\frac{{m}_{2}{r}_{2}^{2}}{{m}_{1}{r}_{1}^{2}}$.
Substituting ${m}_{2}=8{m}_{1}$ and ${r}_{2}=2{r}_{1}$ into the equation, we have:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\left(8{m}_{1}\right)\left(2{r}_{1}{\right)}^{2}}{{m}_{1}{r}_{1}^{2}}=\frac{32{m}_{1}{r}_{1}^{2}}{{m}_{1}{r}_{1}^{2}}=32$.
Therefore, the moment of inertia ${I}_{2}$ of sphere 2 exceeds the moment of inertia ${I}_{1}$ of sphere 1 by a factor of 32.
Don Sumner
We are given two solid steel spheres, where Sphere 2 has twice the radius of Sphere 1. Let's denote the radius of Sphere 1 as ${r}_{1}$ and the radius of Sphere 2 as ${r}_{2}$.
The moment of inertia of a solid sphere is given by the formula:
$I=\frac{2}{5}m{r}^{2}$,
where $m$ is the mass of the sphere and $r$ is its radius.
Since both spheres are made of solid steel, their densities are the same. Hence, their masses will be proportional to their volumes. The volume of a sphere is given by:
$V=\frac{4}{3}\pi {r}^{3}$.
Thus, the mass $m$ of a sphere will be proportional to ${r}^{3}$.
Let's assume the mass of Sphere 1 is ${m}_{1}$, and the mass of Sphere 2 is ${m}_{2}$. We can write the following equation:
$\frac{{m}_{2}}{{m}_{1}}={\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{3}$.
Since Sphere 2 has twice the radius of Sphere 1, we have ${r}_{2}=2{r}_{1}$. Substituting this into the equation, we get:
$\frac{{m}_{2}}{{m}_{1}}={\left(\frac{2{r}_{1}}{{r}_{1}}\right)}^{3}={2}^{3}=8$.
Thus, the mass of Sphere 2 is eight times the mass of Sphere 1.
Now, let's find the ratio of the moments of inertia:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}{m}_{2}{r}_{2}^{2}}{\frac{2}{5}{m}_{1}{r}_{1}^{2}}=\frac{{m}_{2}{r}_{2}^{2}}{{m}_{1}{r}_{1}^{2}}$.
Substituting ${m}_{2}=8{m}_{1}$ and ${r}_{2}=2{r}_{1}$ into the equation, we have:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\left(8{m}_{1}\right)\left(2{r}_{1}{\right)}^{2}}{{m}_{1}{r}_{1}^{2}}=\frac{32{m}_{1}{r}_{1}^{2}}{{m}_{1}{r}_{1}^{2}}=32$.
Therefore, the moment of inertia ${I}_{2}$ of Sphere 2 exceeds the moment of inertia ${I}_{1}$ of Sphere 1 by a factor of 32.
Explanation:
$I=\frac{2}{5}m{r}^{2}$
where $I$ is the moment of inertia, $m$ is the mass of the sphere, and $r$ is the radius of the sphere.
Let's assume the mass of both spheres is the same. Therefore, we can compare the moments of inertia by using the ratio:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}m\left(2{r}_{1}{\right)}^{2}}{\frac{2}{5}m{r}_{1}^{2}}$
Simplifying the equation, we have:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}\left(2{r}_{1}{\right)}^{2}}{\frac{2}{5}{r}_{1}^{2}}$
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}4{r}_{1}^{2}}{\frac{2}{5}{r}_{1}^{2}}$
$\frac{{I}_{2}}{{I}_{1}}=\frac{4{r}_{1}^{2}}{{r}_{1}^{2}}$
$\frac{{I}_{2}}{{I}_{1}}=4$
Therefore, the moment of inertia ${I}_{2}$ of sphere 2 exceeds the moment of inertia ${I}_{1}$ of sphere 1 by a factor of 4.
Do you have a similar question?<|endoftext|>
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# Conditional Probability: Definition & Uses
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Instructor: Joseph Vigil
We know a coin can land on either heads or tails. But what would happen if one coin flip changed the next? In this lesson, we'll look at events that are dependent on each other, and we'll learn how to calculate the probability of two events occurring in a combined manner.
## Probability
What are the chances your favorite baseball team will make it to their league's playoffs? What are the chances that they make the World Series? Or even that they'll win the World Series? The answers to all these questions depend on probability, which is a numerical expression of how likely it is that an event will occur.
Let's flip a coin, for example. It landing will be one of two sample points, or possible outcomes: heads or tails. These two sample points make up our sample space, the complete set of possibilities in an experiment. Since one out of the two sample points is heads, the probability of the coin landing on heads is a 1 out of 2, or 1/2. Likewise, since one out of the two sample points is tails, the probability of the coin landing on tails is 1 out of 2, or 1/2.
That's simple enough. Every time we flip the coin, we have the same sample space. Even if we flipped it a million times, the next flip will still give either heads or tails. But what if the sample space changed every time we ran an experiment?
## Dependent Events
You've got to get to work! You've slept past your alarm, and it's still dark. You need two black dress socks to finish dressing. You have two white socks and two black ones in your drawer. What are your chances of grabbing the right socks?
Let's look at the socks we have now and how that group of socks may change. When you grab your first sock, it can of course be black or white. Let's say you get lucky, and you grab a black sock on your first try. Since 2 out of the 4 socks are white, you have a 2/4, or 1/2, chance of picking a black sock.
Unlike the coin flip, though, the odds of grabbing a certain sock change every time we run this experiment. This is an example of conditional probability, which is the probability of one event happening given that another event has already happened. In other words, the second event is conditional based on the previous event.
Now you have one black sock and two white socks left. The group of socks has changed. So the odds of picking either a white or black sock has also changed. Each time you pick a sock, you change future options. In other words, each selection is dependent on the one before it; so each selection is a dependent event.
Now that there are two white socks and one black sock left, you have a 1/3 chance of picking a black sock. Since you have a 1/2 chance of picking a black sock the first time and a 1/3 chance of picking a black sock the second time, we'll multiply those two probabilities together. 1/2 * 1/3 = 1/6. You have a one out of six chance of picking two black socks right away. Not very good odds. We'd better turn on the lights!
## Notation for Conditional Probability
We have to use a common language so people can be on the same page when talking about probabilities. So, when we write P(A), we mean the probability that some event, A, will happen.
Let's go back to our coin. If A represents the coin landing on heads, then P(A) = 1/2, because there's a 1 out of 2 chance that the coin will land on heads.
If B represents the coin landing on tails, then P(B) also equals 1/2, because there's a 1 out of 2 chance that the coin will land on tails.
When we were dealing with the socks, though, each event was dependent on those before it. So, we write P(B|A), which means the probability that B happens if A has already happened, also known as the conditional probability of B given A.
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# A thin glass plate of thickness is $\dfrac{{2500}}{3}\lambda$ ( $\lambda$ is wavelength of light used) and refractive index $\mu = 1.5$ is inserted between one of the slits and the screen in Young’s double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is:A) $2:1$B) $1:4$C) $4:1$D) $4:3$
Last updated date: 18th Jun 2024
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Hint: Calculate the path difference due to insertion of the thin glass plate. Due to this path difference, there will be a phase difference. Calculate that path difference. Now, intensity can be calculated using the formula of resultant intensities.
Complete step by step solution:
We know that there will be path difference due to the insertion of the glass plate and this path difference will also lead to phase difference.
Now, the path difference when the thin glass plate is inserted is given as $\Delta p$ ,
$\Delta p = (\mu - 1)t$.....................equation 1
Here $t$ is the thickness of the glass plate
$\mu$ is the refractive index of the glass
Now, the phase difference when the thin glass plate is inserted $$\Delta \phi$$ is given as:
$$\Delta \phi = \dfrac{{2\pi }}{\lambda }\Delta p$$
$$\Delta \phi = \dfrac{{2\pi }}{\lambda }\left( {\mu - 1} \right)t$$.............equation 2
Where $\lambda$ is wavelength of light used
The intensity at the centre $${I_c}$$ after the insertion of the glass plate will be given as:
$${I_c} = {I_s} + {I_s} + 2\sqrt {{I_s}^2} \cos \Delta \phi$$
Here, $${I_s}$$ is the intensity of light from each slit
Solving this equation, we get
$${I_c} = 2{I_s}(1 + \cos \Delta \phi )$$
But from trigonometric identities, we know that $$1 + \cos \theta = 2{\cos ^2}\left( {\dfrac{\theta }{2}} \right)$$
$$\Rightarrow {I_c} = 2{I_s}(2{\cos ^2}\left( {\dfrac{{\Delta \phi }}{2}} \right))$$
$$\Rightarrow {I_c} = 4{I_s}({\cos ^2}\left( {\dfrac{{\Delta \phi }}{2}} \right))$$
Before the glass plate was inserted, the phase difference was zero and hence the intensity was
$${I_c} = 4{I_s}({\cos ^2}{0^0})$$
$$\Rightarrow {I_c} = 4{I_s} = {I_0}$$............equation 3
Substituting the value of phase difference, we get
$$\Rightarrow {I_c} = 4{I_s}{\cos ^2}\left( {\dfrac{{2\pi }}{\lambda }\left( {\mu - 1} \right)t} \right)$$
$$\Rightarrow {I_c} = 4{I_s}{\cos ^2}\left( {\dfrac{{2\pi }}{\lambda }\left( {1.5 - 1} \right)\dfrac{{2500}}{3}\lambda } \right)$$
$$\Rightarrow {I_c} = 4{I_s}{\cos ^2}\left( {2\pi \left( {\dfrac{1}{2}} \right)\dfrac{{2500}}{3}} \right)$$
$$\Rightarrow {I_c} = 4{I_s}{\cos ^2}\left( {2500\dfrac{\pi }{3}} \right)$$
But
$${\cos ^2}\left( {2500\dfrac{\pi }{3}} \right) = \dfrac{1}{4}$$
$$\Rightarrow {I_c} = 4{I_s} \times \dfrac{1}{4}$$
$$\Rightarrow {I_c} = {I_s}$$
Comparing this with equation $3$ , we have
$$\dfrac{{{I_0}}}{{{I_C}}} = \dfrac{4}{1}$$
Therefore, the ratio of the intensities before and after the introduction of the glass plate is $$4:1$$
Thus, option C is the correct option.
Note: In Young’s double slit experiment, a pattern of bright and dark fringes is observed. Remember when there is no glass plate, the phase difference and the path difference will be zero. Also remember that due to insertion of a thin glass plate, the intensity decreased for this problem.<|endoftext|>
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# Unit 2 Review - PowerPoint PPT Presentation
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## Unit 2 Review
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Title: Unit 2 Review
1
Unit 2 Review
• Try the practice questions first and check your
• Then, answer the real questions, 1-8. Show
Start!
2
Find the product.
x
3
Multiply the numerator by the numerator,
and denominator by denominator.
Now, simplify
The product is
4
Now try this on your own 1.
x
Show your work on the worksheet.
5
Find the quotient.
2
6
Convert the mixed number to an improper
fraction. Then, multiply by the reciprocal of the
second fraction.
2
• To convert to an improper fraction
• Remember, ask yourself how many thirds are in 2
1/3? There are 7/3s.
• If you are not sure, draw a picture of 2 1/3,
then count the thirds.
• Multiply the denominator by the whole number,
• (3 x 2 1 7)
x
Now, multiply and simplify the fractions.
2
x
7
Solve using a picture. Show 2 1/3.
Then divide into sixths.
Next, number or circle groups of 5/6.
You only have 4 out of 5 sixths left. So you
cannot make a third group. Your remainder is
4/5.
you have numbered 2
8
Now, try this on your own. Use pictures or
numbers to solve. 2. 3
Show your work on the worksheet.
9
Find the product of the following decimals
0.38 X 1.45
10
1.45 100 40 5 x .38
30 8 40 (8 x
5) 320 (8 x 40) 800 (8 x 100)
150 (30 x 5) 1200 (30 x 40) 3000 (30 x
100) .5510 Or 1.45 x .38 1160
435 .5510
11
worksheet. 3. Find the product
0.25 x 2.37
12
Find the quotient 4.67 .25
13
Find the quotient 4.67 .55 .55 4.67
If the divisor is a decimal, make it a
whole number by moving the decimal. Then move
the decimal in the dividend the same number of
places. Then divide. Remember to line up the
decimal point in the quotient. So, if you move
the decimal 2 places to the right in .55, it
becomes 55. You must then move the decimal 2
places in 4.67, to make it 467 The new problem
is 55 467 8. 27/55 8.49
55. 467. -220 4
247 4 -220 27
8 This is the whole
number portion
27/55.
Now convert this fraction to a decimal. Use a
calculator 27 55 .49 Now combine the whole
14
worksheet. 4. 8.4 .32
15
• Here are some word problems. Look for key words
that tell which operation is needed to solve the
problem. Then, find the solution and show your
work on the worksheet.
• Sarah needs 2 ¼ yards of material to make one
dress for her daughter. How many dresses can she
make from 10 ½ yards of material?
• A half pound of raisins was in a jar in Lukes
cabin. He and his cabin mates ate 2/3 of them.
How much of the raisins did they eat?
• Mrs. Melton is making ribbons to give to the
participants in the spelling bee. Her roll of
ribbon contains 8.4 feet of ribbon. She wants to
make ribbons that are .75 feet long. How many
ribbons can she make?
• Jason is helping his mother shop at the grocery
store. He needs to buy granola that costs 2.49
per pound. How much will 3.1 pounds of granola
cost?
16
Congratulations! You have finished the
review! Remember to bring your completed
worksheet to class tomorrow! Good luck on the
assessment!<|endoftext|>
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# Dividing Fractions in 5 Easy Steps
Dealing with fractions can be difficult, and dividing fractions is no different. Let’s learn how to divide fractions in 5 easy steps.
Author
Taylor Hartley
Expert Reviewer
Published: August 24, 2023
# Dividing Fractions in 5 Easy Steps
Dealing with fractions can be difficult, and dividing fractions is no different. Let’s learn how to divide fractions in 5 easy steps.
Author
Taylor Hartley
Expert Reviewer
Published: August 24, 2023
# Dividing Fractions in 5 Easy Steps
Dealing with fractions can be difficult, and dividing fractions is no different. Let’s learn how to divide fractions in 5 easy steps.
Author
Taylor Hartley
Expert Reviewer
Published: August 24, 2023
Key takeaways
• You want to use steps to divide fractions – If you try to do it all at once, it might get confusing. Separate the process into steps.
• Know your terminology – Know the difference between the numerator and the denominator before you start working on your division problems.
Fractions are already parts of a whole number. So, you might be wondering: how can you divide a fraction into an even smaller part—or, for that matter, divide it to turn it into an even bigger part?
Honestly, it works the same way dividing whole numbers works. You can divide a small number into an even smaller number, right? 4 divided by 2 gives you 2. So how can ¼ divided by ½ give you ½? Today, we’re going to walk through how to divide fractions in 5 easy steps. But first, let’s make sure we understand the basics to make it super easy to follow along.
## The parts of a fraction
Unlike a whole number, a fraction has two parts.
• The numerator – This is the number on the top of the fraction. If you think of a fraction as shaded parts of a whole, it is how many sections you would shade in.
• The denominator – This is the number on the bottom of the fraction. So, using the example of a fraction as shaded parts of a whole, it would be how many sections available to shade in.
Visually, 1/4 would look like this, with 1 out of 4 parts shaded in.
And 3/16 would look like this, with 3 out of 16 parts shaded in.
## Dividing fractions, step by step
There are several different ways to divide fractions, but if you try to master them or use them all at once, it might get super confusing. It’s always best to divide complicated ideas into smaller parts to make the ideas easier to understand. Let’s DIVIDE (ha, see what we did there?) dividing fractions into 5 easy steps to help make it easier.
It might sound simple, but first, you need to understand what you’re being asked to do. Let’s look at the following problem:
Basically, you’re being asked to divide five-sixths by seven-eighths.
### 2. Keep the first fraction as is
Next, you need to know what to do with each of the fractions in the problem. Do not do anything to the first fraction. Keep it EXACTLY as it is.
### 3. Change the division sign to a multiplication sign
You mean I have to know how to multiply to know how to divide? Absolutely you do! What you want to do in this step is to change the DIVISION sign to a MULTIPLICATION sign.
So, if we were just rewriting the problem and this was our final step, the problem would look like this:
But it ISN’T the last step. The last step is…
### 4. Flip the second fraction
The last step before you can solve the problem is to flip the numerator with the denominator in the second fraction. So, now your problem looks like this:
### 5. Solve by multiplying the numerators and denominators together, then simplify if necessary
The answer is 16/9, which cannot be simplified, so it remains the same.
Ready to give it a go?
Dividing fractions is pretty simple, right? Just keep that in mind…keep dividing fractions as simple as possible. The only person who can get you confused is you!
Ready to practice on your own? We’ve included some no-risk practice problems to get you ready for the classroom!
## Practice problems
The answer is 15/6, and since both 15 and 6 can be divided by 3, the answer, when simplified, will be 5/2.
The answer is 16/15. This cannot be simplified, as these numbers cannot be divided by a similar number.
The answer is 4/3. This is simplified from 24/18, numbers that can each be divided by 6 to give you 4/3.
The answer is 12/7. This cannot be simplified, as these numbers cannot be divided by a similar number.
The answer is 7/4, simplified from 14/8, which can both be divided by 2 to give you the answer.
## Parent Guide
The answer is 15/6, and since both 15 and 6 can be divided by 3, the answer, when simplified, will be 5/2.
How did we get here?
2. Keep 5/6 as is.
3. Change the division symbol to a multiplication symbol.
4. Flip 1/3 to 3/1.
5. Then, multiply 5 by 3 to get 15 and 6 by 1 to get 6, which equals 15/6. Divide each of these by 3 to get 5/2.
The answer is 16/15. This cannot be simplified, as these numbers cannot be divided by a similar number.
How did we get here?
2. Keep 4/5 as is.
3. Change the division symbol to a multiplication symbol.
4. Flip 3/4 to 4/3.
5. Then, multiply 4 by 4 to get 16 and 5 by 3 to get 15, which equals 16/15.
The answer is 4/3. This is simplified from 24/18, numbers that can each be divided by 6 to give you 4/3.
How did we get here?
2. Keep 8/9 as is.
3. Change the division symbol to a multiplication symbol.
4. Flip 2/3 to 3/2.
5. Then, multiply 8 by 3 to get 24 and 9 by 2 to get 18. Both of these can be divided by 6, which leaves you with 4/3.
The answer is 12/7. This cannot be simplified, as these numbers cannot be divided by a similar number.
How did we get here?
1. Read the problem: 3/7 ➗1/4
2. Keep 3/7 as is.
3. Change the division symbol to a multiplication symbol.
4. Flip 1/4 to 4/1.
5. Then, multiply 3 by 4 to get 12 and 7 by 1 to get 7, which equals 12/7.
The answer is 7/4, simplified from 14/8, which can both be divided by 2 to give you the answer.
How did we get here?
2. Keep 7/8 as is.
3. Change the division symbol to a multiplication symbol.
4. Flip 1/2 to 2/1.
5. Then, multiply 7 by 2 to get 14 and 8 by 1 to get 8, then simplify 14/8 to 7/4.
## FAQs about math strategies for kids
We understand that diving into new information can sometimes be overwhelming, and questions often arise. That’s why we’ve meticulously crafted these FAQs, based on real questions from students and parents. We’ve got you covered!
To divide fractions, you multiply the first fraction by the reciprocal (flipped version) of the second fraction. The reciprocal of a fraction is obtained by swapping the numerator and denominator. For example, to divide 1/2 by 3/4, you multiply 1/2 by 4/3. The result is (1/2) × (4/3) = 4/6, which simplifies to 2/3.
If the divisor is a whole number, you can convert it to a fraction by putting it over 1. For example, if you want to divide 3/4 by 2, you can rewrite it as 3/4 ÷ 2/1. Then, you proceed with the division using the reciprocal: (3/4) × (1/2) = 3/8.
Yes, the dividend can be a whole number! In division, the whole number is treated as a fraction with a denominator of 1. For example, if you want to divide 2 by 3/4, you can rewrite it as 2/1 ÷ 3/4. Then, you proceed with the division using the reciprocal: (2/1) × (4/3) = 8/3.
To simplify the result, reduce the fraction to its simplest form by dividing both the numerator and denominator by their greatest common divisor. For example, if the result of dividing fractions is 12/16, you can simplify it by dividing both the numerator and denominator by 4. The simplified form is 3/4.
## Related Posts
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Lesson credits
Taylor Hartley
Taylor Hartley is an author and an English teacher based in Charlotte, North Carolina. When she's not writing, you can find her on the rowing machine or lost in a good novel.
Jill Padfield has 7 years of experience teaching high school mathematics, ranging from Alegra 1 to AP Calculas. She is currently working as a Business Analyst, working to improve services for Veterans while earning a masters degree in business administration.
Taylor Hartley
Taylor Hartley is an author and an English teacher based in Charlotte, North Carolina. When she's not writing, you can find her on the rowing machine or lost in a good novel.
Jill Padfield has 7 years of experience teaching high school mathematics, ranging from Alegra 1 to AP Calculas. She is currently working as a Business Analyst, working to improve services for Veterans while earning a masters degree in business administration.
# Are you a parent, teacher or student?
Are you a parent or teacher?
## Hi there!
Book a chat with our team<|endoftext|>
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# Tennessee - Grade 1 - Math - Measurement and Data - Ordering Objects by Length - 1.MD.1
### Description
Order three objects by length; compare the lengths of two objects indirectly by using a third object.
• State - Tennessee
• Standard ID - 1.MD.1
• Subjects - Math Common Core
### Keywords
• Math
• Measurement and Data
## More Tennessee Topics
Compare two two-digit numbers based on the meanings of the digits in each place and use the symbols >, =, and < to show the relationship.
Create a composite shape and use the composite shape to make new shapes by using two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, rectangular prisms, cones, and cylinders).
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that partitioning into more equal shares creates smaller shares.
Apply properties of operations (additive identity, commutative, and associative) as strategies to add and subtract.
(4) Understand subtraction as an unknown-addend problem. For example, to solve 10 – 8 = ___, a student can use 8 + ___ = 10. (5) Add and subtract within 20 using strategies such as counting on, counting back, making 10, using fact families and related known facts, and composing/ decomposing numbers with an emphasis on making ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9 or adding 6 + 7 by creating the known equivalent 6 + 4 + 3 = 10 + 3 = 13). (6) Fluently add and subtract within 20 using mental strategies. By the end of 1st grade, know from memory all sums up to 10.<|endoftext|>
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Murmuration refers to the phenomenon that results when hundreds, sometimes thousands, of starlings fly in swooping, intricately coordinated patterns through the sky.
Maybe you've seen a murmuration video before. But this one is especially beautiful. It was shot earlier this month in Wales, at Cosmeston Lakes in the Vale of Glamorgan, and posted on Facebook by the BBC Cymru Wales.
Why do I love this short video so much?
It's all about science. Just how do the starlings manage to fly in such an amazingly coordinated way?
A few years ago, George F. Young and his colleagues investigated starlings' "remarkable ability to maintain cohesion as a group in highly uncertain environments and with limited, noisy information" — a nice description of what goes on in a murmuration.
Going in, Young et al. already knew that starlings pay attention to a fixed number of their neighbors in the flock, regardless of flock density — seven, to be exact. Their new contribution was to figure out that "when uncertainty in sensing is present, interacting with six or seven neighbors optimizes the balance between group cohesiveness and individual effort."
Young et al. analyzed still shots from videos of starlings in flight (flock size ranging from 440 to 2,600), then used a highly mathematical approach and systems theory to reach their conclusion. Focusing on the birds' ability to manage uncertainty while also maintaining consensus, they discovered that birds accomplish this (with the least effort) when each bird attends to seven neighbors.
In following this role of seven, then, the birds are part of a dynamic system in which the parts combine to make a whole with emergent properties — and a murmuration results.
That's just incredibly cool.
Also, starlings are essentially an invasive species in this country. They were famously introduced to North America at New York City's Central Park in the 1890s by Shakespeare enthusiasts who wanted all the bird species ever mentioned by Shakespeare to inhabit this continent.
With starlings, they certainly succeeded: 200 million of these birds now inhabit North America. They aren't welcomed by everyone. As the Cornell Lab of Ornithology puts it, starlings in the U.S. are "sometimes resented for their abundance and aggressiveness." In this Scientific American piece, they're even called a "menace."
The murmuration video invites us to see them with fresh eyes.
Barbara J. King is an anthropology professor emerita at the College of William and Mary. She often writes about the cognition, emotion and welfare of animals, and about biological anthropology, human evolution and gender issues. Barbara's most recent book on animals is titled How Animals Grieve, and her forthcoming book, Personalities on the Plate: The Lives and Minds of Animals We Eat, will be published in March. You can keep up with what she is thinking on Twitter: @bjkingape<|endoftext|>
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Stage of prehistory and early history when copper and bronze (an alloy of tin and copper) became the first metals worked extensively and used for tools and weapons. One of the classifications of the Danish archaeologist Christian Thomsen's Three Age System, it developed out of the Stone Age and generally preceded the Iron Age. It first began in the Far East and may be dated 5000–1200 BC in the Middle East and about 2000–500 BC in Europe.
Mining and metalworking were the first specialized industries, and the invention of the wheel during this time revolutionized transport.
Agricultural productivity (which began during the New Stone Age, or Neolithic period, about 6000 BC) was transformed by the ox-drawn plough, increasing the size of the population that could be supported by farming.
In some areas, including most of Africa, there was no Bronze Age, and ironworking was introduced directly into the Stone Age economy.
Metalworking The earliest use of bronze has been found in the Far East and was originally assumed to have spread, through diffusion of culture, westwards and throughout Europe. However, absolute dating techniques now suggest that metalworking may have been independently invented in other areas, including the Aegean.
Some regions, such as Denmark, became important manufacturing centres although they had no native copper or tin ores, indicating the importance of long-distance trade in the European Bronze Age. Tin ore is particularly restricted in its distribution – found, for example, in Cornwall, southwestern England; Bohemia, Germany; and Anatolia, Turkey – and must have been carried great distances. The other chief metal used during the Bronze Age was gold, which occurs widely in the pure state and was used for ornament.
Culture In Europe the burials of the Bronze Age peoples have produced a great range of pottery vessels and other objects which provide clues to the way of life and ritual beliefs of the ordinary and the elite. This material can also be used in conjunction with metal artefacts to describe, distinguish, and observe the spread of local cultures, such as that of the Beaker people. In general, Bronze Age society in temperate Europe was based on farming, in which stockrearing continued to play a large part, but the Bronze Age also saw the rise of urbanization in the Middle East and palace economies, such as those of the Minoan civilization and Mycenaean civilization, which were based almost as much on manufacturing and service industries as on the rural hinterland. Interrelationships of cultures joined by maritime trade routes are important, in particular with regard to the Mediterranean.
The evidence of Bronze Age weapons in graves suggests the symbolic importance as well as the prestige value of such artefacts. Palace economies, such as Knossos on Crete, epitomized organized systems based on hierarchical structure and the trade in prestige goods, sometimes manufactured on site, as evidenced by workshops. The investigation of the Ulu Burun shipwreck off Caş, Turkey, revealed a cargo of metal (copper) ingots, Egyptian faience, unworked ivory, and amphorae among thousands of objects, such as Mesopotamian seals and Egyptian stone scarabs, which may be traced to a source of raw or manufactured material. This complex system may be seen in prestige burials and trade goods in a network of long-distance exchange. Early Bronze Age burial mounds in areas such as Wessex, southern England, were either simple mounds or elaborate forms with ditches and banks, which may differentiate the ruling elite.
Bronze Age dagger
(3500–800 BCE) Scientists categorize human development according to material culture, social and political institutions, sophistication of...
The cultural phase during which metallurgical technology, based first on copper (in the Chalcolithic period) and then on bronze (copper...
The division of prehistory into the Stone Age, Bronze Age, and Iron Age, first proposed by the Danish archaeologist Christian Thomsen 1816–19. Subseq<|endoftext|>
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Balbharti Maharashtra State Board Class 7 Science Solutions Chapter 7 Motion, Force and Work Notes, Textbook Exercise Important Questions and Answers.
Maharashtra State Board Class 7 Science Solutions Chapter 7 Motion, Force and Work
Class 7 Science Chapter 7 Motion, Force and Work Textbook Questions and Answers
1. Fill ¡n the blanks with the proper words from the brackets.
(stationary, zero, changing, constant, displacement, velocity, speed. acceleration, stationary but not zero. inc reuses)
Question a.
If a body traverses a distance in direct proportion to the time, the speed of the body is ……………… .
constant
Question b.
If a body is moving with a constant velocity, its acceleration is ……………… .
zero
Question c.
……………. is a scalar quantity.
Speed
Question d.
…………….. is the distance traversed by a body in a particular direction in unit time.
Velocity
2. Observe the figure and answer the questions.
Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE’? Can this velocity be called average velocity?
Question a.
Observe the figure and answer the questions
Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE’? Can this velocity be called average velocity?
1. Actual distance = $$\overline{\mathrm{AB}}$$ + $$\overline{\mathrm{BC}}$$ + $$\overline{\mathrm{CD}}$$ + $$\overline{\mathrm{DE}}$$ = 3 + 4 + 5 + 3
Actual distance = 15 km
2. Displacement = $$\overline{\mathrm{AB}}$$ + $$\overline{\mathrm{BD}}$$ + $$\overline{\mathrm{DE}}$$
= 3 + 3 + 3
Displacement = 9 km
3. Speed = $$\frac{\text { Distance travelled }}{\text { Total time }}$$
Distance = 15 km = 15 × 1000 = 15000 m
Time = 1 hr = 1 × 60 × 60 = 3600 sec.
s = $$\frac{15000}{3600}$$ or s = $$\frac{15 \mathrm{~km}}{1 \text { hour }}$$ = 15km/hour
= 4.16 m/sec. or 15 km/hour
4. Velocity = $$\frac{\text { Distance travelled }}{\text { Total time }}$$
Displacement = 9 km = 9 × 1000 = 9000 m
Time = 1 hr = 1 × 60 × 60 = 3600 sec
V = $$\frac{9000}{3600}$$ or V = $$\frac{9 \mathrm{~km}}{1 \text { hour }}$$ = 9 km/hour
= 2.5 m/sec. or 9 km/hour
5. Yes, this velocity can be called as average velocity.
3. From the groups B and C, choose the proper words, for each of the words in group A.
Question a.
From the groups B and C, choose the proper words, for each of the words in group A.
Group ‘A’ Group B’ Group ‘C’ Work Joule erg Force Newton dyne Displacement Metre cm
4. A bird sitting on a wire, flies, circles around and comes back to its perch. Explain the total distance it traversed during its flight and its eventual displacement.
Question a.
The total distance the bird has traversed is the length of the distance covered by circling, but the eventual displacement are the bird is zero as its initial and final position are one and the same.
5. Explain the following concepts in your own words with everyday examples: force, work, displacement, velocity, acceleration, distance.
Question a.
Explain the following concepts in your own words with everyday examples: force, work, displacement, velocity, acceleration, distance.
1. Force: The interaction that brings about the acceleration is called force.
e.g: An ox is pulling a cart, applying brakes to a bicycle, lifting heavy iron object with a crane.
2. Work: When an object is displaced by applying a force on it, work is said to be done.
e.g: A bucketful of water is to be drawn from a well and taken to the home by walking from well to home.
3. Displacement: The minimum distance
traversed by a moving body in one direction from the original point to reach the final point is called displacement.
e.g: A rolling of a ball from point A to point B in the same direction.
4. Velocity: Velocity is the distance traversed by a body in a specific direction in unit time.
e.g: A truck is covering a distance of 40km from A to D in a straight line in 1 hour.
5. Acceleration: It is change in velocity per second. It can be deduced.
Acceleration = $$\frac{\text { Change in velocity }}{\text { Time taken for change }}$$
e.g:
(i) In the above example a truck covered the distance AB at velocity of 60 km/hr, BC at 30 km/hr and CD at 40 km/hr. (ii) It means that the velocity for the distance CD is greater than the velocity for the distance BC. (iii) From the number of seconds required for this change in velocity to take place, the change in velocity per second can be deduced. This is called acceleration (iv) Distance: The length of the route actually traversed by a moving body irrespective of the direction is called distance.
e.g: Ranjit travelled 1km. from his home to school.
