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# (ii) $$\frac{105}{a} < \frac{100}{b}$$
Guest Jan 10, 2016
#2
+91786
+15
If a and b are positive numbers and 100/a =95/b,
prove that:
(i) a > b
$$\frac{100}{a}=\frac{95}{b}\\ 100b=95a\\ 95a=100b\\ a=\frac{100b}{95}\\ a=\frac{95b}{95}+\frac{5b}{95}\\ a=b+more\\ a>b$$
i) prove $$\frac{105}{a}<\frac{100}{b}$$
$$\frac{100}{a}=\frac{95}{b}\\ \frac{100}{a}\times 1.05=\frac{95}{b}\times 1.05\\ \frac{105}{a}=\frac{99.75}{b}\\ \frac{105}{a}=\frac{100}{b}-\frac{0.25}{b}\\ \frac{105}{a}=\frac{100}{b}-\;\;a\; bit\\ so\\ \frac{105}{a}<\frac{100}{b}$$
Melody Jan 10, 2016
Sort:
#1
0
If a and b are positive numbers and 100/a =95/b,
prove that:
(i) a > b
(ii) 105/a <100/b
Guest Jan 10, 2016
#2
+91786
+15
If a and b are positive numbers and 100/a =95/b,
prove that:
(i) a > b
$$\frac{100}{a}=\frac{95}{b}\\ 100b=95a\\ 95a=100b\\ a=\frac{100b}{95}\\ a=\frac{95b}{95}+\frac{5b}{95}\\ a=b+more\\ a>b$$
i) prove $$\frac{105}{a}<\frac{100}{b}$$
$$\frac{100}{a}=\frac{95}{b}\\ \frac{100}{a}\times 1.05=\frac{95}{b}\times 1.05\\ \frac{105}{a}=\frac{99.75}{b}\\ \frac{105}{a}=\frac{100}{b}-\frac{0.25}{b}\\ \frac{105}{a}=\frac{100}{b}-\;\;a\; bit\\ so\\ \frac{105}{a}<\frac{100}{b}$$
Melody Jan 10, 2016
#3
+82839
+15
If a and b are positive numbers and 100/a =95/b,
prove that:
(i) a > b
(ii) 105/a <100/b
(i) if 100/a = 95/b, this implies that 95a = 100b, which implies that a = (100/95)b......Then a must be greater than b since we have to multiply b by a quantity > 1 to get a
(ii) 105/a < 100/b..... Cross-multiplying........
105b < 100a........but, by definition, a =(100/95)b....so....
105b < 100(100/95)b .......divide both sibes by 100 →
(105/100)b < (100/95)b ........reduce the fractions →
(21/20)b < (20/19)b ........divide both sides by b →
(21/20) < (20/19) → cross-multiply, again
19*21 < 20*20 →
399 < 400......and since the left side is less than the right side......then the left side of the original inequality is also less than the right
CPhill Jan 10, 2016
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# Derive the equation of motion for a simple harmonic oscillator
A simple harmonic oscillator is a system that exhibits periodic motion, where the restoring force is proportional to the displacement from the equilibrium position.
Examples of simple harmonic oscillators include a mass attached to a spring, a pendulum, and a vibrating string. In this article, we will derive the equation of motion for a simple harmonic oscillator.
Derive the equation of motion for a simple harmonic oscillator-Consider a mass m attached to a spring with spring constant k. The mass is free to move along a horizontal axis. Let x be the displacement of the mass from the equilibrium position, and let F be the net force acting on the mass. According to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration. Therefore, we have:
• F = ma
The restoring force of the spring is given by Hooke's law, which states that the force exerted by a spring is proportional to its extension or compression. Therefore, the restoring force acting on the mass is given by:
• F = -kx
Derive the equation of motion for a simple harmonic oscillator-where the negative sign indicates that the force is directed towards the equilibrium position. Substituting this expression for F in the equation of motion, we get:
• -kx = ma
Dividing both sides by m, we get:
• a = -kx/m
Derive the equation of motion for a simple harmonic oscillator-This is the equation of motion for a simple harmonic oscillator. It is a second-order differential equation, since it relates the acceleration of the mass to its displacement from the equilibrium position. The solution to this equation is a sinusoidal function, which describes the periodic motion of the mass.
## Explain The Difference Between Kinematics And Kinetics In Mechanics
To solve this equation, we first make the assumption that the displacement of the mass from the equilibrium position is given by:
• x = A cos(ωt + φ)
where A is the amplitude of the motion, ω is the angular frequency, t is the time, and φ is the phase angle. The cosine function represents the oscillatory behavior of the mass, since it varies between -A and A as the mass moves back and forth along the axis. The angular frequency ω is related to the frequency f of the motion by the equation:
• ω = 2πf
where π is the mathematical constant pi. The phase angle φ represents the initial position of the mass at t=0.
Taking the first derivative of x with respect to time, we get:
• v = dx/dt = -Aω sin(ωt + φ)
where v is the velocity of the mass. Taking the second derivative of x with respect to time, we get:
• a = d^2x/dt^2 = -Aω^2 cos(ωt + φ)
Derive the equation of motion for a simple harmonic oscillator-Substituting these expressions for x, v, and a in the equation of motion, we get:
• -mAω^2 cos(ωt + φ) = -kA cos(ωt + φ)
Dividing both sides by -mA, we get:
• ω^2 = k/m
This equation relates the angular frequency of the motion to the spring constant and the mass of the oscillator. It is known as the angular frequency equation, and it describes the natural frequency of the oscillator, which is the frequency at which the oscillator oscillates without any external force acting on it.
Substituting the value of ω from the angular frequency equation into the expression for x, we get:
• x = A cos(ωt + φ) = A cos(√(k/m)t + φ)
This is the solution to the equation of motion for a simple harmonic oscillator. It describes the periodic motion of the mass, where the amplitude A and the phase angle φ depend on the initial conditions of the motion.
Conclusion
The equation of motion for a simple harmonic oscillator is a = -kx/m, where the acceleration of the mass is proportional to its displacement from the equilibrium position.
Derive the equation of motion for a simple harmonic oscillator-This equation is a second-order differential equation, and its solution is a sinusoidal function that describes the periodic motion of the mass. The natural frequency of the oscillator is given by the angular frequency equation ω^2 = k/m, which depends on the spring constant and the mass of the oscillator.
Derive the equation of motion for a simple harmonic oscillator-The amplitude and phase angle of the motion depend on the initial conditions of the motion. Understanding the equation of motion for a simple harmonic oscillator is important in various fields such as physics, engineering, and mathematics, as it helps in the design of machines and structures, and in the understanding of how objects move in the real world.
## FAQ.
Q. What is a simple harmonic oscillator?
Ans. A simple harmonic oscillator is a system that exhibits periodic motion, where the restoring force is proportional to the displacement from the equilibrium position.
Q. What is the equation of motion for a simple harmonic oscillator?
Ans. The equation of motion for a simple harmonic oscillator is a = -kx/m, where the acceleration of the mass is proportional to its displacement from the equilibrium position.
Q. What is the solution to the equation of motion for a simple harmonic oscillator?
Ans. The solution to the equation of motion for a simple harmonic oscillator is a sinusoidal function that describes the periodic motion of the mass.
Q. What is the natural frequency of a simple harmonic oscillator?
Ans. The natural frequency of a simple harmonic oscillator is given by the angular frequency equation ω^2 = k/m, which depends on the spring constant and the mass of the oscillator.
Q. What factors affect the motion of a simple harmonic oscillator?
Ans. The motion of a simple harmonic oscillator is affected by the mass of the oscillator, the spring constant, and the initial conditions of the motion such as the amplitude and phase angle.
Q. What is the importance of understanding the equation of motion for a simple harmonic oscillator?
Ans. Understanding the equation of motion for a simple harmonic oscillator is important in various fields such as physics, engineering, and mathematics, as it helps in the design of machines and structures, and in the understanding of how objects move in the real world.<|endoftext|>
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# What are Equivalent Fractions? - Definition & Examples
Coming up next: What is a Percent? - Definition & Examples
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• 0:00 The Parts Of A Fraction
• 0:30 Multiplication
• 1:55 Division
• 3:35 Testing For Equivalency
• 4:25 Lesson Summary
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Lesson Transcript
Instructor: Joseph Vigil
In this lesson, you'll review some fraction terminology and then learn what equivalent fractions are. Afterward, you can test your new knowledge with a brief quiz.
## The Parts of a Fraction
It will benefit us to review some fraction terminology before we define equivalent fractions.
When we write a fraction, there's always a number above the dividing bar and a number below the dividing bar. The upper number in a fraction is called the numerator. The lower number is called the denominator. So in the fraction 1/2, 1 would be the numerator, and 2 would be the denominator.
An easy trick to remember which part goes where is that denominator and down both start with the letter 'd,' so the denominator is always the number that's down in a fraction.
## Multiplication
Your sly friend has made two pizzas. He said that he cut one pizza into two big slices, and you can have one of them. He's cut the other pizza into six slices and said you can have three of them. You have to decide which would give you more pizza.
Well, three slices are more than one, but let's take a closer look...
Let's consider the fractions of the two pizzas: 1/2 and 3/6. Look at the numerators in both fractions. To get from 1 to 3, we multiply by 3. We do the same thing with the denominators: we multiply 2 by 3 to get 6.
In essence, what we've done is multiply 1/2 by 3/3, and 3/3 is just a fractional form of 1. When we multiply any number by 1, we're not changing its value. So we could multiply 1/2 by any fractional form of 1.
And on and on...
Since we're really multiplying by 1 in each case, we're not changing the value of the original fraction; we're just creating another fraction with the same value. These are called equivalent fractions. 1/2 and 3/6 may look different, but they have the same value.
Let's look at our original example, 1/2 and 3/6, graphically:
Although 1/2 and 3/6 are different in the way they're written, they cover the same portion of the circles. This is because they're equivalent fractions with the same value!
No matter which pizza you chose, you'd get the same amount of pizza. Each of the three slices would just be smaller than the one big slice.
## Division
We can also use division to create equivalent fractions. Just like multiplying a number by 1 doesn't change it's value, the same is true if we divide by 1. So we could take a fraction like 6/8 and divide it by a fractional form of 1, such as 2/2:
Let's use the circles again to see their equivalency:
Although 6/8 and 3/4 are different in the way they're written, they cover the same portion of the above circles. This is because they're equivalent fractions with the same value!
Equivalent fractions are an important tool when adding or subtracting fractions with different denominators. Let's look at an example:
Johnny bought one half of a cake. He didn't know that while he was out, his wife, Lyndon, bought one fourth of a cake. When they got home, how much cake did they have altogether?
1/2 + 1/4
When we're adding fractions, they must have the same denominator, but in this case, they have different denominators (2 and 4). However, we can make their denominators equal by changing the denominator in 1/2.
Remember that we can form equivalent fractions by multiplying the numerator and denominator by the same number. If we multiply the denominator in 1/2 by 2, we'll have a new denominator of 4, and the fractions will have the same denominator.
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Colonialism/Imperialism: The simple way to distinguish these two is to think of colonialism as practice and imperialism as the idea driving the practice. Colonialism is the implanting of settlements on a distant territory.
Colonialism in its modern form first began to take shape about 400 years ago, and it changed the economic landscape of the world forever. For one thing, it enabled Europe to get fabulously rich on the trade it produced. The foundations of what we now think of as free-market capitalism were invented during the colonial era, partly to handle trade.
It’s an undecided question in academic circles (amongst historians for instance) as to whether colonialism is important purely for its economic consequences, or whether cultural factors (such as missionary Christianity or a sense of racial superiority) also plays a part.
Imperialism is a word with a long history. It was first associated with ancient Rome (a fact that is borne out quite emphatically in the first pages of Heart of Darkness, where the presence of the Romans gives a sense of history). It didn’t begin to be used much in the English-speaking world until the late 1800s. Imperialism has a specifically expansionist connotation.
Globalization generally refers to the period since 1989, following the collapse of the Berlin Wall and the break-down of the Cold War system. In contrast to the earlier, colonial era, globalization is characterized by the decay of national boundaries and state institutions in favor of transnational economic activity. The “globalization” era has also been characterized by intensified cross-cultural interactions (facilitated by technology), as well as an explosion in migrations of various peoples in many different directions.
There is considerable debate amongst historians and economists as to when the elements that we now think of as constituting “globalization” first appeared. Some say the most important moment was 1970, when the “gold standard” was dropped. Others see the structure of the global political hierarchy, as well as the patterns of economic growth and development (where some countries have grown fabulously rich while others have largely languished, or been exploited for low-level labor) as roughly similar to that achieved in the high imperialist era of the 1890s. For these scholars, “globalization” is simply “imperialism” with a different name.
1492-1650: Period of exploration and early European colonization of the new world and some African and Asian territories. Birth of the new mercantile commodity economy (driven by “cash crops” such as sugar, tobacco, coffee, tea, textiles, etc.)
1607: British foundation of a colony at Jamestown.
1757: The Battle of Plassey – the beginning of British military superiority in India
1885: Congo Conference. Europeans carve up Africa into slices.
1914: World War I begins (largely a European war in fact). It’s seen by many as a war that puts a halt to the rampant territorial acquisitiveness of the preceding 40 years. European nations are forced to confront the consequences of their penchant for “gobbling” up colonies when they apply the same principles against their own neighbors.
1939: World War II begins, involving nearly the entire world. Not only are there dozens of sites in which battles occur, but people from the colonies fight for the major powers (we’ll see this in The English Patient, where an Indian soldier is in Italy fighting for the British army).
1945: End of World War II – the beginning of the Cold War, which will largely polarize the world into two “camps” (this is the origin of the language of “first world,” “second world,” and “third world”)
1947: Indian/Pakistani independence. The beginning of the steady decline in the British empire.
1960-1963: Most British colonies in Africa and the Caribbean become free nations, generally peacefully. Nigeria, Uganda, Tanzania, Ghana, Jamaica, Trinidad & Tobago are some of the most important on this list. Most become
1961: Marxist Revolution in Cuba; Fidel Castro comes to power. Before the revolution, Cuba had been largely a U.S. protectorate, and was heavily dominated by parasitical industries like gambling, with a strong Mafia presence.
1964: Civil rights legislation passed in the U.S. brings changes to the racial hierarchy of the U.S. economy.
1965: New laws encourage the immigration of skilled laborers from the
Late 1960s: Alongside the emergence of the counter-culture in the U.S., the first moment at which African-Americans develop relationships with newly independent African states.
1970: At an international conference, the major economic powers of the (non-communist) world abolish the “gold standard,” thus initiating a new period in economic speculation.
1979: Islamic revolution in Iran. Before the revolution, Iran had been a kind of U.S. protectorate, ruled by a U.S. supported dictator (the Shah).
1983-4: Beginnings of “glasnost” (opening) in the Soviet Union.
1989: Berlin wall falls – start of the period of contemporary Globalization by most accounts.
1991: Failed Coup in the Soviet Union leads to the break up of the USSR. The collapse of the Cold War system.
1997: Hong Kong becomes independent from England, and is returned to China.
1996-2000: Explosion of the Internet changes the rules and accelerates the pace of global interaction.<|endoftext|>
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# What is the distance between (–3, –2) and (5, 2)?
Apr 2, 2016
$4 \sqrt{5}$
#### Explanation:
The distance, $r$, between two points with coordinates $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by
$r = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$
It is an application of Pythagoras Theorem.
Therefore, the distance between $\left(- 3 , - 2\right)$ and $\left(5 , 2\right)$ is
$\sqrt{{\left(- 3 - 5\right)}^{2} + {\left(- 2 - 2\right)}^{2}} = \sqrt{64 + 16}$
$= \sqrt{80}$
$= 4 \sqrt{5}$
Apr 2, 2016
$4 \sqrt{5}$
#### Explanation:
color(blue)((-3,-2)and(5,2)
Use the distance formula
So,
color(purple)(x_1=-3,x_2=5
color(purple)(y_1=-2,y_2=2
$\therefore d = \sqrt{{\left(- 3 - 5\right)}^{2} + {\left(- 2 - 2\right)}^{2}}$
$\rightarrow \sqrt{{\left(- 8\right)}^{2} + {\left(- 4\right)}^{2}}$
$\rightarrow \sqrt{64 + 16}$
$\rightarrow \sqrt{80}$
$\rightarrow \sqrt{16 \cdot 5}$
color(green)(rArr4sqrt5<|endoftext|>
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 9.3E: Undetermined Coefficients for Higher Order Equations (Exercises)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
In Exercises [exer:9.3.1}– [exer:9.3.59} find a particular solution.