6. A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/s. It took 2 seconds for it to go from B to C. What is the acceleration of the ball as it goes from B to C.
Question a.
A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/s. It took 2 seconds for it to go from B to C. What is the acceleration of the ball as it goes from B to C.
As its initial and final positions are one and the same.
Initial Velocity = 2 cm/s.
Final Velocity = 4 cm/s
Time taken for the change in velocity from B to
D = 4 cm/s – 2 cm/s = 2 cm/s
7. Solve the following problems.
Question a.
A force of 1000 N was applied to stop a car that was moving with a constant velocity. The car stopped after moving through 10m. How much is the work done?
Force (F) = 1000 N
displacement (s) = 10m
work done (W) = ?
W = Fs
= 1000 × 10
W = 10,000 Joule
Question b.
A cart with mass 20 kg went 50 m in a straight line on a plain and smooth road when a force of 2 N was applied to it. How much work was done by the force?
Force (F) = 2 N
Displacement (s) = 50 m
Work done (W) = ?
W = Fs
= 2 × 50
W = 100 Joule
Project:
Question a.
Collect information about the study made by Sir Isaac Newton regarding force and acceleration and discuss it with your teacher.
Class 7 Science Chapter 7 Motion, Force and Work Important Questions and Answers
Fill in blanks:
Question 1.
Displacement is a …………. quantity.
vector
Question 2.
The …………. of an object can change even while it is moving along a straight line.
velocity
Question 3.
The …………. velocity can be different at different times.
instantaneous
Question 4.
Change in velocity per second is called …………. .
acceleration
Question 5.
The interaction that brings about the acceleration is called …………. .
force
Question 6.
The scientist …………. was the first to study force and the resulting acceleration.
Sir Isaac Nezvton
Question 7.
Ability to do work is called …………. .
Energy
Question 8.
W = …………. × S.
F
Question 9.
Unit of work is …………. and …………. .
Joule, erg
Question 10.
Unit of force is …………. and …………. .
Newton, dyne
Question 11.
Force is a …………. quantity.
vector
Question 12.
The velocity at a particular time is called …………. velocity.
instantaneous
Question 13.
The …………. of a body is the distance traversed per unit time.
speed
Question 14.
Unit of acceleration is …………. and …………. .
m/s2 and cm/s2
Question 15.
Force is measured by the …………. that it produces.
acceleration
Question 16.
Work done by a body with no displacement will be …………. .
zero
Say whether True or False, correct the false 1 statements:
Question 1.
Velocity is distance travelled per unit of time.
False. Speed is distance travelled per unit of time
Question 2.
In displacement, both distance and direction are taken into account.
True
Question 3.
Speed = Distance/time.
True
Question 4.
Change in speed per second is acceleration.
False. Change in velocity per second is acceleration
Question 5.
Work done depends on the force and the displacement.
True
Question 6.
C.G.S. unit of acceleration is m/s2.
False. C.G.S. unit of acceleration is cm/s2.
Question 7.
M.K.S. unit of force is dyne.
False. M.K.S. unit of force is Newton
Question 8.
Force is measured by the acceleration that it produces.
True
Write the difference between the following:
Question 1.
Speed and Velocity
Speed Velocity 1. Speed is distance travelled per unit of time. 1. Velocity is the distance traversed by a body in a specific direction in unit time. 2. It is a scalar quantity. 2. It is a vector quantity. 3. Formula: Speed = $$\frac{\text { Distance traversed }}{\text { Total time }}$$ 3. Formula: Velocity = $$\frac{\text { Displacement }}{\text { Total time }}$$
Question 2.
Distance and Displacement
Distance Displacement 1. The length of the route actually traversed by a moving body, irrespective of the direction is called distance. 1. The minimum distance traversed by a moving body in one direction from the original point to reach the final point is called displacement. 2. It is a scalar quantity. 2. It is a vector quantity.
Solve the following problems!
Question 1.
A bus travelled 200 km in the first 3 hours and then 100 kms for the next one and a half hours and then 120 kms for the next one and a half hours. What is the average velocity of the bus if it has moved in a straight line for the whole journey.
Question 2.
See the diagram and calculate the Distance and Displacement travelled by the body from A to I.
Distance travelled =
A → B → C → D → E → F → G → H + I
= 5 + 7 + 6 + 3 + 5 + 4 + 6 + 5
= 41 m
Displacement = A → I in a straight line shortest distance
= 1m<|endoftext|>
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Land of Israel and Jewish heritage
A series from the book written by Naveed Anjum
Jewish Life in the Land of Canaan, cont’d
The Promised Land, cont’d
Late Roman period
In this late Roman period, the Tannaim and Amoraim were active and dominant. They were rabbis who organized and debated the Jewish oral traditions and declared them law. The Tannaim and Amoraim were two related but distinctly and chronologically different groups. The Amoraim came after the Tannaim and debated the earlier group’s codified oral traditions. Teachings of the Tannaim are contained in the Mishnah, Beraita, Tosefta, and various Midrash compilations. In addition to the writings of the Tannaim, the later commentaries of the Amoraim were also compiled into the Jerusalem Talmud, which was completed around 400 CE. It was most likely completed in Tiberius.
First were the Tannaim, which, in Hebrew, is the plural תנאים [tanaˈim] of the singular תנא [taˈna]. Tanna means “studies,” “repeats,” or “teaches,” which, through use, becomes “studied,” “repeated,” and “taught.” The Tannaim era, also called the Mishnaic period, lasted about 210 years. Tannaim were rabbinic sages during the period of approximately 10-220 CE. The first Tannaim were disciples of Shammai and Hillel, fl. c.30 BCE–10 CE, and the last Tannaim were contemporaries of Judah ha-Nasi I. The Tannaim were direct transmitters and teachers of the previously uncodified oral tradition. They codified (wrote down) the oral traditions that they taught. Judah ha-Nasi I is reported to have redacted (compiled) their works into one document: The Mishnah.
(To be continued…)<|endoftext|>
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# Time and Distance
A man in a train notices that he can count 21 telephone posts in one minute.If they are known to be 50 meters apart , then at what speed is the train travelling ?
Number of gaps between 21 telephone posts = 20.
Distance travelled in 1 minute = (50 × 20) m= 1000 m = 1 km.
∴ Speed = 60 km/hr.
An express train travelled at an average speed of 100 km/hr,stopping for 3 minutes after every 75 km.How long did it take to reach its destination 600 km from the starting point ?
Time taken to cover 600 km = $$\left ( \frac{600}{100} \right )$$ hrs = 6 hrs.
Number of stoppages = $$\left ( \frac{600}{75} - 1 \right )$$ = 7
Total time of stoppages = (3 × 7) min= 21 min.
Hence,total time taken = 6 hrs 21 min.
A certain distance is covered by a cyclist at a certain speed. If a jogger covers half the distance in double the time,the ratio of the speed of the jogger to that of the cyclist is :
Let the distance covered by, the cyclist be x and the time taken be y. Then,
Required ratio = $$\frac{\frac{1}{2}x}{2y} : \frac{x}{y} = \frac{1}{4} : 1$$ = 1 : 4
The speed of a car increases by 2 km after every one hour. If the distance travelled in the first one hour was 35 km, what was the total distance travelleld in 12 hours ?
Total distance travelled in 12 hours = (35 + 37 +39 + …… upto 12 terms ).
This is an A.P. with first term, a = 35, number of terms, n= 12, common difference, d = 2.
∴ Required distance = $$\frac{12}{2}\left [ 2 \times 35 + \left ( 12 - 1 \right ) \times 2 \right ]$$
= 6(70 + 22)
= 552 km.
A train covers a distance of 10 km in minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be :
$$Speed = \left ( 10 \times \frac{60}{12} \right )$$ km/hr = 50 km/hr.
New speed = (50 - 5) km/hr = 45 km/hr.
∴ Time taken = $$\frac{10}{45}$$ hr
= $$\left ( \frac{2}{9} \times 60 \right )$$ min
= $$\frac{40}{3}$$ min
= 13 min 20 sec.
A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes.The length of the bridge is :
$$Speed =\left ( 5 \times \frac{5}{18} \right )$$ m/sec = $$\frac{25}{18}$$ m/sec.
Distance covered in 15 minutes = $$\left ( \frac{25}{18} \times 15 \times 60 \right )$$ m = 1250 m.
How long will a boy take to run round a square field of side 35 metres, if he runs at the rate of 9 km/hr ?
Speed = 9 km/hr = $$\left ( 9 \times \frac{5}{18} \right ) m/sec = \frac{5}{2}$$ m/sec
Distance = (35 × 4) m = 140 m
∴ Time taken = $$\left ( 140 \times \frac{2}{5} \right )$$ sec = 56 sec
A car is running at a speed of 108 kmph. What distance will it cover in 15 seconds ?
Speed = 108 kmph = $$\left ( 108 \times \frac{5}{18} \right )$$ m/sec = 30 m/sec.
∴ Distance covered in 15 sec = (30 × 15) m = 450 m.
A truck covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 km in 45 minutes. The ratio of their speeds is :
Ratio of speeds = $$\left ( \frac{550}{60} \times \frac{18}{5} \right ) : \left ( \frac{33}{45} \times 60 \right )$$
= 33 : 44
= 3 : 4.
The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is :
Then, $$8x = \frac{400}{4} = 100$$ ⇒ $$x = \left ( \frac{100}{8} \right ) = 12.5$$.<|endoftext|>
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Post Visit Activities
Post Visit Activities
Encourage the students to search the web for more information on the artist they adopted for their trip to the Museum.
Review the pre-trip Great Expectations! Chart to see how the students expectations relate to their actual experience.
Design an in-class museum display about the museum on a bulletin board by having students share their memories of the trip in words and pictures.
Encourage the use of new vocabulary words on the board. Let the students provide tours of the in-class museum to other classes.Using stiff cardboard or wood, have the students paint their own panels and install them in the classroom like the dining room in the Griswold House.
Have the students create their own “Fox Chase” parody featuring members of the class as the participants. Encourage positive characteristics in identifying each others’ individuality, personality, and talents. Discuss how the class would look in its own version of The Fox Chase.
Use prompts such as: “What would we replace the image of the Griswold House with?” or “What things should we include to represent our class?”
More Activities for Fox Chase & the Painted Panels
Have students adopt an artist from The Fox Chase.
Have them share information by completing the sentence, “The Art Colony at Old Lyme would not have been the same without [insert name of artist, for example “Childe Hassam”] because he/she . . .”
Encourage them to use The Fox Chase website to get specific ideas and information about their artist.
Have students “adopt” one of the painted panels in the In Situ: The Painted Panels section.
Ask them to share their panel with the class after working with the Museum’s What? Why? Wow! worksheet.
More Activities for the Lyme Art Colony
Divide the students into small groups and have them design their own artist colony.
Encourage them to give it a name and imagine where it would be, where the artists would stay, what kind of things would they make, what art style would they choose, etc. Have the various art colonies create examples of their artwork and report out to the class.
Have students create short skits based on ideas that are in The Story of Miss Florence and The Lyme Art Colony.
Have them decide where in the boardinghouse their skit would take place.
Have students design a poster advertising the Lyme Art Colony.
More Writing Activities
Have students write letters to the Museum about their visit.
Letters should be addressed to the Education Department, Florence Griswold Museum, 96 Lyme Street, Old Lyme, Connecticut 06371 or send email to: [email protected].
Have students record their experiences at the Museum in a creative journal project using words, pictures, drawings, etc.
Have students create a find-a-word puzzles using words and ideas that were learned during the field trip.<|endoftext|>
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Inside some whales and dolphins are small bones that show they once had back legs and that their ancestors walked on land. These occasionally reappear as tiny rear flippers.
An artist's impression of what Pakicetus, an early ancestor of the whale, might have looked like
In the sixth week of gestation, the human embryo possesses a tail, complete with several vertebrae. In the next couple weeks of development, however, the tail disappears, and over time the vertebrae fuse to form the coccyx, or tailbone, in the adult. Humans and their ape relatives are distinguished from other groups of primates in part by their taillessness, though it is unclear why apes lost their tails.
On rare occasion, a human infant is born with a vestigial tail. In modern medical literature, such tails lack vertebrae and typically are harmless, though some are associated with spina bifida (failure of the vertebrae to completely enclose the spinal cord). Tails in human infants typically are removed through surgery without complication.<|endoftext|>
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The way students learn today has been greatly changed by the utilization of technology by…
Use of Technology in Classroom to Enhance Teaching and Learning
In the past, classes were conducted using chalk-boards, textbooks, notebooks, pencils and pens. However, today there has been a rise in new technological teaching-tools such as whiteboards, projectors and smart-devices that have actually helped to make learning more efficient and effective than in the past. In fact, technology has provided teachers with numerous tools that can be used inside and outside the classroom and even students can now learn better than in the past. However, new technologies are rinsing at a fast pace and this actually requires teachers and student to adapt to fast growing technologies in-order to gain from them in classroom environment.
Ways of Using Technology in a Classroom to Enhance Teaching and Learning:
- Presentation Software: Today there are several softwares like; PowerPoint that can be used visual-illustrations or presentation while in a classroom-environment. In fact, visual-aid technologies help to complement teaching, stimulate discussions and even allows for out-of-class teaching. However, teachers and students need to learn how to use presentation-software in order to achieve the best results while in a classroom environment.
- Word Processing: Today, students can install word-processing software on their computers and then use it to write while in a classroom environment. In fact, using a computer to write feels more convenient than handwriting. In fact, even teachers can easily create PDFs and handouts for their students using word-processors thus eliminating the need for textbooks.
- Classroom Response Systems: Teachers can also encourage student engagement in a classroom-environment by using electronic-devices that allow students to record their answers to multiple choice questions and then instantly display the results. In fact, the anonymity of these systems helps to encourage student participation which is a great thing. For example, Clickers is a great response-system that can serve as a catalyst for discussions in a classroom environment.
- Class Websites: Today, teachers can easily create Class-websites using softwares like WordPress in order to enhance on learning. In fact, a class-website can feature a message-board where students and teachers can have conversations and solve difficult topics. Additionally, teachers can even post educational-material on certain lessons/topics in-order to allow students learn more about a particular-subject online. ( Learn How To Create a Website …here…)
- Online-Projects and Collaboration-Tools: Today, teachers and students can easily carryout different projects online and there are even several collaboration-tools that help students work together to achieve a deeper understanding of any course material.
- Digital Textbooks: Technology has also allowed teachers to create digital-textbooks that can be easily shared amongst students online thus eliminating the need for physical-textbooks that were used in the past. In fact, digital-textbooks are very easy to create, share and even very convenient since they can easily be accessed online using your computer-device.
- Information Visualization Tools: Technology helps in the creation visualized educational-content in order to make learning more interesting for students. In fact, there are several tools that can be used to transform words into pictures or videos and this actually enhances on learning-process under a classroom-environment.
- Educational Computer Games: using computer-games with an educational-theme under a classrooms-environment actually helps students to interact with educational-content and even learn without realizing that they are learning. In fact, computer-games work perfectly for elementary students that tend to have less attention under a classroom-environment.
- Use of Podcasts: with Podcasts, teachers can easily create digital audio-clips containing information that students may need for initial learning or reviewing. On the other hand, even students can create audio-clips on different topics in-order to help educate their friends. In fact, podcasts can be uploaded on a class-website and then downloaded or shared between students in-order to maximize learning.
- Video Streaming: Using videos in a classroom environment actually helps students to learn efficiently and effectively than never before. For example, YouTube is a great tool for finding great video-clips that can be used to compliment any lesson under a classroom environment.
- Wikis and Blogs: These technologies allow students to publish their ideas and stories while sharing them with friends. In fact, Wikis and Blogs can encourage learning in an interactive way compared to traditional learning methods.
- Video Conferencing: Using Video Conferencing technology allows students to get lessons from different teachers around the world while in a classroom-environment. In fact, this technology allows students to speak with virtual-guest-lecturers and even interact with other students from different countries.
- Teaching with computer-devices: Today, different teachers and students are using computer-devices like tablets, laptops and even Smartphones to carryout lessons under a classroom environment. In fact, schools are only beginning to explore the many possibilities of using computer-devices in classrooms.
- Use of social-media platforms: Although social-media can be a disrupting tool for students in a classroom environment, social-media can actually be used to share thoughts, comments and concerns in real-time making it a perfect for shy students who feel more comfortable conversing online than in a classroom. In fact, social-media is an easy way for students to share resources and help each other out both outside and inside a classroom environment.<|endoftext|>
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The return interval and number of fires vary, depending on the geographical location in interaction with climate, topography and amount of fuel. During recent decades, in northern Europe the number and severity of fires have been insignificant compared with Mediterranean region, in which fire return intervals may be 15–35 years, compared to the average of 60–120 years for boreal forests. This is partly due to the efficient system of fire protection in northern Europe, but is mainly due to the less favourable climate for fire and the smaller human impact on ignition of forest fires. The consequences of fire are related to both site and stand characteristics, site being the most important factor controlling the stability of stands. Dry sites being more flammable and likely to ignite are associated with high risk of fire. In northern Europe, due to the interaction between species and site, the role of species difference in risk of fire damage is not clear. In southern Europe, fire risk cannot be explained by differences between tree species. There, other vegetation (shrubs, etc.) is of major importance for the risk of fire. Management of forests can, to some degree, alter the risk and the occurrence of fire. In northern Europe, logging may have compensated for fire occurrence by decreasing the amount of fuel. In addition, forest roads act as fire-breaks and facilitate fire-fighting. On the contrary, in southern Europe the risk of fire has been found to increase because the traditional forest uses and management have decreased, which increases the accumulation of fuel. However, it is not yet possible to quantify and compare the effect of management in absolute terms. Currently, some tools, such as fire-risk indices, remote sensing and GIS-based techniques, are available for prediction of fire risk in some areas. For example, fire-risk indices are most suitable for areas, like northern Europe, which have a low fire risk. In high-risk areas, such as southern Europe, more sophisticated techniques are needed for assessment of the risk. In the future, assuming global warming at northern latitudes (2 x CO2 climate), the risk of fire damage could also increase in northern Europe. Therefore, to allow the various locational and silvicultural factors to be assessed on the European level, an integrated risk model is needed.<|endoftext|>
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# 2.4 Hypothesis testing of single mean and single proportion: examples (Page 4/6)
Page 4 / 6
Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with theresult that 43 of the households have three cell phones.
Set up the Hypothesis Test:
${H}_{o}$ : $p$ $=0.30\phantom{\rule{20pt}{0ex}}$ ${H}_{a}$ : $p$ $()$ $0.30$
Determine the distribution needed:
The random variable is $P\text{'}$ = proportion of households that have three cell phones.
The distribution for the hypothesis test is $P\text{'}$ ~ $N$ $\left(0.30,\sqrt{\frac{\left(0.30\right)\cdot \left(0.70\right)}{150}}\right)$
The value that helps determine the p-value is $p\text{'}$ . Calculate $p\text{'}$ .
$p\text{'}=\frac{x}{n}\phantom{\rule{20pt}{0ex}}$ where $x$ is the number of successes and $n$ is the total number in the sample.
$x=43$ , $n=150$
p' = $\frac{43}{150}$
What is a success for this problem?
A success is having three cell phones in a household.
What is the level of significance?
The level of significance is the preset $\alpha$ . Since $\alpha$ is not given, assume that $\alpha =0.05$ .
Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.
Calculate the p-value.
p-value = 0.7216
Make a decision. _____________(Reject/Do not reject) ${H}_{0}$ because____________.
Assuming that $\alpha$ = 0.05, $\alpha <\text{p-value}$ . The Decision is do not reject ${H}_{0}$ because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.
The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a singlepopulation proportion. This means that the null and alternate hypotheses use theparameter $p$ . The distribution for the test is normal. The estimated proportion $p\text{'}$ is the proportion of fleas killed to the total fleas found on Fido. This is sample information.The problem gives a preconceived $\alpha =0.01$ , for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!
Hypothesis testing problems consist of multiple steps. To help you do the problems, solution sheets are provided for your use. Look in the Table of Contents Appendix for the topic "Solution Sheets." If you like, use copies of the appropriate solution sheet for homework problems.
My dog has so many fleas, They do not come off with ease.As for shampoo, I have tried many types Even one called Bubble Hype,Which only killed 25% of the fleas, Unfortunately I was not pleased.I've used all kinds of soap, Until I had give up hopeUntil one day I saw An ad that put me in awe.A shampoo used for dogs Called GOOD ENOUGH to Clean a HogGuaranteed to kill more fleas. I gave Fido a bathAnd after doing the math His number of fleasStarted dropping by 3's! Before his shampooI counted 42. At the end of his bath,I redid the math And the new shampoo had killed 17 fleas.So now I was pleased. Now it is time for you to have some funWith the level of significance being .01, You must help me figure outUse the new shampoo or go without?
Set up the Hypothesis Test:
${H}_{o}$ : $p=0.25\phantom{\rule{20pt}{0ex}}$ ${H}_{a}$ : $p>0.25$
Determine the distribution needed:
In words, CLEARLY state what your random variable $\overline{X}$ or $\mathrm{P\text{'}}$ represents.
$\mathrm{P\text{'}}$ = The proportion of fleas that are killed by the new shampoo
State the distribution to use for the test.
Normal: $N\left(0.25,\sqrt{\frac{\left(0.25\right)\left(1-0.25\right)}{42}}\right)$
Test Statistic: $z=2.3163$
Calculate the p-value using the normal distribution for proportions:
$\text{p-value}=$ $0.0103$
In 1 – 2 complete sentences, explain what the p-value means for this problem.
If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 $\left(\frac{17}{42}\right)$ or more.
Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p-value.
Compare $\alpha$ and the p-value:
Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using COMPLETE SENTENCES.
alpha decision reason for decision
0.01 Do not reject ${H}_{o}$ $\alpha$ $()$ $\text{p-value}$
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.
Construct a 95% Confidence Interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the Confidence Interval.
Confidence Interval: $\left(0.26,0.55\right)$ We are 95% confident that the true population proportion $p$ of fleas that are killed by the new shampoo is between 26% and 55%.
This test result is not very definitive since the p-value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Got questions? Join the online conversation and get instant answers!<|endoftext|>
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· major Cold War confrontation between the United States and the Union of Soviet Socialist Republics (USSR) over Soviet-supplied missile installations in Cuba, regarded by many as the world’s closest approach to nuclear war
· after Castro’s revolution on Cuba, Eisenhower’s administrative anulled diplomatic relations with Cuba in 1960, because Cuba abused American soldiers on ambassys, took over American property and supported guerrilas in Latin America
· on April 17, 1962, a group of Cuban refugees, trained and equiped in USA, landed on the Cuban coast, in the Bay of Pigs; Castro’s soldiers were waiting for them and 2,000 Cuban refugees were captured or killed by 20,000 Castro’s army
· in October USA had leaerned that Soviet Union had begun missile shipment and spy planes overflying Cuba had photographed Soviet-managed construction work by August 29; and the first ballistic missile was spotted on October 14.
· after a week of secret consultation with his advisers, during which options of invasion, air strikes, blockade, or diplomacy were discussed, on October 22 US President John F. Kennedy announced his intention of placing a naval blockade around Cuba to prevent the arrival of more missiles.
· Kennedy demanded that the Soviet Union dismantle and remove the weapons, declaring that US naval forces would intercept and inspect ships bound for Cuba to determine whether they were carrying missiles.
· Soviet vessels en route for Cuba turned away to avoid the quarantined zone, while communication between Khrushchev and Kennedy was opened through diplomatic channels. After several days of negotiations during which nuclear war was feared by many to be a possibility, Khrushchev agreed, on October 28, to dismantle the missile sites and return the weapons to the Soviet Union, offering the United States on-site inspection in return for a guarantee not to invade Cuba
· Kennedy halted the blockade and also secretly promised to withdraw US missiles recently stationed in its NATO partner Turkey<|endoftext|>
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Human Rights in Iraq
The concept of human rights has become a well-known and widely accepted term to use. Varying interpretations are possible, with differences usually being based on cultural background. Nonetheless, most of these understandings consciously or subconsciously include the basic rights outlined in the United Nations’ Universal Declaration of Human Rights.
The United Nations General Assembly (GA) adopted the Universal Declaration of Human Rights on December 10th, 1948. It was written in the aftermath of World War II, “… as a common standard of achievement for all peoples and all nations, to the end that every individual and every organ of society, keeping this Declaration constantly in mind…” Thus it was truly meant to be universal, to protect citizens from any type of violation the world had recently experienced, as outlined in the Preamble and 30 Articles.
As such, it includes articles on the right to life in dignity; liberty and security; freedom of movement; right to nationality and education; just treatment of human beings and respect; as well as freedom of expressions and opinions, from torture or inhumane treatment, as well as economic, social and cultural rights.
International Human Rights Law
The Declaration is not legally binding, but is the basis of international human rights law. Two binding UN covenants were formed as a result of the UDHR; the International Covenant on Civil and Political Rights and the International Covenant on Economic, Social and Cultural Rights. Combined, these three documents are often referred to as the “International Bill of Human Rights”.
Over the years other conventions have been written to expand on and add to this fundament, focusing on a variety of topics such as refugees (1951 and 1967), discrimination of women (‘CEDAW’, 1979) and disabled persons (2008), against torture (1987), protection of migrant workers (1990), and against racial discrimination (1969) to name a few.
Additionally, the International Labour Organization has compiled a large number of conventions specifically related to work force and labour standards, of which 8 are considered ‘fundamental conventions’ and relate to freedom of association (1948, C087), collective bargaining (1949, C098), forced labour (1930, C029 and 1957, C105), minimum age (1973, C138), child labour (1999, C182), equal remuneration (1951, C100), equal opportunity and treatment (C111).
The Geneva Conventions are a revision of previously constructed conventions, adjusted after WWII and specifically focus on treatment of persons in time of war. It consists of four Conventions, and three additional protocols. The International Committee of the Red Cross (ICRC) explains that the Conventions “aims at ensuring that, even in the midst of hostilities, the dignity of the human person, universally acknowledged in principle, shall be respected.”
During a series of expert meetings, congregations by Red Cross agencies, and a confluence of government representatives over time, the articles were revised until a draft was represented at The Diplomatic Conference for the Establishment of International Conventions for the Protection of Victims of War in 1949. The Final Act was signed by fifty-nine nations, some of which no longer exist, and has attained more signatories since.
The Cairo Declaration on Human Rights in Islam (CDHRI) was compiled by the Organisation of Islamic Cooperation (OIC) in 1990, during the 19th Islamic Conference of Foreign Ministers in Cairo, and has 57 signatories. This Declaration holds similar – if not identical – principles as the UDHR, but notably also included articles related to ‘jus in bello’ – acceptable wartime conduct, alike the Geneva Conventions. The CDHRI also addresses equality between women and men, rights of the child, freedom, right to medical care, right to self-determination, amongst others. Most notably is that this 25 Article document clearly lists the Sharia as reference point including for punishment. The CDHRI has been adopted by 45 countries, out of the total 57 members of the OIC.
Conventions Signed By Iraq
Iraq signed the Geneva Conventions on 14 February 1956, and the Additional Protocol I (Protection of Victims of International Armed Conflicts) on April 1 2010. Moreover, it became signatory to the Convention for the Rights of the Child (CRC) on 15 June 1994, and signed the Optional Protocol to the CRC on 24 June 2008.
Iraq voted in favor of the UDHR, along with 48 other states, and has been member of the OIC since 1975. Only one of the 8 fundamental ILO conventions has not been ratified by Iraq, being the convention on freedom of association (C087).
The Convention relating to the Status of Refugees is based off Article 14 of the UDHR, and recognizes the right of asylum and protection of refugees. It was approved during the General Assembly meeting of December 14, 1950 and came into force on April 22, 1954. However, the original Convention limited its scope to refugees fleeing prior to 1 January 1951. As such, an additional protocol was compiled in 1967, removing these limitations.
Iraq is not party to the 1951 Convention, nor its Protocol.
The Convention on the Elimination of All Forms of Discrimination against Women – also called CEDAW, was approved during the General Assembly Session on 18 December, 1979 and entered into force on 3 September 1981. Iraq confirmed its accession to the Convention on 13 August, 1986. Through “accession” a state accepts the offer or the opportunity to become a party to a treaty, which has already negotiated and signed by other states. It has the same legal effect as ratification. Countries that have ratified or acceded CEDAW are legally bound to put its provisions into practice, and thereby agree to submit national reports on measures taken to comply with its obligations. Such reports are to be compiled at least every four years.
On the same date, Iraq voiced reservations, saying it is not bound by Article 2 (f) and (g), condemning discrimination against women and taking of appropriate measures to ensure equality in national law, customs, and practices; Article 9 (1) and (2), concerning equal rights to nationality between men and women, as well as children; and Article 16, eliminating discrimination in marriage and family matters and child marriage. Additionally, it entered a reservation to Article 29 (1), discussing dispute between states concerning interpretation or application of the Convention specifically stating it “in no way implies recognition of or entry into any relations with Israel.” It is noteworthy that Article 29 (2) specifically allows for a reservation against Article 29 (1).
Persons with Disabilities
The Convention on the Rights of Persons with Disabilities was approved during the General Assembly session on December 13, 2006 and came into force on May 3, 2008. Simultaneously, the Optional Protocol was approved, giving the Committee on the Rights of Persons with Disabilities (CRPD) competence to examine individual complaints with regard to alleged violations by States parties to the Protocol. The CRPD is the body of independent experts that monitors implementation of the Convention.
Iraq acceded to the Convention on March 20, 2013 without any reservations. It is not party to the Optional Protocol.
The Convention against Torture and Other Cruel, Inhuman, or Degrading Treatment or Punishment, also referred to as just the Convention against Torture, was adopted during the General Assembly session on December 10, 1984. On June 26, 1987 it was registered and thereby came into force. Its implementation is monitored by the Committee Against Torture (CAT), composed of ten individuals of various nationalities. All signatory states are obliged to send regular reports to the CAT, based on which recommendations are made. Iraq confirmed its accession to the Convention on July 7, 2011 with no reservations.
The International Convention on the Protection of the Rights of All Migrant Workers and Members of their Families was approved by the General Assembly on December 18, 1990 and entered into force on July 1, 2003. Iraq is not party to the Convention.
The Convention on the Elimination of All Forms of Racial Discrimination was approved by the General Assembly and accordingly opened for signature on March 7, 1966. It entered into force on January 4, 1969. Despite the obvious as stated in the Convention title, it aims to obliterate hate speech and promote understanding. Implementation of the articles is monitored by the Committee on the Elimination of Racial Discrimination, to which bi-annual reports are submitted by each signatory. It also is responsible for handling inter-state and individual complaints related to non-conformity to the provisions of the Convention, as prescribed in Article 14.
Iraq signed the Convention on February 18, 1969 and ratified it on February 13, 1970. Upon signing, Iraq clearly stated that this in no way signified entry into dealings with Israel and that it does not consider itself bound by Article 22 – pertaining to inter-state dispute resulting from interpretation or implementation of the Convention, stating that all parties to the dispute are to expressly consent to intervention of the International Court of Justice. Both these statements were explicitly repeated upon ratification.<|endoftext|>
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# Beginner Differential Equations
• Oct 20th 2012, 03:31 PM
astuart
Beginner Differential Equations
Hello,
I've just started on separating differential equations and am just having a little trouble on how to proceed with this sort of question.
$\frac {dy}{dx} = y^2 \,e^(2x)$
I understand that the y needs to be bough to the left, and then the formal separation of $\frac {dy}{dx}$ occurs - as my tutor puts it.
$=\frac{1}{y^2}\,dy = e^{2x} dx$
$= \int \frac{1}{y^2}dy = \int e^{2x}$
$= ln\,y^2 = \frac {e^{2x}}{2} + C$
This is where it breaks down a little bit for me...I'd assume that we need to get y to its simplest form, but I don't know how to do that...
$=y^2 = e^{\frac {e^2x}{2}} + C$
That isn't right (I'm assuming), so can someone point me in the direction of the correct step.
Thanks :)
• Oct 20th 2012, 03:55 PM
TheEmptySet
Re: Beginner Differential Equations
$\int \frac{1}{y^2}dy = -\frac{1}{y}$
So you should get
$-\frac{1}{y}=\frac{1}{2}e^{2x}+C$
• Oct 20th 2012, 03:59 PM
MarkFL
Re: Beginner Differential Equations
First, when you separate the variables to get:
$\frac{1}{y^2}\cdot\frac{dy}{dx}=e^{2x}$
you should note that you are losing the trivial solution $y\equiv0$.
Then integrating with respect to $x$, we have:
$\int y^{-2}\,dy=\int e^{2x}\,dx$
As noted above, you have integrated incorrectly.
• Oct 20th 2012, 04:00 PM
astuart
Re: Beginner Differential Equations
Quote:
Originally Posted by TheEmptySet
$\int \frac{1}{y^2}dy = -\frac{1}{y}$
$-\frac{1}{y}=\frac{1}{2}e^{2x}+C$
Ahh, of course...It's basically $y^{-2}$.
I was treating it as $y^{-1}$ or something like that.<|endoftext|>
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During the growth and development stage in children, their brains form stronger and more complex connections that help them gain experience. Research has shown that during the first ten years of a child’s life, they learn things at a very speedy rate. This stage is known as the “Windows of Opportunity” or “Critical Period” stage. The ability to learn a language in children occurs during their first year. Their vocabulary increases fourfold by the time they are two years of age. This is known as the “Vocabulary Explosion” stage.
Unlike children, adults are prone to losing some synaptic connections. The synapses that survive and flourish depend on how often they are used. Let’s take a case in example of a two-year-old child whose parents speak two separate languages. The child is able to recognize some words from both languages if both parents use both languages in front of him or her. However, the ability to speak or understand either of the languages gets lost when the parents stop speaking it.
Therefore, since children grasp and understand things faster than adults do, a parent or guardian can use this opportunity to improve their vocabulary. This can be done in such a way that the child doesn’t feel like they are being tutored or compelled to learn. To do so, parents or guardians can use the following exercises to improve the vocabulary for kids:
1. Teach them to read and write on their own
Research has shown that reading and writing contributes about 20% of our vocabulary. When parents teach their kids how to read and write from an early age, they help promote the correct and regular use of new words. One simple way of teaching your child how to read is by taking them to the library as often as possible and getting them library cards. When they come across new words, help them figure out their meanings and how they can use them in sentences.
2. Create a word-rich environment for them
To improve vocabulary for kids, parents ought to promote word-consciousness. This can be done through an array of games and songs. Playing puzzles and crosswords with your child will allow you to not only bond with them, but also to help enhance their vocabulary. Children can also improve their vocabulary using a number of kids learning programs that can be found online. Most of these kids learning programs are free of charge and easily accessible to parents.
Storytelling is a fun way of teaching your child new words and the meaning of different words as well. Encourage your child to tell you stories and respond to them using synonyms. If, for example, your child says, “the giant was huge”, you could ask, “So he was enormous?”
4. Use proper language while speaking to them
Avoid using ‘baby talk’ when you are talking to your kids. Do not use incongruous sounds or utter words in a monotonous cadence or rhythm while using exaggerated facial expressions. Rather, use real and normal words. Otherwise, you can use songs that mimic different sounds.
5. Talk to them regularly
Holding a conversation can help improve the vocabulary for kids. Research shows that talking to your child can help improve their IQ as well. It is also important for the parent to use non-verbal cues and repetition to help expedite the learning process.<|endoftext|>
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Frisian, people of western Europe whose name survives in that of the mainland province of Friesland and in that of the Frisian Islands off the coast of the Netherlands but who once occupied a much more extensive area.
In prehistoric times the Frisians inhabited the coastal regions from the mouth of the Rhine at Katwijk (north of The Hague) to the mouth of the Ems. Much of the land then was covered by lakes, estuaries, and swamps and exposed to the incursions of the sea, and the inhabitants lived mostly on terpen, or man-made mounds. Slowly the Frisians brought the lower-lying land under cultivation and protected themselves against the sea by building more terpen (dikes were not practicable). Most of these were in the modern provinces of Friesland and Groningen; the east bank of the Rhine itself was almost uninhabited. Excavations in the terpen have thrown some light on Frisian life in the centuries before the Romans came.
Roman and Frankish periods
The Roman general Nero Claudius Drusus encountered the Frisians after his crossing of the Rhine from the south in 12 bce. He then made them tributary to Rome. The fact that they supplied oxhide for the Roman army suggests considerable cattle raising. In 28 ce the Frisians revolted and temporarily freed themselves from Roman rule, but in 47 ce they were once more made tributary by Gnaeus Domitius Corbulo. Thereafter they supplied soldiery for the Roman armies. They took part in the Batavian revolt of 69–70 ce, but relations with the Roman authority were soon restored, and Frisians continued to serve in the Roman legions in many places, including Britain (inscriptions from the 2nd and 3rd centuries for instance at Appleby and at Watermore). Pottery from Trier and from southern Gaul and imported bronzes found in Frisia provide clear evidence of the trade that the Romans brought in return for exports of cattle. This trade developed gradually in the 2nd and 3rd centuries but then declined, though during the 4th and 5th centuries commercial relations with Cologne and the Rhine province still existed, as coins found in Frisia show. Even so, there was little fusion between the Frisian and the Roman cultures, and the native ways held. Frisia had begun to show that spirit of independence which still survives in cultural, social, and political matters.
By the end of the 5th century, Roman rule had disappeared and migrations were changing the face of western Europe. Between 450 and 500 the Angles and Saxons, on their way from the Elbe River to Britain, may well have invaded Frisia and spent some time there. Some of them may have settled permanently in Frisia, while a number of Frisians may have gone on to Britain with the majority of the newcomers. Some finds suggest that after about 500 the Frisians became somewhat mixed with Anglo-Saxon elements and that there was some blending of cultures (language, ceramics, architecture).
After about 600, Frankish influences also made themselves felt. The Franks pushed forward to the north and became neighbours of the Frisians, who in the meantime had expanded outside the coastal belt proper. This larger region, Frisia Magna (up to the lower reaches of the Rhine) was so named because its inhabitants were mainly Frisians from the coast. Its economy included not only agriculture and cattle breeding but also a textile industry. Dorestad (southeast of Utrecht) was the commercial centre. The Frisians, a seafaring people, also traded by river with the regions of the Rhine and of the Moselle and later with the north of France.
The Franks invaded the southern borderland of Frisia Magna and occupied the remains of the Roman fort of Trajectum ad Rhenum (Utrecht) and the trading station of Dorestad about 600. About 40 years later the Frisians recovered this area and penetrated into the region between the Rhine and the Meuse; in Dorestad they even minted coins. In 689, however, the Frisian kingRadbod was dislodged from Dorestad and Trajectum by the Frankish leader Pippin II, who wanted to protect both the Frankish trade on the Rhine and the Christian missions.
Anglo-Saxon missionaries now began to convert the Frisians to Christianity: Willibrord arrived with his followers at Utrecht in 690. In the 8th century came Boniface (who was to be martyred near Dokkum in 754), and the Frisian Liudger continued the work. Utrecht was the centre of this activity, but the ill will of the Frisian kings partly counteracted its effect.
In 734 the Frankish ruler Charles Martel defeated the Frisians and drove them back to the north, thus dissolving Frisia Magna. Charlemagne finally subdued the Frisians, and after this they were converted permanently to Christianity. When the Saxon revolts had been crushed, peace reigned in Frisia. The national laws were codified (Lex Frisionum, 802 ce). Norse invasions temporarily disturbed this peace in the 9th century.
Frisia was little affected by the partition of the Carolingian empire. From 925 the lands occupied by Frisians belonged to the German kingdom. The people’s subsequent history, however, is that of the three regions into which the country was divided: (1) West Frisia, from the mouth of the Rhine to the Vlie and Lake Flevo; (2) Middle Frisia, from the Vlie to Lauwers; and (3) East Frisia, from Lauwers to the Jade estuary, where the region east of the Ems was penetrated by Frisians as colonists between 600 and 1200. The boundary between West and Middle Frisia was soon formed by the floods of Lake Flevo, which developed into the Zuiderzee.
West Frisia fell to the counts of Holland, who occupied first the south and the coastal area and, after 1250, the northeastern part. In East Frisia the city of Groningen, with its surrounding country, was ruled by the bishop of Utrecht, and part of the district between the Ems and the Jade became a countship (under the Cirksena family from 1454 to 1744, when it passed to the kingdom of Prussia). Middle Frisia kept itself free of overlords, whether native or foreign, except for short periods, until the end of the Middle Ages.
Middle Frisia or Friesland
The Middle Frisians repeatedly expelled the counts of Holland, although they remained “immediate” vassals of the Holy Roman emperor throughout the Middle Ages. Feudalism was never accepted there, and this gave rise to the idea of Frisian liberties. These found expression in the so-called Privilege of Freedom, allegedly granted by Charlemagne but in fact a forgery, from the second half of the 13th century.
These Frisians settled all affairs themselves, keeping legislation, administration, and the judiciary in their own hands and themselves maintaining their dikes, roads, and canals. The Union of the Upstalbeam, by which they consolidated themselves politically, included some of the East Frisians. This degree of liberty was rare in medieval Europe; parallels are found only in the Swiss confederacy and in the administration of Dithmarschen.
As late as 1457 the emperor Frederick III acknowledged the "immediate" status of the Frisians, but in 1498 the emperor Maximilian I presented Frisia to Albert, duke of Saxony, with the title gubernator in Frisia. The Saxon dukes, however, could not hold their own; the Frisians, helped by Charles of Gelders, ousted them. In 1524 the Frisian lands between Lauwers and the Zuiderzee fell to the emperor Charles V. Thenceforth they were joined to the Burgundian portion of the Habsburg heritage. Charles was the first foreign ruler to bring the free Frisians a central administration that they accepted.
Friesland adopted the new doctrines at the Reformation and participated in the revolt of the northern Netherlands against Philip II of Spain. It thus became a province of the republic of the Netherlands, as constituted by the Union of Utrecht (1579). A junior branch of the house of Nassau filled the office of stadtholder. In 1815 Friesland was absorbed into the kingdom of the Netherlands.
The Frisian language, which has many dialects, is taught in the schools in Friesland. It is acknowledged as an official language in Friesland, but it is not legally codified as such by the Dutch government. Literary and scientific works are written in it, and there is a Frisian academy (Fryske Akademy) in Leeuwarden. In East and North Frisia the language has been largely supplanted by German. In the early 21st century, there were approximately 470,000 Frisian speakers worldwide.
Learn More in these related Britannica articles:
Frisia…the traditional homeland of the Frisians, a Germanic people who speak a language closely related to English.…
France: The hegemony of NeustriaThe Frisians reached the mouth of the Schelde River and controlled the towns of Utrecht and Dorestat; the attempted conversion of Frisia by Wilfrid of Northumbria had to be abandoned (
c.680). In southern Gaul the duke Lupus changed the status of Aquitaine from a duchy…
Germany: Merovingian GermanyIn the north the Frisians and Saxons remained independent of Frankish control into the 8th century, preserving their own political and social structures and remaining for the most part pagan. In areas under Frankish lordship, Christianity made considerable progress through the efforts of native Raetians in the Alpine regions,…
Netherlands: Ethnic groups…Dutch are a mixture of Frisians, Saxons, and Franks. In fact, research has made plausible the contention that the autochthonous inhabitants of the region were a mixture of pre-Germanic and Germanic population groups who in the course of time had converged on the main deltaic region of western Europe. There…
history of the Low Countries: Frankish ruleThe area occupied by the Frisians in the north was completely outside the Frankish sphere of influence, but the Rhine delta and even what is now Noord-Brabant also appear to have retained the virtually independent status they had possessed during the Roman era.…<|endoftext|>
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Japanese researchers have discovered a number of the earliest stars that emerged in the Universe, just 250 million years after the Big Bang, as reported in a study issued by Nature journal. Based on data gathered by the huge Chilean ALMA telescope, scientists have sighted the faraway galaxy MACS1149-JD1 which formed 550 million years ago. At the time of the observation, the Japanese researchers noticed that this galaxy housed numerous 300 million years old stars.
The “redshift” of the galaxy, a measurement that establishes the age and distance of a space object, was 9.1096 units, the biggest value obtained so far from the study of spectral lines, as revealed by the report for this study.
Scientists have determined this measurement based on lines in the ionized oxygen spectra, rather than employing ionized carbon, which is more commonly used when scanning faraway space objects.
Earliest stars in the Universe formed just 250 million years after the Big Bang
“These results demonstrate the ability of ALMA as a tool to measure the redshift of distant galaxies,” says Richard Bouwens, of the University of Leiden in the Netherlands, in the paper issued in the Nature journal.
The study conducted by Takuya Hashimoto and his staff from the Osaka Sangyo University provides insight into the formation process of early stars and also indicates that upcoming space telescopes like the James Webb, that will substitute Hubble in orbit in early 2020, are likely to provide new insights into early star formation, says Bouwens.
Researchers think that the earliest stars in the Universe were built in areas with a very high density of matter, even though comprehension of this phenomenon is not yet well-defined.
Bowens points out that it is still uncertain if the star activity observed in MACS1149-JD1 also took place in different parts of the early Universe but he claims that the finding “will certainly boost similar studies of other galaxies”.
In the meantime, the discovery that the earliest stars in the Universe formed only 250 million after the Big Bang remains a stunning one.<|endoftext|>
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There are lots of critical social values we work on as teachers. A few of these include: cooperation, freedom, happiness, honesty, humility, love, peace, respect, responsibility, simplicity, tolerance and unity.
They all have their place and their value but for me I always think that humility is everything.
But being humble means different things to different people.
In Factfulness, Hans Rosling et al (2018) say that being humble ‘means being aware of how difficult your instincts can make it to get the facts right. It means being realistic about the extent of your knowledge. It means being happy to say “I don’t know.” It also means, when you do have an opinion, being prepared to change it when you discover new facts.”
I like this definition because we don’t have to feel pressured. It’s okay not to know and we can be relaxed about it. This is not about being a know-it-all and stops us from being too big for our boots.
Being humble is about recognising the contributions of all those around you. Children might experience success and achieve great things but they can’t take all the credit. There is a team around them and acknowledging that is important.
Teaching humility is important because we are letting children know that the most successful people in the world don’t boast, they acknowledge. Successful people, particularly sports ‘stars’, don’t get carried away and start bragging how good they are. They are graceful and respectful in their words and thoughts for others and the people that help them.
Roger Federer is a great example and the perfect role model of how to live and breathe humility. He is confident without being arrogant and respects others and maintains his self-respect. If you have ever seen him being interviewed after winning a tennis match he is humble, gentle and genuine. He is clearly still proud in what he does but he doesn’t rub others’ noses in his success. He always has a good word to say about his opponents.
Federer is a class act and even though he wins plenty he does still suffer losses. His response though is the same and he carries himself with an unbridled humbleness that is a joy to watch. This is when humility meets greatness. He doesn’t think less of himself but simply thinks of himself less. His humility is laced with true modesty.
As a humble athlete, Federer is the total package of everything we want children to learn – sportsmanship, teamwork, confidence and respect for others.
This calm and modest mindset is something we can share with children and help them to consider others ahead of themselves. We can help model a zero-tolerance for rudeness and teach them how to respect everyone. We can help them to see other people’s perspectives and appreciate their feelings.
Being humble is also being open and admitting to mistakes. When we are wrong, how many of us ‘own’ the mistake and apologise? If we face criticism do we see it as a personal attack and go into defend mode? Or do we demonstrate our humility and recognise what we have strengths and weaknesses? Do we teach children that apologising makes us place our pride aside, and it is one of the most humbling things we can do.
Here’s the tricky part: a big part of being humble is understanding and accepting that we don’t always get credit for what we do. Children might find that hard to stomach but humility is tough to swallow when you first start out.
Raymond Tang: Be humble – and other lessons from the philosophy of water.<|endoftext|>
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Bones are semi-rigid, porous, mineralized organs, consisting of cells in a hard matrix, that form part of the endoskeleton of vertebrates. Bones function to move, support, and protect the body, produce red and white blood cells, and store minerals.
Although externally bones may appear to be simple and even solid structures, in reality they are composed of living bone tissue interlaced with blood vessels, nerve fibers, and so forth, and their formation, structure, and function involves stunningly complex coordination.
Bones come in a variety of shapes and have a intricate internal and external structure, allowing them to be lightweight yet strong and hard, while fulfilling their many other functions. One of the types of tissues that make up bones is the mineralized osseous tissue, also called bone tissue, a specialized connective tissue that gives bones their rigidity and honeycomb-like, three-dimensional internal structure. Other tissue types found in bones in their entirety include marrow, the periosteum, nerves, blood vessels, and cartilage.
Because a group of tissues are involved that perform a specific function or group of functions, bones can be referred to as organs, although bone tissue is the dominant tissue, leading bone to often be classified as a specialized type of connective tissue.
All bones consist of living cells embedded in the mineralized organic matrix that makes up the osseous tissue.
The primary tissue of bone, osseous tissue, is a relatively hard and lightweight composite material, formed mostly of calcium phosphate in the chemical arrangement termed calcium hydroxylapatite (this is the osseous tissue that gives bones their rigidity). It has relatively high compressive strength but poor tensile strength, meaning it resists pushing forces well, but not pulling forces. While bone is essentially brittle, it does have a significant degree of elasticity, contributed chiefly by collagen. Collagen is the main protein of connective tissue in animals and involves the harmonization of three polypeptide chains into the form of a triple helix. It is characterized by the regular arrangement of amino acids in each of the three chains; under tension, the triple helix coils tight, resisting stretching, and making collagen valuable for structure and support, while giving bones some elasticity.
Bone is not a uniformly solid material, but rather has some spaces between its hard components. The hard outer layer of bones is called compact bone tissue due to its minimal gaps or spaces. This tissue gives bones their smooth, white, and solid appearance, and accounts for 80 percent of the total bone mass of an adult skeleton. Compact bone may also be referred to as dense bone or cortical bone. Filling the interior of the organ is the hole-filled spongy bone tissue (also called cancellous bone or trabecular bone) which is comprised of a network of flat or needle-shaped trabeculae, which makes the overall organ lighter and allows room for blood vessels and marrow. Spongy bone accounts for the remaining 20 percent of total bone mass, but has nearly ten times the surface area of compact bone.
The exterior of bones (except where they interact with other bones through joints) is covered by the periosteum, which has an external fibrous layer, and an internal osteogenic layer. The periosteum is richly supplied with blood, lymph, and nerve vessels, attaching to the bone itself through Sharpey's fibers.
Bone can also be either woven or lamellar (layered). Woven bone is weak, with a small number of randomly oriented collagen fibers, but forms quickly and without a preexisting structure during periods of repair or growth. Lamellar bone is stronger, formed of numerous stacked layers and filled with many collagen fibers parallel to other fibers in the same layer. The fibers run in opposite directions in alternating layers, assisting in the bone's ability to resist torsion forces. After a break, woven bone quickly forms and is gradually replaced by slow-growing lamellar bone on preexisting, calcified hyaline cartilage through a process known as "bony substitution."
There are seven main functions of bones.
Most bones perform all of these functions to one degree or another, but certain bones are more specialized for certain functions.
There are five types of bones in the human body: long, short, flat, irregular, and sesamoid.
The process of bone resorption releases stored calcium into the systemic circulation and is an important process in regulating calcium balance. As bone formation actively fixes circulating calcium in its mineral form, removing it from the bloodstream, resorption actively unfixes it, thereby increasing circulating calcium levels. These processes occur in tandem at site-specific locations and are known as bone turnover or remodeling. Osteoblasts and osteoclasts, coupled together via paracrine cell signaling, are referred to as bone remodeling units. The iteration of remodeling events at the cellular level is influential on shaping and sculpting the skeleton during growth and in response to stress (such as weight-bearing exercise or bone healing).
The matrix comprises the other major constituent of bone. It has inorganic and organic parts. The inorganic is mainly crystalline mineral salts and calcium, which is present in the form of hydroxyapatite. The matrix is initially laid down as unmineralized osteoid (manufactured by osteoblasts). Mineralization involves osteoblasts secreting vesicles containing alkaline phosphatase. This cleaves the phosphate groups and acts as the foci for calcium and phosphate deposition. The vesicles then rupture and act as a center for crystals to grow on.
The organic part of matrix is mainly Type I collagen. This is made intracellularly as tropocollagen, and then exported. It then associates into fibrils. Also making up the organic part of matrix are various growth factors, the functions of which are not fully known. Other factors present include glycosaminoglycans, osteocalcin, osteonectin, bone sialo protein, and Cell Attachment Factor. One of the main things that distinguishes the matrix of a bone from that of another cell is that the matrix in bone is hard.
The formation of bone during the fetal stage of development (in humans, after the 7th or 8th week until birth) occurs by two methods: Intramembranous and endochondral ossification.
Intramembranous ossification mainly occurs during formation of the flat bones of the skull; the bone is formed from mesenchyme tissue. The steps in intramembranous ossification are:
Endochondral ossification occurs in long bones, such as limbs; the bone is formed from cartilage. The steps in endochondral ossification are:
Endochondral ossification begins with points in the cartilage called "primary ossification centers." They mostly appear during fetal development, though a few short bones begin their primary ossification after birth. They are responsible for the formation of the diaphyses of long bones, short bones, and certain parts of irregular bones. Secondary ossification occurs after birth, and forms the epiphyses of long bones and the extremities of irregular and flat bones. The diaphysis and both epiphyses of a long bone are separated by a growing zone of cartilage (the epiphyseal plate). When the child reaches skeletal maturity (18 to 25 years of age), all of the cartilage is replaced by bone, fusing the diaphysis and both epiphyses together (epiphyseal closure).
Bone marrow can be found in almost any bone that holds cancellous tissue. In newborns, all such bones are filled exclusively with red marrow (or hemopoietic marrow), but as the child ages it is mostly replaced by yellow, or "fatty," marrow. In adults, red marrow is mostly found in the flat bones of the skull, the ribs, the vertebrae, and pelvic bones.
"Remodeling" is the process of resorption followed by replacement of bone with little change in shape and occurs throughout a person's life. Its purpose is the release of calcium and the repair of micro-damaged bones (from everyday stress). Repeated stress results in the bone thickening at the points of maximum stress (Wolff's law).
The study of bones and teeth is referred to as osteology. It is frequently used in anthropology, archaeology, and forensic science for a variety of tasks. This can include determining the nutrition, health, age, or injury status of the individual the bones were taken from. Preparing fleshed bones for these types of studies can involve maceration—boiling fleshed bones to remove large particles, then hand-cleaning.
Anthropologists and archaeologists also study bone tools made by Homo sapiens and Homo neanderthalensis. Bones can serve a variety of uses, such as projectile points or artistic pigments, and can be made from endoskeletal or external bones such as antler or tusk.
There are several alternatives to mammalary bone seen in nature; though they have some similar functions, they are not completely functionally analogous to bone.
Bone penetrating the skin and being exposed to the outside can be both a natural process in some animals, and due to injury:
Several terms are used to refer to features and components of bones throughout the body:
|articular process||A projection that contacts an adjacent bone.|
|articulation||The region where adjacent bones contact each other—a joint.|
|canal||A long, tunnel-like foramen, usually a passage for notable nerves or blood vessels.|
|condyle||A large, rounded articular process.|
|crest||A prominent ridge.|
|eminence||A relatively small projection or bump.|
|epicondyle||A projection near to a condyle but not part of the joint.|
|facet||A small, flattened articular surface.|
|foramen||An opening through a bone.|
|fossa||A broad, shallow depressed area.|
|fovea||A small pit on the head of a bone.|
|labyrinth||A cavity within a bone.|
|line||A long, thin projection, often with a rough surface. Also known as a ridge.|
|malleolus||One of two specific protuberances of bones in the ankle.|
|meatus||A short canal.|
|process||A relatively large projection or prominent bump.(gen.)|
|ramus||An arm-like branch off the body of a bone.|
|sinus||A cavity within a cranial bone.|
|spine||A relatively long, thin projection or bump.|
|suture||Articulation between cranial bones.|
|trochanter||One of two specific tuberosities located on the femur.|
|tubercle||A projection or bump with a roughened surface, generally smaller than a tuberosity.|
|tuberosity||A projection or bump with a roughened surface.|
Several terms are used to refer to specific features of long bones:
|Diaphysis||The long, relatively straight main body of the bone; region of primary ossification. Also known as the shaft.|
|epiphyses||The end regions of the bone; regions of secondary ossification.|
|epiphyseal plate||The thin disc of hyaline cartilage between the diaphysis and epiphyses; disappears by twenty years of age. Also known as the growth plate.|
|head||The proximal articular end of the bone.|
|neck||The region of bone between the head and the shaft.|
All links retrieved June 15, 2016.
New World Encyclopedia writers and editors rewrote and completed the Wikipedia article in accordance with New World Encyclopedia standards. This article abides by terms of the Creative Commons CC-by-sa 3.0 License (CC-by-sa), which may be used and disseminated with proper attribution. Credit is due under the terms of this license that can reference both the New World Encyclopedia contributors and the selfless volunteer contributors of the Wikimedia Foundation. To cite this article click here for a list of acceptable citing formats.The history of earlier contributions by wikipedians is accessible to researchers here:
The history of this article since it was imported to New World Encyclopedia:<|endoftext|>
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How can acceleration change the motion of an object?
It can change the direction or size of the velocity vector or both.
In most cases of an object which is accelerating the size (or magnitude) of the velocity is changed, so the object speeds up or slows down. The aspect that makes the difference is whether the acceleration is in the same direction as the velocity vector or in the opposite direction. If the acceleration is in the same direction as the velocity the size of the velocity will be increased (if it is in the opposite direction the size of the velocity will be decreased).
If the problem is one-dimensional then the acceleration only affects the size of the velocity vector and the direction will not be changed (so you only have to worry about whether the acceleration is in the same / opposite direction as the velocity). In two or three dimensional problems the size and direction of the velocity vector maybe change. Mathematical methods must be used to find the resultant velocity vector. An example of this would be projectile motion.
In some cases the acceleration is perpendicular to the velocity vector, in these cases the size of the velocity will not be affected. However the acceleration will change the direction of the velocity. A good example of this is circular motion with constant speed. During circular motion the velocity vector is always tangential to the circular path and the acceleration always acts towards the centre of the circle, so they are always perpendicular.
(Hence the speed remains constant because the size of the velocity vector does not change.)<|endoftext|>
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Statistics – Combination with replacement
Each of several possible ways in which a set or number of things can be ordered or arranged is called permutation Combination with replacement in probability is selecting an object from an unordered list multiple times.
Combination with replacement is defined and given by the following probability function:
Formula
nCr=(n+r−1)!r!(n−1)!nCr=(n+r−1)!r!(n−1)!
Where −
· nn = number of items which can be selected.
· rr = number of items which are selected.
· nCrnCr = Unordered list of items or combinations
Example
Problem Statement:
There are five kinds of frozen yogurt: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. What number of varieties will there be?
Solution:
Here n = 5 and r = 4. Substitute the values in formula,
nCr=(n+r−1)!r!(n−1)! =(5+3+1)!3!(5−1)! =7!3!4! =50406×24 =35nCr=(n+r−1)!r!(n−1)! =(5+3+1)!3!(5−1)! =7!3!4! =50406×24 =35<|endoftext|>
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Long gone are the days when a generic ‘Well done!’ or ‘Good!’ would suffice as adequate and constructive teacher feedback for pupils’ writing. Today, feedback given to pupils needs to be specific and clear and should give pupils guidance on how to improve their writing. Feedback is often confused with editing, it is much more than just correcting mistakes, it is showing them the way to make the next step in improving.
Feedback comments must be limited to three or four major suggestions and should focus on positive feedback, as well as something to improve on. Writing targets need to be limited to this amount, too. An overwhelming amount of feedback and targets would prevent pupils from acting on your suggestions. Your feedback should help them understand what is most important for them to work on, even if the end result isn’t quite there yet. Your targets should help them succeed in showing improvement and becoming better writers.
Remember, writing is an ongoing process, not a one-time event. Pupils need to be given opportunities to close the gap between their current performance and the desired outcome.
Use these templates to set writing targets for your pupils. Celebrate their success with a certificate and move on to the next set of targets to keep them developing further!
Download the free Writing Targets Sheet and Writing Target Achievement Certificate by clicking on the images below!
Till next time…<|endoftext|>
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# Angles In Parallel Lines
🏆Practice angles in parallel lines
## Angles on Parallel Lines
If we add a third line that intersects the two parallel lines (those lines that could never cross), we will obtain various types of angles.
To classify these angles we must observe if they are:
above the line - the pink part
below the line - the light blue part
to the right of the line - the red part
to the left of the line - the green part
## Test yourself on angles in parallel lines!
If one of two corresponding angles is a right angle, then the other angle will also be a right angle.
## Corresponding angles
The corresponding angles located between parallel lines are equal.
They are called corresponding angles because:
• they are on the same side of the transversal
• they are on the same "floor" in relation to the line
Here are some examples of corresponding angles:
The two angles marked are on the left side and on the ground floor - they are corresponding and equivalent.
## Vertically Opposite Angles
The vertically opposite angles that are located between parallel lines are equal.
They are called vertically opposite angles because:
• They share the same vertex - they are located on the same vertex
• They are opposite each other
Here are some examples of vertically opposite angles:
The two marked angles are located on the same vertex and are opposite each other, therefore, they are vertically opposite angles.
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The sum of adjacent angles located between parallel lines is equal to $180$.
They are called adjacent angles because:
• They are next to each other
• They are on the same line
Here are some examples of adjacent angles:
The two marked angles are on the same line - (diagonal) - and are next to each other, therefore, they are adjacent angles.
## Alternate Angles
The alternate angles that are located between parallel lines are equal.
They are called alternate angles because:
• they are not on the same side of the transversal
• they are not on the same "level" in relation to the line
Here are some examples of alternate angles:
The two marked angles are on different "levels" and sides, therefore, they are alternate angles.
Do you know what the answer is?
## Collateral angles
The sum of consecutive angles located between parallel lines is equal to $180$
They are called consecutive angles because:
• they are on the same side of the transversal
• but they are not on the same "level" in relation to the line
Here are some examples of consecutive angles:
The two marked angles are on the same side of the line, but at a different height, therefore, they are consecutive angles.
Observe: The angles painted in red in the illustration above are external consecutive angles since they are on the outer side of the parallel lines
The internal consecutive angles are on the inner side of the parallel lines:
## Angles on Parallel Lines
### Now to practice!
Give an example according to the illustration of:
• Alternate angles
• Corresponding angles
• Vertically opposite angles
• Consecutive interior angles
• Consecutive exterior angles
Solution:
• Examples of alternate angles $1, 8$
Both are on different sides and levels, therefore, they are alternate.
• Examples of corresponding angles $8,4$
Both are on the same side and at the same level or floor, therefore, they are corresponding.
• Examples of vertically opposite angles $1,4$
Both share the same vertex and are located opposite each other, therefore, they are vertically opposite angles.
• Examples of adjacent angles $7,8$
Both are on the same line and are located next to each other, therefore, they are adjacent.
• Examples of exterior alternate angles $1,7$
Both are on the same side, but not at the same level. In addition, they are located on the outside of the line, therefore, they are exterior alternate angles.
• Examples of interior alternate angles $3,5$
Both are on the same side, but not at the same level. In addition, they are located on the inside of the line, therefore, they are interior alternate angles.
Another exercise:
What are the marked angles called in the illustration?
Solution
The marked angles are alternate
They are located on different sides and heights, therefore, they are alternate.
Another exercise:
What are the angles shown in the illustration called?
Solution
The indicated angles are consecutive
They are on the same side of the line, but at different heights, therefore, they are external consecutive angles.
Another exercise:
What are the angles shown in the illustration called?
Solution
The indicated angles are adjacent
They are on the same blue line and are next to each other, therefore, they are adjacent angles.
## Examples and exercises with solutions of right angles and parallels
### Exercise #1
Which type of angles are shown in the figure below?