[exer:9.3.1] $$y'''-6y''+11y'-6y=-e^{-x}(4+76x-24x^2)$$
[exer:9.3.2] $$y'''-2y''-5y'+6y=e^{-3x}(32-23x+6x^2)$$
[exer:9.3.3] $$4y'''+8y''-y'-2y=-e^x(4+45x+9x^2)$$
[exer:9.3.4] $$y'''+3y''-y'-3y=e^{-2x}(2-17x+3x^2)$$
[exer:9.3.5] $$y'''+3y''-y'-3y=e^x(-1+2x+24x^2+16x^3)$$
[exer:9.3.6] $$y'''+y''-2y=e^x(14+34x+15x^2)$$
[exer:9.3.7] $$4y'''+8y''-y'-2y=-e^{-2x}(1-15x)$$
[exer:9.3.8] $$y'''-y''-y'+y=e^x(7+6x)$$
[exer:9.3.9] $$2y'''-7y''+4y'+4y=e^{2x}(17+30x)$$
[exer:9.3.10] $$y'''-5y''+3y'+9y=2e^{3x}(11-24x^2)$$
[exer:9.3.11] $$y'''-7y''+8y'+16y=2e^{4x}(13+15x)$$
[exer:9.3.12] $$8y'''-12y''+6y'-y=e^{x/2}(1+4x)$$
[exer:9.3.13] $$y^{(4)}+3y'''-3y''-7y'+6y=-e^{-x}(12+8x-8x^2)$$
[exer:9.3.14] $$y^{(4)}+3y'''+y''-3y'-2y=-3e^{2x}(11+12x)$$
[exer:9.3.15] $$y^{(4)}+8y'''+24y''+32y'=-16e^{-2x}(1+x+x^2-x^3)$$
[exer:9.3.16] $$4y^{(4)}-11y''-9y'-2y=-e^x(1-6x)$$
[exer:9.3.17] $$y^{(4)}-2y'''+3y'-y=e^x(3+4x+x^2)$$
[exer:9.3.18] $$y^{(4)}-4y'''+6y''-4y'+2y=e^{2x}(24+x+x^4)$$
[exer:9.3.19] $$2y^{(4)}+5y'''-5y'-2y=18e^x(5+2x)$$
[exer:9.3.20] $$y^{(4)}+y'''-2y''-6y'-4y=-e^{2x}(4+28x+15x^2)$$
[exer:9.3.21] $$2y^{(4)}+y'''-2y'-y=3e^{-x/2}(1-6x)$$
[exer:9.3.22] $$y^{(4)}-5y''+4y=e^x(3+x-3x^2)$$
[exer:9.3.23] $$y^{(4)}-2y'''-3y''+4y'+4y=e^{2x}(13+33x+18x^2)$$
[exer:9.3.24] $$y^{(4)}-3y'''+4y'=e^{2x}(15+26x+12x^2)$$
[exer:9.3.25] $$y^{(4)}-2y'''+2y'-y=e^x(1+x)$$
[exer:9.3.26] $$2y^{(4)}-5y'''+3y''+y'-y=e^x(11+12x)$$
[exer:9.3.27] $$y^{(4)}+3y'''+3y''+y'=e^{-x}(5-24x+10x^2)$$
[exer:9.3.28] $$y^{(4)}-7y'''+18y''-20y'+8y=e^{2x}(3-8x-5x^2)$$
[exer:9.3.29] $$y'''-y''-4y'+4y=e^{-x}\left[(16+10x)\cos x+(30-10x)\sin x\right]$$
[exer:9.3.30] $$y'''+y''-4y'-4y=e^{-x}\left[(1-22x)\cos 2x-(1+6x)\sin2x\right]$$
[exer:9.3.31] $$y'''-y''+2y'-2y=e^{2x}[(27+5x-x^2)\cos x+(2+13x+9x^2)\sin x]$$
[exer:9.3.32] $$y'''-2y''+y'-2y=-e^x[(9-5x+4x^2)\cos 2x-(6-5x-3x^2)\sin2x]$$
[exer:9.3.33] $$y'''+3y''+4y'+12y=8\cos2x-16\sin2x$$
[exer:9.3.34] $$y'''-y''+2y=e^x[(20+4x)\cos x-(12+12x)\sin x]$$
[exer:9.3.35] $$y'''-7y''+20y'-24y=-e^{2x}[(13-8x)\cos 2x-(8-4x)\sin2x]$$
[exer:9.3.36] $$y'''-6y''+18y'=-e^{3x}[(2-3x)\cos 3x-(3+3x)\sin3x]$$
[exer:9.3.37] $$y^{(4)}+2y'''-2y''-8y'-8y=e^x(8\cos x+16\sin x)$$
[exer:9.3.38] $$y^{(4)}-3y'''+2y''+2y'-4y=e^x(2\cos2x -\sin2x)$$
[exer:9.3.39] $$y^{(4)}-8y'''+24y''-32y'+15y=e^{2x}(15x\cos2x+32\sin2x)$$
[exer:9.3.40] $$y^{(4)}+6y'''+13y''+12y'+4y=e^{-x}[(4-x)\cos x-(5+x)\sin x]$$
[exer:9.3.41] $$y^{(4)}+3y'''+2y''-2y'-4y=-e^{-x} (\cos x-\sin x)$$
[exer:9.3.42] $$y^{(4)}-5y'''+13y''-19y'+10y=e^x (\cos2x+\sin2x)$$
[exer:9.3.43] $$y^{(4)}+8y'''+32y''+64y'+39y=e^{-2x}[(4-15x)\cos3x-(4+15x)\sin 3x]$$
[exer:9.3.44] $$y^{(4)}-5y'''+13y''-19y'+10y=e^x[(7+8x)\cos 2x+(8-4x)\sin2x]$$
[exer:9.3.45] $$y^{(4)}+4y'''+8y''+8y'+4y=-2e^{-x} (\cos x-2\sin x)$$
[exer:9.3.46] $$y^{(4)}-8y'''+32y''-64y'+64y=e^{2x} (\cos2x-\sin2x)$$
[exer:9.3.47] $$y^{(4)}-8y'''+26y''-40y'+25y=e^{2x}[3\cos x-(1+3x)\sin x]$$
[exer:9.3.48] $$y'''-4y''+5y'-2y=e^{2x}-4e^x-2\cos x+4\sin x$$
[exer:9.3.49] $$y'''-y''+y'-y=5e^{2x}+2e^x-4\cos x+4\sin x$$
[exer:9.3.50] $$y'''-y'=-2(1+x)+4e^x-6e^{-x}+96e^{3x}$$
[exer:9.3.51] $$y'''-4y''+9y'-10y=10e^{2x}+20e^x\sin2x-10$$
[exer:9.3.52] $$y'''+3y''+3y'+y=12e^{-x}+9\cos2x-13\sin2x$$
[exer:9.3.53] $$y'''+y''-y'-y=4e^{-x}(1-6x)-2x\cos x+2(1+x)\sin x$$
[exer:9.3.54] $$y^{(4)}-5y''+4y=-12e^x+6e^{-x}+10\cos x$$
[exer:9.3.55] $$y^{(4)}-4y'''+11y''-14y'+10y=-e^x(\sin x+2\cos2x)$$
[exer:9.3.56] $$y^{(4)}+2y'''-3y''-4y'+4y=2e^x(1+x)+e^{-2x}$$
[exer:9.3.57] $$y^{(4)}+4y=\sinh x\cos x-\cosh x\sin x$$
[exer:9.3.58] $$y^{(4)}+5y'''+9y''+7y'+2y=e^{-x}(30+24x)-e^{-2x}$$
[exer:9.3.59] $$y^{(4)}-4y'''+7y''-6y'+2y=e^x(12x-2\cos x+2\sin x)$$
[exer:9.3.60] $$y'''-y''-y'+y=e^{2x}(10+3x)$$
[exer:9.3.61] $$y'''+y''-2y=-e^{3x}(9+67x+17x^2)$$
[exer:9.3.62] $$y'''-6y''+11y'-6y=e^{2x}(5-4x-3x^2)$$
[exer:9.3.63] $$y'''+2y''+y'=-2e^{-x}(7-18x+6x^2)$$
[exer:9.3.64] $$y'''-3y''+3y'-y=e^x(1+x)$$
[exer:9.3.65] $$y^{(4)}-2y''+y=-e^{-x}(4-9x+3x^2)$$
[exer:9.3.66] $$y'''+2y''-y'-2y=e^{-2x}\left[(23-2x)\cos x+(8-9x)\sin x\right]$$
[exer:9.3.67] $$y^{(4)}-3y'''+4y''-2y'=e^x\left[(28+6x)\cos 2x+(11-12x)\sin2x\right]$$
[exer:9.3.68] $$y^{(4)}-4y'''+14y''-20y'+25y=e^x\left[(2+6x)\cos 2x+3\sin2x\right]$$
[exer:9.3.69] $$y'''-2y''-5y'+6y=2e^x(1-6x),\quad y(0)=2, \quad y'(0)=7,\quad y''(0)=9$$
[exer:9.3.70] $$y'''-y''-y'+y=-e^{-x}(4-8x),\quad y(0)=2, \quad y'(0)=0,\quad y''(0)=0$$
[exer:9.3.71] $$4y'''-3y'-y=e^{-x/2}(2-3x),\quad y(0)=-1, \quad y'(0)=15,\quad y''(0)=-17$$
[exer:9.3.72] $$y^{(4)}+2y'''+2y''+2y'+y=e^{-x}(20-12x),\, y(0)=3,\; y'(0)=-4,\; y''(0)=7,\; y'''(0)=-22$$
[exer:9.3.73] $$y'''+2y''+y'+2y=30\cos x-10\sin x, \quad y(0)=3,\quad y'(0)=-4,\quad y''(0)=16$$
[exer:9.3.74] $$y^{(4)}-3y'''+5y''-2y'=-2e^x(\cos x-\sin x),\; y(0)=2,\; y'(0)=0,\; y''(0)~=~-1, \; y'''(0)=-5$$
[exer:9.3.75] Prove: A function $$y$$ is a solution of the constant coefficient nonhomogeneous equation
$a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=e^{\alpha x}G(x) \tag{A}$
if and only if $$y=ue^{\alpha x}$$, where $$u$$ satisfies the differential equation
$a_0u^{(n)}+{p^{(n-1)}(\alpha)\over(n-1)!}u^{(n-1)}+ {p^{(n-2)}(\alpha)\over(n-2)!}u^{(n-2)}+\cdots+p(\alpha)u=G(x) \tag{B}$
and
$p(r)=a_0r^n+a_1r^{n-1} + \cdots + a_n\nonumber$
is the characteristic polynomial of the complementary equation
$a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=0.\nonumber$
[exer:9.3.76] Prove:
The equation
$\begin{array}{lcl} a_0u^{(n)}&+&\{p^{(n-1)}(\alpha)\over(n-1)!}u^{(n-1)}+ \{p^{(n-2)}(\alpha)\over(n-2)!}u^{(n-2)}+\cdots+p(\alpha)u\\ &=&\left(p_0+p_1x+\cdots+p_kx^k\right)\cos \omega x\\&&\,+ \left(q_0+q_1x+\cdots+q_kx^k\right)\sin\omega x \end{array} \tag{A}$
has a particular solution of the form
$u_p=x^m\left(u_0+u_1x+\cdots+u_kx^k\right)\cos\omega x+ \left(v_0+v_1x+\cdots+v_kx^k\right)\sin\omega x.\nonumber$
If $$\lambda+i\omega$$ is a zero of $$p$$ with multiplicity $$m\ge1$$, then (A) can be written as
$a(u''+\omega^2 u)= \left(p_0+p_1x+\cdots+p_kx^k\right)\cos\omega x+ \left(q_0+q_1x+\cdots+q_kx^k\right)\sin\omega x,\nonumber$
which has a particular solution of the form
$u_p=U(x)\cos\omega x+V(x)\sin\omega x,\nonumber$
where
$U(x)=u_0x+u_1x^2+\cdots+u_kx^{k+1},\,V(x)=v_0x+v_1x^2+\cdots+v_kx^{k+1}\nonumber$
and
$\begin{array}{rcl} a(U''(x)+2\omega V'(x))&=&p_0+p_1x+\cdots+p_kx^k\\[10pt] a(V''(x)-2\omega U'(x))&=&q_0+q_1x+\cdots+q_kx^k. \end{array}\nonumber$<|endoftext|>
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A verbal art like poetry is reflective; it stops to think. Music is immediate, it goes on to become.” – W.H. Auden. This quote best explains the complex art of music. Music is an elaborate art form that will always remain ever changing. Music developed drastically from it’s beginning in the Prehistoric era to the 14th Century.
The exact origin of music is unknown. It is known that music was used in prehistoric times in magical or spiritual rituals but no other use is known. This knowledge is borne out of the fact that music still forms a vital part of most religious ceremonies today.
The history of Greek music is problematic. Although there are frequent references to musical performance in Greek manuscripts, there are less than twelve fragments of actual Greek music, including both vocal and instrumental music, that have survived. It is impossible to fully understand the notation to make an authentic performance.
For the Greeks, music was of divine origin. According to Greek mythology, the gods themselves invented music and it’s instruments. Many of the early myths told of the powerful effects of music. Music played an important part in both the public and private lives of the Greeks. They believed it could deeply affect human behavior. Greek music was built up of a series of distinct modes, each with it’s own name. According to the doctrine of ethos, each mode was so powerful that it gave music the ability to influence human actions in a precise way. The Phrygian mode expressed passionate and intimate emotions, where as the Dorian mode produced forceful, rigid feelings.
In later Greek history the doctrine of ethos was widely argued by the most philosophical of men. Plato and Aristotle both had broadly different views on the power and importance of music. The persocratic philosopher Pythagoras was even interested enough in music to develop the numerical octave system that we still use today. The Classical Greeks used music in much of their drama and by the time Greece was made a Roman province, music dominated dramatic performances and social activities.
There is not a great deal of original Roman music. Most of the music that did come out of the Roman era was derived from the Greeks. Despite this, there was definite musical activity in the later Roman Empire. An ample amount of evidence survived for instruments and a good deal of theory also. But by in large Greek music remained the most popular in the Roman Empire.
Early Christian music drew off of Jewish sources. The custom of singing sacred verses at services was an ancient Jewish tradition that goes back to Mesopotamian sources. As the Church grew the music fell more into the care of professionals and it became greatly complex. Soon the church officials became fearful that the music was overpowering the worship and music was regulated in worship services.
The beginnings of Byzantine music was mainly based on Syraic and Hebrew music. Most music of this time was written for religious purposes and was strictly regulated by church officials. By 386 AD Saint Ambrose of Milian began the use of vernacular hymns in the church worship services.
The development of the music of the Early Middle Ages was intertwined with the grow the of the Christian church. Chanting of scriptures and prayers was practiced earlier. By the sixth century AD modalchant, known as plainchant, had increased so greatly that Pope Gregory I had it collected and organized, and it came to be called Gregorian chant. The chant did not have a regular rhythm but was fitted to the natural accents of the Latin words. Like all previous music, each chant consisted of a single melody, and all the singers sang the same notes. This type of music is called monophonic, or one-voiced.
Nonreligious, or secular, music was composed by wandering poets who sang of chivalry and courtly love in the twelfth and thirteenth centuries. In France they were either jongleurs, itinerant minstrels who made a living from their songs, or troubadour and troueres, aristocrats who sang for the love of music. In Germany the poet-musicians were called minnesingers. Some two thousand minnstrel melodies are preserved in old manuscripts.
The discovery that two voices could sing two separate melodies at the same time and still produce a pleasing sound occurred sometime around the ninth century. This discovery was called Polyphony. The genesis of polyphony occurred in France, first in very basic notation lacking precise pitch. By the twelfth century, polyphony was developed into elaborate forms in two centers: Paris and St. Martial de Limoges, the latter preceding the former. By this time, precise pitch notation is given, and so the footing is fairly firm.
The first experiments in polyphony were called organum. A secong voice or voices sang the chant melody at perhaps an interval of a fourth or fifth above the original, or tenor. Sometimes the two moved in opposite directions. Above the tenor a more elaborate part might be sung. As the two parts become more independent, often two distinct melodies proceeded at the same time. When the third and fourth parts were added, the music became truly polyphonic.
Sometime after the mid-twelfth century, a new Notre Dame Cathedral was being built in Paris, and with it grew a school of composers. Two names have been preserved from that school- Leonin and Perotin. They stretched the organum to unheard-of lengths and embellished it with flourishes of long melismas, or many notes sung to one syllable. New rhythmic patterns developed, as did repetitions of motifs, sequential patterns, and imitation.
Out of this developed the motet, originally in Latin on a sacred text. Unlike the organum, the text was sung in the upper voices as well as the tenor. Bilingual motets (French-Latin, English-Latin) arose, and secular texts or combinations of sacred and secular texts were used. Tenors were sometimes chosen from French popular songs instead of from plainchant. Instruments played lower parts, making the motet an accompanied solo song.
The period culminated in the works of Guillaume de Machaut. He left 23 motets, more than 100 secular songs, and a mass. They are characterized by excellent craftsmanship with colorful melodic and harmonic inflections and constantly shifting rhythms.
The later fourteenth century was a period during which the French style dominated secular composition throughout Europe. It modified to reflect local tastes in Italy and England, but remained largely French in inspiration for some decades. However, Italian composers continued to develop a more native idiom, combining French Ars Nova ideas with indigenous genres.
Music as a whole progressed slowly through the many years it’s been around, taking it’s time to perfect itself. It can be seen that in just the last few hundred years,1300- 1500AD, that the styles in music took a dramatic leap towards the future. It will be interesting to see how swiftly music will accustom itself to the next thousand years.
Cunningham, Lawrence S., Reich, John J. Culture and Values; A Survey of Western Humanties. New York: Harcourt Brace College Publishers, 1994.
McComb, Todd Michel. ;a href=”http://www.music.indiana.edu”;http://www.music.indiana.edu The Origin of Polyphony: 1996
McComb, Todd M. ;a href=”http://www.music.indiana.edu”;http://www.music.indiana.edu A Selection of Medieval Music: 1996
McComb, Todd M. ;a href=”http://www.music.indiana.edu”;http://www.music.indiana.edu Early Music: 1996
“Music,Classical.” Comptom’s Interactive Encyclopedia, Inc. 1995.
Stinson, John music14.html @ <a href=”http://www.lib.latrobe.edu.au”>http://www.lib.latrobe.edu.au The Music of the 14th century: 1997<|endoftext|>
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Grade 5 level
To provide background information about school supplies.
Students will be able to describe and introduce classroom objects.
Students will be able to name the classroom objects.
Students will be able to recognize the school supplies.
Procedure (33-42 minutes)
- T asks Ss what comes to their minds when they hear 'school' -Ss do brainstorming. -T writes the answers on the board and creates a mindmap.
-T distributes the reading text to the students. -Ss read the text. -After reading the text, Ss circles the picture of objects mentioned in the text. -T gives the question sheet about the text and ask them to answer the questions according to text. -After finishing the exercise, teacher elicits answers from the class and whole class check together.
-Teacher distributes the reaading text "What am I?" to the students. - Students reads the short paragraphs and decide which classroom object it is. -Students match the objects and their definitions individually.
-After matching the paragraph and objects, a student reads a paragraph and other students tries to find the correct object.<|endoftext|>
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LCM of 40 and 50 is the the smallest number among all usual multiples that 40 and 50. The first few multiples the 40 and 50 are (40, 80, 120, 160, 200, 240, 280, . . . ) and (50, 100, 150, 200, 250, 300, . . . ) respectively. There room 3 typically used methods to uncover LCM the 40 and also 50 - by element factorization, by division method, and by listing multiples.
You are watching: What is the least common multiple of 40 and 50
1 LCM that 40 and 50 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM of 40 and also 50 is 200.
Explanation:
The LCM of two non-zero integers, x(40) and also y(50), is the smallest hopeful integer m(200) the is divisible through both x(40) and y(50) without any kind of remainder.