### Step-by-Step Solution
Alternate angles are a pair of angles that can be found on the opposite side of a line that cuts two parallel lines.
Furthermore, these angles are located on the opposite level of the corresponding line that they belong to.
Alternate
### Exercise #2
$a$ is parallel to
$b$
Determine which of the statements is correct.
### Step-by-Step Solution
Let's review the definition of adjacent angles:
Adjacent angles are angles formed where there are two straight lines that intersect. These angles are formed at the point where the intersection occurs, one next to the other, and hence their name.
Now let's review the definition of collateral angles:
Two angles formed when two or more parallel lines are intersected by a third line. The collateral angles are on the same side of the intersecting line and even are at different heights in relation to the parallel line to which they are adjacent.
Therefore, answer C is correct for this definition.
$\beta,\gamma$ Colaterales$\gamma,\delta$ Adjacent
### Exercise #3
Is it possible to have two adjacent angles, one of which is obtuse and the other right?
### Step-by-Step Solution
Remember the definition of adjacent angles:
Adjacent angles always complement each other up to one hundred eighty degrees, that is, their sum is 180 degrees.
This situation is impossible since a right angle equals 90 degrees, an obtuse angle is greater than 90 degrees.
Therefore, together their sum will be greater than 180 degrees.
No
### Exercise #4
In which of the diagrams are the angles $\alpha,\beta\text{ }$ vertically opposite?
### Step-by-Step Solution
Remember the definition of angles opposite by the vertex:
Angles opposite by the vertex are angles whose formation is possible when two lines cross, and they are formed at the point of intersection, one facing the other. The acute angles are equal in size.
The drawing in answer A corresponds to this definition.
### Exercise #5
The lines a and b are parallel.
What are the corresponding angles?
### Step-by-Step Solution
Given that line a is parallel to line b, let's remember the definition of corresponding angles between parallel lines:
Corresponding angles are angles located on the same side of the line that intersects the two parallels and are also situated at the same level with respect to the parallel line to which they are adjacent.
Corresponding angles are equal in size.
According to this definition $\alpha=\beta$and therefore the corresponding angles
$\alpha,\beta$<|endoftext|>
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# Combining vectors
https://arbital.com/p/combining_vectors
by Adele Lopez Dec 3 2016 updated Jan 1 2018
One of the most useful things we can do with vectors is to combine them!
[summary: We can combine vectors to get other vectors, which is called a linear combination of vectors. A set of vectors that can be combined to reach all of the vectors in the vector space exactly once is called a basis. The number of elements in the basis for a vector space is called the dimension of the vector space because it matches our intuitive notion of dimension.
By choosing a basis, we can break down any vector into different components, each one a scaled basis vector. These scalars are called the coordinates of the vector. For convenience, we will usually use a standardized basis with special properties, but other bases are equally valid.]
Now that we know how vectors work, we want to know how to work with them. In this page, we'll study some fundamental concepts in working with vectors. This will make using them and thinking about them easier. We'll start by combining vectors to make new vectors, and we will use this tool to help us describe and understand vector spaces, and to make it easier to talk about different vectors. In particular, we will learn how we can communicate a particular vector algebraically, without needing to share a picture.
# Linear combinations
Here are some vectors:
We can use these vectors to make other vectors, by taking different amounts of each one, adding all the parts together:
This is called a linear combination.
Sometimes, there are different ways to make the same vector from a certain set of ingredient vectors:
Sometimes, there are no ways to make a particular vector. This is easiest to see in three dimensions. In this example, by rotating our point of view, we can see that there is no way of combining $\mathbf u$, $\mathbf v$, and $\mathbf w$ to make the vector $\mathbf s$:
We call the set of all the vectors that are reachable by our set of vectors the span of those vectors. In the example above, the span of $\mathbf u$, $\mathbf v$, and $\mathbf w$ is the set of all vectors that could lie within the plane highlighted in gray.
# Basis
How many vectors do we need to span the whole vector space? It depends on the space!
For a flat space, we can use the vectors $\mathbf x$ and $\mathbf y$ to reach the whole space:
What happens if we remove one or the other? What vectors can we reach with a linear combination of just $\mathbf x$, or just $\mathbf y$?
%%hidden(Show solution):
Linear combinations of just $\mathbf x$ are just $\mathbf x$ scaled by different amounts. So the only vectors that will be reachable are the ones that could lie in the same line as $\mathbf x$. Similarly, with just $\mathbf y$, we can only reach $\mathbf y$ scaled by different amounts with linear combinations. %% So we might guess that we need at least two vectors to span a flat space.
To span a space that also has some depth, we can't just use these two vectors though:
Not even if we just add any other vector in a different direction:
This doesn't add anything new since $\mathbf d$ is a linear combination of $\mathbf x$ and $\mathbf y$!
We say that $\mathbf d$ is linearly dependent on $\mathbf x$ and $\mathbf y$.
To get the full depth of the space, we need to add a vector that is linearly independent of $\mathbf x$ and $\mathbf y$, that is, not a linear combination of $\mathbf x$ and $\mathbf y$.
A linearly independent set of vectors that span the whole space is called a basis for that space.
## Dimension
What does a vector space with just one vector $\mathbf v$ in its basis look like?
Linear combinations of $\mathbf v$ are just $\mathbf v$ scaled by different numbers. So the vector space is just any vector that fits in a line containing $\mathbf v$. A basis with one vector can reach a one-dimensional space.
Now if we have two vectors $\mathbf x$ and $\mathbf y$ in our basis, we can reach any vector within the plane containing both $\mathbf x$ and $\mathbf y$:
So a basis with two vectors spans a two-dimensional space.
You can probably guess that a basis with three vectors reaches a three-dimensional space.
The dimension of a vector space is the number of vectors in the basis!
This allows us to accurately imagine higher dimensional spaces, even though we cannot visualize them directly. Even though four linearly independent vectors do not exist in physical space, the same mathematical rules can be applied to determine different properties of such a world.
In this way, math allows us to explore all kinds of strange, exotic worlds beyond our own, including vector spaces of any dimension. It also turns out that many of these higher dimensional vector spaces are very useful for real world situations!
### Zero dimensions
What is a linear combination of nothing? Just as zero is the amount of things we have if we have nothing, the zero vector is the vector of a linear combination of nothing. So a zero-dimensional vector space is a vector space that has practically nothing - just the zero vector!
## Standard basis
It's convenient to work in a standardized basis, with vectors that have convenient properties. In many spaces, there's a standard set of vectors with particular magnitudes and directions that are conventionally used.
In two dimensional geometric space, the standard basis is $\mathbf x$, $\mathbf y$, where $\mathbf x$ and $\mathbf y$ both have magnitude one.
Similarly, in three dimensional geometric space, a basis with vectors $\mathbf x$, $\mathbf y$, and $\mathbf z$, all unit vectors, is commonly used.
However this one is less standardized, and it is not unusual to see things that use a different convention for three dimensional geometric space.
One property that is very often used is to have a basis where all the basis vectors are normalized to have unit magnitude. This is called a normal basis. %note: It's also useful to have a basis where all the vectors are in as different directions as possible. However, this only makes sense in vector spaces with a concept of angle. We'll learn about angle in geometric vector spaces soon. A basis with this property is called an orthogonal basis (Orthogonal is Greek for right angular). A basis that is both orthogonal and normal is called an orthonormal basis. %
Although using a standardized basis is convenient, it's important to remember that standardized bases are not mathematically fundamental. The math works equally well when using any basis. For example, we could also use $\mathbf r$ and $\mathbf s$ as a basis for the plane of your screen:
# Coordinates
If we have a basis, we can describe any vector as a linear combination of those basis elements in exactly one way. Here we have $\mathbf x$ and $\mathbf y$ as our basis, and we want to describe the vector $\mathbf v$:
We can write $\mathbf v = 3\mathbf {x} + 4 \mathbf {y}$. The scaled basis vectors we use to reach our vector $v$ are called the components of $v$. In this example, the $\mathbf x$-component of $\mathbf v$ is $3$, and the $\mathbf y$-component of $\mathbf v$ is $-1$. We can also refer to this vector by the list of the scalars for each of the components – these are called the coordinates of the vector. So the coordinates of $\mathbf v$ are $(3,4)$ in this case. %note: We also need to know the order of the basis vectors in order to use coordinates unambiguously. In this case, the implicit order is x, y.%
It's important to remember that if we use a different basis, the same vector will have different coordinates:
Coordinates give us a handy way to describe different vectors in our vector space:
This lets us describe specific vectors in an algebraic way, as long as the basis in use is understood. So we can share a particular vector with our friend by sharing the coordinates of the vector using the standard basis, all without needing to draw a picture!
## Reference point
We can do a similar thing with points. We choose a particular point as a reference point, called the origin, which we usually write with a capital $O$:
We can now describe points by the vector will take us from the origin to that point. And that vector can be described in terms of the basis vectors, using coordinates!
In this example, $p = O + 2\mathbf x + 3\mathbf y$, and $q = O - 3\mathbf x + \mathbf y$. We call these the coordinates of the point. For short, we can write these just as $p = (2, 3)$ and $q = (-3,1)$ – but it's really important to make sure that we are clear on what basis is being used, and what the origin is. In the $\mathbf s, \mathbf t$ basis we used earlier, and with the same origin, our points will have different coordinates:
This time, $p$ can be described as $(2,\frac{1}{2})$, and $q = (-3,2)$.
Next, we will start exploring the things that make geometric vectors geometric, by thinking about how we can compare different vectors to each other in a geometric vector space.<|endoftext|>
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The Iron Age is the last of the Three Ages of British later prehistory. It begins with the arrival of the new metal, iron, around 800 BC and ends with Roman troops landing on the shores of Kent, in AD 43. The Romans gave the British writing and with writing came recorded history – which is why prehistory is said to cease with their arrival. People in Iron Age Britain are sometimes described as Celts and they spoke Celtic languages, which survive today in Breton, Welsh, Gallic (Scotland) and Gaelic (Ireland). The working of iron requires greater control of very high temperatures which led to improvements in pottery firing and less regionalised pottery styles. The Iron Age saw the appearance of ditched enclosed farmstead-type settlements as at Itter Crescent, open settlements characterised by roundhouses and pits as at Fengate, and the building of the hillforts like the earthworks at Newborough. Societies were hierarchically organised in this period, having moved from the extended clan to the chiefdoms and the earliest named rulers. These are the tribes the Romans encountered when they came to Britain in the first century. The best known of these rulers was Queen Boudicca of the Iceni tribe/kingdom. She led a popular rebellion against Roman rule, in AD 60-1.
Environmentally, the Iron Age sees increased flooding and higher groundwater levels in the fens.<|endoftext|>
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Question
# y’’ -4y’ +4y = $12x^2 -6x$e^2x Y$0$= 1 Y’$0$=0 Y$x$=c1y1+c2y2+yp
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## Answer to a math question y’’ -4y’ +4y = $12x^2 -6x$e^2x Y$0$= 1 Y’$0$=0 Y$x$=c1y1+c2y2+yp
Corbin
4.6
To solve the given second-order linear homogeneous ordinary differential equation:
y'' - 4y' + 4y = $12x^2 - 6x$e^{2x}
We first find the complementary function $CF$ by solving the associated homogeneous equation:
y'' - 4y' + 4y = 0
The auxiliary equation is obtained by substituting y = e^{rx} into the homogeneous equation:
r^2e^{rx} - 4re^{rx} + 4e^{rx} = 0
Factoring out e^{rx}:
e^{rx}$r^2 - 4r + 4$ = 0
r^2 - 4r + 4 = $r-2$^2 = 0
This implies a repeated root at r = 2, so the complementary function $CF$ is:
y_{CF} = $c_1 + c_2x$e^{2x}
To find the particular integral $PI$, we use the method of undetermined coefficients. Let's assume the particular solution is of the form:
y_{PI} = Ax^2e^{2x} + Bxe^{2x}
Now, let's find the first and second derivatives of y_{PI}:
y'_{PI} = $2Ax^2e^{2x} + 2Axe^{2x}$ + $Be^{2x} + 2Bxe^{2x}$
y''_{PI} = $4Ax^2e^{2x} + 8Axe^{2x} + 2Ae^{2x}$ + $2Be^{2x} + 4Be^{2x} + 2Bxe^{2x}$
Substituting these derivatives into the original differential equation and simplifying, we get:
$4Ax^2 + 12Bx + 12A - 6B$e^{2x} = $12x^2 - 6x$e^{2x}
Comparing coefficients, we have:
4Ax^2 + 12Bx + 12A - 6B = 12x^2 - 6x
Equating the coefficients of like powers of x:
4A = 12 \quad \text{$coefficient of }x^2$
12B + 12A - 6B = -6 \quad \text{$coefficient of }x$
Solving the equations, we find:
A = 3
B = -1
Therefore, the particular solution $PI$ is:
y_{PI} = 3x^2e^{2x} - xe^{2x}
The general solution $GS$ is the sum of the complementary function $CF$ and the particular integral $PI$:
y_{GS} = y_{CF} + y_{PI} = $c_1 + c_2x$e^{2x} + 3x^2e^{2x} - xe^{2x}
Using the initial conditions, we can find the values of c_1 and c_2. Given: y$0$ = 1 and y'$0$ = 0.
Substituting x = 0 and y = 1 into the general solution $GS$:
y_{GS}$0$ = $c_1 + c_2 \cdot 0$e^{2 \cdot 0} + 3 \cdot 0^2 e^{2 \cdot 0} - 0 \cdot e^{2 \cdot 0} = c_1 = 1
Substituting x = 0 and y' = 0 into the general solution $GS$:
y'_{GS}$0$ = $c_2$e^{2 \cdot 0} + 0 - 1 \cdot e^{2 \cdot 0} = c_2 - 1 = 0
Solving for c_2, we get:
c_2 = 1
Therefore, the solution to the differential equation is:
y$x$ = $1 + x$e^{2x} + 3x^2e^{2x} - xe^{2x}
Answer: y$x$ = $1 + x + 3x^2 - x$e^{2x} = $1 + 2x + 3x^2$e^{2x}
Frequently asked questions $FAQs$
Math question: What is the period of the sine function f$x$ = sin$x$?
+
What is the indefinite integral of sin$x$?
+
What is the maximum value of the function f$x$ = 2x^2 - 3x + 4 in the interval [-1, 3]?
+<|endoftext|>
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Having always been intrigued by historical geology with its eras of hundreds of millions of years, I was interested to know to what extent it could be, and has been, tested by experiment. Some years ago, I renewed my studies on the subject, paying particular attention to the principles of stratigraphy which are used to obtain a relative chronology from superposition of strata.
First some definitions. A stratum is a single layer of homogeneous or gradational lithology deposited parallel to the original dip of the formation. It is separated from adjacent strata or cross-strata by surfaces of erosion, non-deposition, or abrupt change in character. The term stratum includes "bed" and "lamination" which strata vary in thickness from less than a millimeter for micro-strata or laminae to more than a meter. A stratum often shows evidence of sorting of the particles of which it is composed, with the size decreasing from bottom to top of the stratum.
A facies is a recognizable series of superposed strata.
Superposed strata and, on a larger scale, superposed facies, are considered as successive layers or isochronously-deposited sediments. That is to say, the lower one formed first, the one above it second, and so on. The principles of stratigraphy arose from this belief of strata and facies being successive layers.
In 1667, Nicholas Steno, a naturalist from Tuscany, wrote in his book, Canis Carchariae, that "the layers of subsoil are strata of ancient successive sediments." From this assertion he deduced the principles of stratigraphy defined in his 1669 book, Prodromus. These include:
A. The principle of superposition:
". . . at the time when any given stratum was being formed, all the matter resting upon it was fluid, and, therefore, at the time when the lower stratum was being formed, none of the upper strata existed."
B. The principle of initial horizontality:
"Strata either perpendicular to the horizon or inclined to the horizon were at one time parallel to the horizon."
C. The principle of strata continuity:
"Material forming any stratum were continuous over the surface of the Earth unless some other solid bodies stood in the way."
Implication of these definitions
De Luc, at the beginning of the nineteenth century, and later Brongniart, considered that if the principle of superposition indicated a time sequence, the difference in fossil content in superposed strata would show a change of species in time. Subsequently, the presence of index fossils in superposed strata was used as evidence of a succession of time periods. Later, these principles of stratigraphy and biostratigraphy provided the basis upon which nineteenth century geologists established the geological column.
Examination of existing data
At the end of the nineteenth century, Johannes Walther1 studied the formation of contemporaneous sedimentary deposits in a delta which prograded, or developed from the coast towards the open sea. He observed the same succession of facies from the surface downwards, as from the coast towards the sea. This was clear evidence that facies in sequences, when superposed and juxtaposed at the same time, do not always follow the principles of superposition and continuity.
In the 1970s and 80s, Glomar Challenger borings into the Pacific Ocean bottom showed that Walther's discovery of superposed and juxtaposed facies also applied to deep sea sediments. Walther's observations can in fact be shown to be a natural consequence of sedimentary mechanics.
American geologist Edwin McKee2 reported his observations of sediments deposited in 1965 when Bijou Creek in Colorado overflowed its banks due to 48 hours of torrential rain. The stratified deposits, reaching a thickness of 12 feet, exhibited particle sorting and bedding planes.
Bedding planes are generally interpreted as the result of interruptions in sedimentation, with hardening of the surface of the last-formed layer prior to further sedimentation. It was impossible by simple observation to tell whether the Bijou Creek strata had formed successively one on top of the other or were co-temporaneous. In any case, 48 hours gave no time for the surface of any stratum to harden before subsequent sediment was deposited on top of it. The partings had to be due to some other mechanism.
My Own Experiments
Searching the geological and sedimentological literature, I was surprised to find that there was little experimental data on the formation of strata. It had apparently been assumed that all strata form in accordance with the principles of stratigraphy. My objective was, therefore, to test by experiment this assumed mechanism for strata building.
Examination of conditions under which strata form
I started by examining how sedimentary particles are deposited in both dry and wet conditions. Sand particles of differing size produced micro-strata or laminae when poured into a flask. The micro-strata formed from the particles of sand sorting themselves out according to size, with the larger particles at the bottom grading up to the smaller ones at the top. The process repeated itself, producing multiple laminae.
This was fundamental because it showed that micro-strata or laminae formed from particle sorting, irrespective of the speed of sedimentation, and not by one layer forming first and then the next one forming on top of it. My results were published by the French Academy of Science,3,4 which encouraged me to continue my experiments, but on a much larger scale.
Experiments in the Colorado State University
Large scale experiments required the resources of a laboratory with the latest technology. Having read their reports on sedimentology, I contacted Colorado State University in the USA. This led to a series of experiments conducted in their modern hydraulics laboratory at Fort Collins. Pierre Julien, a sedimentologist, was in charge of the experiments. They took place in large glass-walled flumes, which allowed observation and filming from above and through the sides of the tanks.
Different sized particles of sand were poured into water circulating in the flume. Variations in current velocity caused the particles to be sorted according to size. At 1 m/s superposed laminae formed laterally in the direction of the current. A reduction of velocity to 0.5 m/s caused larger particles to collect on the previous laminae, always migrating in the direction of the current. An increase in velocity back to 1 m/s caused laminae similar to the previous ones to form, mainly due to friction, on top of the stratum of larger particles. The accumulation of sediment produced a deposit consisting of the downstream part of the lower laminae, part of the sloping stratum of larger particles, and the upstream part of the upper laminae. Each individual deposit formed successively downstream and was therefore younger than the one before it. Variations in current velocity, as found in rivers and oceans, could thus cause deposits to form both vertically and laterally at the same time in the direction of the current.
The flume experiments further demonstrated the mechanical nature of stratification, whereby: (1) Particles segregated according to their size when transported by a current of variable velocity; (2) Desiccation, or drying out, of deposits caused bedding partings; (3) Stratification of the deposit, under both dry and wet conditions, formed parallel to the slope of the deposit, which could exceed 30°.5
It was discovered that where there is a current:
1. Strata can form laterally and vertically at the same time;
2. Strata can form in the same way as sequences of facies;
3. Strata are not always a measure of chronology.
These highlighted experimental facts show clearly:
a. Superposed strata do not always result, according to Steno's beliefs, from successive layers of sediment; consequently the principle of superposition does not always apply to strata formed in a current;
b. Stratification formed parallel to a slope exceeding an angle of 30°, can invalidate the principle of original horizontally. Inclined strata are not necessarily, therefore, the result of subsidence or uplift.
Correspondence between experimental results and geological formations
The experiments demonstrated that current deposited strata can form in the same way as sequences of facies. This is consistent with sequence stratigraphy. The experiments show that bedding planes, considered as resulting from interruptions of sedimentation, can result from desiccation of sediments. Moreover, recent submarine observations such as Rubin,6 the flume experiments summarized by Southard,7 and river studies initiated by Hjulström8 and developed by several other scientists, have shown the relationships between contemporaneous hydraulic conditions and sedimentary structures, particularly between critical speed of sedimentation and particular size. Such relationships correspond to those measured in our experiments. These relationships can be used to determine the minimum paleohydraulic conditions (velocity of current, depth of water, discharge and speed of accumulation of sediments) from sedimentary rock structures.
1 Walther J., 1893-1894, Einleitung in die Geologie als historische Wissenschaft: Jena Verlag von Gustav Fisher, Sud. 1055p.
2 McKee, E.D., Crosby, E.J. & Berryhill, H.L. Jr. 1967, Flood deposits, Bijou Creek, Colorado, 1965, Journal of Sedimentary Petrology, 37, 829-851.
3 Berthault G. 1986, Sedimentology—experiments on lamination of sediments, C.R. Acad. Sc. Paris, 303 II, 17, 1569-1574.
4 Berthault G. 1988, Sedimentation of heterogranular mixture—experimental lamination in still and running water, C.R. Acad. Sc. Paris, 306, II, 717-724.
5 Julien P, Lany, Berthault G., 1993, Experiments on stratification of heterogeneous sand mixtures, Bulletin of the Geological Society, France, 164-5, 649-660.
6 Rubin D.M. and McCulloch D.S. 1980, Single and superposed bedforms: a synthesis of San Francisco Bay and flume observations, Journal of Sedimentary Petrology, 26:207-231.
7 Southard J. and Boguchwal J.A. 1990, Bed configuration in steady unidirectional waterflows, part 2, Synthesis of flume data, Journal of Sedimentary Petrology 60(5) : 658-679.
8 Hjulström F. 1935, The morphological activity of rivers as illustrated by river fyris, Bulletin of the Geological Institute Uppsala, 25, chapter 3.
For further study, see the video describing the Berthault and Julien investigation "Experiments in Stratification," 1999.
* Guy Berthault is an independant researcher from France.
Cite this article: Berthault, G. 2000. Experiments in Stratification. Acts & Facts. 29 (10).<|endoftext|>
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Last updated on Apr 5, 2024
Solve these Decimal MCQs Quiz and improve your speed and accuracy of solving questions. ‘Decimals’ is one of the most commonly appearing topics in competitive exams and candidates must know how to stay on their toes while solving these questions. Decimal objective questions are actually an opportunity for candidates to save time for more difficult questions in their exams. Check out the explanations and tips to solve Decimal question answers with the least effort in the least amount of time.
## Latest Decimals MCQ Objective Questions
#### Decimals Question 1:
0.123 + 12.3 + 0.0123 + 1.23 = ?
1. 13.776
2. 13.6653
3. 13.6563
4. 13.5663
Option 2 : 13.6653
#### Decimals Question 1 Detailed Solution
Given Data:
First number = 0.123
Second number = 12.3
Third number = 0.0123
Fourth number = 1.23
Concept:
Sum up all given numbers to get the total.
Solution:
⇒ Total = 0.123 + 12.3 + 0.0123 + 1.23
⇒ Total = 13.6653
Therefore, the sum of the numbers is 13.6653.
#### Decimals Question 2:
The value of
$$\frac{{5.169 \times 5.169 \times 5.169 - 64 \times {{\left( {0.888} \right)}^3}}}{{{{\left( {5.169} \right)}^2} + {{\left( {3.552} \right)}^2} + 4\left( {0.888} \right)\left( {5.169} \right)}}$$is (1 + k), where k = _____.
1. 0.617
2. 0.623
3. 0.317
4. More than one of the above
5. None of the above
Option 1 : 0.617
#### Decimals Question 2 Detailed Solution
Given:
$$\frac{{5.169 \times 5.169 \times 5.169 - 64 \times {{\left( {0.888} \right)}^3}}}{{{{\left( {5.169} \right)}^2} + {{\left( {3.552} \right)}^2} + 4\left( {0.888} \right)\left( {5.169} \right)}} = 1 + k$$
Formula used:
(a3 - b3) = (a - b)(a2 + ab + b2) ---- (1)
Calculations:
⇒ $$\frac{{5.169 \times 5.169 \times 5.169 - 64 \times {{\left( {0.888} \right)}^3}}}{{{{\left( {5.169} \right)}^2} + {{\left( {3.552} \right)}^2} + 4\left( {0.888} \right)\left( {5.169} \right)}} = 1 + k$$
⇒ $$\frac{{5.169^3 - (4 \times {{\left( {0.888}) \right)}^3}}}{{{{\left( {5.169} \right)}^2} + {{\left( {3.552} \right)}^2} + 4\left( {0.888} \right)\left( {5.169} \right)}} = 1 + k$$
Using equation (1), we get
⇒ $$\frac{(5.169 - 3.552)((5.169)^2 + (3.552)(5.169) +(3.552)^2)}{(5.169)^2 + (3.552)(5.169) +(3.552)^2)} = 1 + k$$
⇒ 5.169 - 3.552 = 1 + k
⇒ 1.617 = 1 + k
⇒ k = 0.617
∴ The value of k is 0.617.
#### Decimals Question 3:
Simplify: (0.3 × 0.3 + 0.01) × (0.1 × 0.1 + 0.09)
1. 1/100
2. 1/10
3. 1
4. 9
Option 1 : 1/100
#### Decimals Question 3 Detailed Solution
Given:
(0.3 × 0.3 + 0.01) × (0.1 × 0.1 + 0.09)
Calculation:
The expression is = (0.3 × 0.3 + 0.01) × (0.1 × 0.1 + 0.09)
⇒ (0.3 × 0.3 + 0.01) × (0.1 × 0.1 + 0.09) = (0.09 + 0.01) x (0.01 + 0.09)
⇒ (0.3 × 0.3 + 0.01) × (0.1 × 0.1 + 0.09) = (0.10) x (0.10)
⇒ (0.3 × 0.3 + 0.01) × (0.1 × 0.1 + 0.09) = 0.01 = (1/100)
∴ The value of (0.3 × 0.3 + 0.01) × (0.1 × 0.1 + 0.09) is 1/100.
#### Decimals Question 4:
Find the value of the following expression:
$$\frac{(7.03)^{3}-0.027}{(7.03)^{2}+2.109+(0.3)^{2}}$$
1. 7.06
2. 6.73
3. 7
4. 7.33
5. Not Attempted
Option 2 : 6.73
#### Decimals Question 4 Detailed Solution
Givan:
( 7.03 )3 - 0.027 / ( 7.03 )2 +2.109 + ( 0.3 )2
Formula used:
( a3 - b3 ) = (a - b ) ( a2 + ab + b2
Calculation:
According to the question
⇒ ( 7.03 )3 - ( 0.3 )3 /( 7.03 )2 + 2.109 + ( 0.3 )2
According to the formula
⇒ ( 7.03 - 0.3 ) { ( 7.03 )2 + 7.03 × 0.3 + ( 0.3 )2 } / ( 7.03 )2 + 2.109 +( 0.3 )2
⇒ (7.03 - 0.3 ) = 6.73
∴ ( 7.03)3 - 0.027 / ( 7.03 )2 + 2.109 + ( 0.3 ) is 6.73.
#### Decimals Question 5:
What is the value of x, if [0.9 {1.1 - 0.3(0.3 - 0.2 ÷ 0.5) + 0.4}] ÷ 0.81 = x+1 ?
1. -0.8
2. 0.7
3. 1
4. -0.5
5. Not Attempted
Option 2 : 0.7
#### Decimals Question 5 Detailed Solution
Given:
[0.9 {1.1 - 0.3(0.3 - 0.2 ÷ 0.5) + 0.4}] ÷ 0.81 = x + 1
Concept used:
Calculation:
[0.9 {1.1 - 0.3(0.3 - 0.2 ÷ 0.5) + 0.4}] ÷ 0.81
⇒ [0.9 {1.1 - 0.3(0.3 - 0.4) + 0.4}] ÷ 0.81
⇒ [0.9 {1.1 + 0.03 + 0.4}] ÷ 0.81
⇒ [0.9 × 1.53] ÷ 0.81
⇒ 1.377 ÷ 0.81
⇒ 1.7
Here 1.7 = x + 1
So, X = 1.7 - 1 = 0.7
## Top Decimals MCQ Objective Questions
#### Decimals Question 6
On simplification $$\sqrt {{{\left( {0.65} \right)}^2} - {{\left( {0.16} \right)}^2}}$$ reduces to
1. 0.63
2. 0.65
3. 0.54
4. None of these
Option 1 : 0.63
#### Decimals Question 6 Detailed Solution
$$\sqrt {{{\left( {0.65} \right)}^2} - {{\left( {0.16} \right)}^2}}$$
Since,
a2 - b2 = (a - b) ( a + b)
$$\begin{array}{l} \Rightarrow \sqrt {\left( {0.65 + 0.16} \right)\left( {0.65 - 0.16} \right)} \\ \Rightarrow \sqrt {\left( {0.81} \right)\left( {0.49} \right)} \\ \Rightarrow \sqrt {\left( {0.9} \right)\left( {0.9} \right) \times \left( {0.7} \right)\left( {0.7} \right)} \end{array}$$
⇒ 0.9 × 0.7 = 0.63
#### Decimals Question 7
Solve:
(81.84 + 118.16) ÷ 53 = 1.2 × 2 + ?
1. 0.8
2. -0.8
3. 0.6
4. -0.6
Option 2 : -0.8
#### Decimals Question 7 Detailed Solution
Given expression,
(81.84 + 118.16) ÷ 53 = 1.2 × 2 + ?
⇒ 200 ÷ 53 = 1.2 × 2 + ?
⇒ 200 ÷ 125 = 1.2 × 2 +?
⇒ 1.6 = 2.4 + ?
⇒ ? = -0.8
#### Decimals Question 8
Solve:
(?)3 × 10 = 13230 ÷ (9.261 ÷ 7)
1. 10
2. 25
3. 125
4. 0.5
Option 1 : 10
#### Decimals Question 8 Detailed Solution
Concept used:
Given expression is,
⇒ (?)3 × 10 = 13230 ÷ (9.261 ÷ 7)
⇒ (?)3 × 10 = 13230 ÷ (1.323)
⇒ (?)3 × 10 = 10000
⇒ (?)3 = 1000
⇒ (?)3 = (10)3
⇒ ? = 10
#### Decimals Question 9
What approximate value should come in place of question mark ‘?’ in the following question?
1132.757 – 2315.996 – 1753.829 + 2 × 2846.639 = ?
1. 2746
2. 2757
3. 2656
4. 2646
5. None of these
Option 2 : 2757
#### Decimals Question 9 Detailed Solution
1132.757 – 2315.996 – 1753.829 + 2 × 2846.639
= 1133 – 2316 – 1754 + 5694
= (1133 + 5694) – (2316 + 1754 )
= 6827 – 4070
= 2757
#### Decimals Question 10
$$\frac{6.75~\times ~6.75~\times ~6.75-4.25~\times ~4.25~\times ~4.25}{67.5~\times ~67.5~+~42.5~\times ~42.5~+~67.5~\times ~42.5}=?$$
1. 2.5
2. 0.25
3. 0.0025
4. 0.025
Option 4 : 0.025
#### Decimals Question 10 Detailed Solution
Mistake Points
In this type of question, in the numerator, there are 6.75 and 4.25, while in the denominator there is 67.5, and 42.5 is present in multiplication.
From the denominator, we can write 67.5 × 67.5 as 6.75 × 10 × 6.75 × 10 = 100 × (6.75), and similarly for other terms.
Therefore we take 100 as common from the denominator.