Let's look at the different methods for finding the LCM of 40 and also 50.
By prime Factorization MethodBy Listing MultiplesBy division Method
### LCM of 40 and 50 by prime Factorization
Prime administrate of 40 and 50 is (2 × 2 × 2 × 5) = 23 × 51 and also (2 × 5 × 5) = 21 × 52 respectively. LCM of 40 and 50 deserve to be obtained by multiply prime determinants raised to their respective highest possible power, i.e. 23 × 52 = 200.Hence, the LCM of 40 and 50 by prime factorization is 200.
### LCM the 40 and 50 by Listing Multiples
To calculation the LCM the 40 and also 50 by listing the end the common multiples, we can follow the given below steps:
Step 1: perform a few multiples of 40 (40, 80, 120, 160, 200, 240, 280, . . . ) and 50 (50, 100, 150, 200, 250, 300, . . . . )Step 2: The usual multiples indigenous the multiples of 40 and 50 space 200, 400, . . .Step 3: The smallest common multiple that 40 and 50 is 200.
∴ The least usual multiple the 40 and 50 = 200.
### LCM that 40 and also 50 by department Method
To calculate the LCM that 40 and also 50 through the department method, we will divide the numbers(40, 50) by your prime components (preferably common). The product of this divisors gives the LCM the 40 and 50.
Step 3: proceed the measures until just 1s space left in the critical row.
The LCM the 40 and also 50 is the product of every prime number on the left, i.e. LCM(40, 50) by department method = 2 × 2 × 2 × 5 × 5 = 200.
☛ likewise Check:
Example 2: Verify the relationship between GCF and LCM that 40 and 50.
Solution:
The relation in between GCF and LCM of 40 and also 50 is provided as,LCM(40, 50) × GCF(40, 50) = Product that 40, 50Prime administrate of 40 and 50 is given as, 40 = (2 × 2 × 2 × 5) = 23 × 51 and also 50 = (2 × 5 × 5) = 21 × 52LCM(40, 50) = 200GCF(40, 50) = 10LHS = LCM(40, 50) × GCF(40, 50) = 200 × 10 = 2000RHS = Product that 40, 50 = 40 × 50 = 2000⇒ LHS = RHS = 2000Hence, verified.
## FAQs top top LCM the 40 and 50
### What is the LCM of 40 and also 50?
The LCM the 40 and 50 is 200. To discover the least common multiple the 40 and 50, we need to discover the multiples that 40 and also 50 (multiples of 40 = 40, 80, 120, 160 . . . . 200; multiples of 50 = 50, 100, 150, 200) and choose the the smallest multiple the is exactly divisible through 40 and also 50, i.e., 200.
### How to find the LCM of 40 and also 50 by element Factorization?
To discover the LCM that 40 and also 50 using prime factorization, us will uncover the element factors, (40 = 2 × 2 × 2 × 5) and (50 = 2 × 5 × 5). LCM of 40 and also 50 is the product the prime determinants raised to your respective highest possible exponent among the numbers 40 and 50.⇒ LCM the 40, 50 = 23 × 52 = 200.
### What is the the very least Perfect Square Divisible by 40 and also 50?
The least number divisible through 40 and also 50 = LCM(40, 50)LCM of 40 and 50 = 2 × 2 × 2 × 5 × 5 ⇒ least perfect square divisible by each 40 and also 50 = LCM(40, 50) × 2 = 400 Therefore, 400 is the forced number.
### Which of the following is the LCM that 40 and also 50? 10, 200, 40, 36
The value of LCM that 40, 50 is the smallest common multiple that 40 and also 50. The number to solve the given problem is 200.
### If the LCM the 50 and also 40 is 200, find its GCF.See more: The Cell Plate Is Formed During A Anaphase Class 9 Biology Cbse
LCM(50, 40) × GCF(50, 40) = 50 × 40Since the LCM of 50 and 40 = 200⇒ 200 × GCF(50, 40) = 2000Therefore, the greatest usual factor (GCF) = 2000/200 = 10.<|endoftext|>
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# How many ways can I color 8 tiles with colors? (easy)
I'm trying to learn the very basics of permutations and combinations and since I'm completely new to this subject I'd like to get a good explanation to the following problems. My task is this:
1) Suppose we have 8 tiles and that we're going to color exactly two of them with blue and exactly two with green. How many way can we do that?
2) Suppose we have eight tiles, but this time we're going to color exactly two tiles blue such that that they're not adjacent. Can a genereral formula be derived from this?
My work so far:
1) The ordered selection of eight tiles is $8!$ and we have to divide this by the number of permutations of colored and non-colored tiles. The number of permutations of the non-colored tiles are $4!$ and $2!$ for each color. If this is correct we should get the following: $$\dfrac{8!}{4!2!2!}=\dfrac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot1}{4\cdot 3\cdot 2\cdot 1\cdot 2\cdot 1\cdot 2\cdot 1}=7\cdot 6\cdot 5\cdot 2=420$$
I'm not sure how to tackle problem 2). So to sum up, verification of problem 1) and help on problem two, atleast good hints. Also if you have good, relatively short guides (pages/channels/notes etc.) on combinatorics/permutations, feel free to post it. Thanks in advance.
• A more intuitive way (at least for way) of understanding the solution to the first problem is $\binom{8}{2}\binom{6}{2},$ i.e., how many ways can we choose two out of first eight, then six. – Bobson Dugnutt Oct 27 '16 at 7:21
1) If it is so, we subtract from 420 the colorings which have the two blue tiles adjacent. With 8 tiles there are 7 ways to choose 2 adjacent tiles. Then, among the remaining 6 tiles we choose the two tiles to be colored green in $\frac{6!}{4!2!}=15$ ways.
2) If the color green is not used in part 2, we simply subtract from $\frac{8!}{6!2!}=28$ the 7 ways to choose 2 adjacent tiles.<|endoftext|>
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## Mathematical operation and symbol notation
#### Mathematical operation and symbol notation
Direction: In each of the following questions, two rows of numbers are given. The resultant number in each row is to be worked out separately based on the following rules and the questions below the rows of number are to be answered. The operations of numbers progress from left to the right.
Rules
(i) If a two-digit number is followed by another even number, the first one is to be divided by the second one.
(ii) If an even number is followed by a prime number, the two are to be multiplied.
(iii) If an odd number is followed by another odd number, the two are to be added.
(iv) if a three-digit number is followed by a two-digit number which is a perfect square, the second number is to be subtracted from the first number.
(v) If a three-digit number is followed by a two-digit number which is not a perfect square, the first number is to be divided by the second one.
1. 97 45 71
48 8 11
What is the sum of the resultants of the two rows ?
1. Rules
(i) (2-digit even number) ÷ ( Even number)
(ii) (Even number) x (Prime number)
(iii) (Odd number) + (Odd number)
(iv) (3-digit number) - (2-digit perfect square number)
(v) (3-digit number) ÷ (2-digit non-perfect square number)
Hint
1st Row 97 45 71 ⇒ 97 + 45 = 142 [rule (iii)]
##### Correct Option: A
Rules
(i) (2-digit even number) ÷ ( Even number)
(ii) (Even number) x (Prime number)
(iii) (Odd number) + (Odd number)
(iv) (3-digit number) - (2-digit perfect square number)
(v) (3-digit number) ÷ (2-digit non-perfect square number)
Solution
1st Row 97 45 71 ⇒ 97 + 45 = 142 [rule (iii)]
⇒ 142 ÷ 71 = 2 [rule (v)]
2nd Row 48 8 11 ⇒ 48 ÷ 8 = 6 [rule (i)]
⇒ 6 x 11 = 66 [rule (ii)]
∴ Required sum = 2 + 66 = 68
1. 125 64 33
282 x 39
If 'x' is the resultant of the first row, what will be the resultant of the second row ?
1. Rules
(i) (2-digit even number) ÷ ( Even number)
(ii) (Even number) x (Prime number)
(iii) (Odd number) + (Odd number)
(iv) (3-digit number) - (2-digit perfect square number)
(v) (3-digit number) ÷ (2-digit non-perfect square number)
Hint
1st Row 125 64 33 ⇒ 125 - 64 = 61 [rule (iv)]
##### Correct Option: B
Rules
(i) (2-digit even number) ÷ ( Even number)
(ii) (Even number) x (Prime number)
(iii) (Odd number) + (Odd number)
(iv) (3-digit number) - (2-digit perfect square number)
(v) (3-digit number) ÷ (2-digit non-perfect square number)
Solution
1st Row 125 64 33 ⇒ 125 - 64 = 61 [rule (iv)]
⇒ 61 + 33 = 94 [rule (iii)]
= x
2nd Row 282 x 39 ⇒ 282 94 39
⇒ 282 ÷ 94 = 3 [rule (v)]
⇒ 3 + 39 = 42 [rule (iii)]
∴ Resultant of second row = 42
1. 24 7 81
x 27 19
If 'x' is the resultant of the first row, what will be the resultant of the second row ?
1. Rules
(i) (2-digit even number) ÷ ( Even number)
(ii) (Even number) x (Prime number)
(iii) (Odd number) + (Odd number)
(iv) (3-digit number) - (2-digit perfect square number)
(v) (3-digit number) ÷ (2-digit non-perfect square number)
Hint
1st Row 24 7 81 ⇒ 24 x 7 = 168 [rule (ii)]
##### Correct Option: D
Rules
(i) (2-digit even number) ÷ ( Even number)
(ii) (Even number) x (Prime number)
(iii) (Odd number) + (Odd number)
(iv) (3-digit number) - (2-digit perfect square number)
(v) (3-digit number) ÷ (2-digit non-perfect square number)
Solution
1st Row 24 7 81 ⇒ 24 x 7 = 168 [rule (ii)]
⇒ 168 - 81 = 87 [rule (iv)]
= x
2nd Row x 27 19 ⇒ 87 27 19
⇒ 87 + 27 = 114 [rule (iii)]
⇒ 114 ÷ 19 = 6 [rule (v)]
∴ Resultant of second row = 6
1. Which of the following sets of operation with the usual notations replacing the starts in the order given makes the statements valid ?
100 * √16 * √225 * √1
1. 100 + √16
= √225 -√1
##### Correct Option: B
100 + √16
= √225 -√1
10 + 4 = 15 - 1
⇒ 14 = 14
1. Which of the following interchange of signs would make the given equation correct ?
5 + 3 x 8 - 12 ÷ 4 = 3
1. Apply All option one by one
5 + 3 x 8 - 12 ÷ 4 = 3
Now, after changing the sign for option C
5 + 3 x 8 ÷ 12 - 4 = 3
Apply the BODMAS Rule
5 + 3 x 8/12 - 4 = 3
##### Correct Option: C
Apply All option one by one
5 + 3 x 8 - 12 ÷ 4 = 3
Now, after changing the sign for option C
5 + 3 x 8 ÷ 12 - 4 = 3
Apply the BODMAS Rule
5 + 3 x 8/12 - 4 = 3
5 + 2 - 4 = 3
3 = 3<|endoftext|>
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Glacier Boundaries over
North America (Wisconsin Era)
Long before European explorers reached the Red River valley, around the time that ancient Greece was rising to power, Lockport and St. Clements were part of a bountiful region of hunting, fishing, and trading. “For the 3000 years before recorded history, native peoples came here to camp, hunt, and fish” (Kenosewun – Signage).
From 70,000 to 10,000 years ago, 97% of Canada was covered with ice. This continental glacier, known as the Wisconsin glacier era, made life in most areas of North America uninhabitable. However, around 10,000 years ago, when the glacier began to recede, much rich and fertile land was uncovered. With the melting of this glacier, endless amounts of water were deposited on the earth. By 8000 BC, current day St. Clements was covered by the waters of Glacial Lake Agassiz, the largest freshwater lake on the continent. Although first explorers began arriving in Manitoba as early as 11,000 years ago, it was not until 6300-6000 BC that St. Clements’ region dried enough to become habitable.
How do we know this? Archeology provides the information. Archeology is the study of previous ways of life by analyzing all types of natural surroundings, and preserved objects, and mentally recreating a cultural lifestyle. Through archeology, it is estimated that First Nations explorers may have traveled the Selkirk area anywhere from 6500-5500 BC. However, more established bands were not present in the area until near 1000 BC, when they arrived in present day Lockport.
Archeology at Lockport Site
Archeological surveys done in the 1980s revealed much historic evidence of aboriginal settlement at the Lockport area. But why was Lockport such a favourable site? Before the Lockport dam was completed in 1910, this area along the Red River was home to the St. Andrews Rapids. These rapids, falling over 15 feet in a 16-kilometer distance, attracted hunters over 300 years ago. Huge herds of bison came to cross the river at the shallow waters. The fast moving waters over the rapids was also an “ideal spawning and feeding ground for fish [… and the bedrock of the rapids] provided people with large quantities of stone for making tools” (Bringing It All Together – Signage). Spring floods, which were more common at the time, spilled nutrients onto the surrounding soil, making the land very fertile.
Before 200 BC, the climate of the area was wetter and cloudier than today, with cooler summers, and more severe winter storms and spring flooding. The present-day forests had not reached our municipality, and the landscape was completely grassland.
Prehistory of this region is marked by specific cultural groups arriving/leaving the land. The first included the Larter Culture, which existed from 1000 – 200 BC. They were named for the family whose property held evidence of the early peoples, discovered in 1951. They were the first group of people to occupy the Lockport area. Being nomadic people, their tools and houses were easily assembled/disassembled and transferred. The main purpose for coming here was because they “followed the bison to the shelter of the Red River Valley in the winter months. Bison bones and corner-notched dart points at the site show this” (Larter Culture – Signage).
Bison were very important to the Larter Culture. Early European accounts of bison sightings describe herds of thousands, which took three days to pass. The bison had many uses to the First Nations peoples. Its hide “was used for blankets, garments, boats, ropes and housing. The hoofs produced glue, and its sinews were used for bowstrings and twine. The meat was cut in strips and dried, and mixed with marrow and suet to create pemmican. The bones could be used for weapons and tools […]. Even the bison’s droppings were used as fuel for fires” (Silent Thunder – Signage). To the First Nations peoples, the buffalo was also more than just a resource. “The buffalo was a part of us, his flesh and blood being absorbed by us until it became our own flesh and blood. Our clothing, our tipis, everything we needed for life came from the buffalo’s body. It was hard to say where the animal ended and the man began” – John [Fire] Lame Deer, Sioux medicine man, 1972 (Silent Thunder – Signage).
Following the Larter Culture, was the arrival of the Laurel Culture. The Laurel Culture was present on the Lockport site from 200 BC – 1000 AD. They were named for the Minnesota town where fragments of their pottery were found. As the Lockport grasslands gradually turned into mixed forests, Great Lakes First Nations peoples began arriving. With them came new technologies, like the “bow and arrow, birchbark canoes, and a knowledge of ceramics” (Laurel Culture – Signage). They harvested wild rice, hunted and fished. The Laurel Culture also commonly developed campsites, allowing them to stay longer than the previous nomadic Larters. The Lockport area contained extensive historic evidence of these people. Several burial mounds were found near Lockport, decorated with items that may have been of value in the afterlife. It is evident that the cultures lived in numerous locations, with seasonal camps; their pottery fragments have been discovered as far north as St. Peter’s Church in East Selkirk, Manitoba.
To understand the Laurel community one must also pay attention to their most valuable resource — fishing. The same species of fish were present in the Red River Valley for centuries. The channel catfish, pickerel, goldeye, lake sturgeon, suckers, freshwater drum and northern pike, were all used as a staple in their diet. [The fish were boiled or dried before eating.] Fishing technology amongst early peoples reached a peak after A.D. 1000. One technique was to spear larger fish such as the giant sturgeon and channel catfish, using harpoons tipped with barbed bone points (An Account of Prehistoric Fishing at Lockport – Signage).
Later, some netting was also used; these were gill nets with stones attached at the corners. It was around 500 AD that the climate changed dramatically in the Red River region, developing into one similar to today. Then, in 1160, came a severe drought. The people of the Laurel Culture left without any further trace; not much is known of what happened to them.
In the 12th century, people in the Dakotas were experiencing severe droughts along with a population explosion. The people living there traveled north to Lockport, attracted by its ample moisture for horticulture; they arrived between 1000 and 1300 AD. Archeological investigations conducted at Lockport in the 1980s unearthed charred corn kernels, hoes made from the shoulder-blades of bison, and underground storage pits which were up to two meters deep. The pottery styles discovered originated among Native farming cultures of the upper Mississippi and Missouri river valleys. “Radiocarbon-dating charcoal and bone recovered at Lockport […] indicate that aboriginal people were planting corn there in 1400 AD” (Aboriginal Horticulture – Signage). They were the earliest pre-European farmers in Manitoba and may have had a population anywhere from 300 to 2000 people. They also hunted “bison, rabbit, beaver, muskrat, moose, grouse, pelicans, and cranes. They fished for catfish, walleye, sturgeon, and pike in the fast moving waters of the Red River. They gathered hazelnuts, wild cherries, raspberries, and strawberries as well as a variety of seeds” (Corn Planting – Signage). It is believed that they kept their gardens on levees within the marsh. Although only evidence of corn has been found, it is assumed they farmed beans, squash, and sunflowers as well.
Some of the ways that corn was prepared included:
- ash balls for seasoning – shelled corn husks were burned and the ashes were cooled and molded to make a ball which could be used for seasoning
- Mapi’ Nakapa’ which was in a stew with meat
- Ma’nakapa which was corn meal and beans with spring salt
- corn balls
- corn bread
- hard yellow corn parched in sand – better known as popcorn. (Corn Planting – Signage)
This farming by the aboriginal peoples, although limited, “represents the earliest known evidence of farming on the Canadian prairies, and the northernmost expressions of pre-European horticulture on the North American Continent” (First Farmers in the Red River Valley – Signage).
The last group of prehistoric peoples to live along the northern Red River was called the Selkirk Culture. They are the ancestors of the modern Cree Indians. They were dominant along the Red River all the way from Lockport to north of Selkirk. Archeological evidence concerning these peoples was discovered during a restoration project of St. Peter’s Church in East Selkirk, Manitoba. One stage of the restoration project involved the removal of the deteriorating flooring within the church. While removing the flooring, traces of early aboriginal lifestyles were found dating back nearly 800 years ago. After digging a mere one metre below the surface, they found fragments of pottery, stone tools, bison bones and traces of a hearth or fire pit. There were also fragments of Selkirk Culture pottery found dating back between 800 and 1750 A.D. The Selkirk Culture lived in the region until around 300 years ago, when the European fur trade began in this region. Kenosewun (ke-no’-se-wun), which is a Cree word meaning There are many fishes, is a fitting name to the Lockport site which was settled for its abundance of fish and wildlife. It is because of these resources that our municipal area was a wonder of activity, long before history was ever recorded.
Article written by Jared Laberge St. Clements Heritage Advisory Committee – 08/01/2007
Archives of Manitoba. Kenosewun Museum. Museum Signage. Lockport, 2007.
Archives of Manitoba. Kenosewun Park Trail. Historical Signage. Lockport, 2007.
Manitoba Culture, Heritage and Citizenship. First Farmers In The Red River Valley. Manitoba: Historic Resources, 1994.
Manitoba Culture, Heritage and Citizenship. Manitoba’s First Explorers. Manitoba: Historic Resources, 1992.
Manitoba Culture, Heritage and Citizenship. The Prehistory of the Lockport Site. Manitoba: Historic Resources, 1985.
McLeod, K. David. Land Below The Forks, Archeology, Prehistory and History of the Selkirk and District Planning Area. Manitoba: Historic Resources, 1987.
Petch, Virginia. St. Peter’s Dynevor Heritage Resources Impact Assessment. Personal E-mail. 2002.<|endoftext|>
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# Multiples of 15
The multiples of 15 are the numbers which are n-times of 15, where n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and so on. So, basically, multiples are n-times of any number, where n is the list of natural numbers. Thus, number 15 has multiples in the form of 15n, such that:
• 15 x 1 = 15
• 15 x 2 = 30
• 15 x 3 = 45
• 15 x 4 = 60
and so on.
Here is the list of all multiples:
15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375, 390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, 735, …
All numbers can be divided or the product of 15 is a multiple of 15. The multiples are different from the factors of 15. The factors are the numbers which when multiplied in a pair gives the original number such as 3 x 5 = 15, so here 3 and 5 are not the multiples but the factors. Multiples do not produce the original number instead they are divisible from the original number, for example, 45 ÷ 15 = 3.
## What is a multiple of 15?
Any number that can be denoted in the form 15n where n is an integer is a multiple of 15. So if two values p and q, we say that q is a multiple of p if q = np for some integer n.
For example, 60, 45, 150 and 120 are all multiples of 15 for the following reasons
60 = 15 x 4 45 = 15 x 3 150 = 15 x 10 120 = 15 x 8
These values are called the multiples as these values are obtained by adding or subtracting the original value many times.
### What is the fifth multiple of 15?
The fifth multiple of 15 could be found by multiplying 15 by 5, such that;
15 x 5 = 75
Thus, the fifth multiple of 15 is 75.
### Multiples of 15 Chart
Here is the table of all the multiples of integer 15 from 1 to 20. Students can try writing the multiples till 100 by just multiplying 15 with natural numbers up to 100 to have practice.
Multiplication: Multiples of 15 15 x 1 15 15 x 2 30 15 x 3 45 15 x 4 60 15 x 5 75 15 x 6 90 15 x 7 105 15 x 8 120 15 x 9 135 15 x 10 150 15 x 11 165 15 x 12 180 15 x 13 195 15 x 14 210 15 x 15 225 15 x 16 240 15 x 17 255 15 x 18 270 15 x 19 285 15 x 20 300
### Facts about Multiples of 15
Some interesting facts about multiples of 15 are given below:
15 × z = multiple of 15; where z is an integer Nth multiple of 15 = 15 × N For example, 3rd multiple of 15 = 3 × 15 = 45 Multiple of 15 is always a multiple of 3 and 5 since 3 and 5 are the factors for the number 15 The common multiples of 3 and 5 are the multiples of 15
Also, check the multiples of some numbers given here.
• The first ten multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100
• The first ten multiples of 20 are 20, 40, 60, 80, 100, 120, 140, 160, 180, 200
• The first ten multiples of 25 are 25, 50, 75, 100, 125, 150, 175, 200, 225, 250
• The first ten multiples of 30 are 30, 60, 90, 120, 150, 180, 210, 240, 270, 300
## Solved Examples on Multiples of 15
Question 1:
Dena visits the pool for swimming once every thirty days and Rachael once in fifteen days. If we assume that they go swimming on the first day of May, when will their next possible meet together at the pool?
Solution:
The common multiple of the numbers 15 and 30 are given below.
The multiples of 15 are 15, 30, 45,…..
The multiples of 30 are 30, 60, 90,…..
The multiples in common to 30 and 15 are 30, 60, 90, …..
The smallest common multiple of the numbers 30 and 15 is 30.
They visited the pool and swam together on the first day of May.
So, their next possible meet is 1 + 30 = 31 May
Question 2:
Reena can eat 15 candies in a day. Find the maximum number of candies that she could eat in 14 days.
Solution:
Number of candies that Reena can eat in a day = 15
The maximum number of candies that she could eat in 14 days = 14th multiple of 15
= 15 × 14
= 210<|endoftext|>
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# Find Zeros of a Polynomial Function
Related Topics:
More Lessons for PreCalculus
Math Worksheets
Videos, worksheets, examples, solutions, and activities to help PreCalculus students learn how to find the zeros or roots of a polynomial function.
The following figure show how to find the zeros or roots of a polynomial function Graphically or using the Rational Zeros Theorem. Scroll down the page for more examples and solutions.
Find the Zeros of a Polynomial Function - Integer Zeros
This video provides an introductory example of how to find the zeros of a degree 3 polynomial function.
Example:
Find all the zeros or roots of the given function graphically and using the Rational Zeros Theorem.
f(x) = x3 - 3x2 - 13x + 15 Find the Zeros of a Polynomial Function - Real Rational Zeros
This video provides an example of how to find the zeros of a degree 3 polynomial function with the help of a graph of the function.
Example:
Find all the zeros or roots of the given function.
f(x) = 6x3 - 11x2 - 26x + 15
Find the Zeros of a Polynomial Function with Irrational Zeros
This video provides an example of how to find the zeros of a degree 3 polynomial function with the help of a graph of the function. The function as 1 real rational zero and 2 irrational zeros.
Example:
Find all the zeros or roots of the given function.
f(x) = x3 - 4x2 - 11x + 2 Zeros of Polynomial Functions
How to use the graphing calculator to find real zeros of polynomial functions?
Example:
Find all the zeros or roots of the given function.
f(x) = 3x4 - 4x3 - 11x2 + 16x - 4 Finding the Zeros of a Polynomial Function
A couple of examples on finding the zeros of a polynomial function.
Example:
Find all the zeros or roots of the given functions.
f(x) = 3x3 - 19x2 + 33x - 9
f(x) = x3 - 2x2 - 11x + 52
Graphing polynomials in factored form
Graphing polynomials in factored form taking into account multiplicities of zeros.
Example:
Graph y = -3.5x2(x - 4)3(x + 3)
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.<|endoftext|>
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# A Puzzle From The MathsJam Shout (With Generating Functions)
In a recent MathsJam Shout, courtesy of Bristol MathsJam, we were given a situation, which I paraphrase:
• Cards bearing the letters A to E are shuffled and placed face-down on the table.
• You predict which of the cards bears which letter
• (You make all of your guesses before anything is revealed, and must predict each letter exactly once).
What’s the probability of getting all five correct? How about the probabilities for 4, 3, 2, 1 and 0?
### An easy start
The probability of getting five correct is simply $\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}$, or $\frac{1}{120}$.
The probability of getting four correct is even more simple: it’s zero, because if you have four correct, the fifth must also be correct.
You might conceivably know that the number of derangements of five objects is 44 – I had a vague recollection – but let’s pretend we don’t know that. Instead, let’s use generating functions.
### Genewhotothewhatnow?
Generating functions have many uses, one of which is to turn problems involving distributions of integers into problems of algebra.
There’s a fairly natural bijective mapping between the possible probability distributions on the integers and the polynomials with non-negative coefficients $p(x)$ such that $p(1) = 1$: suppose $X$ is a random variable that takes integer values, there is a polynomial such that the $x^k$ term has coefficient $p_k$ if and only if $P(x=k) = p_k$.
That looks like gobbledygook, so let’s take an example: suppose $X$ is the number of sixes you get from rolling two fair dice. You can work out that $P(X=0) = \frac{25}{36}$, $P(X=2) = \frac{1}{36}$ and $P(X=1) = \frac{10}{36}$. The corresponding generating function is $\frac{25}{36} + \frac{10}{36}x + \frac{1}{36}x^2$.
This turns out to be $\br{ \frac{5}{6} - \frac{1}{6}x}^2$, which you might expand binomially. It’s not a coincidence that the probabilities come from the binomial distribution!
### So how does it help here?
Let’s start with an easier problem: suppose you have only the letters A and B to deal with, rather than A to E. You have even chances of getting them both right, or of getting neither right, which corresponds to a generating function of $\frac{1}{2} + \frac{1}{2}x^2$. (I’m going to call this $g_{(2,0)}(x)$ - the generating function for when you have two cards, neither of which is a dud).
Suppose instead you have cards A and C to match with cards A and B. Here, you’ll get one correct match half the time, and neither correct the other half, making the generating function for this case $\frac{1}{2} + \frac{1}{2}x$. (This is $g_{(2,1)}(x)$: two cards, one of which is a dud.)
### The three-card case
We can use this to extend the problem to three cards, A, B and C.
Without loss of generality, suppose you predict the first card to be A. You will be correct one time in three, leaving yourself with two cards and no duds; the remaining two-thirds, you will leave yourself with a possible match and a dud.
You would score a point in the first case, and thus need to increase all of the powers in the corresponding generating function by one - which you can do by multiplying by $x$.
That makes the generating function for the three-card case $\frac{1}{3}x g_{(2,0)}(x) + \frac{2}{3} g_{(2,1)}(x)$, or $\frac{1}{3}x\br{ \frac{1}{2} + \frac{1}{2}x^2} + \frac{2}{3}\br{\frac{1}{2}+\frac{1}{2}x}$, making $\frac{1}{3} + \frac{1}{2}x + \frac{1}{6}x^3$.
Even without counting the cases, that’s clearly plausible: the coefficients (and therefore probabilitities) add up to 1; there’s no way to get exactly two correct; and the probability of getting all three correct is $\frac{1}{6}$, as we’d have worked out.
### More generally
Suppose you have $n$ cards, none of which are duds.
You have a probability of $\frac{1}{n}$ of getting the first card correct, in which case you now have $n-1$ cards, none of which are duds.
You have a probability of $1 - \frac{1}{n}$ of getting the first card wrong, in which case you now have one dud out of $n-1$ cards.
That makes a recursively-defined generating function of $g_{(n,0)}(x) = \frac{1}{n}x g_{(n-1,0)}(x) + \br{1 - \frac{1}{n}} g_{(n-1,1)}(x)$, with base cases still to come.
But we haven’t really considered dud cases yet!
### A clever strategy for dealing with duds
There's a clever way to handle how duds to ensure you only ever have one in your hand at a time: if you know one of the cards you have to match is a dud, check that one next. Obviously, it will be wrong, but there are two flavours of wrong: either it matches a card you still have to match (the bad wrong, as it leaves you with another, different dud), or it matches a card you've already tried to match (the good wrong, as you now have no duds.)
For example, suppose I predict the ordering BDACE and the correct (but face-down) order is ABCDE.
I begin by checking my prediction for A - it's wrong, as I turn over a C.
The next card I check is my C - which is bad-wrong, as I turn over a D.
(Now I have B, D and E still to match; A, B and E are still face down. D is my only dud.)
I check my D, turning over a B - this is also bad-wrong (I now have B and E in my hand, and A and E face-down).
I check B, and turn over A, which is good-wrong - it leaves me with E to match with E, so I've matched one card correctly.
### Dud scoring
In general, if you have a single dud card out of a total of $n$, you have a $\frac{1}{n}$ probability of the next match involving $n-1$ non-duds, and a $1 - \frac{1}{n}$ chance of having a dud in your $n-1$ cards.
Logically, $g_{(n,1)} = \frac{1}{n} g_{(n-1,0)}(x) + \br{ 1- \frac{1}{n}} g_{(n-1,1)}(x)$.
The only difference between this and the $g_{(n,0)}$ function is that the first term is multiplied by $x$ in the other version. We could even write:
$g_{(n, d)}(x) = \frac{1}{n}x^{1-d} g_{(n-1,0)}(x) + \br{ 1- \frac{1}{n}} g_{(n-1,1)}(x)$ for $n \gt 0$ and $d \in \{0,1\}$.
We also need to work on the base cases: I think all we need to state is that $g_{(0,0)}(x) = 1$.
### So how about $g_{(5,0)}(x)$?
It’s simplest to build the functions up from the base, although it’s possible to work downwards if you’re really careful with your fractions. Let’s go from the very beginning:
$g_{(0,0)}(x) = 1$.
That means $g_{(1,0)}(x) = x$ and $g_{(1,1)}(x) = 1$, as we would hope.
For two cards, we have $g_{(2,0)}(x) = \frac{1}{2}x g_{(1,0)}(x) + \frac{1}{2} g_{(1,1)}(x) = 1 + x^2$, as before; also, $g_{(2,1)}(x) = \frac{1}{2}g_{(1,0)}(x) + \frac{1}{2}g_{(1,1)}(x) = 1+x$. It’s agreeing so far!
Keeping going, $g_{(3,0)}(x) = \frac{1}{3}x g_{(2,0)}(x) + \frac{2}{3} g_{(2,1)}(x) = \frac{1}{3} + \frac{1}{2}x + \frac{1}{6}x^2$, as before; $g_{(3,1)} = \frac{1}{3} g_{(2,0)}(x) + \frac{2}{3} g_{(2,1)}(x) = \frac{1}{2} + \frac{1}{3}x + \frac{1}{6}x^2$. This remains plausible.
Penultimately, $g_{(4,0)}(x) = \frac{1}{4}x g_{(3,0)}(x) + \frac{3}{4} g_{(3,1)}(x) = \frac{3}{8} + \frac{1}{3}x + \frac{1}{4}x^2 + \frac{1}{24}x^4$ - which is again plausible; $g_{(4,1)}(x) = \frac{1}{4} g_{(3,0)}(x) + \frac{3}{4} g_{(3,1)}(x) = \frac{11}{24} + \frac{3}{8}x + \frac{1}{8}x^2 + \frac{1}{24}x^3$.
Last push for a final answer - and this time, we only need $g_{(5,0)}(x)$. Working it out, $g_{(5,0)}(x) = \frac{1}{5}x g_{(4,0)}(x) + \frac{4}{5} g_{(4,1)}(x) = \frac{11}{30} + \frac{3}{8}x + \frac{1}{6}x^2 + \frac{1}{12}x^3 + \frac{1}{120}x^5$.
This gives the probabilities of the results at $\frac{1}{120}$ for 5, zero for 4, $\frac{1}{12}$ for 3, $\frac{1}{6}$ for 2, $\frac{3}{8}$ for 1 and $\frac{11}{30}$ for 0.
### Special bonus!
It turns out that for a generating function $G(x)$, the expected value is given by $G’(1)$. So, by differentiating, we get $g’_{(5,0)}(x) = \frac{3}{8} + \frac{1}{3}x + \frac{1}{4}x^2 + \frac{1}{24}x^3$.
Then $g’_{(5,0)}(1) = \frac{3}{8} + \frac{1}{3} + \frac{1}{4} + \frac{1}{24} = 1$: on average, we expect to get one card correct!
If your eyes are especially sharp, you might spot that $g’_{(5,0)} = g_{(4,0)}$. Does this pattern continue? Can you prove it?
A lovely puzzle, that – I’m sure there are less involved ways to solve it, but I like this one, dagnabbit.
* Thanks to Bristol MathsJam for the puzzle, and to Barney Maunder-Taylor and Adam Atkinson for helpful comments.
## Colin
Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.
### One comment on “A Puzzle From The MathsJam Shout (With Generating Functions)”
• ##### Christopher Allan
Long ago I made some notes on this problem, known as “The Matching Problem”. As you say, a derangement means no matches. I did it with two numbered packs of cards. You have done it for 5 cards. For 6 there are 720 arrangements: 265 give zero matches, 264 give 1 match, 135 give 2 matches, 40 give 3 matches, 15 give 4 matches, zero give 5 and 1 gives 6, of course.
I have the general formula which resembles yours. Somewhere along this derangements thing there is a term known as ‘subfactorial N’. Subfactorial N can be expressed in terms of N! and associated lower factorials, but I forget the details. (You sometimes see it written with the exclamation mark upside down). Subfac 4 = 9, subfac 5 = 44, subfac 6 = 265, etc.
Also, as N gets large, it can be shown that probability of no matches (i.e. all the cards differing from their original positions) approximates to 1/e. This means it becomes very close to 0.36788 or about a 37% chance of zero matches.
Have just discovered I extended it to 8 cards:
N = 7 gives no. of derangements as 1854
N = 8 gives 14,833
I gave up at this point. The approx. to 1/e is quite good for N = 8.
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# How to decompose function like x^2+x-6?
I was trying to wrap my brain around this but could not think of anything. I am very poor at maths but trying to learn of it.
Edit: Exact question is If x^2+x-6 is a composite function f(g(x)); then figure out f(x) and g(x) ?
To me information looks incomplete to figure out such details.