Identity used:
a3 - b3 = (a - b) (a2 + b2 + ab)
Calculation:
$$\Rightarrow \text{ }\!\!~\!\!\text{ }\frac{6.75~\times ~6.75~\times ~6.75-4.25~\times ~4.25~\times ~4.25}{67.5~\times ~67.5~+~42.5~\times ~42.5~+~67.5~\times ~42.5}$$
$$\Rightarrow {\rm{}}\frac{{\left[ {\left( {6.75{\rm{\;}}-{\rm{\;}}4.25} \right)\left( {6.75{\rm{\;}} \times {\rm{\;}}6.75{\rm{\;}} + {\rm{\;}}4.25{\rm{\;}} \times {\rm{\;}}4.25{\rm{\;}} + {\rm{\;}}6.75{\rm{\;}} \times {\rm{\;}}4.25} \right)} \right]}}{{[100{\rm{\;}} \times {\rm{\;}}\left( {6.75{\rm{\;}} \times {\rm{\;}}6.75{\rm{\;}} + {\rm{\;}}4.25{\rm{\;}} \times {\rm{\;}}4.25{\rm{\;}} + {\rm{\;}}6.75{\rm{\;}} \times {\rm{\;}}4.25} \right)}}$$
⇒ ? = 2.5/100
⇒ ? = 0.025
∴ ? = 0.025
#### Decimals Question 11
What should come in place of question mark ‘?’ in the following question?
0.36 of 52.5 + ? of 35 = 57.4
1. 1.01
2. 1.21
3. 1.16
4. 1.10
5. None of these
Option 4 : 1.10
#### Decimals Question 11 Detailed Solution
Calculations:
0.36 of 52.5 + ? of 35 = 57.4
⇒ 18.9 + ? of 35 = 57.4
⇒ ? of 35 = (57.4 − 18.9)
⇒ ? = 38.5/35
⇒ ? = 1.1
#### Decimals Question 12
Find the value of {(.98)3 + (0.02)3 + 3 × 0.98 × 0.02 – 1}
1. 1.98
2. 1.09
3. 1.562
4. 0
Option 4 : 0
#### Decimals Question 12 Detailed Solution
Given:
{(.98)3 + (0.02)3 + 3 × 0.98 × 0.02 – 1}
Concept used:
(a + b)3 = a3 + b3 + 3ab(a + b)
Calculation:
{(.98)3 + (0.02)3 + 3 × 0.98 × 0.02 – 1
⇒ [(.98)3 + (0.02)3 + 3 × 0.98 × 0.02(0.98 + 0.02)} – 1]
Now, first term is of the form (a + b)3 where a = .98 and b = 0.02
⇒ (0.98 + 0.02)3 – 1 = 1 – 1 = 0
∴ The value is 0.
#### Decimals Question 13
Correct expression of $$0.0\overline {18} = ?$$
1. 1/55
2. 18/100
3. 18/1000
4. 1/66
Option 1 : 1/55
#### Decimals Question 13 Detailed Solution
Calculation:
Let x = 0.0181818....
⇒ 10x = 0.1818.... ----eq (1)
⇒ 1000x = 18.1818.... ----eq (2)
Now, eq (2) - eq (1)
⇒ 1000x - 10x = 18.1818... - 0.1818...
⇒ 990x = 18
⇒ x = 18/990
⇒ x = 1/55
∴ The fraction of $$0.0\overline {18}$$ is 1/55
Shortcut Trick
$$0.\overline {ab} = ab/99$$
ab¯" role="presentation" style="display: inline; position: relative;" tabindex="0">
⇒ ab¯" role="presentation" style="display: inline; position: relative;" tabindex="0">$$0.0\overline {18} = 18/990$$
⇒ 2/110 = 1/55
∴ The fraction of $$0.0\overline {18}$$ is 1/55.
ab¯" role="presentation" style="display: inline; position: relative;" tabindex="0">ab" id="MathJax-Element-5-Frame" role="presentation" style="position: relative;" tabindex="0">ab = ab/99
#### Decimals Question 14
Simplify:
(9.6 × 3.6 ÷ 7.2 + 10.8 of $$\frac{1}{{18}}$$ - $$\frac{1}{{10}}$$).
1. 5.2
2. 5.4
3. 5.3
4. 5
Option 3 : 5.3
#### Decimals Question 14 Detailed Solution
Concept used:
BODMAS
Calculation:
(9.6 × 3.6 ÷ 7.2 + 10.8 of $$\frac{1}{{18}}$$ - $$\frac{1}{{10}}$$)
= (9.6 × 3.6 ÷ 7.2 + 0.6 - $$\frac{1}{{10}}$$)
= (9.6 × $$\frac{1}{2}$$ + 0.6 - $$\frac{1}{{10}}$$)
= (4.8 + 0.6 - $$\frac{1}{{10}}$$)
$$\frac{48\ +\ 6\ - \ 1}{10}$$
$$\frac{53}{10}$$
= 5.3
#### Decimals Question 15
The value of $$1.\overline{23}-0.\overline{86}+0.\overline{29}$$ is equal to:
1. $$0.65\overline{65}$$
2. $$0.\overline{65}$$
3. $$0.6\overline{56}$$
4. $$0.6\overline{65}$$
Option 2 : $$0.\overline{65}$$
#### Decimals Question 15 Detailed Solution
Shortcut TrickFormula used:
$$a.\overline {bc} = \frac{abc - 1}{99}$$
$$0.\overline {ab} = \frac{ab}{99}$$
Calculation:
$$1.\overline{23}-0.\overline{86}+0.\overline{29}$$
By using the above formula
⇒ $$\frac{123-1}{99}-\frac{86}{99}+\frac{29}{99}$$
⇒ $$\frac{122-86+29}{99} = \frac{65}{99}$$
∴ $$1.\overline{23}-0.\overline{86}+0.\overline{29}$$ = $$0.\overline{65}$$
Alternate Method
Calculation:
Let x = $$1.\overline{23}$$ ----eq.1
⇒ 100x = $$123.\overline{23}$$ ----eq.2
By subtracting eq.1 from eq.2:
99x = 122
⇒ x = 122/99
Let y = $$0.\overline{86}$$ ----eq.3
⇒ 100y = $$86.\overline{86}$$ ----eq.4
By subtracting eq.3 from eq.4:
99y = 86
⇒ y = 86/99
Let z = $$0.\overline{29}$$ ----eq.5
⇒ 100z = $$29.\overline{29}$$ ----eq.6
By subtracting eq.5 from eq.6:
99z = 29
⇒ z = 29/99
$$1.\overline{23}-0.\overline{86}+0.\overline{29}$$ = x - y + z
⇒ (122/99) - (86/99) + (29/99) = 65/99
∴ $$1.\overline{23}-0.\overline{86}+0.\overline{29}$$ = $$0.\overline{65}$$<|endoftext|>
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Algebra II : Simplifying Expressions
Example Questions
← Previous 1 3 4 5 6 7 8
Example Question #1 : Simplifying Expressions
Simplify x(4 – x) – x(3 – x).
1
x2
3x
0
x
x
Explanation:
You must multiply out the first set of parenthesis (distribute) and you get 4x – x2. Then multiply out the second set and you get –3x + x2. Combine like terms and you get x.
x(4 – x) – x(3 – x)
4x – x2 – x(3 – x)
4x – x2 – (3x – x2)
4x – x2 – 3x + x2 = x
Example Question #1 : How To Divide Trinomials
Divide:
Explanation:
Factor the numerator and denominator:
Cancel the factors that appear in both the numerator and the denominator:
Example Question #2 : Simplifying Expressions
Simplify:
Explanation:
and cancel out, leaving in the numerator. 5 and 25 cancel out, leaving 5 in the denominator
Example Question #4 : How To Divide Monomial Quotients
Simplify the following:
Explanation:
First, let us factor the numerator:
Example Question #3 : Simplifying Expressions
Find the product:
Explanation:
First, mulitply the mononomial by the first term of the polynomial:
Second, multiply the monomial by the second term of the polynomial:
Example Question #1 : Simplifying Expressions
Multiply, expressing the product in simplest form:
Explanation:
Cross-cancel the coefficients by dividing both 15 and 25 by 5, and both 14 and 21 by 7:
Now use the quotient rule on the variables by subtracting exponents:
Example Question #5 : Simplifying Expressions
Simplify the following:
Explanation:
In this problem, you have two fractions being multiplied. You can first simplify the coefficients in the numerators and denominators. You can divide and cancel the 2 and 14 each by 2, and the 3 and 15 each by 3:
You can multiply the two numerators and two denominators, keeping in mind that when multiplying like variables with exponents, you simplify by adding the exponents together:
Any variables that are both in the numerator and denominator can be simplified by subtracting the numerator's exponent by the denominator's exponent. If you end up with a negative exponent in the numerator, you can move the variable to the denominator to keep the exponent positive:
Example Question #5 : Simplifying Expressions
Factor the expression:
Explanation:
To find the greatest common factor, we need to break each term into its prime factors:
Looking at which terms all three expressions have in common; thus, the GCF is . We then factor this out:
Example Question #4 : Simplifying Expressions
Expand:
Explanation:
To expand, multiply 8x by both terms in the expression (3x + 7).
8x multiplied by 3x is 24x².
8x multiplied by 7 is 56x.
Therefore, 8x(3x + 7) = 24x² + 56x.
Example Question #1 : How To Multiply Binomials With The Distributive Property
Simplify:
None of the other answers are correct.
Explanation:
First, distribute –5 through the parentheses by multiplying both terms by –5.
Then, combine the like-termed variables (–5x and –3x).
← Previous 1 3 4 5 6 7 8<|endoftext|>
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Picture Cube Puzzle
What is a Picture Cube Puzzle? Six Puzzles Turning the Cubes Similar Puzzles Comments Picture Cube Puzzle on the Internet
To the Main Page "Mathematische Basteleien"
What is a Picture Cube Puzzle?
...... The picture cube puzzle is a puzzle of - in this case - nine cubes. Every cube has parts of a picture on its sides. The nine cubes form a picture if you lay them properly side by side. It is special that you find five more pictures, if you turn over the cubes systematically.
Six Puzzles top
When you buy this puzzle, you often get six sheets of paper as templates of the six puzzles.
A B C
D E F
Turning the Cubes top
After you have solved one puzzle, you can easily get the remaining pictures.
...... In order to descibe the way to the pictures, I call them A, B, C, D, E, and F. The parts of one picture may have the names 1 to 9. Every cube has the same order. The names A to F stand on the sides and a number.
...... Thus it is sufficient to look at one cube only.
...... You see more if you draw a net and show all the sides of the cube. You can see that you bring the pictures B and C on top by a quarter-turn (from the front side to the top side), starting with A above. E doesn't fit in this sequence. Starting again with A you can get picture F by a turn from the top side to the right side.
...... If you say "turn" then you mean all nine cubes. You turn three cubes side by-side at the same time. Here you can see how to turn it from the front to the top side for the first row. The other rows will follow. If you want to turn the columns, you grasp three cubes one below the other in the same way. Each cube stays on its place inside a mat.
...... If you turn the cube so that F is at the top and you draw the net, you see: You get the pictures D and E by quarter turns. B doesn't fit in this sequence.
There are always three pictures, which you get by turning from the front to the top side. You change from one three-sequence-picture to the next one by A/F.
You pass all pictures along the sequence A-B-C-E(mixed)-A----F-B(mixed)-D-E-F.
This is a clever order of the pictures on the cubes.
This no-name-3x3-puzzle is from China and doesn't have flowers as pictures, but ugly, humanized animals. I plan to replace them by my photos.
If you want to know the names of the flowers from my garden, here they are:
Elecampane, Mittagsblume, Clematis, Geranium, Ringelblume, and Crocus.
German: Alant, Mittagsblume, Klematis, Geranium, Ringelblume und Krokus.
Similar Puzzles top
Ravensburger Puzzle
This is the name of the puzzles of the Ravensburger Spieleverlag for the good old cube puzzle.
The motives are nice and tasteful. They are made of plastic material and carefully fabricated. They aren't cheap.
There are different sizes like 3x2 or 4x3.
It isn't possible to bring all pictures to the top only by turning. You must change rows or columns.
If you have found a picture, you have a second one on the reverse side.
Many designs are possible. I am sure there is also a picture cube puzzle with six pictures all isolated, so that you can't find a next picture by turning. Then it is a real 6-in-1-puzzle.
Escher-Block
...... The Escher block has eight cubes, which form a 2x2x2 cube together. If you put them together, you must find a picture by Escher on each side. There is the difficulty that a single cube shows pictures several times and that you must hide them inside the cube. It is not easy to find the solution. Therefore, and because of the Escher picture, it is a puzzle for grown-ups. The fish-bird-picture has the title "Sky and Water (1938)".
...... You can find the manufacturer's name on the description: Made by PUSSICAT, Bad Salzuflen, Germany ©1990 M.C.Escher, Art.No.80 5050 I live in Bad Salzuflen. You see the former factory at the Lemgoer Straße. The factory was there until 2000. As I see the building is empty now. The main emphasis of the company was sliding puzzles with artists' drawings. They are still "in". I made a German page (not yet translated) about the chain puzzle Elefant.
Jigsaw Puzzle
The jigsaw puzzles fit on this puzzle page. I think they are the reason that the picture cube puzzle came out of fashion.
LÜK
Then I remember LÜK. I see this equipment for math exercises together with tortured pupils and unnerved parents. LÜK is the abbreviation of the German words Lerne-Übe-Kontolliere (learn-exercise-control).
You had to solve problems written on tiles and put them in a certain place. If the solutions were correct, the pieces altogether showed a nice pattern on the reverse side. This was a version of the 1970s.
But no bad words. Of course I know that LÜK can be a nice toy that gives fun.
I know the puzzle from my childhold. The cubes were from wood and the pictures were from fairy tales. You could easily find four different pictures by turns, but for the last two ones you had to put the single cubes together as I remember.
Picture Cube Puzzle on the Internet top
German
Wikipedia
Puzzle
Gail from Oregon coast thank you for supporting me in my translation.
Feedback: Email address on my main page<|endoftext|>
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Module 1: Understanding Behaviour
In order to teach new behaviours or to decrease challenging ones, you need to underst and that is anything a person says or does (i.e., breathing, talking, hitting, eating, crying, etc.) There are lots of things in our environment (stimuli) that can affect our behaviour and we may do things more or less often depending on certain things or people around us.
We know that events that occur before and after behaviour have a big impact on the probability of them happening again. Think of this as the ABCs of behaviour (a behavioural contingency). A st ands for Antecedent (i.e., things that happen before behaviour). B st ands for Behaviour. C st ands for Consequence (i.e. events that occur right after a behaviour has happened). People tend to think that a consequence is a negative thing, but when talking about behaviour in general, consequences are simply whatever occurs after the behaviour.
If we’re going to increase or decrease a behaviour, then we need to know exactly what that behaviour is and define it clearly. This is referred to as an operational definition. The definition helps caregivers, parents or teachers know exactly when the target behaviour has or has not occurred. You should be able to easily observe and measure if a behaviour has happened. For example, if you’re talking about a tantrum, what is considered a tantrum? The more clear the definition, the easier it is to document.
Let’s consider an example. If you’re trying to decrease throwing, you know that you want to eliminate the target behaviour. However, when has throwing occurred and when do we track it? If we simply say, anytime the child throws, you may get questions like, “What if he threw something on the floor and not at a person” or “what if he threw a soft object that can’t hurt anybody?” If you have a clear operational definition right from the get go, you can avoid the confusion. In this case, you could define throwing as: “whenever the child throws any object in the direction of people, other objects, the wall or the floor.”
Remember the 3 ABCs of behaviour (Antecedent – Behaviour – Consequence). When you’re trying to increase or decrease a behaviour, you need to have a clear concise definition of the target behaviour.<|endoftext|>
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# introductory algebra glossary - PowerPoint PPT Presentation
Introductory Algebra Glossary. Unit One of Nine Units. Introduction. WELCOME Using the Introductory Algebra Glossary is simple. Click thru the slide show and check your knowledge of definitions before you display them. natural numbers . The numbers used for counting:
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### Introductory AlgebraGlossary
Unit One of Nine Units
• WELCOME
• Using the Introductory Algebra Glossary is simple.
• Click thru the slide show and check your knowledge of definitions before you display them.
• The numbers used for counting:
• {1, 2, 3, 4, ...}.
• The set of whole numbers is:
• {0, 1, 2, 3, 4, 5, ...}.
• The number above the fraction bar that shows how many equivalent parts are being considered.
• The number below the fraction bar in a fraction. It shows the number of equal parts in a whole.
• Any number that divides evenly (without remainder) into the given number:
• 1, 2, 3 and 6 are factors of 6.
• The answer to a multiplication problem.
• 6 is the product of 2 times 3.
• A number is factored by writing it as the product of two or more numbers.
• 6 is factored as 2 times 3.
• A natural number (except one) that has only one and itself as factors.
• 2, 3, 5, 7, 11, 13, and 17 are prime numbers.
• A composite number has at least one factor other than itself and one.
• The largest common factor of a list of integers or the largest term that is a factor of all terms in the polynomial.
• A fraction is in lowest terms when there are no common factors in the numerator and denominator (except 1).
• Pairs of numbers whose product is 1:
• 1/3 and 3 are reciprocals.
• The answer to a division problem.
• The answer to an addition problem.
• Given several denominators, the smallest expression that is divisible by all the denominators is called the least common denominator.
• A whole number and a fraction written together and understood to be their sum.
• The answer to a subtraction problem.
• A number that indicates how many times a factor is repeated:
• Given 23 the exponent is three.
• The number that is a repeated factor when written with an exponent:
• Given 23 the base is two.
• A number or letter (variable) written with an exponent:
• Examples: 23 or x6.
• Parentheses, ( ), square brackets, [ ], or fraction bars.
• A variable is a symbol used to represent an unknown number:
• In the term 3x the variable is x.
• Any collection of numbers or variables joined by the basic operations of addition, subtraction, multiplication, or division (except by zero), or the operation of taking roots.
• A statement that two algebraic expressions are equal:
• Example: 4x = 5y.
• Any replacement for the variable that makes the equation true.
• A collection of objects.
• The objects that belong to a set.
• A line with a scale that is used to show how numbers relate to each other.
• A number located to the left of zero on a number line.
• A number located to the right of zero on the number line.
• Numbers that can be written with a positive or negative sign.
• The set of integers is:
• {...-3, -2, -1, 0, 1, 2, 3,...}.
• The point on a number line that corresponds to a number is its graph.
• Rational numbers can be written as the quotient of two integers, with denominator not zero.
• Set-builder notation is used to describe a set of numbers without actually having to list all of the elements.
• Irrational numbers cannot be written as the quotient of two integers but can be represented by points on the number line.
• All numbers that can be represented by points on the number line, that is, all rational and irrational numbers.
• Two numbers that are the same distance from zero on a number line but on opposite sides of zero. The sum of two additive inverses equals zero.
• The distance between zero and a number on a number line.
• The multiplicative inverse of a nonzero real number a is 1/a. The product of multiplicative inverses is one.
• The order of numbers in an addition problem can be changed without changing the sum:
• 6 + 4 + 3 = 3 + 6 + 4
• The product in a multiplication problem remains the same regardless of the order of the factors:
• 6 • 4 = 4 • 6
• The way in which numbers being added are grouped does not change the sum:
• 6 + (3 + 2) = (6 + 3) + 2
• The way in which numbers being multiplied are grouped does not change the product:
• 6 • (2 • 3) = (6 • 2) • 3
• The sum of zero and any number equals the number, and the product of one and any number equals the number:
• X + 0 = x x • 1 = x
• A number added to its opposite is zero and a number multiplied by its reciprocal is one:
• 1 + (-1) = 0 1 • (1/2) = 1
• For any real numbers a, b, and c, the distributive property states that:
• a (b + c) = ab + ac.
• A number, a variable, or the product or quotient of a number and one or more variables raised to powers.
• The numerical factor in a term. In the term 6x2 the numerical coefficient is 6.
• The same variables raised to exactly the same powers. The terms 2x2 and 7x2 are like terms.
• Terms that do not have the same variable or the variables are not raised to the same powers. The terms 2x2 and 7x3 are unlike terms.
• A method of adding or subtracting like terms by using the properties of real numbers:
• 2x2 + 7x2 = 9x2.<|endoftext|>
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# Free fall
Free fall is the motion of an object under the influence of gravity alone, without any other forces acting on it, such as air resistance. In a vacuum, where air resistance is negligible, all objects fall at the same rate regardless of their mass. For example, if you drop a feather and a rock from the same height in a vacuum, they will hit the ground simultaneously. On Earth, free fall can be observed when you drop an object from a height and it accelerates downward due to gravity, typically at a rate of 9.8 meters per second squared. Free fall is an important concept in physics as it helps explain the motion of objects under the influence of gravity and is a fundamental aspect of understanding gravitational forces.
Contents
## Practice problems
### Problem #1
A feather is released from the top of a building, free-falling for a time of 20 seconds. Calculate the distance traveled by the feather during this time and determine its velocity. Assume a gravitational acceleration of g = 9.81 m/s2.
Solution
Given data:
• Time taken by a feather, t = 20 s
• Distance traveled by a feather, h = ?
• Velocity of a feather, v = ?
• Gravitational acceleration, g = 9.81 m/s2
Applying the formula of free fall distance, in terms of time:
• h = ½ × g t2
• h = ½ × 9.81 × (20)2
• h = ½ × 9.81 × 400
• h = 1962 m
Therefore, the distance traveled by a feather is 1962 m.
Applying the formula of free fall velocity, in terms of time:
• v = g t
• v = 9.81 × 20
• v = 196.2 m/s
Therefore, the velocity of a feather is 196.2 m/s.
### Problem #2
A ball is dropped from the top of a tower and falls a distance of 600 meters. Determine the time it takes for the ball to reach the ground and calculate its velocity upon reaching the ground. Use a gravitational acceleration of g = 9.81 m/s2.
Solution
Given data:
• Distance traveled by a ball, h = 600 m
• Time taken by a ball to reach the ground, t = ?
• Velocity of a ball, v = ?
• Gravitational acceleration, g = 9.81 m/s2
Applying the formula of free fall time, in terms of distance:
• t = √2 h/g
• t = √(2 × 600)/9.81
• t = √1200/9.81
• t = √122.3241
• t = 11.06 s
Therefore, the time taken by a ball to reach the ground is 11.06 s.
Applying the formula of free fall velocity, in terms of distance:
• v = √2 g h
• v = √2 × 9.81 × 600
• v = √11772
• v = 108.49 m/s
Therefore, the velocity of a ball is 108.49 m/s.
### Problem #3
A small piece of paper is dropped from the 10th floor of a building and falls with an initial velocity of 26 m/s. Determine the time it takes for the paper to reach the ground and calculate the distance traveled by the paper. Use a gravitational acceleration of g = 9.81 m/s2.
Solution
Given data:
• Velocity of a paper, v = 26 m/s
• Time taken by a paper to reach the ground, t = ?
• Distance traveled by a paper, h = ?
• Gravitational acceleration, g = 9.81 m/s2
Applying the formula of free fall time, in terms of velocity:
• t = v/g
• t = 26/9.81
• t = 2.65 s
Therefore, the time taken by a piece of paper to reach the ground is 2.65 s.
Applying the formula of free fall distance, in terms of velocity:
• h = v2/2 g
• h = (26)2/(2 × 9.81)
• h = 676/19.62
• h = 34.45 m
Therefore, the distance traveled by a piece of paper is 34.45 m.
### Problem #4
A box is released from a tower and takes 4 seconds to reach the ground. Calculate the distance traveled by the box during its free fall. Assume a gravitational acceleration of g = 9.81 m/s2.
Solution
Given data:
• Time taken by a box to reach the ground, t = 4 s
• Distance traveled by a box, h = ?
• Gravitational acceleration, g = 9.81 m/s2
Applying the formula of free fall distance, in terms of time:
• h = ½ × g t2
• h = ½ × (9.81) × (4)2
• h = ½ × 9.81 × 16
• h = 78.48 m
Therefore, the distance travelled by a box is 78.48 m.<|endoftext|>
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Focus on Developing the Virtue of Detachment in Children.
Use this check-off style worksheet to help children recognize five common activities that require the virtue of detachment in some way. Learners will also be encouraged to come up with five additional practical applications of their own.
Two versions of this virtue worksheet are included - color and grayscale.
Focus: Virtue study
Theme: Virtue of detachment
This printable resource is only one small part of Our Virtue Lesson on Detachment.
Help Kids Learn About Detachment This lesson celebrates the virtue of detachment and is intended to help Catholic kids practice and live their faith in a real way. It is designed to be used daily, over the course of a month. With a weekly core lesson,...
|File Size||3.74 MB|
|Register For Access|<|endoftext|>
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Winter in the Arctic is not only cold and dark, it is also storm season when hurricane-like cyclones traverse the northern waters from Iceland to Alaska. These cyclones are characterized by strong localized drops in sea level pressure, and as Arctic-wide decreases in sea level pressure are one of the expected results of climate change, this could increase extreme Arctic cyclone activity, including powerful storms in the spring and fall.
A new study in Geophysical Research Letters uses historical climate model simulations to demonstrate that there has been an Arctic-wide decrease in sea level pressure since the 1800’s.
“This research shows that the Arctic appears to be expressing symptoms expected from ongoing climate change,” said Dr. Stephen Vavrus from the University of Wisconsin-Madison. “The long-term decline in atmospheric pressure over most of the Arctic is consistent with the response typically simulated by climate models to greenhouse warming, and this study finds a general corresponding increase in the frequency of extreme Arctic cyclones since the middle 19th century.”
Tracking changes in Arctic cyclone activity through time, Vavrus calculated a statistically significant, though minor, increase in extreme Arctic cyclone frequency over the study period, with increases strongest near the Aleutian Islands and Iceland. Dr. Vavrus suggests that, as of yet, the effect of climate change on Arctic cyclone activity has been minimal, but that future changes in polar climate will drive stronger shifts.
“One societally relevant implication is that more storminess probably means more erosion of Arctic coastlines, especially in tandem with declines in buffering sea ice cover and increases in thawing coastal permafrost,” concluded Dr. Varnus. “Erosion of Arctic coastlines has already been growing more severe during recent decades, and this study points to a contributing factor that will likely become an even more recognizable culprit in the future.”<|endoftext|>
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# Arithmetic Sequences (the Nth Term)
In this worksheet, students find the Nth term in simple linear sequences.
Key stage: KS 3
Curriculum topic: Algebra
Curriculum subtopic: Use Sequences to Find the nth Term
Difficulty level:
### QUESTION 1 of 10
In an Arithmetic sequence like this:
4, 10, 16, 22, 28, 34 ...
the nth term is found as follows:
Step 1
Find the common difference between the terms
Here it is 6.
Step 2
The common difference is the coefficient (this just means the number in front.) of n
Here we get 6n
Step 3
Compare this to the actual sequence and adjust the formula by adding or subtracting a fixed number.
Here 6n would give us the terms 6, 12, 18, 24, 30 etc.... which is 2 more than our sequence of 4, 10, 16, 22, 28 etc.
So we must subtract 2.
So the nth term is 6n - 2
We can test this by putting in n = 5 to check that the 5th term is really 28.
It is, because 6 x 5 - 2 = 30 - 2 = 28
For this worksheet, It is important that you do not put spaces in your answer (For example, you would need to put in 5n-3 instead of 5n - 3. If you put in spaces, the system will mark it as wrong.
Work out the nth term in this sequence:
1, 2, 3, 4, 5, 6 ...
Work out the nth term in this sequence:
2, 4, 6, 8, 10, 12 ...
Work out the nth term in this sequence:
5, 10, 15, 20, 25, 30 ...
Work out the nth term in this sequence:
2, 3, 4, 5, 6, 7 ...
Work out the nth term in this sequence:
1, 3, 5, 7, 9, 11 ...
Work out the nth term in this sequence:
1, 4, 7, 10, 13, 16 ...
Work out the nth term in this sequence:
4, 7, 10, 13, 16, 19 ...
Work out the nth term in this sequence:
1, 6, 11, 16, 21, 26 ...
Work out the nth term in this sequence:
10, 17, 24, 31, 38, 45 ...
Work out the nth term in this sequence:
9, 13, 17, 21, 25, 29 ...
• Question 1
Work out the nth term in this sequence:
1, 2, 3, 4, 5, 6 ...
CORRECT ANSWER
n
EDDIE SAYS
Common difference is 1.
No need to adjust.
• Question 2
Work out the nth term in this sequence:
2, 4, 6, 8, 10, 12 ...
CORRECT ANSWER
2n
EDDIE SAYS
Common difference is 2.
No need to adjust.
• Question 3
Work out the nth term in this sequence:
5, 10, 15, 20, 25, 30 ...
CORRECT ANSWER
5n
EDDIE SAYS
Common difference is 5.
No need to adjust.
• Question 4
Work out the nth term in this sequence:
2, 3, 4, 5, 6, 7 ...
CORRECT ANSWER
n+1
EDDIE SAYS
Common difference is 1.
Add 1 to adjust.
• Question 5
Work out the nth term in this sequence:
1, 3, 5, 7, 9, 11 ...
CORRECT ANSWER
2n-1
EDDIE SAYS
Common difference is 2.
Subtract 1 to adjust.
• Question 6
Work out the nth term in this sequence:
1, 4, 7, 10, 13, 16 ...
CORRECT ANSWER
3n-2
EDDIE SAYS
Common difference is 3.
Subtract 2 to adjust.
• Question 7
Work out the nth term in this sequence:
4, 7, 10, 13, 16, 19 ...
CORRECT ANSWER
3n+1
EDDIE SAYS
Common difference is 3.
Add 1 to adjust.
• Question 8
Work out the nth term in this sequence:
1, 6, 11, 16, 21, 26 ...
CORRECT ANSWER
5n-4
EDDIE SAYS
Common difference is 5.
Subtract 4 to adjust.
• Question 9
Work out the nth term in this sequence:
10, 17, 24, 31, 38, 45 ...
CORRECT ANSWER
7n+3
EDDIE SAYS
Common difference is 7.
Add 3 to adjust.
• Question 10
Work out the nth term in this sequence:
9, 13, 17, 21, 25, 29 ...
CORRECT ANSWER
4n+5
EDDIE SAYS
Common difference is 4.
Add 5 to adjust.
---- OR ----
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Goats (Capra hircus) were introduced to New Zealand in the 1770s. They were first liberated as a food source and to clear weeds, and later to provide fibre for commercial industries. Goats were easily domesticated, and as a result were moved throughout the country as land was cleared for farming and settlement. Populations of wild goats have largely been a result of escapees from farms and deliberate releases.
Unlike domestic goats, feral goats don't have any identification or branding, and are not contained.
Wild goats are browsers rather than grazers. They cause considerable damage to under-storey vegetation up to two metres above the ground, damaging young trees in exotic forests and along replanted soil-conservation areas. The effect of goats destroying undergrowth, coupled with the effects of possums browsing the canopy, results in significant and often permanent damage to native vegetation.
Goats are agile animals, able to exploit steep hill slope areas unsuitable to other animals. The loss of vegetation in these areas may lead to increased erosion. Goats were recognised as a threat to New Zealand’s native vegetation from the 1890s and were identified as a major pest in the 1930s.
Male goats stand around 70 cm high at the shoulder and can grow to 1.5 metres in length, weighing between 50-60 kg. Adult females are considerably smaller. Both sexes may be white, black, brown or a combination and have horns. Male goats have chin beards and a pungent smell. Both sexes have a flat tail that is bare on the underside.
Feral goats are widespread throughout the Waikato region. Populations are generally highest on reverting farmland on steep hill country, but they are also found in exotic and indigenous forests, and scrub-lands.
They are hardy animals and will adapt to most environmental conditions. Their characteristic habitat is forest or scrub covered hill slopes, but they will move to grasslands where the opportunity arises.
The Department of Conservation (DOC) has the primary responsibility for goat management under the Wild Animal Control Act. Current government policy is to control goats in areas of highest conservation priority.
DOC has prioritised goat control in 24 locations in the region, covering over 14,000 hectares. The 24 locations are listed in the table below:
|Lakeshore reserves (Taupo)||Tongariro/Waiotaka Rivers (Turangi)|
|Tirohanga reserves (North Taupo)||Moehau (Coromandel)|
|Mt Pirongia||Mt Karioi|
Goat control is generally undertaken in conjunction with possum control in these areas. Land owners/occupiers adjacent to the above priority goat control areas may be directed by an authorised person to control wild goats on their land.
A constant problem with goat management is their dual identity as both farm resource and pest.
Our Regional Pest Management Plan aims to prevent the escape of farmed goats into areas of regional ecological significance, and to reduce populations of wild goats in key areas to a level where significant environmental damage does not occur.
One of the best methods of control for private land owners/occupiers is shooting. Every person shooting must either hold a firearms licence or be under supervision of a person who holds a firearms licence and is over 20 years of age. You should inform your neighbours where and when you intend to shoot. This may be an opportunity to co-ordinate your efforts with neighbours.
Contact Waikato Regional Council or the Department of Conservation to discuss shooting techniques for goats.
For additional advice and information on effective methods of control contact Council's Pest Animal staff on 0800 BIOSEC or the Department of Conservation(external link).<|endoftext|>
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Succeed with maths: part 2
Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.
# 3 Volume
Here the units that are used in everyday life differ from the SI units for volume. In the SI, units of volume are based upon the metre, with a cubic metre being standard. However, this is a very large volume and not of much use to us when measuring everyday objects. To get an idea of how large 1 cubic metre is, imagine a box that is 1 metre high by 1 metre wide by 1 metre long – that is 1 cubic metre and would be a lot of milk!
The base unit for volume that is used in everyday situations is the litre (abbreviated as ‘l’ – lower case ‘L’, not upper case ‘i'). This is from the metric system of measurement on which the SI is based. Adding the prefixes as before gives us the related units of millilitre (ml) and centilitre (cl). Note this could also be continued to include a kilolitre, but larger volumes are usually measured in cubic metres, where 1 cubic metre is the same as 1000 litres (or a kilolitre).
Again, using the knowledge of prefixes it can be deduced that a millilitre is a thousandth of a litre, and a centilitre is a hundredth of a litre. From this the following can be stated:
• 1 litre = 1000 ml
• 1 litre = 100 cl.
These relationships can also be displayed in a diagram, showing how to convert between the different units, as shown below:
Figure _unit3.3.1 Figure 1 Converting between units of volume
Now you’ve had lots of practice with converting between units, see how you get on with this next activity. Think back to the start of Week 1 – do you feel that the activities are now becoming more straightforward?
## Activity _unit3.3.1 Activity 5 Volume
Timing: Allow approximately 5 minutes
• a.You fill your car with 35.6 litres of fuel one day and four days later, another 15.2 litres. How much fuel in total have you put in your car this week in centilitres? Remember, clicking on ‘reveal comment’ will give you additional hints and tips.
Converting from a physically larger to a physically smaller unit, means you need to multiply.
• b.A bottle contains 14 cl of medicine. The dose is 5 ml. How many doses can be given from this bottle?
Click on reveal comment if you need a hint.
### Discussion
You are dealing with different units here, so you need to start by converting 14 cl to ml or 5 ml to cl.
You may find it easier to convert 14 cl to ml.
To find the number of doses, the volume of the bottle and the dose need to be in the same units.
There are 10 ml in 1 cl.
Therefore,
You have probably guessed by now that the next topic will be the imperial units for volume. Some of these will no doubt prove more familiar than others, as they are still used in some everyday situations, such as buying drinks.<|endoftext|>
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The onset of cold and flu season serves as a reminder of the critical importance of maintaining adequate stores of the mineral zinc.
While best known for its role in fighting colds and flus, zinc has far-ranging effects on human health, from strengthening the body’s immune defenses to promoting optimal growth and development.1 The tenth most common element in the human body, zinc is vital to the functioning of more than 300 hormones and enzymes.2 One of the most important of these is copper/zinc superoxide dismutase, an antioxidant enzyme associated with longevity and protection against oxidative stress.3-5
Zinc deficiency is associated with numerous disease states and even with DNA damage, which may contribute to cancer.6 With a preponderance of scientific evidence suggesting that zinc deficiency is common in the United States and throughout the world, it is all the more important that health-conscious adults take steps to ensure optimal zinc status.
Boosting the Body’s Antioxidant Defenses
Zinc is an essential component of one of the body’s most important natural antioxidants, the enzyme known as copper/zinc superoxide dismutase (Cu/Zn SOD).3,4 Cu/Zn SOD is one of a family of antioxidant enzymes that disarm dangerous superoxide radicals, which can wreak havoc on cellular compounds such as lipids, proteins, and DNA.5,7 Cu/Zn SOD rapidly breaks these compounds down into hydrogen peroxide, which is quickly converted to harmless water and ordinary oxygen.8
Research suggests that maintaining adequate levels of zinc and SOD is especially important to increasing longevity. Cumulative oxidative damage caused by reactive oxygen species is correlated with aging. In the laboratory, organisms that were manipulated to produce higher levels of SOD demonstrated extended life spans.5 Conversely, mutations in the gene that encodes Cu/Zn SOD have been associated with familial amyotrophic lateral sclerosis (Lou Gehrig’s disease), a fatal neurodegenerative disorder.7 Oxidative damage to the enzyme itself has recently been implicated in a variety of pathologies, including Alzheimer’s and Parkinson’s diseases.9
Oral supplementation with a bioavailable SOD preparation provides a way to greatly augment the protective benefits conferred by the body’s own supply of superoxide dismutase.10
A Cornerstone of Healthy Immune Function
In addition to bolstering the body’s antioxidant defenses, zinc plays a crucial role in supporting proper immune system function.11-15 T-lymphocytes are white blood cells that help fight infection and depend on zinc for their development and activation. In humans, zinc deficiency can result in a decreased number of T-lymphocytes and a diminished ability to fight infection and heal wounds.16-18 Because supplemental zinc may help fight infection and heal wounds,17 zinc status is especially important for patients with conditions such as HIV infection.18
Maintaining adequate levels of zinc is equally important to aging adults. As people grow older, their immune function declines, partly due to the decreasing size and function of the thymus gland. New evidence suggests that zinc may help to maintain healthy function of the thymus gland in elderly people.19 Scientists have also found evidence that a zinc deficiency in elderly adults may contribute to impaired immune function with aging, or immunosenescence.20 These findings suggest that zinc plays a critical role in supporting healthy immune function with age.
Zinc Fights Symptoms of the Common Cold
One of zinc’s most important uses in recent years is reducing the severity and duration of colds. The common cold is caused by any one of more than 200 distinct viruses that target the respiratory tract.23 Zinc interferes with the viruses’ ability to attach to the surface of respiratory tract cells and reproduce, which may help prevent infections from taking hold and causing symptoms.24
Numerous clinical trials, involving hundreds of child and adult patients, support the effectiveness of zinc lozenges in mitigating cold symptoms.23-29 One study concluded that lozenge use within 24 hours of the first onset of cold symptoms reduces the severity of symptoms, the duration of the illness, subsequent use of antibiotics to treat secondary symptoms, and overall incidence of colds per year.29
Colds can also lead to even more serious secondary conditions, such as sinusitis in adults and middle ear infections in children. Indeed, acute upper-respiratory infection is the single most common reason people seek medical treatment in the United States.25 Vanderbilt University scientists discovered that zinc inhibits the activity of a common respiratory virus that routinely threatens children’s health. The doctors added zinc salts, such as zinc sulfate and zinc lactate, to cell cultures infected with a common pediatric respiratory disease known as respiratory syncytial virus. Zinc salts prevented the harmful virus from replicating freely.30
To be most effective, zinc lozenges should be taken within 24 hours of the appearance of initial cold symptoms, and should be taken continuously throughout the course of the illness. Although generally well tolerated, some patients have reported mild nausea when taking zinc lozenges, and temporary alterations in taste sensation are not uncommon. Taking zinc with food may help to alleviate these effects.
Supporting Human Growth and Maturation
Zinc is vital to human growth and maturation. Studies have shown that zinc deficiency results in cognitive deficits and growth retardation, including delays in sexual maturation, among both males and females.36-39 Scientists in California report that even moderate zinc deficiency results in disruptions in normal growth and maturation among rhesus monkeys, and behavioral changes, including lethargy, may precede physical manifestations of deficiency.40,41 The same appears to hold true for humans. Zinc deficiency is linked to deficits in healthy bone mineralization, and has been linked to short stature and developmental delays.36-39
According to researchers in California who conducted a meta-analysis of clinical trials examining the effects of zinc supplementation on children’s growth, “Zinc supplementation produced highly significant, positive responses in height and weight increments.”42 Researchers in Turkey recently reported that zinc supplementation stimulates the production of insulin-like growth factor 1 (IGF-1) and other proteins and polypeptides involved in bone and overall growth.43 IGF-1 responds directly to growth hormone and is responsible for regulating the growth and development of a wide range of cells, as well as DNA synthesis.
Zinc also plays a key role in spermatogenesis, the process by which the male testes produce sperm capable of fertilizing an egg, or ovum. Numerous studies comparing fertile and infertile men have demonstrated that zinc is abnormally low in the blood serum and seminal fluid of infertile men.44-47 Replenishing zinc levels may thus be helpful in optimizing male fertility.<|endoftext|>
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When a child compares himself to others
Do you have a child who is always comparing himself to others in your group? Certain children tend to constantly compare the way they play, their results, their board game wins, and even the size of their father's muscles... Each comment hides an indirect message.
Comparisons can hide certain desires or emotions: to be like others, to stand out, a fear of having less, and even jealousy. Often, comparisons can be associated with low self-esteem. You most likely have noticed children who are constantly observing others and comparing their performances.
In childhood development, a certain level of comparison is normal. At a very young age, when children are still in the parallel play stage and self-centered, they show little interest in comparing themselves to others. Around their third birthday, children may begin to make general comparisons, mainly on a physical level. It's only after pre-k, around 5 years old, that children begin to make more serious comparisons. Certain interventions can help increase children's sense of competence and therefore reduce their need to compare themselves to others.
- Avoid comparing children among themselves. Highlight their individual successes.
- Build children's self-esteem gradually, daily even. Help children discover their abilities, recognize their successes, and celebrate their victories.
- Offer plenty of positive reinforcement. Notice, encourage, and congratulate children's efforts.
- Celebrate children's differences. Highlight each child's strengths.
- Speak positively to the children in your group and invite them to do the same. Keep in mind that words impact our thoughts and feelings.
- Aim to put a STOP to comparisons and take advantage of discussion periods to point out children's accomplishments.
In conclusion, don't forget that children learn through imitation. Make sure you are a positive role model and avoid comparing yourself to others. Be proud of your personal accomplishments and who you are. Children will learn from your behavior.<|endoftext|>
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Magnetic Resonance Imaging: What is it?
Magnetic Resonance Imaging (MRI) scanners have been around for almost 40 years, but their inner workings can still seem mysterious. And, to a patient coming into the scanning room for the first time, they can also be large and intimidating.
WHAT ARE THE DIFFERENT PARTS OF AN MRI MACHINE?
The machine itself comes in many shapes and sizes (we use the Optima MR450w with GEM Suite from GE Healthcare), but the major components are basically the same. One of the most important parts is the large magnet, which is thousands of times more powerful than the earth’s magnetic field. A horizontal tube, called the bore, runs through the centre of this magnet, which is where the patient enters while lying on a bed.
The magnet consists of many coils of wire through which a current of electricity is passed, creating a magnetic field of 1.5 tesla (the unit of measure for magnetic field strength).To generate such a strong magnetic field, very large amounts of electricity are needed. Normally, this would cause the wires to heat up and burn, since wires tend to “resist” electrical current, but superconductivity, or cooling of the wires, reduces this resistance to almost zero.
The wires are continually bathed in liquid helium at very low temperatures (-450˚F) and insulated by a vacuum. An added benefit of this superconductive magnet is that once the initial electrical current is sent through the wires, it flows continuously (since there is no resistance) and additional current is not necessary.
The machine also contains three much lower strength magnets, called gradient magnets. The purpose of the main magnet is to create an intense, stable magnetic field around the patient, while the gradient magnets create fields of varying strengths, which allow the system to target specific areas of the body.
Next, we have the radiofrequency coils. These transmit radio waves into the patient's body; there are usually different coils which conform to the contour of different parts of the body: knees, shoulders, wrists, head, neck, and so on.
Finally, a very powerful computer system controls various aspects of the scanning process, including the gradient coils, the radiofrequency pulse, as well as interpreting all the information gathered and constructing high quality images.
|Axial head image||Coronal knee image|
HOW DOES THE MACHINE “READ” MY BODY?
Our body is made up of billions of atoms, including an abundance of hydrogen atoms. The MRI machine is specifically concerned with these atoms, because within them the hydrogen protons are randomly spinning on their axis in various directions - except when they are placed in a magnetic field.
When they are exposed to a magnetic field, hydrogen protons line up in the direction of that field. Since the magnetic field runs straight down the centre of an MRI machine, the protons line up to point at either the head or the feet of the patient. If we match up the head-facing protons with the feet-facing protons, we are left with only a few unmatched protons out of every million. It’s doesn’t sound like much, but it is enough to create detailed images.
In addition to the magnetic field, the MRI machine emits a radio frequency (RF) pulse specific to the “resonance” of hydrogen. That pulse is directed toward the area of the body we want to examine, where the unmatched protons absorb the energy and are forced to spin at a particular frequency, in a particular direction. This specific frequency of resonance (the Larmour frequency) depends on the tissue being imaged and the strength of the magnetic field.
While the RF pulse is being emitted, the gradient magnets are turned off and on rapidly, in a specific way, to alter the magnetic field on a small, but localized level. This allows the system to target a specific area for imaging. Images can be taken of any part of the body with incredible detail, providing doctors with scans of the tissue from a variety of angles and directions.
|Head and neck image|
When the RF pulse is stopped, the hydrogen protons slowly return to their previous alignment and release the energy they had absorbed. In the process, they give off signals that the coils pick up and send to the computer system. The computer interprets these signals as mathematical data, which it then converts into a picture.
Sometimes an MRI scan involves an injectable contrast, or dye, that alters the local magnetic field in the tissue being examined. Normal and abnormal tissue respond differently to this slight alteration, allowing doctors to visualize different types of tissue abnormalities.
IS IT DANGEROUS?
There are no known health risks attributed to MRI scans. However, due to the nature of superconductivity, the magnet is always on, so it’s important to take care around the machine. In particular, it’s important to remove all metal objects before entering the room.
MRI machines make a loud knocking noise during a scan, which can be disconcerting. It is caused when the gradient coils work to alter the main magnetic field. Due to the huge magnetic forces involved, they move slightly (vibrate) and produce sound. These sounds are normal. At Mayfair, we will provide you with headphones to listen to music and help muffle the noise. For patients with sensitive hearing our machines also offer Acoustic Reduction Technology to reduce noise for brain, spine, and MSK exams.
For more information about MRI scanning, please visit the services page for a list of scans and our procedures.
Tilakaratna, P. How Magnetic Resonance Imaging works explained simply. Taken from www.howequipmentworks.com<|endoftext|>
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Most of us use the word ‘heat’ to mean something that feels warm, but science defines heat as the flow of energy from a warm object to a cooler object.
Actually, heat energy is all around us – in volcanoes, in icebergs and in your body. All matter contains heat energy.
Heat energy is the result of the movement of tiny particles called atoms, molecules or ions in solids, liquids and gases. Heat energy can be transferred from one object to another. The transfer or flow due to the difference in temperature between the two objects is called heat.
For example, an ice cube has heat energy and so does a glass of lemonade. If you put the ice in the lemonade, the lemonade (which is warmer) will transfer some of its heat energy to the ice. In other words, it will heat up the ice. Eventually, the ice will melt and the lemonade and water from the ice will be the same temperature. This is known as reaching a state of thermal equilibrium.
Matter is all around you. It is everything in the universe – anything that has both mass and volume and takes up space is matter. Matter exists in different physical forms – solids, liquids and gases.
All matter is made of tiny particles called atoms, molecules and ions. These tiny particles are always in motion – either bumping into each other or vibrating back and forth. It is the motion of particles that creates a form of energy called heat (or thermal) energy that is present in all matter.
The particles in solids are tightly packed and can only vibrate. The particles in liquids also vibrate but are able to move around by rolling over each other and sliding around. In gases, the particles move freely with rapid, random motion.
Transferring heat energy – particles in collision
At higher temperatures, particles have more energy. Some of this energy can be transmitted to other particles that are at a lower temperature. For example, in the gas state, when a fast moving particle collides with a slower moving particle, it transfers some of its energy to the slower moving particle, increasing the speed of that particle.
With billions of moving particles colliding into each other, an area of high energy will slowly transfer across the material until thermal equilibrium is reached (the temperature is the same across the material).
Changing states by heat transfer
Faster moving particles ‘excite’ nearby particles. If heated sufficiently, the movement of particles in a solid increases and overcomes the bonds that hold the particles together. The substance changes its state from a solid to a liquid (melting). If the movement of the particles increases further in the liquid, then a stage is reached where the substance changes into a gas (evaporation).
Three ways of transferring heat energy
All heat energy, including heat generated by fire, is transferred in different ways:
Convection transfers heat energy through gases and liquids. As air is heated, the particles gain heat energy allowing them to move faster and further apart, carrying the heat energy with them. Warm air is less dense than cold air and will rise. Cooler air moves in below to replace the air that has risen. It heats up, rises, and is again replaced by cooler air, creating a circular flow called a convection current. These currents circle and heat the room.
Conduction transfers heat energy in solids. The moving particles of a warm soild material can increase the heat energy of the particles in a cooler solid material by transferring it directly from one particle to the next. Since particles are closer together, solids conduct heat better than liquids or gases.
Radiation is a method of heat transfer that does not require particles to carry the heat energy. Instead, heat is transferred in infrared waves (part of the electromagnetic spectrum). Heat waves radiate out from hot objects in all directions, travelling at the speed of light, until they hit another object. When this happens, the heat energy carried by the waves can be either absorbed or reflected.
Fire illustrates the three different methods of heat transfer. For example, the firebox will heat up due to convection. The air above the fire will be warm due to convection. You can warm your hands near to the flames due to radiant heat transfer.
An effect of heat – expansion
When gases, liquids and solids are heated, they expand. As they cool, they contract or get smaller. The expansion of the gases and liquids is because the particles are moving around very fast when they are heated and are able to move further apart so they take up more room. If the gas or liquid is heated in a closed container, the particles collide with the sides of the container, and this causes pressure. The greater the number of collisions, the greater the pressure.
Sometimes when a house is on fire, the windows will explode outwards. This is because the air in the house has been heated and the excited molecules are moving at high speed around the room. They are pushing against the walls, ceiling, floor and windows. Because the windows are the weakest part of the house structure, they break and burst open, releasing the increased pressure.<|endoftext|>
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What were the causes and effects of new immigration during the Gilded Age? Why did it result in a surge in nativism?
Explain how the Gilded Age fostered the consolidation of business, the government, and disadvantaged economic and social classes.
What strategies enabled big business to eliminate competition and limit the power of labor unions?
Give examples of how the Progressive movement partially succeeded in improving life for average Americans by curbing big business, making the government more responsive to the will of the people, and enacting social welfare legislation.
Rank the Presidents/Parties in order from Most to Least Progressive. Explain each of your rankings with specific evidence.
What were the most successful Progressive reforms? What were the limitations of the Progressive movement?
Unit 2 – Imperialism and Isolationism
Trace how the United States became increasingly active and aggressive in world affairs from 1890 to 1918.
What three factors spurred the new American Imperialism? Which was the strongest?
Explain the causes and effects of the Spanish-American War.
In which situations was the U.S. most/least justified in the desire for acquisition or intervention? In which situations did the U.S. use methods that would be considered acceptable/unacceptable by today’s standards?
Identify the long-term and immediate causes of WWI.
Why did the U.S. get involved in a world war? Should we have gotten involved sooner?
How did the United States prepare for and “sell” the war at home? How did the war impact citizens on the homefront?
Explain the debate over the Treaty of Versailles and the League of Nations.
What were the major international consequences of WWI?
Analyze nativism, racism, and the fear of communism after World War I.
Explain how WWI resulted in isolationist foreign policy and conservative domestic policy in the 1920s. Give specific examples of each.
Compare Reagan and Clinton – What are the similarities and differences between Conservative Republicans and New Democrats?
Explain the public’s disillusionment over social issues and democratic ideals coupled with its recognition of the limitations of the federal government’s abilities to solve social problems in the 1960s and early 1970s. Why did this lead to a resurgence of conservative ideology in the 1980s?
What were the policies and events that brought about an end to the Cold War?
What U. S. actions and world events reflected the identity of America as “The superpower” of the world during this time period? What events and trends challenged that notion of U. S. superiority and power?
Following the breakup of the Soviet Union, why has the US struggled to find a cohesive foreign policy that both minimizes our intervention and protects our global and national interests?
Trace the War on Terrorism from September 11th through the War in Afghanistan.
Explain how U.S. interests in Israel and oil have resulted in increasing involvement in the Middle East. How has the U.S. approached this problem over time?
Compare and contrast the Gulf War with the Iraq War. Include both causes and effects.
How did technological developments between 1950 and 2000 radically alter the economic, social, and moral fiber of the nation?
What led to the prosperity of the 1990s and the recession in the new 21st century?
How are energy consumption and sources of energy at odds with environmental concerns?
How did population trends, including the “graying” of the baby boomers and the growth of minority populations in the U. S., affect U. S. culture and policies during the era?<|endoftext|>
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The Quantitative Reasoning section is often considered the most difficult part of the ISEE. You will have 35 minutes to answer 19 word problems and 18 quantitative comparison questions, and you are not allowed to use a calculator. This section is very similar to Mathematics Achievement, except that it contains quantitative comparison questions.
To learn about what level of math you should know for the ISEE and for quick tutorials on the 48 most tested concepts, click on the button below.
Each word problem is followed by four answer choices. Choose the answer choice that best answers the word problem. Several of the word problems may be solved without actually doing any math on paper.
Systematic Approach to Mathematics Word Problems
General 5-Step Approach
|1||Do I understand the question?||Write out the necessary information to the side of the problem.||SKIP IT! Circle the question in the test booklet and come back to it if time permits.|
|2||Do I know which formulas I should use?||Quickly write out the necessary formulas to the side of the problem.||SKIP IT! Circle the question in the test booklet and come back to it if time permits.|
|3||Will the question take me a long time to complete?||SKIP IT! Circle the question in the test booklet and come back to it if time permits.||See if you can solve the problem without performing any math.|
|4||Can I input values to solve the question?||Use simple values such as -1, 0, and 1. If the answer chances, then choose D.||Keep in mind all properties of algebraic and geometric concepts.|
|5||Can I make an educated guess with what I know?||Eliminate portions of each value that are the same in both columns.||DO NOT leave it blank. Choose the answer that makes the most sense.|
Each question contains two quantities that you must compare. Some questions will include extra information to help answer the question. Each answer choice designates a particular relationship between the two quantities:
- The quantity in Column A is greater.
- The quantity in Column B is greater.
- The two quantities are equal.
- The relationship cannot be determined from the information given.
Systematic Approach to Quantitative Comparisons
The hardest part about Quantitative Comparison questions is where answer choice D might be the correct answer. Don’t automatically assume something is unsolvable!<|endoftext|>
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# Mental Subtraction Strategies - grade 1 Mental Math for Class 1 Math
Subtraction is a process of finding the difference between two or more items together. Mental Subtraction is used to find the difference between large numbers in the simplest way.
This concept introduces mental math strategies for subtraction for students.
In this learning concept, the students will learn to
• Apply the Subtraction tricks
• Use of forward counting and backward counting in subtraction.
Each concept is explained to class 1 maths students with illustrations and examples, and a concept map is given to summarize the idea. At the end of the page, two printable worksheets with solutions are attached for students to practice. Download the worksheets with solutions and assess your knowledge.
#### What Is Subtraction?
• To find the difference between any numbers.
#### How to Subtract Mentally?
We will learn some easy methods to do subtraction mentally.
• #### Subtraction by Forward Counting
• Consider the second number (Subtrahend) and count the number of steps forward till getting first number (Minuend).
#### Method:
• In the above example, we start counting from number 5 till getting the number 8.
• Then count the number of steps that we have counted forward.
So, we get:
• #### Subtraction by Backward Counting
• Consider the first number (Minuend) and count the number of steps backwards equal to the second number (Subtrahend).
#### Method:
• In the above example, we start counting backwards from the number 7.
• We have subtrahend as 3, so we count 3 steps backwards.
• We reached at number 4. It is our answer.
• This method will make our subtraction easier.
So, we get:
• #### Subtract any 1-digit number from 10 using number bonds of 10.
• We use here number bonds of 10.
#### Method:
• In the above example; we know that, if we add 3 to 7, we get 10. (Number bond of 10: 7 + 3 = 10)
• So, here 7 is 3 less than 10.
So, we get:
#### Method:
• In the above example; we know that, if we add 8 to 2, we get 10. (Number bond of 10: 2 + 8 = 10)
• So, here 2 is 8 less than 10.
So, we get:
• #### Subtraction By Splitting of 2-digit Number Into Tens and Ones
• Split 2-digit number into Tens and Ones, make subtraction of Ones.
#### Method:
• In the above example, we split 28 into 20 Tens and 8 Ones.
• Then subtract 5 from 8 Ones. We get 3.
• Adding 20 Tens and 3 Ones, we get 23.<|endoftext|>
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The Sangre de Cristo National Heritage Area is a melting pot of settlers from around the world. Traditional holiday customs from this diverse group of ethnicities has been handed down over generations. Several traditions have been adopted and integrated between cultures. The Sangre de Cristo National Heritage Area would like to share just a few “holiday traditions” or “seasonal traditions” from each group known to have made settlement in the Sangre de Cristo National Heritage Area. These include: Native Tribes, Spanish, German, Dutch, Swedes, Japanese and Amish.
The tribes that came through the San Luis Valley were transient and did not set up permanent residence. Many tribes believed traditions had a “season” upon which to be administered. Tribes passed down their culture and heritage from generation to generation through storytelling and oral traditions. The season for storytelling in the Ute Tribe, is during the winter months, when they went into their lodge.
The Pueblo Tribe also came to the San Luis Valley seasonally. Their permanent pueblo was established in modern day Taos, in the Southern end of the San Luis Valley and is part of the Northern Rio Grande National Heritage Area. When the Spanish arrived in the 16th century, they began converting the Pueblo peoples to Catholicism. The missionaries’ attempts resulted in a merging of traditional feast days with Christian saints’ days. As a result, holiday celebrations on the Pueblos reflect the blend of Spanish culture and the Roman Catholic influence with indigenous Native American Indian religious ceremony. The fusion of the two spiritual traditions can be observed in the festivities, which include the Matachines Dance. (See Matachina Dance under Hispanic Holiday Traditions)
The moment the Sun reaches its southernmost point in the sky marks the December solstice, the official beginning of winter in the Northern Hemisphere (where it is called the winter solstice) and a time of great celebration in many northern cultures. The word solstice comes from the Latin solstitium — sol meaning “sun” and -stitium “stoppage.” The winter solstice marks the shortest day and longest night of the year. From now on, the days begin to grow longer and the nights shorter. In native and ancient cultures, the winter solstice was an auspicious moment.To many tribes it is a time where they offer gratitude, honor family and ancestors, and follow a ritual observance of beliefs.
There are culinary traditions from the Mayans and Aztecs that have crossed over to become traditional foods eaten and made around the Christmas season in the Sangre de Cristo National Heritage Area.
Posole, translated as “hominy” is a traditional stew that has a long ritual history among the Aztecs. Since Maize was a sacred plant for the Aztecs, posole was made to be consumed on special occasions to celebrate the creation of man. Meso-Americans believed the god Quetzalcoati made man from masa (corn-meal dough). As a result, corn took on a religious significance. The kernels were soaked in a mixture of ground limestone (farther north, they used ashes to cure the corn) and water, soaked for several days then dried. Processing the corn in this way allowed the corn to be preserved for several years while keeping fresh taste and free from vermin, and allowed the release of a multitude of important nutrients to become accessible for digestion. In the “General History of Things from New Spain” Fray Bernardino de Sahagun mentioned during the festivities to honor the god Xipe, the Emperor was served a massive dish of pozolli-crowned with the thigh of a sacrificed prisoner. That’s right. Curiously, chili was left out of that recipe. After the conquest of the Aztec Empire cannibalism was banned and pork became the new meat added to the dish. Posole didn’t arrive in the Rio Grande Valley and the San Luis Valley until was brought by the Spanish in the 1600s. These cultures took the posole recipe and adapted it to make their own, endowing it with a life-affirming significance.
1 pound dried posole
1 quart water, or more
2 pounds pork, steak or roast, cut into ½ inch cubes
1 Tbsp salt or more to taste
2 cloves garlic, minced
Pinch of Mexican oregano
1 Tbsp cumin, or to taste
½ Caribe chili, or to taste
Simmer the posole in unseasoned water until it becomes soft and the kernels have burst open; it usually requires 1 ½ to 2 hours.
Brown the pork in a cold, well-seasoned frying pan, adding no fat or oil. Sautes until very browned then add the polole. Deglaze the frying pan with 1 cup water, stirring to loosen the brown bits sticking to the pan. Also add these to the posole.
Add remaining ingredients, adding one-half the cumin and cook the stew for 1 or more hours, to blend the flavors. Just before serving, add the remaining cumin.
Hispanic Holiday Traditions
The tradition of luminarias in the Sangre de Cristo National Heritage Area dates back more than 300 years. The tradition began when the Spanish villages along the Rio Grande displayed unique and easy to make Christmas lanterns, called luminarias, to guide the spirit of Christ along their paths.
Luminarias today are often made of a brown paper bag, which has been folded at the top and filled with a few cups of sand and a votive candle illuminating the bag from within.
These lanterns have not always been made out of paper bags, the early versions were actually small bonfires of crisscrossed piñon branches which were built in three-foot high squares. Instead of making lanterns that would hang in a tree or from a roof, which would become damaged by the wind, small bags were made and placed on the ground, rooftops and along pathways. Luminarias are said to originate from Spaniard merchants who were impressed with the paper lanterns from the Chinese culture and decided to make their own version when they migrated to New Spain.
The name of the decoration is the subject of a long-running item of contention. Some insisting the correct term is farolito which translates as “little lantern”. While others are sure the correct name is luminaria which means “festival light”. Historically luminaria referred not to a paper lantern, but to a small festival or vigil bonfire; however, this distinction is not commonly made outside of northern New Mexico. Farolitos may be referred to as “luminarias” by some, but on Christmas Eve, when the farolitos(illuminated paper bags)are lit in streets of Santa Fe, luminarias (Posada vigil fires) are burning in the small mountain villages of Northern New Mexico. Luminaria bonfires made of square, stacked piñon and juniper wood can often can be seen in towns and pueblos across northern New Mexico. In the mountain villages and by the roadways they are built by local residents to welcome visitors and to commemorate holiday activities. No matter what you call them they are beautiful to see this time of year.
Las Posadas takes place on each of the nine nights leading up to Christmas, from December 16 to 24th. The word posada means “inn” or “shelter” in Spanish, and in this tradition, the Bible story of Mary and Joseph’s journey to Bethlehem and their search for a place to stay is re-enacted. The celebration begins with a procession in which the participants hold candles and sing Christmas carols. Sometimes there will be individuals who play the parts of Mary and Joseph who lead the way. The procession will make its way to a particular home (a different one each night), where a special song (La Cancion Para Pedir Posada) is sung. There are two parts to the traditional posada song. Those outside the house sing the part of Joseph asking for shelter and the family inside responds singing the part of the innkeeper saying that there is no room. The song switches back and forth a few times until finally the innkeeper agrees to let them in. The hosts open the door and everyone goes inside.The nine nights of posadas leading up to Christmas are said to represent the nine months that Jesus spent in Mary’s womb, or alternatively, to represent nine days journey that it took Mary and Joseph to get from Nazareth (where they lived) to Bethlehem (where Jesus was born).