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I suppose you mean, how to factor the polynomial? – Adrián Barquero Nov 23 '10 at 23:45
high school algebra $\neq$ linear-algebra. retagged as polynomials. – Willie Wong Nov 23 '10 at 23:52
Hmm well that changes things a little bit. But now your problem does not have a unique solution. – Adrián Barquero Nov 23 '10 at 23:58
There are potentially many ways of doing this in general. For e.g. take $f(x)=x-6$ and $g(x)=x^2+x$. – Timothy Wagner Nov 24 '10 at 0:08
It may be awfully cheap, but consider $f(x)=x^2+x-6$, and $g(x)=x$. Then $f(g(x))=f(x)=x^2+x-6$. – yunone Nov 24 '10 at 0:13
We always have
(x+a)(x+b) = (x(x+b))+(a*(x+b)) = (x*x)+(x*b)+(a*x)+(a*b) = (x^2)+(b*x)+(a*x)+(a*b) (x^2)+(b*x)+(a*x)+(a*b) = (x^2)+(a*x)+(b*x)+(a*b) = (x^2)+((a+b)*x)+(a*b) (x+a)*(x+b) = (x^2)+((a+b)*x)+(a*b)
Converting your equation to that form gives
x^2+x-6 = (x^2)+(1*x)+(-6)
where we want
(x^2)+(1*x)+(-6) = (x^2)+((a+b)*x)+(a*b)
So, we're looking for numbers whose sum is 1 and whose product is -6.
a+b = 1 and a*b = -6
That should be enough for you to figure it out, but in case it's not,
if you just want the answer you can put
"n vf rdhny gb artngvir gjb naq o vf rdhny gb cbfvgvir guerr"
into rot13.com
-
You've misread the question. – Qiaochu Yuan Nov 24 '10 at 0:20
@Qiaochu: I'd be more generous and assume he missed the update (the question was updated while Ricky was posting the answer). Two minutes isn't that long a gap. – J. M. Nov 24 '10 at 0:29
The 'simplest' possibility is f(x) := x-3 and g(x) := x^2+x-3.
It is the more natural of the 2 pairs of integer polynomials of degree at most 2 that minimize the sum of squares of their coefficients while composing to give x^2+x-6.
...
and apparently it's not this easy to get back the account I created on another computer.
-
The context of this question suggests that $f=f(x)$ and $g=g(x)$ are going to be polynomials (with either real, rational or integer coefficients). If this is the case then...
Hint: While there's going to be an infinite number of functions $f,g$ for which $f(g(x))=x^2+x-6$, there will only be two situations when this can occur (consider the degrees of $f$ and $g$, and how that effects $f(g(x))$). These can be classified by setting $f$ and $g$ as arbitrary polynomials of the appropriate degrees, composing them, then equating coefficients with $f(g(x))=x^2+x-6$.
If $f$ and $g$ are not restricted to polynomials, there's going to be all sorts of functions $f$ and $g$ for which $f(g(x))=x^2+x-6$. For example: Let $A=\left(\begin{matrix}-2 & 1 \\ 4 & 1 \end{matrix}\right)$. Then the characteristic polynomial $\det(A-xI)=(-2-x)(1-x)-4=x^2+x-6$.
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Since I don't think anyone else has covered it and because it can be useful for thinking about inverse functions/relations, consider completing the square: \begin{align} f(x)&=x^2+x-6 \\ &=x^2+x+\frac{1}{4}-\frac{25}{4} \\ &=\left(x+\frac{1}{2}\right)^2-\frac{25}{4} \end{align} so $f(x)=a(b(c(x)))$ where $a(x)=x-\frac{25}{4}$, $b(x)=x^2$, and $c(x)=x+\frac{1}{2}$. The advantage here (though probably not significant for your purpose) is that each of these functions can be thought of as a single simple operation on their input (that the input appears only once in each function also makes it easier to think of this way).
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sempteim245
2022-03-31
I have to simplify and evaluate this :
${\mathrm{cos}70}^{\circ }+4{\mathrm{cos}70}^{\circ }$
### Answer & Explanation
zalutaloj9a0f
The easy way :
$\frac{{\mathrm{cos}70}^{\circ }+4{\mathrm{cos}70}^{\circ }{\mathrm{sin}70}^{\circ }}{{\mathrm{sin}70}^{\circ }}=\frac{{\mathrm{sin}20}^{\circ }+2{\mathrm{sin}40}^{\circ }}{{\mathrm{cos}20}^{\circ }}$
$=\frac{{\mathrm{sin}20}^{\circ }+{\mathrm{sin}40}^{\circ }+{\mathrm{cos}50}^{\circ }}{{\mathrm{cos}20}^{\circ }}$
Sum to product formula gives :
$=\frac{{\mathrm{cos}10}^{\circ }+{\mathrm{cos}50}^{\circ }}{{\mathrm{cos}20}^{\circ }}$
Again using the formula gives $2{\mathrm{cos}30}^{\circ }=\sqrt{3}\approx 1.732$
Leonardo Mcpherson
First $70=90-20$
We can express all in terms of $\mathrm{cos}\left(20\right)$ and use that $\frac{1}{2}=\mathrm{cos}\left(60\right)=4{\mathrm{cos}}^{3}\left(20\right)-3\mathrm{cos}\left(20\right)$
Let's write $x=\mathrm{cos}\left(20\right),y=\mathrm{sin}\left(20\right)$ to write less.
So, $4{x}^{3}-3x-\frac{1}{2}=0$
We square your expression such that we don't have to write radicals, but we can go without it too if we wanted.
${\left(\mathrm{cot}\left(70\right)+4\mathrm{cos}\left(70\right)\right)}^{2}={\left(\frac{\mathrm{cos}\left(90-20\right)+4\mathrm{cos}\left(90-20\right)\mathrm{sin}\left(90-20\right)}{\mathrm{sin}\left(90-20\right)}\right)}^{2}$
$={\left(\frac{y+4xy}{x}\right)}^{2}$
$={y}^{2}\frac{16{x}^{2}+8x+1}{{x}^{2}}$
But
$\frac{1}{x}=8{x}^{2}-6$
$=\left(1-{x}^{2}\right)\left(16{x}^{2}+8x+1\right){\left(8{x}^{2}-6\right)}^{2}$
$=-1024{x}^{8}-512{x}^{7}+2496{x}^{6}+1280{x}^{5}-1952{x}^{4}-1056{x}^{3}+444{x}^{2}+288x+36$
Now divide this polynomial by $4{x}^{3}-3x-\frac{1}{2}$, which is zero.
$=\left(-256{x}^{5}-128{x}^{4}+432{x}^{3}+192{x}^{2}-180x-66\right)\cdot \left(4{x}^{3}-3x-\frac{1}{2}\right)+3$
The remainder gives you the value 3.
Do you have a similar question?
Recalculate according to your conditions!<|endoftext|>
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## Number Theory
### 1. Introduction
Numbers are used to measure and count. The number system which is globally in use is the decimal number system, where the base is $10$. This just means that there are $10$ digits in this number system. They are $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$. Unless specified otherwise, the base 10 number system will be used for all calculations.
The base $9$ number system will use the first $9$ digits mentioned above which are $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$ and $8$. Likewise, the base $8$ number system will not include the digits $8$ and $9$. So, the base $2$ number system will include only the digits $0$ and $1$ - called the binary number system.
As number systems of base higher than $10$ have never been widely used, additional digits used have widely varied. The widely agreed form of hexadecimal (base $16$) representation is to use the letters A, B, C, D, E and F as the digits after $9$. In the management exams, typically these letters in that order have been used to represent digits for number systems with base higher than $10$. We will discuss these in detail in the $Number \space Systems$ lesson.
#### 1.1 Decimal Number System
A number in the decimal number system, for example $94.25$,
$94.25 = 9 \times 10^1 + 4 \times 10^0 + 2 \times 10^-1 + 5 \times 10^-2$
$94.25 = 9 \times 10 + 4 \times 1 + \dfrac{2}{10} + \dfrac{5}{100}$
Note that digits to the left of the decimal point are multiplied with $10^0$, $10^1$, $10^2$ and so on, while the digits to the right of the decimal point are multiplies with $10^{-1}$, $10^{-2}$, and so on, in that order.
Therefore, digits in the places to the left of the decimal point are units digit, tens digit, hundreds digit, thousands digit and so on. The digits in places to the right of the decimal point are called tenths digit, hundredths digit, thousandths digit and so on.
In the above example, $9$ is tens digit, $4$ is the units digit, $2$ is the tenths digit and $5$ is the hundredths digit.
#### 1.2 Application in MBA tests
Questions in MBA tests might require us to find $2$-digit or $3$-digit numbers with certain conditions around the digits. In these questions, we assume the digits to be variables and form the numbers.
A $2$-digit number can be written as $10a + b$, where $a$ and $b$ are the tens and units digit respectively.
Likewise, a $3$-digit number can be written as $100a + 10b + c$ where $a$, $b$ and $c$ are the hundreds, tens and units digit respectively.
Or, a $4$-digit numbers can be written $1000a + 100b + 10c + d$ , where $a$, $b$, $c$ and $d$ are the thousands, hundreds, tens and units digit respectively.
### Example 1
When a $3$-digit number is reversed, we get another $3$-digit numbers. If the difference between these two numbers is $297$, then what is the difference between the hundreds digit and the units digit of the number?
### Solution
Let $x$, $y$ and $z$ be the hundreds, tens and units digit of the number. Therefore, the number is $\bm{100a} + \bm{10b} + \bm{c}$.
When the digits are reversed, the number formed is $\bm{100c} + \bm{10b} + \bm{a}$.
$100 a + 10 b + c - (100 c + 10 b + a) = 297$
$\implies 99 a - 99c = 297$
$\implies a - c = 3$
Answer: $3$<|endoftext|>
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# How do you find the 7th term in the geometric sequence 2, 6, 18, 54, ...?
Nov 10, 2015
$4374$
#### Explanation:
In a geometric sequence, you choose a first term $a$, a ratio $r$, and then you obtain every term multiplying the previous one by the ratio. Let's compute some terms:
• ${a}_{0} = a$
• ${a}_{1} = a \cdot r$
• ${a}_{2} = \left(a \cdot r\right) r = a \cdot {r}^{2}$
• ${a}_{3} = \left(a \cdot {r}^{2}\right) r = a \cdot {r}^{3}$
and so on. We can see that the relation is ${a}_{n} = a \cdot {r}^{n}$, which means that the seventh term is ${a}_{7} = a \cdot {r}^{7}$.
In your case, the first term $a$ is $2$, and the ratio can be easily computed: if ${a}_{0} = 2$ and ${a}_{1} = 6$, then ${a}_{1} = {a}_{0} \cdot r = 6$, which means $2 \cdot r = 6$, and finally $r = 3$.
So, the seventh term will be $2 \cdot {3}^{7} = 2 \cdot 2187 = 4374$<|endoftext|>
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# What is value of Cos Pi?
Table of Contents
## What is value of Cos Pi?
-1
The value of cos pi is -1. Cos pi can also be expressed using the equivalent of the given angle (pi) in degrees (180°). Since the cosine function is a periodic function, we can represent cos pi as, cos pi = cos(pi + n × 2pi), n ∈ Z.
### How do you find Cos 3pi?
The value of cos 3pi can be calculated by constructing an angle of 3π radians with the x-axis, and then finding the coordinates of the corresponding point (-1, 0) on the unit circle. The value of cos 3pi is equal to the x-coordinate (-1). ∴ cos 3pi = -1.
How do you simplify Cos x π?
cos(x+π)=−cosx .
What is pi in sin and cos?
Summary: The value of sin π is 0 and cos π is -1.
## What is cos pi by 2?
0
The value of cos pi/2 can be calculated by constructing an angle of π/2 radians with the x-axis, and then finding the coordinates of the corresponding point (0, 1) on the unit circle. The value of cos pi/2 is equal to the x-coordinate (0). ∴ cos pi/2 = 0.
### What does a COS function graph look like?
The graphs of sine and cosine have the same shape: a repeating “hill and valley” pattern over an interval on the horizontal axis that has a length of . The sine and cosine functions have the same domain—the real numbers—and the same range—the interval of values .
What is the COS curve?
Cos Graph. The cosine graph or the cos graph is an up-down graph just like the sine graph. The only difference between the sine graph and the cos graph is that sine graph starts from 0 while the cos graph starts from 90 (or π/2). The cos graph given below starts from 1 and falls till -1 and then starts rising again.
What is cos at pi 3?
Answer. cos (pi/3) = 1/2.
## What is Cos pi by 4?
The actual value of cos pi/4 is √2/2 in radical form and 0.7071 in decimal form.
### What is the derivative of Cos pi X?
Here, we see that the derivative of the outside function, cos(x) , is −sin(x) . So, we will write −sin(x) but keep the inside function intact, giving us a −sin(πx) . We then multiply that by the derivative of πx , which is just π , giving the full derivative of −πsin(πx) .
What is 2cosx equal to?
2cos x is twice the cosine of angle x. It lies between −2 and 2. cos2x is the cosine of angle 2x. It is two times the angle x.
What is cos 2x equal to?
Cos2x is one of the double angle trigonometric identities as the angle in consideration is a multiple of 2, that is, the double of x. Let us write the cos2x identity in different forms: cos2x = cos2x – sin2x. cos2x = 2cos2x – 1. cos2x = 1 – 2sin2x.
## What is cos π theta?
It is defined as a ratio of adjacent (base) sides to the hypotenuse. As we know that, cos(π) = cos 180º cos 180º lies in the second quadrant where cos has a negative value. Therefore, cos (π) or cos 180º = -1. By using trigonometry identity, cos(-θ) = cos(θ)
### Does pi equal 180 degrees in trigonometry?
180 degrees equals pi radians, so to get one degree divide both sides by 180. One degree is pi over 180 and if you want a decimal value for this you can use your calculator pi divided by 180, this is approximately .<|endoftext|>
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Alec Jeffreys invented genetic fingerprinting, a powerful technology that provided new insights for genetic researchers, and resolved life-or-death questions for law enforcement. It also provided a simple way to establish family relationships in paternity and immigration disputes, screen for mutations, identify human remains, and improve animal breeding.
Human DNA contains long strands of molecules that correlate to each individual's genetic traits. The amount of information coded into DNA is so vast that it is difficult to examine DNA samples. Jeffreys discovered a process that detects extremely variable DNA regions, proving that each human has its own genetic fingerprint. Users of genetic fingerprinting were able to pursue and identify criminals who left behind samples of blood, semen, skin or hair after a crime. Subjecting these DNA samples to genetic fingerprinting made it possible to conclusively link suspects to the scene of a crime. It also exonerated numerous people falsely convicted before genetic fingerprinting was invented.
Born in Oxford, England in 1950, Jeffreys studied at Merton College in Oxford. He conducts his research at the University of Leicester, where he made his well-known discovery. His work has earned him numerous honors, including knighthood in 1994.<|endoftext|>
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# Factors of 52
## |
Understanding factors and their role in mathematics is crucial for building a solid foundation in number theory. In this blog, we embark on a journey to uncover the factors of 52 and delve into the fascinating world of divisibility. So, grab your mathematical thinking caps and join us as we unlock the secrets hidden within the number 52.
Recommended Reading: Factors of 12
### The Basics of Factors
Before we dive into the factors of 52, let’s quickly recap what factors are. Factors are the numbers that divide a given number evenly without leaving any remainder. In simpler terms, they are the building blocks that make up a number. By exploring the factors of a particular number, we gain insight into its divisibility and mathematical properties.
### Discovering the Factors of 52
To find the factors of 52, we need to determine which numbers can divide it without leaving a remainder. Let’s embark on this mathematical journey:
#### Number 1
Every whole number is divisible by 1, including 52. When we divide 52 by 1, the result is 52. Therefore, 1 is a factor of 52.
#### Number 2
Dividing 52 by 2 yields a quotient of 26. Since there is no remainder, we conclude that 2 is also a factor of 52.
#### Number 4
We are continuing our exploration, dividing 52 by 4 results in a quotient of 13, once again with no remainder. Hence, 4 is another factor of 52.
#### Number 13
Our mathematical journey takes us further as we divide 52 by 13, resulting in a quotient of 4. Therefore, 13 is a factor of 52.
### The Complete Set of Factors
Having traversed the realm of divisibility, we have discovered all the factors of 52. They are 1, 2, 4, 13, and 52. These numbers showcase the diverse ways in which 52 can be divided evenly.
### Applications and Significance
The factors of 52 find relevance in various mathematical concepts and practical applications. One important application is in determining whether a number is prime or composite. If a number has only two factors, 1 and itself, it is considered prime. However, if it has more than two factors, it is classified as composite. In the case of 52, since it has multiple factors beyond 1 and 52, it falls into the category of a composite number.
Additionally, the knowledge of factors assists in simplifying fractions, finding common denominators, and solving equations. By understanding the factors of a given number, we can unravel its properties and relationships within the mathematical landscape.
In exploring factors 52, we have gained valuable insights into its divisibility and mathematical nature. Factors 1, 2, 4, 13, and 52 contribute to the intricate web of relationships and patterns within the realm of numbers. Remember, understanding factors not only aids in problem-solving but also nurtures a deeper appreciation for the elegance and beauty of mathematics. So, continue exploring, questioning, and unlocking the secrets hidden within the vast numerical landscape.
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Register for a free 60-minute Advanced Math Workshop today!
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What is Parity of Esteem? - Glossary of Good
Parity of Esteem is most often used, in the UK at least, in reference to the way mental health is treated by health and social care services.
The Royal College of Psychiatrists define it as “Valuing mental health equally with physical health”.
So, put simply, it means equality.
How are mental health problems treated differently from physical health?
The picture is complex, but a few clear example of how mental health is treated unequally to physical health include:
- Mental health services receive less funding. It accounts for 28% of the burden of disease yet it only receives 13% of NHS spending. (Source: Centre for Mental Health)
- You are more likely to be asked to leave your area for mental health care than for a physical health problem. (Source: Guardian)
- You will wait longer for treatment if you have a mental health problem. (Source: Mind)
- If you have a mental health problem, you can be treated against your will, a practice illegal in all other health contexts. (Source: NHS)
- If you have a mental health problem you are more likely to be left until you are at crisis point and need emergency care. (Source: Centre for Mental Health)
- You are more likely to receive no treatment at all. (Source: Guardian)
- If you have a severe mental health problem, your physical health also becomes devalued, to the extent that you will die 10-20 years earlier. (Source Oxford University)
What is being done around Parity of Esteem?
The Mental Health Taskforce
The Mental Health Taskforce was a group of experts, including those with direct experience of mental health problems, who drew up a national strategy to improve mental health services. These included not just the NHS, but all services involved in supporting people with mental health problems and it had ‘parity of esteem’ as one of its core principles.
The strategy they drew up is called theFive Year Forward View which, although a little unwieldy for the casual reader, is the first time there has really been a strategic approach to improving mental health care, with full involvement from patients and service users.
We have seen high profile anti-stigma campaigns over the past decade which have prompted real improvements in public attitudes.
However, research related to the anti-stigma campaign Time to Change shows that some professionals still have discriminatory beliefs and attitudes. This is worrying as this is possibly where it most matters in getting people the healthcare support they need.
Health and Social Care Act 2012
The Health and Social Care Act 2012 included a legal requirement for the NHS to treat mental and physical health equally, however this is yet to have much impact.
More funding (?)
Perhaps as testament to how important mental health care now is in the public consciousness, every few months the Conservative government, and the coalition government before it, announce a new chunk of money for mental health.
Unfortunately, this is often just an accountant’s illusion. That money is either taken from another sorely underfunded, often mental health-related budget, or never emerges.
The King’s Fund has been watching closely, and a report late last year showed that 40% of mental health care trusts (who provide 80% of mental health care) actually had their budgets cut.
What needs to be done around Parity of Esteem?
Most commentators agree that, while the NHS remains so chronically and devastatingly underfunded, there is not much chance of real equality being achieved.
Likewise, while the attitudes of professionals, particularly commissioners, remain behind the rest of the public we cannot reach equality.
We must also not forget that mental health problems are often the result of, or at least made worse by, things like unemployment, poor living conditions, isolation and discrimination.
If we are fully committed to helping people recover from or live well with mental health problems we must listen to the research and address these issues as well.
Parity of Esteem – Further reading
Parity of Esteem – what are we trying to achieve? by Martin McShane for NHS England
Parity of Esteem by Centre for Mental Health
‘The chronic underfunding of mental health care is a stigma proving hard to reverse’ by Alex Langford in Community Care
Mental health trust funding down 8% from 2010 despite coalition’s drive for parity of esteem by Andy McNicoll in Community Care
NHS mental health funding falls in England – FoI figures by Sarah Bloch for BBC News
Trust finances raise concerns about the future of the Mental Health Taskforce recommendations by Helen Gilburt for The King’s Fund
Is there ‘parity of esteem’ between mental and physical health? by Chris Naylor for The King’s Fund
Explore more of Montfort's Glossary of Good: https://montfort.io/glossary-of-good/<|endoftext|>
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# How do you solve the triangle NPQ given P=109^circ, Q=57^circ, n=22?
Jan 28, 2017
$m \angle N = {44}^{\circ}$, $m \angle P = {109}^{\circ}$, $m \angle Q = {27}^{\circ}$
and $n = 22$, $p = 29.94$ and $q = 14.38$
#### Explanation:
Let the triangle be as shown below:
Observe that as $m \angle P = {109}^{\circ}$ and $m \angle Q = {27}^{\circ}$, we have $m \angle N = {180}^{\circ} - {109}^{\circ} - {27}^{\circ} = {44}^{\circ}$
According to sine formula , we will have here,
$\frac{n}{\sin} N = \frac{p}{\sin} P = \frac{q}{\sin} Q$
Hence, $\frac{p}{\sin {109}^{\circ}} = \frac{q}{\sin {27}^{\circ}} = \frac{22}{\sin} {44}^{\circ} = \frac{22}{0.6947}$
Hence $p = \frac{22}{0.6947} \times \sin {109}^{\circ} = \frac{22}{0.6947} \times 0.9455 = 29.94$
and $q = \frac{22}{0.6947} \times \sin {27}^{\circ} = \frac{22}{0.6947} \times 0.454 = 14.38$<|endoftext|>
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# Lesson 13
Multiplying Complex Numbers
## 13.1: $i$ Squared (5 minutes)
### Warm-up
The purpose of this activity is to elicit strategies and understandings students have for multiplying imaginary numbers. Later in this lesson, students will multiply complex numbers and write them in the form $$a+bi$$, so it will be helpful for students to see various strategies for multiplying imaginary numbers.
### Student Facing
Write each expression in the form $$a+bi$$, where $$a$$ and $$b$$ are real numbers.
1. $$4i \boldcdot 3i$$
2. $$4i \boldcdot \text-3i$$
3. $$\text- 2i \boldcdot \text- 5i$$
4. $$\text- 5i \boldcdot 5i$$
5. $$(\text-5i)^2$$
### Activity Synthesis
Ask students to share their strategies for each problem. Record and display their responses for all to see. Highlight ways that students regrouped factors of $$i$$ or $$\text-1$$ in order to simplify their answers. If no student mentions the fact that squaring an imaginary number always results in a real number, ask students to discuss this idea.
## 13.2: Multiplying Imaginary Numbers (15 minutes)
### Activity
In this partner activity, students take turns matching equivalent expressions using the fact that $$i^2 = \text-1$$. They build on a previous activity in which they multiplied imaginary numbers and saw how this changed the numbers’ representation on the complex plane. Fluency with strategically using the fact that $$i^2=\text-1$$ will be an important skill in the next activity when students multiply numbers that have both real and imaginary parts. As students trade roles explaining their thinking and listening, they have opportunities to explain their reasoning and critique the reasoning of others (MP3).
### Launch
Arrange students in groups of 2. Tell students that for each expression in column A, one partner finds an equivalent expression in column B and explains why they think it is equivalent. (One item in column B will not be used.) The partner's job is to listen and make sure they agree. If they don't agree, the partners discuss until they come to an agreement. Students then swap roles. If necessary, demonstrate this protocol before students start working.
Conversing: MLR8 Discussion Supports. As students take turns finding a match and explaining their reasoning to their partner, display the following sentence frames for all to see: “ _____ and _____ are equal because . . .” and “I noticed _____, so I matched . . .” Encourage students to challenge each other when they disagree. This will help students clarify their reasoning about equivalent expressions that involve imaginary numbers.
Design Principle(s): Support sense-making; Maximize meta-awareness
Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer collaboration. When students share their work with a partner, display sentence frames to support conversation such as: “First, I _____ because . . .”, “I noticed _____ so I . . .”, “Why did you . . .?”, “I agree/disagree because . . . .”
Supports accessibility for: Language; Social-emotional skills
### Student Facing
Take turns with your partner to match an expression in column A with an equivalent expression in column B.
• For each match that you find, explain to your partner how you know it’s a match.
• For each match that your partner finds, listen carefully to their explanation. If you disagree, discuss your thinking and work to reach an agreement.
A B
$$5 \boldcdot 7i$$ -9
$$5i \boldcdot 7i$$ $$35i$$
$$3i^2$$ -35
$$(3i)^2$$ 1
$$8i^3$$ 9
$$i^4$$ -3
$$\text- i^2$$ -1
$$(\text- i)^2$$ $$\text- 8i$$
### Anticipated Misconceptions
If students feel stuck with the expressions $$i^3$$ or $$i^4$$, suggest replacing exponents with repeated factors ($$i \boldcdot i \boldcdot i$$) and rewriting the expressions a pair at a time.
### Activity Synthesis
Once all groups have completed the matching, here are some questions for discussion:
• “Which matches were tricky? Explain why.” ($$(\text- i)^2$$ was tricky because there were several negative signs to keep track of and I had to double-check whether the final answer was negative or positive.)
• “Did you need to make adjustments in your matches? What might have caused an error? What adjustments were made?” (I thought $$3i^2$$ was -9 because I thought the 3 was squared too. I had to go back and just square the $$i$$ and then multiply it by 3.)
Ask groups to explain their reasoning for several matches, especially why $$i^4=1$$. If not brought up by students, make sure to discuss that it’s possible to use the fact that $$i^2=\text-1$$ to make equivalent expressions.
## 13.3: Multiplying Complex Numbers (15 minutes)
### Activity
The purpose of this activity is for students to build fluency expressing the product of two complex numbers in the form $$a+bi$$, where $$a$$ and $$b$$ are real numbers. In order to do this, students must use the fact that $$i^2=\text-1$$.
Look for students who answer the last question as $$13+0i$$ as opposed to 13 to highlight during discussion.
Representation: Internalize Comprehension. Activate or supply background knowledge. Display the table provided in the launch. As students share their explanations, use color-coding and annotations to highlight the connections between the expression, the values in the table, and the final product after combining like terms. Students can refer back to the chart throughout the activity.
Supports accessibility for: Visual-spatial processing
### Launch
Tell students that they are now going to multiply complex numbers together. Display the expression $$(3 + 2i)(\text- 4 - 5i)$$ for all to see and give students 1 minute of quiet think time to consider how they would find the product. Then, display this table for all to see:
3 $$\boldsymbol{2i}$$
-4 -12 $$\text- 8i$$
$$\boldsymbol{\text- 5i}$$ $$\text- 15i$$ $$\text- 10i^2$$
After a brief time to consider the diagram, select students to explain where each of the values came from. Ask students, “Now that we have $$\text- 12 - 8i - 15i - 10i^2$$, what do we do in order to write the number in the form $$a+bi$$?” (We know $$i^2 = \text-1$$, so the sum of these is $$\text- 12 - 8i - 15i +10 = \text- 2 - 23i$$.)
### Student Facing
Write each product in the form $$a + bi$$, where $$a$$ and $$b$$ are real numbers.
1. $$(\text- 3 + 9i)(5i)$$
2. $$(8 + i)(\text- 5 + 3i)$$
3. $$(3 + 2i)^2$$
4. $$(3 + 2i)(3 - 2i)$$
### Student Facing
#### Are you ready for more?
On October 16, 1843, while walking across the Broom Bridge in Dublin, Ireland, Sir William Rowan Hamilton came up with an idea for numbers that would work sort of like complex numbers. Instead of just the number $$i$$ (and its opposite $$\text- i$$) squaring to give -1, he imagined three numbers $$i$$, $$j$$, and $$k$$ (each with an opposite) that squared to give -1.
The way these numbers multiplied with each other was very interesting. $$i$$ times $$j$$ would give $$k$$, $$j$$ times $$k$$ would give $$i$$, and $$k$$ times $$i$$ would give $$j$$. But the multiplication he imagined did not have a commutative property. When those numbers were multiplied in the opposite order, they’d give the opposite number. So $$j$$ times $$i$$ would give $$\text- k$$, $$k$$ times $$j$$ would give $$\text- i$$, and $$i$$ times $$k$$ would give $$\text- j$$. A quaternion is a number that can be written in the form $$a+bi+cj+dk$$ where $$a$$, $$b$$, $$c$$, and $$d$$ are real numbers.
Let $$w=2+3i-j$$ and $$z=2i+3k$$. Write each given expression in the form $$a+bi+cj+dk$$.
1. $$w+z$$
2. $$wz$$
3. $$zw$$
### Activity Synthesis
The key takeaway is that the product of complex numbers is another complex number, and we can see this by using usual arithmetic along with the fact that $$i^2=\text-1$$ to write products in the form $$a+bi$$, where $$a$$ and $$b$$ are real numbers.
Select previously identified students to share their responses to the last question, for which $$13+0i$$ is the technically correct response. Discuss the idea that numbers like 13 or $$5i$$ don’t need to be written as $$13+0i$$ and $$0+5i$$ in order to be recognizable as complex numbers. Writing complex numbers as a single term is okay; it’s something they did for a long time before they knew that all real numbers are complex numbers of the form $$a+bi$$ where $$b=0$$.
## Lesson Synthesis
### Lesson Synthesis
In this lesson, students learned about multiplication with complex numbers. Display this equation for all to see, leaving room to record student thinking for all to see during the discussion:
$$\displaystyle (2+2i)(2-3i) = (3+2i) + (7+4i)$$
Ask students, “Mentally, without going through all the trouble of writing each side in the form $$a+bi$$, what about the structure of the expressions on each side of the equals sign tells you that the equation is true or false?” (Looking at the left hand side, we can make the imaginary part by adding the product $$2i \boldcdot 2$$ with the product $$2 \boldcdot \text-3i$$, which gives $$4i-6i=\text-2i$$. On the right hand side, however, the imaginary part is $$2i+4i=6i$$, so the equation is false. The two complex numbers don’t have equal imaginary parts, so they can’t be equal.)
Then, display this equation for all to see:
$$\displaystyle (3-5i)(\text-2+3i) = (\text-10+6i) + (4+13i)$$
Again, ask students to use arguments about the structure of the expressions on each side of the equation to mentally determine whether or not it is true. (The imaginary parts match because on the left hand side, $$3 \boldcdot 3i + \text-5i \boldcdot \text-2 = 9i+10i=19i$$, and on the right hand side, $$6i+13i=19i$$. The real parts don’t match, however. On the right hand side, the real part is $$\text-10+4=\text-6$$. On the left hand side, the real parts of the two factors also multiply to -6, but there are multiples of $$i^2$$ that are real that haven’t been accounted for.)
If time allows, ask students to verify the arguments for one or both equations by writing each side in the form $$a+bi$$, where $$a$$ and $$b$$ are real numbers.
## Student Lesson Summary
### Student Facing
To multiply two complex numbers, we use the distributive property:
$$\displaystyle (2 + 3i )(4 + 5i) = 8 + 10i + 12i + 15i^2$$
Remember that $$i^2 = \text- 1$$, so:
$$\displaystyle (2 + 3i )(4 + 5i) = 8 + 10i + 12i - 15$$
When we add the real parts together and the imaginary parts together, we get:
$$\displaystyle (2 + 3i )(4 + 5i) = \text- 7 + 22i$$<|endoftext|>
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A reading gap fill is one task you may get in the IELTS test. You have to fill in the gaps of a summary of part of the text using words from a box.
There may be more words than you need to use so you need to find the part of the reading that refers to the summary and make sure that you work out which word will fit.
You also need to think about the grammar as the word you put in the reading gap fill must fit grammatically as well. Here are some general strategies
Strategies for Reading Gap Fill Tasks
Looking at question 1, these are the strategies you can try following:
Read through the summary carefully to make sure you understand it.
Work out which section of the reading the summary comes from (in this example, the whole of the text is summarized but in the real test you'll need to look through the reading to find the right paragraphs).
Carefully read the sentence with the first gap and think about what form will fit i.e. should it be an adjective, noun, infinitive, present participle etc? And what type of word is needed i.e. is it an amount, a change, an action?
You should have worked out that for questions 1 you are looking for a noun because 'an' comes before it.
Then look at the words that are in the box - which ones have the right form to fit and the right type? There are several nouns.
Look at the correct part of the full reading that refers to the reading gap fill section you are looking at and decide what happened for the first time to do with air rage in the 1940s?
Use this information to help you choose the correct word for the reading gap fill..
Reading Gap Fill Practice
The first recorded case of an airline passenger turning seriously violent during a flight, a phenomenon now widely known as “air rage”, happened in 1947 on a flight from Havana to Miami. A drunk man assaulted another passenger and bit a flight attendant. However, the man escaped punishment because it was not then clear under whose legal control a crime committed on plane was, the country where the plane was registered or the country where the crime was committed. In 1963, at the Tokyo convention, it was decided that the laws of the country where the plane is registered take precedence.
The frequency of air rage has expanded out of proportion to the growth of air travel. Until recently few statistic were gathered about air rage, but those that have been indicate that passengers are increasingly likely to cause trouble or engage in violent acts. For example, in 1998 there were 266 air rage incidents out of approximately four million passengers, a 400% increase from 1995. In the same period American Airlines showed a 200% rise. Air travel is predicted to rise by 5% internationally by 2010 leading to increased airport congestion. This, coupled with the flying public’s increased aggression, means that air rage may become a major issue in coming years.
Aside from discomfort and disruption, air rage poses some very real dangers to flying. The most extreme of these is when out of control passengers enter the cockpit. This has actually happened on a number of occasions, the worst of which have resulted in the death and injury of pilots or the intruder taking control of the plane, almost resulting in crashes. In addition, berserk passengers sometimes attempt to open the emergency doors while in flight, putting the whole aircraft in danger. These are extreme examples and cases of air rage more commonly result in physical assaults on fellow passengers and crew such as throwing objects, punching, stabbing or scalding with hot coffee.
Look at the words in the table and decide which word will fit in the reading gap fill summary. Type the word into the gap (when you have completed it you can click below to reveal and check your answers).
The first time that an (1)
of air rage was recorded was in the 1940’s, but the passenger was never actually charged for an offence because there were no clear rules in place to specify where to prosecute. It was later
that it would be the country where the plane is registered. Air rage has
significantly since this time, growing by a staggering 400% from 1995 to 1998. Air rage is
to be a major problem in the future as air travel increases, as do levels of aggression. Angry (5)
can put everyone in danger including the pilots, the crew and the other passengers, with some form of
being the most common consequence.
The first time that an incident of air rage was recorded was in the 1940’s, but the passenger was never actually charged for an offence because there were no clear rules in place to specify where to prosecute. It was later established that it would be the country where the plane is registered. Air rage has increased significantly since this time, growing by a staggering 400% from 1995 to 1998. Air rage is predicted to be a major problem in the future as air travel increases, as do levels of aggression. Angry passengers can put everyone in danger including the pilots, the crew and the other passengers, with some form of assault being the most common consequence.
Reading Gap Fill Answer Discussion
You should have worked out the this is a synonym for 'case'. The other nouns in the box would not fit here.
'Establish' can mean to set up something up such as a system of rules. So this word fits here. You may have thought it was 'found', but this means to discover something. The new legislation wasn't 'discovered'.
If you refer to the reading you can see that this is the trend referred to. 'Rose' does not fit grammatically.
The reading and the summary show that the future is being discussed, so this word fits.
It must be the plural as it is being used as a general noun. For the singular, an article would need to have been used.
If you put 'injury' this is wrong as the reading does not say people are commonly injured, but it does refer to assaults. You can get assaulted without getting injured.
Do you want to know what some of the latest IELTS writing topics have been? Here you can view real recent IELTS exam writing questions. They are from the Academic and General Test. You can also post t…<|endoftext|>
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|Posted by Gilbert Stack on November 9, 2018 at 7:00 AM|
On this day (November 9) in 1938 the Germans began a night of violence against German Jews called Kristallnacht or the Night of Broken Glass. At least 91 Jews were murdered and 30,000 were arrested and shipped to concentration camps where many more died. Houses, homes and schools were demolished, 267 synagogues were destroyed and more than 7000 business were damaged or destroyed. Kristallnacht was followed by additional economic and political persecution of Jews and is widely seen as the beginning of the NAZIs “Final Solution” and the Holocaust.