The Matachines dance is one of the few traditional dances observed by both the Pueblo and Hispano-Mexicano peoples. The dance is a ritual performance that tells the drama of the Spanish military conquest over the Native Tribes and shows the final result where they convert to Christianity. The Matachines dancers were originally brought over from Spain, the music and dance depicts the coming together of the Christians and the Moores. Matachin in Spanish means “sword dancer in costume”. The Matachines dancers,(danzantes) are distinctively costumed. Their mitrelike headdresses, or cupiles, with ribbons streaming down the back and fringe in the front, are their signature symbol. The mask consists of the band of fringe (fleco) over the eyes and a folded kerchief over the lower face. Each danzante carries the palma in his left hand and the guaje in his right. Large colorful scarves hang like capes from the backs of their shoulders. They move in two parallel rows of five or six dancers each.The most basic or universal dramatic elements of the Rio Grande Matachines performance involve several dance sets by the characters El Monarca and La Malinche, an exchange of trident (palma) and rattle (guaje) between them, a variable combination of choreographic interweavings, crossovers, and reversals between the two columns of dancers, a movement involving El Toro-the bull-and his ultimate demise, and processional and recessional marchas at the beginning and end. The clowns, known as Los Abuelos (the grand-fathers), function as conductors and provide comic relief throughout the proceedings.The music for the Matachines dance is typically performed by a fiddler and a guitarist.
Mis Oremos and Mis Crismas
Mis Oremos seems to be a mixture of Las Posadas with Matachines. Traditionally it begins with the lighting of one luminaria (bonfire) on Dec 16th. On the 17th two luminarias are lit. On the 18th three, so that on the 24th there will be 9 luminarias in each village. On the evening of the 24th, Christmas Eve, a few of the older teenagers don masks and costumes as not to be recognized. They are called abuelos (grandfathers). They make the younger kids sing, march in formation, etcetera-all to the beat of a chicote (a horse whip) around the bonfires. Finally the whole group goes to the homes of their neighbors in the villages to ask for Oremos. In front of the homes the farolitos (candles in bags) are lit. At each door the children, guided by the abuelos, recite this poem.“Oremos, oremos Angelitos semos. Del cielo venemos a pedir oremos. Si no nos dan oremos Puertas y ventanas quebraremos.”English Translation:“We pray, we pray. We are little angels. From heaven we come to ask for oremos. If you do not give us oremos,Doors and windows will be broken.”
Traditionally, on Christmas day, a similar event of going house to house followed, called Mis Crismas, where kids go door to door getting treats from neighbors.
These festivities are held around Christmas time and relate to Christian tradition, but parts are reminiscent of Judaism. The luminarias are lit, increasing by one fire a day, similar to the lighting of the candles on Hanukkah. It is possible this was a way that Jewish people, who were avoiding persecution in Europe, could have secretly celebrated Judaism in this area. They would have avoided questioning, since they tied it into Christmas tradition. Occasionally the lighting of the luminarias corresponds exactly to the nights of Hanukkah.
These crisp cultural cookies are flavored with cinnamon and anise, and are a yearly tradition for many Hispanic families in the Sangre de Cristo National Heritage Area. These cookies are the result of centuries of influence brought by local and indigenous customs. These biscuits originated in Spain, where they were called mantecados, and date back to the 16th century, a time when the region of Andalusia experienced a surplus of grains and pork products. The roots of this cookie took on greater significance during Mexico’s Battle of Puebla in 1862, when Mexicans overthrew the French-backed Emperor Maximilian (Celebrated today as Cinco de Mayo). It is said that Mexican women wanted a commemorative cookie and used tin cans to cut the cookies to symbolize stamping out the French. The cookie symbolizes freedom and victory and has also become strongly associated with the Christmas season. They are often offered to the posadistas—the people who participate in Las Posadas, a nine-day celebration where a nightly procession re-enacts Joseph and Mary’s search for lodging in Bethlehem. Traditionally, the procession is always refused “lodging”, though the hosts often provide refreshments and biscochitos. Luminaria bonfires were lit to provide light and warmth for the posadistas.
Makes approximately 5 dozen thin cookies
The traditional shapes are stars, circles and crescents
1 cup lard
1 cup sugar
2 teaspoons vanilla extract
1/4 cup brandy or rum
1 teaspoon anise seeds
3 cups all purpose flour
1 teaspoon salt
1 teaspoon baking powder
1/4 teaspoon cinnamon
1/2 cup sugar
In a stand mixer, cream together lard and sugar on medium speed until fluffy. With mixer running on low, add egg, vanilla extract, brandy, and anise seeds and mix until homogeneous. Add flour, salt, and baking powder and mix just until dough forms into cohesive ball.
Form dough into two cylinders about 1 1/2 to 2 inches in diameter. Chill for 2 hours or wrap tightly in plastic wrap and store in refrigerator for up to 3 days. For longer storage, freeze logs tightly wrapped for several months. Defrost in refrigerator for 1 day before using.
3.Adjust oven rack to middle position and preheat oven to 350°F. Combine sugar and cinnamon in small bowl and set aside. Cut cylinders into 1/4 inch disks and on ungreased cookie sheets leaving 1/2-inch gap between cookies. Bake until golden brown, about 12 minutes.
4.Dip rim of each cookie into cinnamon sugar mixture. Let cool on racks and store at room temperature in airtight container for up to 5 days.
1 lb Pork or Beef (Optional)
3 tbsp lard or bacon grease
1/4 cup garlic (chopped finely)
1/2 cup onion (chopped finely)
2 tbsp Flour
1 cup Water
1 cup Diced Green Chile (roasted, peeled, and seeded)
1/2 Tbsp Salt* to taste
In oil in a heavy saucepan, saute’ garlic and onion.
Blend in flour with wooden spoon.
Add water, salt, and green chile.
Bring to a boil, then drop temperature and simmer, stirring frequently, for 30-50 minutes.
Serve with everything. (you may also refrigerate for 24 hrs for flavors to develop better)
Swedish Christmas Traditions
Saint Lucia Day
In Sweden, Christmas begins with the annual Saint Lucia Day on December 13. The date commemorates Saint Lucy. The saint was a 3rd century martyr who brought food and aid to Christians hiding in the catacombs using a candle-lit wreath to light her way. Her feast once coincided with the winter solstice, the shortest day of the year, it is why her feast day has become known as the Christmas festival of light.Usually, the eldest girl in the family portrays St Lucia. She puts on a white robe in the morning and is allowed to wear a crown full of candles. Personifying St. Lucia, she serves her parents buns, cookies, coffee, or mulled wine
End of the Christmas Season
Christmastime does not end in December for Sweden, but continues on until January 6th, the date of the Epiphany. It is also called trettondedag jul, or “13th-day yule,” as January 6 is the 13th day after Christmas Eve.
Rounding out the end of the Christmas season is Hilarymas, also called Knut’s Day or Tjugondag jul on January 13. Christmas trees are taken down on this day, which is the “20th-day yule,” the 20th day after Christmas Eve. Candies and cookies that decorated the tree are eaten. The feast held during this event is called Knut’s party. Knut, spelled Canute in Danish, was the patron saint of Denmark who was assassinated and canonized for his efforts to secure Denmark from usurpers.
After the festive Christmas Eve dinner, someone dresses up as Tomte. Tomte is a Christmas gnome, who according to Swedish myth, lives on a farm or in the forest. Tomte looks a little like Santa Claus and hands out gifts to the family while saying funny rhymes.
A popular Christmas tradition in Sweden is to serve, a special rice porridge with one almond in it. Traditionally, the person who finds the almond gets to make a wish or is believed to get married the coming year. They also feast on Stollen Bread and Mulled Wine(you can find these recipes in the German Traditions in this article.).
1 cup rice
1 1/4 cups water
1 tbsp butter
pinch of salt
1 small cinnamon stick
3 cups 2% milk
1/3 – 1/2 cup sugar
1 tsp vanilla
1/2 tsp salt
1/4 – 1/2 cup half and half cream
Measure rice into a sieve and rinse under cold water to remove all starch until water runs clear. Add rice, water, butter, pinch of salt and one small cinnamon stick to a medium-sized pot and put on high. Once the rice mixture boils, turn to simmer and finish cooking until all the water is absorbed. This takes about 10 minutes.
In a medium-sized bowl, measure out milk and eggs and whisk together. Whisk in sugar, vanilla and salt to blend.
Once rice is cooked, pour the milk mixture into the pot. Bring back to a simmer then cook for an additional 40 minute or so, until the rice pudding is thickened. Add 1/4 – 1/2 cup half and half cream, this makes the pudding whiter in colour which is so much prettier!
Remove from heat and stir in 1 almond.
German Holiday Traditions
In almost every German city, people celebrate the holiday season with at least one (or a dozen) trips to a traditional Christmas market(Weihnachtsmärkte). These seasonal events, which date back to the 15th century. Their original purpose was to provided food and practical supplies for the cold winter season. Soon the markets became a beloved holiday tradition and a great way to get into the Christmas spirit. The city of Dresden is proud to have the oldest Christmas market in Germany, which goes back to 1434.
Since the Middle Ages, Germans brought into their homes evergreen boughs, yew, juniper, mistletoe, holly, or any plant that maintained its green color through the hard winter months. Even in parts of Northern Germany, where forests were sparse, a similar tradition took hold. The people in Northern Germany used Christmas pyramids (Weihnachtspyramiden). The pyramid form was created using sticks that were then decorated with fir branches.
The first known official Christmas tree(Tannenbaum) was set up in 1419 in Freiburg by the town bakers. They decorated the tree with fruits, nuts, and baked goods, which the children were allowed to remove and eat on New Year’s Day. By 1800, the custom of bringing a tree into the home was firmly established as a holiday tradition in many German-speaking regions and spread throughout Europe, and eventually, around the world. The German Tannenbaum is usually put up and decorated on Christmas Eve, though some families opt to erect their tree during the Advent season. Germans also usually continue to use real lit candles instead of electric lights on the tree.
The Advent Wreath
The Advent wreath (Adventskranz) is adorned with four candles, one of which is lit on each of the four Sundays preceding Christmas. The first Advent wreath, which appeared in the mid-19th century, had 4 larger candles and 19 smaller ones. The Advent wreath was invented by Johann Hinrich Wichern, a German pastor, who founded an orphanage in Hamburg in 1833. During the weeks leading up to Christmas, the children would ask him daily if Christmas had arrived. To make the wait easier, Wichern came up with the first Advent wreath to mark the Christmas countdown. He created his first Advent wreath out of an old wagon wheel and candles. A variation of the tradition of a ring of light existed among the Germanic tribes centuries before the celebration of modern Advent. It is believed that in this tradition fewer candles were lit with each progressive lighting to represent the shortening of the days until the solstice, at which time the Julfest celebrated the return of light. (The English word yule is a cognate with the Germanic Jul).
The Advent Calendar
Originally, families would mark the 24 days of December preceding Christmas with a chalk line on the wall. The first hand-crafted Advent calendars were produced in the mid-19th century; the first printed calendar appeared in Munich in 1903. Eventually the custom was exported all over the world.
The Advent calendar (Adventskalender) is a German invention that was originally designed to involve children in the festivities leading up to Christmas. Advent calendars are usually made of cardboard and have 24 small flaps, one of which is opened on each day leading up to Christmas. A Christmas scene or motif is found behind the window to help tell the story of Christmas. Modern Advent calendars may contain chocolate or candy behind each window, or even small toys. The Advent calendar is a more recent invention of modern capitalism.
Gingerbread makers established their own trade guild in Nuremberg in 1643, and this famous Christmas treat made its first holiday appearance in 1893. Gingerbread houses became part of German Christmas traditions after one was featured in the famous Grimm Brothers’ story of Hansel and Gretel. German families create gingerbread houses, complete with frosting and gumdrops, every December.
A great treat for cold winter days is mulled wine, a traditional Christmas drink from Germany. This hot, spiced wine, is called Glühwein in German (literally “glowing wine”); you can get it
at every German Christmas market.
whole black peppercorns
cinnamon stick (plus 4 for garnish, optional)
fruity red wine
1. With a fine grater, zest, then juice the orange.
2. With the flat side of a knife, press firmly on the cardamom pods to bruise them. In a large pot (not aluminum), combine zest, juice, cardamom, cloves, allspice, peppercorns, cinnamon, wine, sugar, and brandy. Cook over medium heat, stirring, until sugar dissolves, 1 to 2 minutes.
3. Reduce heat to low; simmer until flavors have melded, about 30 minutes. Pour through a fine-mesh sieve; garnish with cinnamon stick, if desired. Serve immediately.
German Stollen Bread
German Stollen, a loaf-shaped fruitcake made of yeast, water and flour, is traditionally eaten around Christmas time in Germany. The treat, which was first baked in Dresden in the 14th century, is filled with nuts, raisins, candied citrus, and spices, and its form is said to represent Baby Jesus in swaddling clothes.The Stollen Festival presents the world’s biggest stollen at 3,429 kilograms, 3.65 meters long, 1.75 meters wide and over a meter high. It is carried through the crowds by a team of horses and surrounded by the pastry chefs who completed the feat. You can purchase a piece of this giant stollen, with the profits going to charity.
1 package regular or quick active dry yeast
¾ cup warm water
½ cup granulated sugar
½ teaspoon salt
1 egg, separated
½ cup butter or margarine
3 ½ cup flour
½ cup blanched almonds
¼ cup chopped candied citron
¼ cup chopped candied cherries(optional)
¼ cup raisins
1 tbsp. melted butter
1 tbsp. grated lemon peel
1 tbsp. water
1 ⅓ cup powdered sugar
2 Tbsp milk
1. In large bowl, dissolve yeast in 3/4 cup warm water. Beat in granulated sugar, the salt, eggs, egg yolk, 1/2 cup butter and 1 3/4 cups of the flour with electric mixer on medium speed 10 minutes, scraping bowl constantly. Stir in remaining flour, the almonds, citron, cherries, raisins and lemon peel. Scrape batter from side of bowl. Cover and let rise in warm place 1 1/2 to 2 hours or until double. (Dough is ready if indentation remains when touched.) Cover and refrigerate egg white.
2. Stir down batter by beating about 25 strokes. Cover tightly and refrigerate at least 8 hours or overnight.
3. Grease cookie sheet. Place dough on well-floured surface; turn to coat with flour. Divide in half; press each half into 10×7-inch oval. Spread with 1 tablespoon butter. Fold ovals lengthwise in half; press only folded edge firmly. Place on cookie sheet. Beat egg white and 1 tablespoon water; brush over folded ovals. Cover and let rise in warm place 45 to 60 minutes or until double.
4. Heat oven to 375°F. Bake 20 to 25 minutes or until golden brown.
5. In medium bowl, mix powdered sugar and milk until smooth. Spread glaze over warm stollen.
Dutch Holiday Traditions
Dutch children in Holland, or the Netherlands celebrate St. Nicholas Day on December 6th. Sinterklass is a kind bishop that wears a red robe and a tall pointed mitre on his head. In dutch tradition, Sinterclass lives in Spain and travels to Amsterdam’s harbor every winter with his white horse and a huge sack full of gifts. He travels with his servants called ‘Zwarte Pieten’. On the evening that Sinterklaas arrives, children leave a shoe out by the fireplace or sometimes a windowsill and sing Sinterklaas songs. They believe that if they leave some hay and carrots in their shoes, for Sinterklaas’s horse, they will be left some sweets or small presents. They’re told that, during the night, Sinterklaas rides on the roofs on his horse and that a ‘Zwarte Piet’ will then climb down the chimney (or through a window) and put the presents and/or candy in their shoes.
A long-time tradition among farmers in the rural east is the ‘mid-winter horn blowing’. This custom begins on Advent Sunday (the fourth Sunday before Christmas) and continues until Christmas Eve.Farmers use long horns made from the wood of elder trees, and everyday at dawn they blow the horn while standing over a well to announce the coming of Christ.
Dutch Apple Beignets(apple fritters)
1 teaspoon sugar (white)
4 1/2 teaspoons dry yeast (instant)
4 cups/400 grams flour (all-purpose)
1/4 cup/50 grams sugar (white)
2 large eggs
2 cups/475 milliliters milk (whole)
1 teaspoon salt
6 medium apples (peeled, cored, and sliced into thick rounds)
8 cups /2 liters oil (vegetable or sunflower)
1. In a small bowl, mix 1 teaspoon of white sugar into 1/2 cup of lukewarm water. Sprinkle the yeast on top and allow to stand for 10 minutes (if the yeast doesn’t bubble, discard and buy new yeast as it means the yeast is no longer active). Stir to combine and set aside.
2. Mix together the flour and 1/4 cup/50 grams white sugar in a large bowl and make a well in the middle. Add the eggs as well as the yeast mixture.
3. Warm up milk in the microwave (it should be lukewarm). Add half of the milk to the well in the flour and mix until all ingredients are combined. Add the rest of the milk and whisk until smooth.
4. Cover the bowl with a damp dish towel and allow to rise in a warm area for about 1 hour. Once the dough has doubled, stir in the salt. The dough should be very slack and have an almost thick batter consistency.
5. Heat the oil in a deep pan or in a deep-fat fryer to 350 F/180 C. To check whether the oil is at the right temperature, stand the handle of a wooden spoon in the oil. If little bubbles form around it, the oil is ready.
6. Using your fingers or tweezers, dip the apple slices into the dough mixture. Gently drop each apple round into the hot oil.
7. The fritters will sink to the bottom of the pan and then pop right back up. You should be able to fry at least 6 fritters at a time. Fry until golden brown on both sides, carefully flipping when required. Drain on a tray lined with paper towels.
8. Sieve confectioners’ sugar over the apple beignets along with a dusting of ground cinnamon, and serve warm.
Garnish: confectioners’ sugar and ground cinnamon
Japanese Holiday Traditions
The majority of the population in Japan are Buddhist or Shinto. When Japanese settled in the San Luis Valley, they brought Buddhism to the area. There is a popular Buddhist holiday, Bodhi Day, which is celebrated in December and goes on for 30 days. Bodhi day is December 8th, the day Buddha achieved enlightenment on his 35th birthday nearly 2,500 years ago (‘Bodhi’ means enlightenment, so ‘Buddha’ means enlightened one).
In many Mahayana Buddhist countries, they decorate their homes or ficus tree with multi-colored lights to signify the many pathways to enlightenment. Buddhists who put up a “Christmas Tree”, do so on the 8th of December and keep it up for 30 days. They use Bodhi day is to not only recreate some of the physical and symbolic things that helped the Buddha achieve enlightenment, but to plant and grow those seeds of enlightenment within themselves.
Some Western Buddhists do celebrate Christmas, because they have families that are Christian and do not take offense to cross celebrating the holiday. Buddhists believe that Jesus exemplified what they call a “Bodhisattva”. A Bodhisattva is one that forgoes their own benefit to help others and has compassion, kindness and love for all beings. Jesus helped others in ways that are still experienced today, by showing the world immense compassion, love, kindness, and beauty and how to incorporate those values into their lives. Buddhists see Jesus as a blessing to this Earth.
Buddha is not a “God” but rather a respected teacher. He stated he was just a man who had found the meaning of life and end to suffering(enlightenment). When Buddhists bow to a statue of Buddha it is out of respect for him as a teacher, not as idol worship. In Asian countries, students everywhere bow to their teachers out of respect and humility, and the Buddha is no different.
Many Buddhists will partake in eating a meal of Milk Rice in on Bodhi Day. The maiden Sujata offered this dish to the future Buddha when he was weak from not eating, while he sat under the tree seeking enlightenment.
1 ¼ cup brown rice or white rice
Dash of salt
3 cups water
1 cup coconut milk (may use cow’s milk instead)
Wash the rice. Pick through it to make sure there are no tiny stones or bits of debris, then run cool water over it to rinse it entirely. Place it in a medium-sized pot.
Add the water and salt. Pour it over the rice, then cover the pot.
Cook the rice over medium-low heat. Continue cooking the rice, covered, until it is soft and plump, and the water is completely absorbed. This should take about 15 minutes. Take care not to burn the rice.
You can also cook the rice in a rice cooker
Reduce the heat to low and add the milk. Pour the milk in slowly and use a spoon to stir it into the rice. Make sure the heat is reduced so that the mixture simmers; if you keep it too high, the texture won’t come out right.
Simmer the rice and milk for ten minutes. Check to make sure it isn’t cooking too quickly; if so, reduce the heat.
Remove the pot from heat. The dish will have the consistency of creamy porridge. Let it cool for about five minutes.
Transfer the rice to a shallow dish. A wide, flat baking dish has the right shape. Use a spoon to scoop all of the rice into the dish and spread it evenly. Use a nonstick dish if you have one available, since rice tends to stick to the bottom of the pan. If you don’t have a nonstick dish, oil the bottom of a glass or metal baking dish.
Flatten the rice. Use the back of a wooden spoon to press the rice flat into the dish. You could also use a spatula or a piece of oiled waxed paper.
Score the rice. Use a knife to score the rice diagonally in one direction, then score it again diagonally in the other direction. This creates the classic diamond shape that Sri Lankans use to serve milk rice.
Cut the rice. After the rice has cooled a stiffened a bit, use the knife to cut it into diamond shapes. Lift the rice from the pan using a spatula and place it on a plate for serving. You can make the dish taste more milky by drizzling the top with more coconut milk after scoring it. Milk rice is traditionally served with curry.
Amish Holiday Traditions
Amish celebrate Christmas, minus the frills and commercialism most mainstream americans have adopted. Their homes will not be covered in decorations or a tree. Faith, family and friends are central to this celebration and to the Amish lifestyle. Hospitality, preparing food, and spending quality time together is part of what enables the Amish culture to thrive and grow. Amish spend time around Christmas reading the Christmas Story from the Bible and can exchange simple gifts, christmas cards and holiday foods like fudge and cookies.
Amish Never Fail Fudge
1/3 C. butter
4 1/2 C. sugar
14 1/2 oz. can evaporated milk
1 C. marshmallow cream
2 t. vanilla
1 bar (13 oz.) sweet chocolate, grated
12 oz. semisweet chocolate chips
2 C. walnuts, chopped
Combine butter, sugar and milk and boil 5 1/2 minutes. Remove from heat and add remaining ingredients except nuts. Beat until well mixed and add nuts. Spoon into buttered pan. Cool until firm then cut. This will make 5 lbs.
Amish Sour-Cream Sugar Cookies
1 cup (2 sticks) butter, softened
1½ cups sugar
3 large eggs, beaten
1 cup sour cream
2 tablespoons vanilla extract
3½ to 4 cups all-purpose flour
2 teaspoons baking powder
1 teaspoon baking soda
¹⁄3 cup shortening
1 teaspoon vanilla
4 cups powdered sugar
½ cup milk
Food coloring (optional)
Colored sprinkles, for decorating (optional)
Chocolate chips, for decorating (optional)
To make the frosting: Cream the shortening with the vanilla and 1 cup of the powdered sugar. Gradually add the milk and the rest of the powdered sugar, beating constantly. More powdered
sugar can be added to give you your desired thickness. Food coloring can also be added if you like. Spread the frosting on the cookies and decorate with colored sprinkles or chocolate chips. Let the frosting set before storing.
Preheat the oven to 350°F. Lightly grease a baking sheet.
Cream the butter and sugar together in a large bowl.
Stir in the eggs, sour cream, and vanilla.
Combine the flour, baking powder, and baking soda in a medium bowl and stir with a whisk to blend.
Add the dry ingredients to the wet ingredients and stir until a soft, firm dough is formed.
Roll the dough out to a ½-inch thickness on a floured surface.
Use your favorite shaped cookie cutters to cut out the dough.
Place the shapes on the prepared pan.
Bake until golden brown around the edges, about 10 minutes.
Remove from the oven and let cool on the pan for 5 minutes, then transfer to wire racks to cool completely.<|endoftext|>
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A consecutive fraction is a number written as a series of alternating multiplicative inverses and integer addition operators. Consecutive fractions are studied in the number theory branch of mathematics. Consecutive fractions are also known as continued fractions and extended fractions.
Consecutive fractions are any number written in the form a(0) + 1/(a(1) + 1/(a(2) + ...))) where a(0), a(1), a(2) and so on are integer constants. The consecutive fraction can continue indefinitely or finitely. Any real number can be written as a finite or infinite consecutive fraction.
Rational numbers can be written in the form p/q where p and q are both integers. Rational numbers are one of the two categories of real numbers. Any rational number can be written as a finite consecutive fraction in the form a(0) + 1/(a(1) + 1/(a(2) + ... 1/a(n))) where a(0), a(1) ... a(n) are integer constants as well.
Sciencing Video Vault
Irrational numbers cannot be written in the form p/q where "p" and "q" are two integers. Common irrational numbers include the √2, pi and e. Irrational numbers cannot be written as finite consecutive fractions, but they can be written as infinite consecutive fractions.
Calculating Finite Consecutive Fractions
To calculate the value of a finite consecutive fraction in the form a(0) + 1/(a(1) + 1/(a(2) + ...1/a(n))), where a(0), a(1) ... a(n) are integers, start from the bottom of the fraction. Solve 1/a(n), add a(n-1), divide 1 by this number and repeat until you solve the fraction. For example, consider 1 + 1/(2 + 1/(3 + 1/4)) = 1 + 1/(2 + 1/(13/4)) = 1 + 1/(2 + 4/13) = 1 + 1/(30/13) = 1 + (13/30) = 43/30.<|endoftext|>
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What is the domain and range of g(x) =(5x)/(x^2-36)?
Feb 10, 2018
$x \in \mathbb{R} , x \ne \pm 6$
$y \in \mathbb{R} , y \ne 0$
Explanation:
The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be.
$\text{solve } {x}^{2} - 36 = 0 \Rightarrow \left(x - 6\right) \left(x + 6\right) = 0$
$\Rightarrow x = \pm 6 \leftarrow \textcolor{red}{\text{ are excluded values}}$
$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne \pm 6$
$\text{or in interval notation as}$
$\left(- \infty , - 6\right) \cup \left(- 6 , 6\right) \cup \left(6 , + \infty\right)$
$\text{for range divide terms on numerator/denominator by the}$
$\text{highest power of x that is } {x}^{2}$
$g \left(x\right) = \frac{\frac{5 x}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{36}{x} ^ 2} = \frac{\frac{5}{x}}{1 - \frac{36}{x} ^ 2}$
$\text{as } x \to \pm \infty , g \left(x\right) \to \frac{0}{1 - 0}$
$\Rightarrow y = 0 \leftarrow \textcolor{red}{\text{is an excluded value}}$
$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne 0$
$\left(- \infty , 0\right) \cup \left(0 , + \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$
graph{(5x)/(x^2-36) [-10, 10, -5, 5]}<|endoftext|>
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A concussion is any injury to the brain that disrupts normal brain function typically on a temporary basis. Concussions are usually caused by a blow or jolt to the head but may occur from a hit to the body that produces a reflex force to the head.
The following is information from the American Academy of Pediatrics about concussions, including guidance on treatment and prevention.
When do concussions occur?
Concussions can happen in any sport but more often occur in collision sports, such as football, rugby, or ice hockey. They also are common in contact sports that don't require helmets, such as soccer, basketball, wrestling, cheerleading, and lacrosse. Many concussions also occur outside organized sports. For example, a child riding a bike or skateboard can fall down and hit their head on the street or an obstacle.
The symptoms of a concussion range from subtle to obvious and usually happen right after the injury but some may take hours to a few days to develop. Athletes who have had concussions may report feeling normal before their brain has fully recovered. With most concussions, the player is not knocked out or unconscious.
Symptoms of a concussion include the following:
Nausea or vomiting
Dizziness or balance problems
Double or blurry vision
Sensitivity to light
Sensitivity to noise
Feeling dazed or stunned
Feeling mentally "foggy"
Confused or forgetful about recent events
Slow to answer questions
Changes in mood—irritable, sad, emotional, nervous
Sleeping more or less than usual
Trouble falling asleep
What to do if you suspect a concussion
Concussions should be taken seriously and all athletes with suspected concussions should not return to full sports practice or participation until cleared by a doctor. A doctor can confirm the diagnosis of concussion and decide when it is OK for the athlete to return to play. Athletes who continue to play after their injury are at increased risk for worse symptoms, a more prolonged recovery and more serious injury to the brain if additional head injuries occur while recovering.
No one knows how many concussions are too many before permanent damage occurs. Repeated concussions are of greater concern, especially if each one takes longer to resolve or if a repeat concussion occurs from a light blow. The evaluating doctor needs to know about all prior concussions, including those that occurred outside of a sports setting, in order to make proper recommendations regarding return to play and future sports participation.
The best treatment for a concussion is reducing physical and mental activity. Children should be monitored often, but there is no need for wake-up checks during sleep. School attendance and work may need to be modified with tests and projects postponed. Students need to be excused from gym class or recess activities. Light physical exertion may be recommended to improve recovery. Physical therapy may be initiated as some problems associated with concussion may need active rehabilitation. Any progressive worsening of concussion symptoms or changes in behavior should be immediately reported to your doctor.
Returning to physical activity
Recovery time from concussion is variable with each injury but may be prolonged in athletes who continue to play immediately after their concussion. An athlete may feel better and want to return to play before their brain has completely recovered. Given the uncertain and unpredictable time frame for recovery, all formal sports activity should be suspended until symptoms have completely resolved at rest. A stepwise return to physical activity can begin once the athlete has clearance from their doctor. Having an athletic trainer, if available, involved in monitoring this plan can be very helpful. It is important to pay close attention to worsening symptoms (like increasing headache, nausea, or dizziness) while completing the activity progression. Any concussion-related symptoms that return with exertion are a clear indicator that the concussion has not healed. Final clearance to return to full activity should also be at the direction of a physician.
Not all concussions can be prevented, but some may be avoided. Helmets should be worn for any riding activities (like horseback, all-terrain vehicle [ATV], motorbike, bike, skateboard, or snowboard) or contact sports (like football, hockey, or lacrosse). Helmets should fit appropriately and be in good condition. Athletes should be taught safe playing techniques and to follow the rules of the game. Most importantly, every athlete needs to know how crucial it is to let their coach, athletic trainer, or parent know if they have hit their head or have symptoms of a head injury—even if it means stopping play. Never ignore a head injury, no matter how minor.
"When in Doubt, Sit Them Out!"<|endoftext|>
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a < b means a is less than b (so b is greater than a)
a £ b means a is less than or equal to b (so b is greater than or equal to a)
a ³ b means a is greater than or equal to b etc.
a > b means a is greater than b etc.
If you have an inequality, you can add or subtract numbers from each side of the inequality, as with an equation. You can also multiply or divide by a constant. However, if you multiply or divide by a negative number, the inequality sign is reversed.
Solve 3(x + 4) < 5x + 9
3x + 12 < 5x + 9
\ -2x < -3
\ x > 3/2 (note: sign reversed because we divided by -2)
Inequalities can be used to describe what range of values a variable can be.
E.g. 4 £ x < 10, means x is greater than or equal to 4 but less than 10.
Inequalities are represented on graphs using shading. For example, if y > 4x, the graph of y = 4x would be drawn. Then either all of the points greater than 4x would be shaded or all of the points less than or equal to 4x would be shaded.
x + y < 7
and 1 < x < 4 (NB: this is the same as the two inequalities 1 < x and x < 4)
Represent these inequalities on a graph by leaving unshaded the required regions (i.e. do not shade the points which satisfy the inequalities, but shade everywhere else).
Inequalities can also be represented on number lines. Draw a number line and above the line draw a line for each inequality, over the numbers for which it is true. At the end of these lines, draw a circle. The circle should be filled in if the inequality can equal that number and left unfilled if it cannot.
On the number line below show the solution to these inequalities.
-7 £ 2x – 3 < 3
This can be split into the two inequalities:
-7 £ 2x – 3 and 2x – 3 < 3
\ -4 £ 2x and 2x < 6
\ -2 £ x and x < 3
The circle is filled in at –2 because the first inequality specifies that x can equal –2, whereas x is less than (and not equal to) 3 and so the circle is not filled in at 3.
The solution to the inequalities occurs where the two lines overlap, i.e. for -2 £ x < 3<|endoftext|>
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# The Jacobian Matrix: Introducing Vector Calculus
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We learn how to construct and apply a matrix of partial derivatives known as the Jacobian matrix. In the process, we also introduce vector calculus.
The Jacobian matrix is a matrix containing the first-order partial derivatives of a function. It gives us the slope of the function along multiple dimensions.
Previously, we’ve discussed how to take the partial derivative of a function with several variables. We can differentiate the function with respect to each variable while keeping the remaining variables constant. Remember that the derivative is nothing but the slope of a function at a particular point. If we take the multivariate function
f(x, y) = x^2 + 3y
The derivative with respect to one variable x will give us the slope along the x dimension.
\frac{\partial{f(x,y)}}{\partial{x}} = 2x
Since x^2 is an exponential term, the slope becomes steeper as we move away from the origin.
Taking the partial derivative with respect to y gives us the slope along the y dimension.
\frac{\partial{f(x,y)}}{\partial{y}} = 3
3y is a linear term, therefore the slope along the y axis remains constant.
## Gradient Definition: Jacobian Row Vector
What about the total derivative with respect to x and y? Since we are differentiating with respect to x and y, it is the slope along both dimensions.
We can express this as a row vector.
df(x,y) =
\begin{bmatrix}
\frac{\partial{f(x,y)}}{\partial{x}} &
\frac{\partial{f(x,y)}}{\partial{y}}
\end{bmatrix}
This is known as the Jacobian matrix. In this simple case with a scalar-valued function, the Jacobian is a vector of partial derivatives with respect to the variables of that function. The length of the vector is equivalent to the number of independent variables in the function.
In our particular example, we can easily “assemble” the Jacobian since we already calculated the partial derivatives.
df(x,y) =
\begin{bmatrix}
2x &
3
\end{bmatrix}
As you’ll hopefully see, the Jacobian can be expressed as a vector. The vector points in the direction of the greatest slope, while its magnitude is proportional to the steepness of the slope at that particular point. This is also known as the gradient of a function. Remember that, unless you are dealing with linear functions and constant slopes, the Jacobian will differ from point to point. To be precise, we’d have to say that the Jacobian is the best linear approximation of the true gradient at a specific point.
Note: if you are unfamiliar with vectors and matrices or need a refresher, I’d suggest checking out my series on linear algebra for machine learning.
## Jacobian Matrix
As we’ve seen, the Jacobian of a function of real numbers is a vector. We can expand the definition of the Jacobian to vector-valued functions.
f(x) =
\begin{bmatrix}
f_1(x) \\
...\\
f_m(x)
\end{bmatrix}
Our function vector has m entries. The resulting Jacobian will be an m \times n matrix, where n is the number of partial derivatives. Each row m in the matrix contains the partial derivatives corresponding to the equivalent row m in the function vector.
\frac{df(x)}{dx} =
\begin{bmatrix}
\frac{\partial f_1(x)}{\partial x_1} & ... & \frac{\partial f_1(x)}{\partial x_n} \\
...\\
\frac{\partial f_m(x)}{\partial x_1} & ... & \frac{\partial f_m(x)}{\partial x_n}
\end{bmatrix}
### What’s the use of the Jacobian?
We can use the Jacobian matrix to transform from one vector space to another. Furthermore, if the matrix is square, we can obtain the determinant. The value of the Jacobian determinant gives us the factor by which the area or volume described by our function changes when we perform the transformation. Confused? Let’s do an example to make this clearer.
We take our standard vector space spanned by the vectors x and y
x =\begin{bmatrix}
1 \\
0
\end{bmatrix}\;
y =\begin{bmatrix}
0 \\
1
\end{bmatrix}
Assume we have another vector space spanned by u and v, where u and v are functions of x and y
u = 3x - y\\
v = x - 2y
Now our Jacobian matrix J can be obtained.
J =
\begin{bmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{bmatrix}
=
\begin{bmatrix}
3 & -1 \\
1 & -2
\end{bmatrix}
If you want to transform any vector from the xy vector space to the uv vector space, You multiply by the Jacobian.
You might say: “This is easy. You only took the coefficients of the functions u and v, and put them in a matrix.”
These are just linear functions. Accordingly, the gradient expressed by the Jacobian will remain constant. No calculus is necessary here. But if you are dealing with non-linear transformations, the pure linear algebra approach isn’t enough anymore. The Jacobian itself will not only contain constants because the transformation will differ from point to point.
We can still calculate the value of the determinant.
det(J) = 3 \times (-2) - (-1) \times (1) = -5
This gives us the factor by which an area is scaled when transforming from the xy to the uv space. Again, here we are dealing with a constant because functions describing the transformations are linear.
## Summary
The Jacobian points us in the direction of the highest local point. In the next post, we learn about the Hessian matrix, which helps us determine whether that point constitutes a global maximum.
This post is part of a series on Calculus for Machine Learning. To read the other posts, go to the index.
Sharing is caring<|endoftext|>
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# Hollow Circular Cylinder
Some examples of a hollow cylinder are pipes, circular buildings and bearing bushes. If $R$ is the outside radius of the cylinder and $r$ is the inside radius of the cylinder, then
(i) $V = \pi {R^2}h – \pi {r^2}h = \pi \left( {{R^2} – {r^2}} \right)h$
(ii) $V = \frac{\pi }{4}\left( {{D^2} – {d^2}} \right)$
with $D$ and $d$ being the outer and inner diameter, and $V$ is volume.
Example:
A well with a 10m inside diameter is dug 14m deep. The earth taken out of it is spread all around to a width of 5m to form an embankment. Find the height of the embankment.
Solution:
Volume of the dug out earth $= \pi {r^2}h$
$= \frac{{22}}{7} \times 5 \times 5 \times 14$
$1100\,{\text{cu}}{\text{.}}\,{\text{m}}$
Area of the embankment (shaded)
$= \pi \left( {{R^2} – {r^2}} \right)$
$= \pi \left( {{{10}^2} – {5^2}} \right)$ $= 75 \times \frac{{22}}{7}\,{\text{sq}}{\text{.m}}$
$\therefore$ Height of the embankment
$= \frac{{{\text{Volume of the earth dug out}}}}{{{\text{Area of the embankment}}}}$
$= \frac{{1100}}{{75 \times \frac{{22}}{7}}}\,\, = \frac{{14}}{3}\,\,\, = 4\frac{2}{3}\,\,{\text{m}}$
Example:
A hollow cylinder copper pipe is 21dm long. Its outer and inner diameters are 10cm and 6cm respectively. Find the volume of the copper used in making the pipe.
Solution:
Given that:
The height of the cylindrical pipe is $h = 21{\text{dm}}\,\,\, = \,210{\text{cm}}$
$\therefore$ External radius, $R = \frac{{10}}{2}\,\, = \,5{\text{cm}}$
Internal radius, $R = \frac{6}{2}\,\, = \,3{\text{cm}}$
Volume of the copper used in making the pipe
$= {\text{ Volume of external cylinder }} – {\text{ volume of internal cylinder}}$
$= \pi {R^2}h – \pi {r^2}h\,\,\,\,\,\, = \,\pi \left( {{R^2} – {r^2}} \right)h\,\,\,\, = \frac{{22}}{7}\left[ {{5^2} – {3^2}} \right] \times 210$
$= \frac{{22}}{7} \times 16 \times 210\,\,\,\, = 22 \times 16 \times 30\,\,\,\, = 10560\,{\text{cu}}{\text{.cm}}$<|endoftext|>
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Table of Contents
Comet 73P/Schwassmann-Wachmann 3NASA's Hubble Space Telescope is providing astronomers with extraordinary views of comet 73P/Schwassmann-Wachmann 3, which is falling apart right before our eyes. Recent Hubble images have uncovered many more fragments than have been reported by ground-based observers. These observations provide an unprecedented opportunity to study the demise of a comet nucleus.
Amateur and professional astronomers around the world have been tracking for years the spectacular disintegration of 73P/Schwassmann-Wachmann 3. As it plunges toward a June 6th swing around the Sun, the comet will pass Earth on May 12th, at a distance of 7.3 million miles, or 30 times the distance between Earth and the Moon.
The comet is currently comprised of a chain of over three dozen separate fragments, named alphabetically, stretching across several degrees on the sky. (The Sun and Moon each have an apparent diameter of about 1/2 of a degree.) Ground-based observers have noted dramatic brightening events associated with some of the fragments (as shown in the bottom frame) indicating that they are continuing to break-up and that some may disappear altogether.
Hubble caught two of the fragments, B and G, (top frames) shortly after large outbursts in activity. Hubble also photographed fragment C (not shown), which was less active. The resulting images reveal that a hierarchical destruction process is taking place, in which fragments are continuing to break into smaller chunks. Several dozen "mini-fragments" are found trailing behind each main fragment, probably associated with the ejection of house-sized chunks of surface material that can only be detected in these very sensitive and high-resolution Hubble images.
Sequential Hubble images of the B fragment, taken a few days apart, suggest that the chunks are pushed down the tail by outgassing from the icy, sunward-facing surfaces of the chunks, much like space-walking astronauts are propelled by their jetpacks. The smaller chunks have the lowest mass, and so are accelerated away from the parent nucleus faster than the larger chunks. Some of the chunks seem to dissipate completely over the course of several days.
Deep-freeze relics of the early solar system, cometary nuclei are porous and fragile mixes of dust and ices. They can be broken apart by gravitational tidal forces when they pass near large bodies (for example, Comet Shoemaker-Levy 9 was torn to pieces when it skirted near Jupiter in 1992, prior to plunging into Jupiter's atmosphere two years later). They can also fly apart from rapid rotation of the nucleus, break apart because of thermal stresses as they pass near the Sun, or explosively pop apart like corks from champagne bottles due to the outburst of trapped volatile gases.
"Catastrophic breakups may be the ultimate fate of most comets," says planetary astronomer Hal Weaver of the Johns Hopkins University Applied Physics Laboratory, who led the team that made the recent Hubble observations and who used Hubble previously to study the fragmentations of comets Shoemaker-Levy 9 in 1993-1994, Hyakutake in 1996, and 1999 S4 (LINEAR) in 2000. Analysis of the new Hubble data, and data taken by other observatories as the comet approaches the Earth and Sun, may reveal which of these breakup mechanisms are contributing to the disintegration of 73P/Schwassmann-Wachmann 3.
German astronomers Arnold Schwassmann and Arno Arthur Wachmann discovered this comet during a photographic search for asteroids in 1930, when the comet passed within 5.8 million miles of the Earth (only 24 times the Earth-Moon distance). The comet orbits the Sun every 5.4 years, but it was not seen again until 1979. The comet was missed again in 1985 but has been observed every return since then.
During the fall of 1995, the comet had a huge outburst in activity and shortly afterwards four separate nuclei were identified and labeled "A", "B", "C", and "D", with "C" being the largest and the presumed principal remnant of the original nucleus. Only the C and B fragments were definitively observed during the next return, possibly because of the poor geometry for the 2000-2001 apparition. The much better observing circumstances during this year's return may be partly responsible for the detection of so many new fragments, but it is also likely that the disintegration of the comet is now accelerating. Whether any of the many fragments will survive the trip around the Sun remains to be seen.
Besides Weaver, the other members of the Hubble observing team are: Carey Lisse (JHU/APL), Philippe Lamy (Laboratoire d'Astronomie Spatiale, France), Imre Toth (Hungarian Academy of Sciences), William Reach (IPAC/Caltech), and Max Mutchler (STScI). Z. Levay (STScI)
A Million Comet Pieces
This infrared image from NASA's Spitzer Space Telescope shows the broken Comet 73P/Schwassman-Wachmann 3 skimming along a trail of debris left during its multiple trips around the sun. The flame-like objects are the comet's fragments and their tails, while the dusty comet trail is the line bridging the fragments.
Comet 73P /Schwassman-Wachmann 3 began to splinter apart in 1995 during one of its voyages around the sweltering sun. Since then, the comet has continued to disintegrate into dozens of fragments, at least 36 of which can be seen here. Astronomers believe the icy comet cracked due the thermal stress from the sun.
The Spitzer image provides the best look yet at the trail of debris left in the comet's wake after its 1995 breakup. The observatory's infrared eyes were able to see the dusty comet bits and pieces, which are warmed by sunlight and glow at infrared wavelengths. This comet debris ranges in size from pebbles to large boulders. When Earth passes near this rocky trail every year, the comet rubble burns up in our atmosphere, lighting up the sky in meteor showers. In 2022, Earth is expected to cross close to the comet's trail, producing a noticeable meteor shower.
Astronomers are studying the Spitzer image for clues to the comet's
composition and how it fell apart. Like NASA's Deep Impact experiment, in
which a probe smashed into comet Tempel 1, the cracked Comet
73P/Schwassman-Wachmann 3 provides a perfect laboratory for studying the
pristine interior of a comet.
Hubble Provides Spectacular Detail of a Comet's Breakup
Hubble Space Telescope is providing astronomers with extraordinary views of Comet 73P/Schwassmann-Wachmann 3. The fragile comet is rapidly disintegrating as it approaches the Sun. Hubble images have uncovered many more fragments than have been reported by ground-based observers. These observations provide an unprecedented opportunity to study the demise of a comet nucleus. The comet is currently a chain of over three dozen separate fragments, named alphabetically, stretching across the sky by several times the angular diameter of the Moon. Hubble caught two of the fragments (B and G) shortly after large outbursts in activity. Hubble shows several dozen "mini-comets" trailing behind each main fragment, probably associated with the ejection of house-sized chunks of surface material. Deep-freeze relics of the early solar system, cometary nuclei are porous and fragile mixes of dust and ices that can break apart due to the thermal, gravitational, and dynamical stresses of approaching the Sun. Whether any of the many fragments survive the trip around the Sun remains to be seen in the weeks ahead. (Courtesy NASA, ESA, JHU/APL, STScI)
Comet 73P/Schwassmann-Wachmann 3 - Fragment B: Apr. 18, 2006
Hubble Space Telescope Advanced Camera for Surveys image of Comet 73P/Schwassmann-Wachmann 3 fragment B on April 18, 2006. (Courtesy NASA, ESA, JHU/APL, STScI)
Comet 73P/Schwassmann-Wachmann 3 - Fragment B: Apr. 19, 2006
Hubble Space Telescope Advanced Camera for Surveys image of Comet 73P/Schwassmann-Wachmann 3 fragment B on April 19, 2006. (Courtesy NASA, ESA, JHU/APL, STScI)
Comet 73P/Schwassmann-Wachmann 3 - Fragment B: Apr. 20, 2006
Hubble Space Telescope Advanced Camera for Surveys image of Comet 73P/Schwassmann-Wachmann 3 fragment B on April 20, 2006. (Courtesy NASA, ESA, JHU/APL, STScI)
Comet 73P/Schwassmann-Wachmann 3 - Fragments B, G
Ground-based color composite image of Comet 73P/Schwassmann-Wachmann 3 fragments B and G on April 21, 2006 made with a 8" f/1.5 Schmidt Camera. (Courtesy M. Jäger and G. Rhemann)<|endoftext|>
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# Question #221ff
Feb 23, 2017
$\sqrt{72 {a}^{19}} = 6 \left\mid {a}^{9} \right\mid \sqrt{2 a}$
#### Explanation:
To find the square root of the given algebraic expression we
$\text{ }$
should find the prime factorization.
$\text{ }$
$72 = 2 \times 36$
$72 = 2 \times 2 \times 18$
$72 = 2 \times 2 \times 2 \times 9$
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$72 = {2}^{3} \times {3}^{2}$
$\text{ }$
$72 {a}^{19} = {2}^{3} \times {3}^{2} \times {\left({a}^{9}\right)}^{2} \times a$
$\text{ }$
$\sqrt{72 {a}^{19}} = \sqrt{{2}^{3} \times {3}^{2} \times {\left({a}^{9}\right)}^{2} \times a}$
$\text{ }$
$\sqrt{72 {a}^{19}} = \sqrt{{2}^{2} \times 2 \times {3}^{2} \times {\left({a}^{9}\right)}^{2} \times a}$
$\text{ }$
$\sqrt{72 {a}^{19}} = \sqrt{{\left(2 \times 3 \times {a}^{9}\right)}^{2} \times 2 \times a}$
$\text{ }$
$\sqrt{72 {a}^{19}} = \left\mid 2 \times 3 \times {a}^{9} \right\mid \sqrt{2 \times a}$
$\text{ }$
$\sqrt{72 {a}^{19}} = 6 \left\mid {a}^{9} \right\mid \sqrt{2 a}$<|endoftext|>
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# Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Additional Questions
## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Additional Questions
EXERCISE 9.1
Question 1.
An unbiased die is thrown. What is the probability of getting
(i) an even number or a multiple of 3.
(ii) a number between 3 and 6.
Solution:
(i) Probability of getting an even number $$\frac{3}{6}=\frac{1}{2}$$
Probability of getting a multiple of 3 = $$\frac{2}{6}$$
Probability of getting an even multiple of 3 = $$\frac{1}{6}$$
Probability of getting an even number or
Question 2.
Two unbiased coins are tossed simultaneously find the probability of getting
Solution:
S = {HH, HT, TH, TT}
(i) probability of two heads = $$\frac{1}{4}$$
(ii) probability of one head = $$\frac{1}{2}$$
(iii) probability of at least one head = $$\frac{3}{4}$$
(iv) probability of at most one head = $$\frac{3}{4}$$
Question 3.
Find the probability that a leap year selected at random will contain 53 Sundays.
Solution:
S = {Sunday Monday, Monday Tuesday, Tuesday Wednesday, Wednesday Thursday, Thursday Friday, Friday Saturday, Saturday Sunday}
n (S) = 7; n (A) = 2; P(A) = $$\frac{2}{7}$$
Question 4.
What is the probability that a number selected from the numbers 1, 2, 3, 25 is a prime number when each of the given numbers is equally likely to be selected?
Solution:
A = {2, 3, 5, 7, 11, 13, 17, 19,23}
P(A) = $$\frac{9}{25}$$
EXERCISE 9.2
Question 1.
Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?
Solution:
A = {3, 6, 9, 12, 15, 18, 7, 14}
Question 2.
One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is
(i) an ace,
(ii) either red card or king.
Solution:
Question 3.
A bag contains 3 red and 2 blue marbles. A marble is drawn at random. What is the probability of drawing a blue marble?
Solution:
Question 4.
Two dice are thrown simultaneously. Find the probability of getting
(i) an even number as the sum.
(ii) a total of at least 10
(iii) a doublet of even number.
Solution:
n(S) = 36
Question 5.
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Solution:<|endoftext|>
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Likert Scales and Data Analyses
by I. Elaine Allen and Christopher A. Seaman
Surveys are consistently used to measure quality. For example, surveys might be used to gauge customer perception of product quality or quality performance in service delivery.
Likert scales are a common ratings format for surveys. Respondents rank quality from high to low or best to worst using five or seven levels.
Statisticians have generally grouped data collected from these surveys into a hierarchy of four levels of measurement:
- Nominal data: The weakest level of measurement representing categories without numerical representation.
- Ordinal data: Data in which an ordering or ranking of responses is possible but no measure of distance is possible.
- Interval data: Generally integer data in which ordering and distance measurement are possible.
- Ratio data: Data in which meaningful ordering, distance, decimals and fractions between variables are possible.
Data analyses using nominal, interval and ratio data are generally straightforward and transparent. Analyses of ordinal data, particularly as it relates to Likert or other scales in surveys, are not. This is not a new issue. The adequacy of treating ordinal data as interval data continues to be controversial in survey analyses in a variety of applied fields.1,2
An underlying reason for analyzing ordinal data as interval data might be the contention that parametric statistical tests (based on the central limit theorem) are more powerful than nonparametric alternatives. Also, conclusions and interpretations of parametric tests might be considered easier to interpret and provide more information than nonparametric alternatives.
However, treating ordinal data as interval (or even ratio) data without examining the values of the dataset and the objectives of the analysis can both mislead and misrepresent the findings of a survey. To examine the appropriate analyses of scalar data and when its preferable to treat ordinal data as interval data, we will concentrate on Likert scales.
Basics of Likert Scales
Likert scales were developed in 1932 as the familiar five-point bipolar response that most people are familiar with today.3 These scales range from a group of categories—least to most—asking people to indicate how much they agree or disagree, approve or disapprove, or believe to be true or false. There’s really no wrong way to build a Likert scale. The most important consideration is to include at least five response categories. Some examples of category groups appear in Table 1.
The ends of the scale often are increased to create a seven-point scale by adding “very” to the respective top and bottom of the five-point scales. The seven-point scale has been shown to reach the upper limits of the scale’s reliability.4 As a general rule, Likert and others recommend that it is best to use as wide a scale as possible. You can always collapse the responses into condensed categories, if appropriate, for analysis.
With that in mind, scales are sometimes truncated to an even number of categories (typically four) to eliminate the “neutral” option in a “forced choice” survey scale. Rensis Likert’s original paper clearly identifies there might be an underlying continuous variable whose value characterizes the respondents’ opinions or attitudes and this underlying variable is interval level, at best.5
Analysis, Generalization To Continuous Indexes
As a general rule, mean and standard deviation are invalid parameters for descriptive statistics whenever data are on ordinal scales, as are any parametric analyses based on the normal distribution. Nonparametric procedures—based on the rank, median or range—are appropriate for analyzing these data, as are distribution free methods such as tabulations, frequencies, contingency tables and chi-squared statistics.
Kruskall-Wallis models can provide the same type of results as an analysis of variance, but based on the ranks and not the means of the responses. Given these scales are representative of an underlying continuous measure, one recommendation is to analyze them as interval data as a pilot prior to gathering the continuous measure.
Table 2 includes an example of misleading conclusions, showing the results from the annual Alfred P. Sloan Foundation survey of the quality and extent of online learning in the United States. Respondents used a Likert scale to evaluate the quality of online learning compared to face-to-face learning.
While 60%-plus of the respondents perceived online learning as equal to or better than face-to-face, there is a persistent minority that perceived online learning as at least somewhat inferior. If these data were analyzed using means, with a scale from 1 to 5 from inferior to superior, this separation would be lost, giving means of 2.7, 2.6 and 2.7 for these three years, respectively. This would indicate a slightly lower than average agreement rather than the actual distribution of the responses.
A more extreme example would be to place all the respondents at the extremes of the scale, yielding a mean of “same” but a completely different interpretation from the ac-tual responses.
Under what circumstances might Likert scales be used with interval procedures? Suppose the rank data included a survey of income measuring $0, $25,000, $50,000, $75,000 or $100,000 exactly, and these were measured as “low,” “medium” and “high.”
The “intervalness” here is an attribute of the data, not of the labels. Also, the scale item should be at least five and preferably seven categories.
Another example of analyzing Likert scales as interval values is when the sets of Likert items can be combined to form indexes. However, there is a strong caveat to this approach: Most researchers insist such combinations of scales pass the Cronbach’s alpha or the Kappa test of intercorrelation and validity.
Also, the combination of scales to form an interval level index assumes this combination forms an underlying characteristic or variable.
Alternative Continuous Measures for Scales
Alternatives to using a formal Likert scale can be the use of a continuous line or track bar. For pain measurement, a 100 mm line can be used on a paper survey to measure from worst ever to best ever, yielding a continuous interval measure.
In the advent of many online surveys, this can be done with track bars similar to those illustrated in Figure 1. The respondents here can calibrate their responses to continuous intervals that can be captured by survey software as continuous values.
Your initial analysis of Likert scalar data should not involve parametric statistics but should rely on the ordinal nature of the data. While Likert scale variables usually represent an underlying continuous measure, analysis of individual items should use parametric procedures only as a pilot analysis.
Combining Likert scales into indexes adds values and variability to the data. If the assumptions of normality are met, analysis with parametric procedure can be followed. Finally, converting a five or seven category instrument to a continuous variable is possible with a calibrated line or track bar.
- Gideon Vigderhous, “The Level of Measurement and ‘Permissible’ Statistical Analysis in Social Research,” Pacific Sociological Review, Vol. 20, No. 1, 1977, pp. 61-72.
- Ulf Jakobsson, “Statistical Presentation and Analysis of Ordinal Data in Nursing Research,” Scandinavian Journal of Caring Sciences, Vol. 18, 2004, pp. 437-440.
- Rensis Likert, “A Technique for the Measurement of Attitudes,” Archives of Psychology, 1932, Vol. 140, No. 55.
- Jum C. Nunnally, Psychometric Theory, McGraw Hill, 1978.
- Dennis L. Clasen and Thomas J. Dormody, “Analyzing Data Measured by Individual Likert-Type Items,” Journal of Agricultural Education, Vol. 35, No. 4, 1994.
- Jacoby, Jacob, and Michael S. Matell, “Three-Point Likert Scales Are Good Enough,” Journal of Marketing Research, Vol. 8, No. 4, 1971, pp. 495-500.
- Jamieson, Susan, “Likert Scales: How to (Ab)use Them,” Medical Education, Vol. 38, No. 12), 2004, pp. 1,217-1,218.
I. ELAINE ALLEN is an associate professor of
statistics and entrepreneurship at Babson College in Babson Park,
MA. She has a doctorate in statistics from Cornell University in
Ithaca, NY. Allen is a senior member of ASQ.
CHRISTOPHER A. SEAMAN is a doctoral student in mathematics at the Graduate Center of City University of New York.<|endoftext|>
| 3.703125 |
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To COOK up planets, stars and people, the universe had to be preheated. In the first slivers of a second after the big bang, the universe expanded exponentially in a process called inflation, leaving the cosmos empty and cold. "Inflation is a flash-freezer," says John Giblin of Kenyon College in Gambier, Ohio. A few minutes later, energetic photons were zipping through a dense cauldron of radiation at a searing 1010 kelvin, the first stage in a cascade of events that gave rise to the matter we have today. So where did all the heat come from?
The energy that drove inflation is one option, but it would have been trapped in a field when the process ended. In the 1980s, physicists suggested that this "inflaton field" decayed into radiation, infusing the universe with heat. But models show the process would have been too slow to make all the matter we see.Read the Original Post<|endoftext|>
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Looking Back and Moving Forward
Pre-K-2
1
This final lesson reviews the work of the previous lessons and suggests a framework for summative assessment. Students will self-select a solution strategy for subtraction from the models introduced in this unit. An extension activity is suggested in which students use the mathematical knowledge and skills developed in the previous lessons to demonstrate understanding and ability to apply that knowledge to playing a new game.
With students in their seats, ask them to name the models used during this unit. Prompt them if they forget any of them. Then have them illustrate with links a fact family of their choice. Have a few students share their fact family.
Give students a story problem in which the number left is provided and they have to find the number taken away. For example, on large chart paper, write:
Kim had 8 stickers. She gave some to her friend Sara. Now Kim has 5 stickers. How many stickers did Kim give to Sara?
Provide students access to paper, crayons, links, pan balances, and number lines. Allow them to choose their solution strategies and materials. After students have solved the problem, have them share their strategies and discuss which strategies worked the best.
Next have the children choose five or six facts that they need to learn from their set of flash cards. Put the children into pairs to practice those facts using the cards they selected.
• Teacher-Generated Subtraction Problem(s)
• Links or Connecting Cubes (in two or more colors)
• Paper
• Crayons
• Pan Balances
• Number Lines
Assessment Option
Give students a subtraction sentence. Have students draw how they would solve this problem. Then have students write the rest of the fact family for this subtraction problem.
Extension
Questions for Students
1. How can you subtract 1 from a number?
[Count backward.]
2. How many weights would you need to take away from the right side to balance a scale with 6 links on the right side and 4 on the left side?
3. What will you land on if you start at 10 and take 5 hops backward on the number line?
[You will land on 5.]
4. What happens if you subtract 0 from a number?
[The number stays the same.]
5. What are the addition facts and the subtraction facts in one family where the difference is 6?
[One possible answer: 8 – 2 = 6, 8 – 6 = 2, 6 + 2 = 8, 2 + 6 = 8.]
6. What activity did you like most? Which was hardest for you? Why?
Teacher Reflection
• Can students demonstrate understanding of the terms difference, take away, and equals?
• What models were the majority of the students most comfortable with?
• Can students explain how to find differences?
• Do the students recognize the facts they know and those they have yet to learn?
• What were the greatest challenges for the students?
• What other situations would extend their experiences with subtraction?
• How might I connect the essential ideas of this unit with lessons about similar mathematics content?
Counting Back and Counting On
Pre-K-2
This lesson, which focuses on the counting model for subtraction, begins with reading a counting book. The students model the numbers as the book is read. Then they make a chain of links and write in vertical and horizontal format the differences suggested by adding and subtracting one link at a time from their chains. Finally, they draw a chain showing one link being taken away and write in two formats the difference it represents.
Taking Away Sets
Pre-K-2
This lesson encourages students to explore another model for subtraction, the familiar set model. Reading one of the many books that feature subtraction set the stage for this lesson in which the students write story problems, find differences using sets, and present results in a table. In the discussion of the table, they focus on the effects of subtracting all and subtracting 0.
Hopping Backward on the Number Line
Pre-K-2
In this lesson, students generate differences using a number line model. Because this model highlights the measurement aspect of subtraction, it is a distinctly different representation from the models presented in the previous lessons of this unit. The order property for subtraction is investigated. At the end of the lesson, children are encouraged to predict differences and solve puzzles involving subtraction.
Finding the Balance
Pre-K-2, 6-8
This lesson encourages students to explore another model of subtraction, the balance. Students will use real and virtual balances. Students also explore recording the modeled subtraction facts in equation form.
Finding Fact Families
Pre-K-2
In this lesson, students explore the relation of addition to subtraction with books and links. Then the children search for related addition and subtraction facts for a given number. They also investigate fact families, including those where one addend is 0 and where the addends are alike.
Practice Makes Perfect
Pre-K-2
During this lesson, students use what they know about fact families to play a concentration game. They will also identify subtraction facts they need to learn.
Learning Objectives
Students will be able to:
• Review the models for subtraction.
• Justify use of specific strategies for solving a subtraction problem.
• Recall subtraction facts.
NCTM Standards and Expectations
• Recognize, describe, and extend patterns such as sequences of sounds and shapes or simple numeric patterns and translate from one representation to another.
• Use concrete, pictorial, and verbal representations to develop an understanding of invented and conventional symbolic notations.
Common Core State Standards – Mathematics
-Kindergarten, Algebraic Thinking
• CCSS.Math.Content.K.OA.A.1
Represent addition and subtraction with objects, fingers, mental images, drawings1, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
-Kindergarten, Algebraic Thinking
• CCSS.Math.Content.K.OA.A.2
Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
-Kindergarten, Algebraic Thinking
• CCSS.Math.Content.K.OA.A.5
Fluently add and subtract within 5.
• CCSS.Math.Content.1.OA.A.1
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
• CCSS.Math.Content.1.OA.B.4
Understand subtraction as an unknown-addend problem. For example, subtract 10 - 8 by finding the number that makes 10 when added to 8.
• CCSS.Math.Content.1.OA.C.6
Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
• CCSS.Math.Content.2.OA.A.1
Use addition and subtraction within 100 to solve one- and two-step word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
• CCSS.Math.Content.2.OA.B.2
Fluently add and subtract within 20 using mental strategies. By end of Grade 2, know from memory all sums of two one-digit numbers.<|endoftext|>
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Also found in: Dictionary, Encyclopedia, Wikipedia.
chi-squared testa statistical routine which is a test of SIGNIFICANCE, comparing the observed results of an experiment or sample against the numbers expected from a theory or prediction. The test produces a value called chi-squared (χ2) which is:
The χ2 number is then converted to a probability value (P) using an χ2 table. If the P value is larger than 5% we can conclude that there is ‘no significant difference’ between the observed results and those expected, any deviation being due to chance. If, however, the probability is less than 5%, it must be concluded that there is a ‘significant difference’ between the observed results and those expected from theory. Note that the χ2test can only be used with data that fall into discrete categories, e.g. heads or tails, long or short, yellow or orange. Take, for example, a sample of 100 plants arising from a cross between two hybrid red parents. Three quarters of the offspring are expected to be red-flowered, one quarter white. The χ2analysis is shown in the table below.
In this example, there are two classes of data (n = 2) so there is one ‘degree of freedom’ (n - 1). Using the Table of χ2 shows that, with one degree of freedom, a χ2 value of 2.61 indicates a greater than 5% chance that the deviation between observed and expected numbers was due to chance alone, i.e. there is no significant difference between the numbers observed and those expected.<|endoftext|>
| 4.40625 |
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