Categories: Today in History<|endoftext|>
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On earth, the equator receives more sunshine than do the poles. This is due to simple geometry of the earth’s curvature, a given amount of sunshine in a beam falling on the equator, which points directly at the sun, has a much more intense effect than the glancing rays spread over a much larger area of the curving surface near the poles. In addition, extensive ice and snow at the poles reflects back to space some of the sun’s energy that reaches the earth. Much more sunshine is absorbed to heat the earth at the equator. This means the land at the equator becomes hotter than the poles. If we had no atmosphere or oceans, the equator would become too hot for life as we know it, and the poles too cold. However, the atmosphere and oceans take some of the excess heat from the equator to the poles, making both habitable to humans. An interesting connection to make is that if the earth were heated evenly at all latitudes there would be no winds or ocean currents.
Many cultures thrive in warm equatorial regions. The Fang people of Gabon, for instance, are successful farmers who take advantage of the warm temperature and long rainy season to cultivate crops such as corn, yams, and plantains. The Fang also raise livestock that have adapted to the climate, such as goats and chickens.
Not all equatorial regions are hot and humid, however. Mount Kilimanjaro, Tanzania, is only 330 kilometers (205 miles) from the Equator, but its elevation creates a climate with cool, dry weather and even alpine glaciers.
The Andes are another equatorial region lacking the hot, humid climate often associated with the Equator. The mountain range includes a desert with almost no rain (the Atacama), as well as some of the tallest peaks on Earth. Here, too, cultures have thrived for thousands of years. The Aymara people of the Altiplano of Bolivia, Peru, and Chile, are primarily an urban people who identify strongly with the innovative navigational successes of their ancestors. In the 20th century, the Aymara helped build railroads through the high, equatorial Andes.
Many plant and animal species thrive in equatorial climates. The Amazon and Congo rain forest ecosystems, for example, are amazingly rich in biodiversity. A single hectare (2.47) of rain forest in Brazil may contain 750 species of trees and twice that many species of insects. The equatorial savanna of Kenya includes mammals such as lions, cheetahs, and elephants. The chilly equatorial Andes are famous for its camelid species: llamas, alpacas, vicunas, and guanacos.
The Sun and planets to scale Diameters: Sun – 1,392,684km Jupiter – 142,984km Saturn – 120,660km Uranus – 51,800km Neptune – 49,500km Earth – 12,750km Venus – 12,100km Mars – 6,800km Mercury – 4,800km.
shadeism.com | This documentary short is an introduction to the issue of shadeism, the discrimination that exists between the lighter-skinned and darker-skinned members of the same community. This documentary short looks specifically at how it affects young women within the African, Caribbean, and South Asian diasporas. Through the eyes and words of 5 young women and 1 little girl – all females of colour – the film takes us into the thoughts and experiences of each. Overall, ‘Shadeism’ explores where shadeism comes from, how it directly affects us as women of colour, and ultimately, begins to explore how we can move forward through dialogue and discussion.<|endoftext|>
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Late last year, the country of Nepal proudly shared the good news that their tiger population had doubled since 2009. Less than a decade ago, there were just 121 wild tigers living within the country’s borders. As of now, that number has increased to over 235.
So, how did Nepal achieve such a great feat? What lessons can we all learn from their actions to help increase wild animal populations around the globe?
Their success is largely contributed to a push for greater conservation efforts and the creation of Nepal’s Tiger Conservation Action Plan, which included the addition of innovative tools to help wildlife survive and thrive.
The World Wildlife Fund notes the country’s political commitment to conservation, something that has clearly paid off.
Nepal banded together as a nation –citizens, government, wildlife conservation funds, and even Leonardo DiCaprio, all worked together to protect natural habitats and wildlife.
As part of Nepal’s Tiger Conservation Action Plan, these efforts included the construction of passageways to connect different protected habitats so that tigers have more freedom to move between interconnected safe spaces.
In addition, they identified prey species and then worked to improve their habitats.
The country created incentives to promote citizens to report any poaching activity. That’s, in part, how they achieved 365 days with zero rhino poaching in May of last year.
“Our commitment to the Global Tiger Recovery Program gains new ground with Nepal’s growing tiger numbers and a successful implementation of Nepal’s Tiger Conservation Action Plan,” said Bishwa Nath Oli, Secretary of the Ministry of Forests and Environment.
Animals were carefully tracked and counted using cameras and occupancy surveys.
“Nepal conducted its national tiger survey between November 2017 and April 2018 in the transboundary Terai Arc Landscape (TAL), a vast area of diverse ecosystems shared with India,” states the World Wildlife Fund press release.
A variety of surveys and surveillance measures were implemented to determine the number and density of tigers. Line transfer surveys were used to calculate prey density.
The survey was led by the Government of Nepal’s Department of National Parks and Wildlife Conservation and Department of Forests. In addition, they worked with the WWF-Nepal, Zoological Society of London, and the National Trust for Nature Conservation.
“This significant increase in Nepal’s tiger population is proof that when we work together, we can save the planet’s wildlife – even species facing extinction,” said Leonardo DiCaprio.
The Hollywood hunk is a WWF-US board member and chairman. He also runs his own tiger conservation fund, the Leonardo DiCaprio Foundation, which funds tiger conservation in Nepal, as well as other parts of the world.
“Nepal has been a leader in efforts to double tigers within its own borders and serves as a model for conservation for all of Asia and the world. I am proud of my foundation’s partnership with WWF to support Nepal and local communities in doubling the population of wild tigers,” DiCaprio added.
At the St. Petersburg Summit in 2010, a goal called “TX2” was put in place to double the world’s tiger population by 2022. Nepal is well on its way to helping the world achieve this record!
“While Nepal is but a few tigers away from our goal to double tiger numbers by 2022, it also underscores the continued need to ensure protection, and improved and contiguous habitats for the long-term survival of the species,” said Dr. Ghana S Gurung, Country Representative, WWF-Nepal.
The WWF estimates that there are 5,000 tigers living in captivity in the US. That’s more than all the wild tigers living across Asia.
It is estimated that 3,890 wild tigers live in the world – most of them in Asia. This doesn’t sound like much, but it’s an increase from 2010, when the WWF estimated that there were 3,200 wild tigers living in the world.
Wild tigers are considered an endangered animal. Up until now, populations have been in decline every year since the 1900s. In 1900, it was estimated that there were 100,000 tigers living wild and free.
This is the first time in a long time that we’ve seen an increase in tiger populations. That alone is promising.<|endoftext|>
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The Space Race refers to the 20th-century competition between two Cold War rivals, the Soviet Union (USSR) and the United States (US), for dominance in spaceflight capability. It had its origins in the missile-based nuclear arms race between the two nations that occurred following World War II, aided by captured German missile technology and personnel from the Aggregat program. The technological superiority required for such dominance was seen as necessary for national security, and symbolic of ideological superiority. The Space Race spawned pioneering efforts to launch artificial satellites, uncrewed space probes of the Moon, Venus, and Mars, and human spaceflight in low Earth orbit and to the Moon.
The Space Race began on August 2, 1955, when the Soviet Union responded to the US announcement four days earlier of intent to launch artificial satellites for the International Geophysical Year, by declaring they would also launch a satellite "in the near future". The Soviet Union beat the US to the first successful launch, with the October 4, 1957, orbiting of Sputnik 1, and later beat the US to have the first human in earth orbit, Yuri Gagarin, on April 12, 1961. The "race" peaked with the July 20, 1969, US landing of the first humans on the Moon with Apollo 11. The USSR attempted several crewed lunar missions but eventually canceled them and concentrated on Earth orbital space stations.
A period of détente followed with the April 1972 agreement on a co-operative Apollo–Soyuz Test Project, resulting in the July 1975 rendezvous in Earth orbit of a US astronaut crew with a Soviet cosmonaut crew. The end of the Space Race is harder to pinpoint than its beginning, but it was over by the December 1991 dissolution of the Soviet Union, after which spaceflight cooperation between the US and Russia flourished.
The Space Race has left a legacy of Earth communications and weather satellites, and continuing human space presence on the International Space Station. It has also sparked increases in spending on education and research and development, which led to beneficial spin-off technologies.
- 1 Early rocket development
- 2 The Race begins
- 3 Uncrewed lunar probes
- 4 First human in space
- 5 First American in space
- 6 Kennedy directs the Race toward the Moon
- 7 Completion of Vostok and Mercury programs
- 8 Kennedy proposes a joint US-USSR program
- 9 Gemini and Voskhod
- 10 Soviet crewed Moon programs
- 11 Outer space treaty
- 12 Disaster strikes both sides
- 13 Onward to the Moon
- 14 Apollo 11
- 15 The Race winds down
- 16 Legacy
- 17 See also
- 18 Notes
- 19 References
- 20 External links
Early rocket development
Germany during World War II
The origins of the Space Race can be traced to Germany, beginning in the 1930s and continuing during World War II when Nazi Germany researched and built operational ballistic missiles capable of sub-orbital spaceflight. Starting in the early 1930s, during the last stages of the Weimar Republic, German aerospace engineers experimented with liquid-fueled rockets, with the goal that one day they would be capable of reaching high altitudes and traversing long distances. The head of the German Army's Ballistics and Munitions Branch, Lieutenant Colonel Karl Emil Becker, gathered a small team of engineers that included Walter Dornberger and Leo Zanssen, to figure out how to use rockets as long-range artillery in order to get around the Treaty of Versailles' ban on research and development of long-range cannons. Wernher von Braun, a young engineering prodigy, was recruited by Becker and Dornberger to join their secret army program at Kummersdorf-West in 1932. Von Braun dreamed of conquering outer space with rockets and did not initially see the military value in missile technology.
During the Second World War, General Dornberger was the military head of the army's rocket program, Zanssen became the commandant of the Peenemünde army rocket center, and von Braun was the technical director of the ballistic missile program. They led the team that built the Aggregat-4 (A-4) rocket, which became the first vehicle to reach outer space during its test flight program in 1942 and 1943. By 1943, Germany began mass-producing the A-4 as the Vergeltungswaffe 2 ("Vengeance Weapon" 2, or more commonly, V2), a ballistic missile with a 320 kilometers (200 mi) range carrying a 1,130 kilograms (2,490 lb) warhead at 4,000 kilometers per hour (2,500 mph). Its supersonic speed meant there was no defense against it, and radar detection provided little warning. Germany used the weapon to bombard southern England and parts of Allied-liberated western Europe from 1944 until 1945. After the war, the V-2 became the basis of early American and Soviet rocket designs.
At war's end, American, British, and Soviet scientific intelligence teams competed to capture Germany's rocket engineers along with the German rockets themselves and the designs on which they were based. Each of the Allies captured a share of the available members of the German rocket team, but the United States benefited the most with Operation Paperclip, recruiting von Braun and most of his engineering team, who later helped develop the American missile and space exploration programs. The United States also acquired a large number of complete V2 rockets.
Soviet rocket development
The German rocket center in Peenemünde was located in the eastern part of Germany, which became the Soviet zone of occupation. On Stalin's orders, the Soviet Union sent its best rocket engineers to this region to see what they could salvage for future weapons systems. The Soviet rocket engineers were led by Sergei Korolev. He had been involved in space clubs and early Soviet rocket design in the 1930s, but was arrested in 1938 during Joseph Stalin's Great Purge and imprisoned for six years in Gulag. After the war, he became the USSR's chief rocket and spacecraft engineer, essentially the Soviet counterpart to von Braun. His identity was kept a state secret throughout the Cold War, and he was identified publicly only as "the Chief Designer." In the West, his name was only officially revealed when he died in 1966.
After almost a year in the area around Peenemünde, Soviet officials conducted Operation Osoaviakhim and later moved more than 170 of the top captured German rocket specialists to Gorodomlya Island on Lake Seliger, about 240 kilometers (150 mi) northwest of Moscow. They were not allowed to participate in final Soviet missile design, but were used as problem-solving consultants to the Soviet engineers. They helped in the following areas: the creation of a Soviet version of the A-4; work on "organizational schemes"; research in improving the A-4 main engine; development of a 100-ton engine; assistance in the "layout" of plant production rooms; and preparation of rocket assembly using German components. With their help, particularly Helmut Gröttrup's group, Korolev reverse-engineered the A-4 and built his own version of the rocket, the R-1, in 1948. Later, he developed his own distinct designs, though many of these designs were influenced by the Gröttrup Group's G4-R10 design from 1949. The Germans were eventually repatriated in 1951–53.
American rocket development
The American professor Robert H. Goddard had worked on developing solid-propellant rockets since 1914, and demonstrated a light battlefield rocket to the US Army Signal Corps only five days before the signing of the armistice that ended World War I. He also started developing liquid-propellant rockets in 1921, yet he had not been taken seriously by the public.
Von Braun and his team were sent to the United States Army's White Sands Proving Ground, located in New Mexico, in 1945. They set about assembling the captured V2s and began a program of launching them and instructing American engineers in their operation. These tests led to the first rocket to take photos from outer space, and the first two-stage rocket, the WAC Corporal-V2 combination, in 1949. The German rocket team was moved from Fort Bliss to the Army's new Redstone Arsenal, located in Huntsville, Alabama, in 1950. From here, von Braun and his team developed the Army's first operational medium-range ballistic missile, the Redstone rocket, that in slightly modified versions, launched both America's first satellite, and the first piloted Mercury space missions. It became the basis for both the Jupiter and Saturn family of rockets.
Cold War missile race
The Cold War (1947–1991) developed between two former allies, the Soviet Union and the United States, soon after the end of the Second World War. It involved a continuing state of political conflict, military tension, proxy wars, and economic competition, primarily between the Soviet Union and its satellite states (often referred to as the Eastern Bloc) and the powers of the Western world, particularly the United States. The primary participants' military forces never clashed directly, but expressed this conflict through military coalitions, strategic conventional force deployments, extensive aid to states deemed vulnerable, proxy wars, espionage, propaganda, a nuclear arms race, and economic and technological competitions, such as the Space Race.
In simple terms, the Cold War could be viewed as an expression of the ideological struggle between communism and capitalism. The United States faced a new uncertainty beginning in September 1949, when it lost its monopoly on the atomic bomb. American intelligence agencies discovered that the Soviet Union had exploded its first atomic bomb, with the consequence that the United States potentially could face a future nuclear war that, for the first time, might devastate its cities. Given this new danger, the United States participated in an arms race with the Soviet Union that included development of the hydrogen bomb, as well as intercontinental strategic bombers and intercontinental ballistic missiles (ICBMs) capable of delivering nuclear weapons. A new fear of communism and its sympathizers swept the United States during the 1950s, which devolved into paranoid McCarthyism. With communism spreading in China, Korea, and Eastern Europe, Americans came to feel so threatened that popular and political culture condoned extensive "witch-hunts" to expose communist spies. Part of the American reaction to the Soviet atomic and hydrogen bomb tests included maintaining a large Air Force, under the control of the Strategic Air Command (SAC). SAC employed intercontinental strategic bombers, as well as medium-bombers based close to Soviet airspace (in western Europe and in Turkey) that were capable of delivering nuclear payloads.
For its part, the Soviet Union harbored fears of invasion. Having suffered at least 27 million casualties during World War II after being invaded by Nazi Germany in 1941, the Soviet Union was wary of its former ally, the United States, which until late 1949 was the sole possessor of atomic weapons. The United States had used these weapons operationally during World War II, and it could use them again against the Soviet Union, laying waste to its cities and military centers. Since the Americans had a much larger air force than the Soviet Union, and the United States maintained advance air bases near Soviet territory, in 1947 Stalin ordered the development of intercontinental ballistic missiles (ICBMs) in order to counter the perceived American threat.
In 1953, Korolev was given the go-ahead to develop the R-7 Semyorka rocket, which represented a major advance from the German design. Although some of its components (notably boosters) still resembled the German G-4, the new rocket incorporated staged design, a completely new control system, and a new fuel. It was successfully tested on August 21, 1957, and became the world's first fully operational ICBM the following month. It was later used to launch the first satellite into space, and derivatives launched all piloted Soviet spacecraft.
The United States had multiple rocket programs divided among the different branches of the American armed services, which meant that each force developed its own ICBM program. The Air Force initiated ICBM research in 1945 with the MX-774. However, its funding was cancelled and only three partially successful launches were conducted in 1947. In 1950, von Braun began testing the Air Force PGM-11 Redstone rocket family at Cape Canaveral. In 1951, the Air Force began a new ICBM program called MX-1593, and by 1955 this program was receiving top-priority funding. The MX-1593 program evolved to become the Atlas-A, with its maiden launch occurring June 11, 1957, becoming the first successful American ICBM. Its upgraded version, the Atlas-D rocket, later served as a nuclear ICBM and as the orbital launch vehicle for Project Mercury and the remote-controlled Agena Target Vehicle used in Project Gemini.
With the Cold War as an engine for change in the ideological competition between the United States and the Soviet Union, a coherent space policy began to take shape in the United States during the late 1950s. Korolev took inspiration from the competition as well, achieving many firsts to counter the possibility that the United States might prevail.
The Race begins
First artificial satellite
In 1955, with both the United States and the Soviet Union building ballistic missiles that could be utilized to launch objects into space, the "starting line" was drawn for the Space Race. In separate announcements four days apart, both nations publicly announced that they would launch artificial Earth satellites by 1957 or 1958. On July 29, 1955, James C. Hagerty, President Dwight D. Eisenhower's press secretary, announced that the United States intended to launch "small Earth circling satellites" between July 1, 1957, and December 31, 1958, as part of their contribution to the International Geophysical Year (IGY). Four days later, at the Sixth Congress of International Astronautical Federation in Copenhagen, scientist Leonid I. Sedov spoke to international reporters at the Soviet embassy and announced his country's intention to launch a satellite as well, in the "near future". On August 30, 1955, Korolev managed to get the Soviet Academy of Sciences to create a commission whose purpose was to beat the Americans into Earth orbit: this was the de facto start date for the Space Race. The Council of Ministers of the Soviet Union began a policy of treating development of its space program as a classified state secret.
Initially, President Eisenhower was worried that a satellite passing above a nation at over 100 kilometers (62 mi), might be construed as violating that nation's sovereign airspace. He was concerned that the Soviet Union would accuse the Americans of an illegal overflight, thereby scoring a propaganda victory at his expense. Eisenhower and his advisors believed that a nation's airspace sovereignty did not extend into outer space, acknowledged as the Kármán line, and he used the 1957–58 International Geophysical Year launches to establish this principle in international law. Eisenhower also feared that he might cause an international incident and be called a "warmonger" if he were to use military missiles as launchers. Therefore, he selected the untried Naval Research Laboratory's Vanguard rocket, which was a research-only booster. This meant that von Braun's team was not allowed to put a satellite into orbit with their Jupiter-C rocket, because of its intended use as a future military vehicle. On September 20, 1956, von Braun and his team did launch a Jupiter-C that was capable of putting a satellite into orbit, but the launch was used only as a suborbital test of nose cone reentry technology.
Korolev received word about von Braun's 1956 Jupiter-C test, but thinking it was a satellite mission that failed, he expedited plans to get his own satellite in orbit. Since his R-7 was substantially more powerful than any of the American boosters, he made sure to take full advantage of this capability by designing Object D as his primary satellite. It was given the designation 'D', to distinguish it from other R-7 payload designations 'A', 'B', 'V', and 'G' which were nuclear weapon payloads. Object D dwarfed the proposed American satellites, by having a weight of 1,400 kilograms (3,100 lb), of which 300 kilograms (660 lb) would be composed of scientific instruments that would photograph the Earth, take readings on radiation levels, and check on the planet's magnetic field. However, things were not going along well with the design and manufacturing of the satellite, so in February 1957, Korolev sought and received permission from the Council of Ministers to create a Prosteishy Sputnik (PS-1), or simple satellite. The Council also decreed that Object D be postponed until April 1958. The new Sputnik was a shiny sphere that would be a much lighter craft, weighing 83.8 kilograms (185 lb) and having a 58-centimeter (23 in) diameter. The satellite would not contain the complex instrumentation that Object D had, but had two radio transmitters operating on different short wave radio frequencies, the ability to detect if a meteoroid were to penetrate its pressure hull, and the ability to detect the density of the Earth's thermosphere.
Korolev was buoyed by the first successful launches of his R-7 rocket in August and September, which paved the way for him to launch his sputnik. Word came that the Americans were planning to announce a major breakthrough at an International Geophysical Year conference at the National Academy of Sciences in Washington D.C., with a paper entitled "Satellite Over the Planet", on October 6, 1957. Korolev anticipated that von Braun might launch a Jupiter-C with a satellite payload on or around October 4 or 5, in conjunction with the paper. He hastened the launch, moving it to October 4. The launch vehicle for PS-1, was a modified R-7 – vehicle 8K71PS number M1-PS– without much of the test equipment and radio gear that was present in the previous launches. It arrived at the Soviet missile base Tyura-Tam in September and was prepared for its mission at launch site number one. On Friday, October 4, 1957, at exactly 10:28:34 pm Moscow time, the R-7, with the now named Sputnik 1 satellite, lifted off the launch pad, and placed this artificial "moon" into an orbit a few minutes later. This "fellow traveler," as the name is translated in English, was a small, beeping ball, less than two feet in diameter and weighing less than 200 pounds. But the celebrations were muted at the launch control center until the down-range far east tracking station at Kamchatka received the first distinctive beep ... beep ... beep sounds from Sputnik 1's radio transmitters, indicating that it was on its way to completing its first orbit. About 95 minutes after launch, the satellite flew over its launch site, and its radio signals were picked up by the engineers and military personnel at Tyura-Tam: that's when Korolev and his team celebrated the first successful artificial satellite placed into Earth-orbit.
The Soviet success raised a great deal of concern in the United States. For example, economist Bernard Baruch wrote in an open letter titled "The Lessons of Defeat" to the New York Herald Tribune: "While we devote our industrial and technological power to producing new model automobiles and more gadgets, the Soviet Union is conquering space. ... It is Russia, not the United States, who has had the imagination to hitch its wagon to the stars and the skill to reach for the moon and all but grasp it. America is worried. It should be."
Eisenhower ordered project Vanguard to move up its timetable and launch its satellite much sooner than originally planned. The December 6, 1957 Project Vanguard launch failure occurred at Cape Canaveral Air Force Station in Florida, broadcast live in front of a US television audience. It was a monumental failure, exploding a few seconds after launch, and it became an international joke. The satellite appeared in newspapers under the names Flopnik, Stayputnik, Kaputnik, and Dudnik. In the United Nations, the Soviet delegate offered the US representative aid "under the Soviet program of technical assistance to backwards nations." Only in the wake of this very public failure did von Braun's Redstone team get the go-ahead to launch their Jupiter-C rocket as soon as they could. In Britain, the US's Western Cold War ally, the reaction was mixed: some celebrated the fact that the Soviets had reached space first, while others feared the destructive potential that military uses of spacecraft might bring.
On January 31, 1958, nearly four months after the launch of Sputnik 1, von Braun and the United States successfully launched its first satellite on a four-stage Juno I rocket derived from the US Army's Redstone missile, at Cape Canaveral. The satellite Explorer 1 was 30.66 pounds (13.91 kg) in mass. The payload of Explorer 1 weighed 18.35 pounds (8.32 kg). It carried a micrometeorite gauge and a Geiger-Müller tube. It passed in and out of the Earth-encompassing radiation belt with its 194-by-1,368-nautical-mile (360 by 2,534 km) orbit, therefore saturating the tube's capacity and proving what Dr. James Van Allen, a space scientist at the University of Iowa, had theorized. The belt, named the Van Allen radiation belt, is a doughnut-shaped zone of high-level radiation intensity around the Earth above the magnetic equator. Van Allen was also the man who designed and built the satellite instrumentation of Explorer 1. The satellite measured three phenomena: cosmic ray and radiation levels, the temperature in the spacecraft, and the frequency of collisions with micrometeorites. The satellite had no memory for data storage, therefore it had to transmit continuously. In March 1958 a second satellite was sent into orbit with augmented cosmic ray instruments.
On April 2, 1958, President Eisenhower reacted to the Soviet space lead in launching the first satellite by recommending to the US Congress that a civilian agency be established to direct nonmilitary space activities. Congress, led by Senate Majority Leader Lyndon B. Johnson, responded by passing the National Aeronautics and Space Act, which Eisenhower signed into law on July 29, 1958. This law turned the National Advisory Committee on Aeronautics into the National Aeronautics and Space Administration (NASA). It also created a Civilian-Military Liaison Committee, chaired by the President, responsible for coordinating the nation's civilian and military space programs.
On October 21, 1959, Eisenhower approved the transfer of the Army's remaining space-related activities to NASA. On July 1, 1960, the Redstone Arsenal became NASA's George C. Marshall Space Flight Center, with von Braun as its first director. Development of the Saturn rocket family, which when mature gave the US parity with the Soviets in terms of lifting capability, was thus transferred to NASA.
Uncrewed lunar probes
In 1958, Korolev upgraded the R-7 to be able to launch a 400-kilogram (880 lb) payload to the Moon. Three secret 1958 attempts to launch Luna E-1-class impactor probes failed. The fourth attempt, Luna 1, launched successfully on January 2, 1959, but missed the Moon. The fifth attempt on June 18 also failed at launch. The 390-kilogram (860 lb) Luna 2 successfully impacted the Moon on September 14, 1959. The 278.5-kilogram (614 lb) Luna 3 successfully flew by the Moon and sent back pictures of its far side on October 6, 1959.
The US reacted to the Luna program by embarking on the Ranger program in 1959, managed by NASA's Jet Propulsion Laboratory. The Block I Ranger 1 and Ranger 2 suffered Atlas-Agena launch failures in August and November 1961. The 727-pound (330 kg) Block II Ranger 3 launched successfully on January 26, 1962, but missed the Moon. The 730-pound (330 kg) Ranger 4 became the first US spacecraft to reach the Moon, but its solar panels and navigational system failed near the Moon and it impacted the far side without returning any scientific data. Ranger 5 ran out of power and missed the Moon by 725 kilometers (391 nmi) on October 21, 1962. The first successful Ranger mission was the 806-pound (366 kg) Block III Ranger 7 which impacted on July 31, 1964.
First human in space
By 1959, American observers believed that the Soviet Union would be the first to get a human into space, because of the time needed to prepare for Mercury's first launch. On April 12, 1961, the USSR surprised the world again by launching Yuri Gagarin into a single orbit around the Earth in a craft they called Vostok 1. They dubbed Gagarin the first cosmonaut, roughly translated from Russian and Greek as "sailor of the universe". Although he had the ability to take over manual control of his capsule in an emergency by opening an envelope he had in the cabin that contained a code that could be typed into the computer, it was flown in an automatic mode as a precaution; medical science at that time did not know what would happen to a human in the weightlessness of space. Vostok 1 orbited the Earth for 108 minutes and made its reentry over the Soviet Union, with Gagarin ejecting from the spacecraft at 7,000 meters (23,000 ft), and landing by parachute. The Fédération Aéronautique Internationale (International Federation of Aeronautics) credited Gagarin with the world's first human space flight, although their qualifying rules for aeronautical records at the time required pilots to take off and land with their craft. For this reason, the Soviet Union omitted from their FAI submission the fact that Gagarin did not land with his capsule. When the FAI filing for Gherman Titov's second Vostok flight in August 1961 disclosed the ejection landing technique, the FAI committee decided to investigate, and concluded that the technological accomplishment of human spaceflight lay in the safe launch, orbiting, and return, rather than the manner of landing, and revised their rules, keeping Gagarin's and Titov's records intact.
Gagarin became a national hero of the Soviet Union and the Eastern Bloc, and a worldwide celebrity. Moscow and other cities in the USSR held mass demonstrations, the scale of which was second only to the World War II Victory Parade of 1945. April 12 was declared Cosmonautics Day in the USSR, and is celebrated today in Russia as one of the official "Commemorative Dates of Russia." In 2011, it was declared the International Day of Human Space Flight by the United Nations.
The radio communication between the launch control room and Gagarin included the following dialogue at the moment of rocket launch:
Korolev: "Preliminary stage..... intermediate..... main..... lift off! We wish you a good flight. Everything is all right."
Gagarin: "Поехали!" (Poyekhali! - Let's go!).
First American in space
The US Air Force had been developing a program to launch the first man in space, named Man in Space Soonest. This program studied several different types of one-man space vehicles, settling on a ballistic re-entry capsule launched on a derivative Atlas missile, and selecting a group of nine candidate pilots. After NASA's creation, the program was transferred over to the civilian agency and renamed Project Mercury on November 26, 1958. NASA selected a new group of astronaut (from the Greek for "star sailor") candidates from Navy, Air Force and Marine test pilots, and narrowed this down to a group of seven for the program. Capsule design and astronaut training began immediately, working toward preliminary suborbital flights on the Redstone missile, followed by orbital flights on the Atlas. Each flight series would first start uncrewed, then carry a non-human primate, then finally humans.
On May 5, 1961, Alan Shepard became the first American in space, launched in a ballistic trajectory on Mercury-Redstone 3, in a spacecraft he named Freedom 7. Though he did not achieve orbit like Gagarin, he was the first person to exercise manual control over his spacecraft's attitude and retro-rocket firing. After his successful return, Shepard was celebrated as a national hero, honored with parades in Washington, New York and Los Angeles, and received the NASA Distinguished Service Medal from President John F. Kennedy.
Kennedy directs the Race toward the Moon
Before Gagarin's flight, US President John F. Kennedy's support for America's crewed space program was lukewarm. Jerome Wiesner of MIT, who served as a science advisor to presidents Eisenhower and Kennedy, and himself an opponent of crewed space exploration, remarked, "If Kennedy could have opted out of a big space program without hurting the country in his judgment, he would have." As late as March 1961, when NASA administrator James E. Webb submitted a budget request to fund a Moon landing before 1970, Kennedy rejected it because it was simply too expensive. Some were surprised by Kennedy's eventual support of NASA and the space program because of how often he had attacked the Eisenhower administration's inefficiency during the election.
Gagarin's flight changed this; now Kennedy sensed the humiliation and fear on the part of the American public over the Soviet lead. Additionally, the Bay of Pigs invasion, planned before his term began but executed during it, was an embarrassment to his administration due to the colossal failure of the American forces. Looking for something to save political face, he sent a memo dated April 20, 1961, to Vice President Lyndon B. Johnson, asking him to look into the state of America's space program, and into programs that could offer NASA the opportunity to catch up. The two major options at the time seemed to be, either establishment of an Earth orbital space station, or a crewed landing on the Moon. Johnson, in turn, consulted with von Braun, who answered Kennedy's questions based on his estimates of US and Soviet rocket lifting capability. Based on this, Johnson responded to Kennedy, concluding that much more was needed to reach a position of leadership, and recommending that the crewed Moon landing was far enough in the future that the US had a fighting chance to achieve it first.
Kennedy ultimately decided to pursue what became the Apollo program, and on May 25 took the opportunity to ask for Congressional support in a Cold War speech titled "Special Message on Urgent National Needs". Full text He justified the program in terms of its importance to national security, and its focus of the nation's energies on other scientific and social fields. He rallied popular support for the program in his "We choose to go to the Moon" speech, on September 12, 1962, before a large crowd at Rice University Stadium, in Houston, Texas, near the construction site of the new Manned Spacecraft Center facility. Full text Khrushchev responded to Kennedy's implicit challenge with silence, refusing to publicly confirm or deny the Soviets were pursuing a "Moon race". As later disclosed, the Soviet Union secretly pursued a crewed lunar program until 1974.
Completion of Vostok and Mercury programs
American Virgil "Gus" Grissom repeated Shepard's suborbital flight in Liberty Bell 7 on July 21, 1961. Almost a year after the Soviet Union put a human into orbit, astronaut John Glenn became the first American to orbit the Earth, on February 20, 1962. His Mercury-Atlas 6 mission completed three orbits in the Friendship 7 spacecraft, and splashed down safely in the Atlantic Ocean, after a tense reentry, due to what falsely appeared from the telemetry data to be a loose heat-shield. As the first American in orbit, Glenn became a national hero, and received a ticker-tape parade in New York City, reminiscent of that given for Charles Lindbergh. On February 23, 1962, President Kennedy escorted him in a parade at Cape Canaveral Air Force Station, where he awarded Glenn with the NASA service medal.
The United States launched three more Mercury flights after Glenn's: Aurora 7 on May 24, 1962, duplicated Glenn's three orbits; Sigma 7 on October 3, 1962, six orbits; and Faith 7 on May 15, 1963, 22 orbits (32.4 hours), the maximum capability of the spacecraft. NASA at first intended to launch one more mission, extending the spacecraft's endurance to three days, but since this would not beat the Soviet record, it was decided instead to concentrate on developing Project Gemini.
Gherman Titov became the first Soviet cosmonaut to exercise manual control of his Vostok 2 craft on August 6, 1961. The Soviet Union demonstrated 24-hour launch pad turnaround and the capability to launch two piloted spacecraft, Vostok 3 and Vostok 4, in essentially identical orbits, on August 11 and 12, 1962. The two spacecraft came within approximately 6.5 kilometers (4.0 mi) of one another, close enough for radio communication. Vostok 4 also set a record of nearly four days in space. Though the two craft's orbits were as nearly identical as possible given the accuracy of the launch rocket's guidance system, slight variations still existed which drew the two craft at first as close to each other as 6.5 kilometers (3.5 nautical miles), then as far apart as 2,850 kilometers (1,540 nautical miles). There were no maneuvering rockets on the Vostok to permit space rendezvous, required to keep two spacecraft a controlled distance apart.
The Soviet Union duplicated its dual-launch feat with Vostok 5 and Vostok 6 (June 16, 1963). This time they launched the first woman (also the first civilian), Valentina Tereshkova, into space on Vostok 6. Launching a woman was reportedly Korolev's idea, and it was accomplished purely for propaganda value. Tereshkova was one of a small corps of female cosmonauts who were amateur parachutists, but Tereshkova was the only one to fly. The USSR didn't again open its cosmonaut corps to women until 1980, two years after the United States opened its astronaut corps to women.
The Soviets kept the details and true appearance of the Vostok capsule secret until the April 1965 Moscow Economic Exhibition, where it was first displayed without its aerodynamic nose cone concealing the spherical capsule. The "Vostok spaceship" had been first displayed at the July 1961 Tushino air show, mounted on its launch vehicle's third stage, with the nose cone in place. A tail section with eight fins was also added, in an apparent attempt to confuse western observers. This spurious tail section also appeared on official commemorative stamps and a documentary.
Kennedy proposes a joint US-USSR program
On September 20, 1963, in a speech before the United Nations General Assembly, President Kennedy proposed that the United States and the Soviet Union join forces in an effort to reach the Moon. Kennedy thus changed his mind regarding the desirability of the space race, preferring instead to ease tensions with the Soviet Union by cooperating on projects such as a joint lunar landing. Soviet Premier Nikita Khrushchev initially rejected Kennedy's proposal. However, on October 2, 1997, it was reported that Khrushchev's son Sergei claimed Khrushchev was poised to accept Kennedy's proposal at the time of Kennedy's assassination on November 22, 1963. During the next few weeks he reportedly concluded that both nations might realize cost benefits and technological gains from a joint venture, and decided to accept Kennedy's offer based on a measure of rapport during their years as leaders of the world's two superpowers, but changed his mind and dropped the idea since he did not have the same trust for Kennedy's successor, Lyndon Johnson.
As President, Johnson steadfastly pursued the Gemini and Apollo programs, promoting them as Kennedy's legacy to the American public. One week after Kennedy's death, he issued an executive order renaming the Cape Canaveral and Apollo launch facilities after Kennedy.
Gemini and Voskhod
Focused by the commitment to a Moon landing, in January 1962 the US announced Project Gemini, a two-man spacecraft that would support the later three-man Apollo by developing the key spaceflight technologies of space rendezvous and docking of two craft, flight durations of sufficient length to simulate going to the Moon and back, and extra-vehicular activity to accomplish useful work outside the spacecraft.
Meanwhile, Korolev had planned further, long-term missions for the Vostok spacecraft, and had four Vostoks in various stages of fabrication in late 1963 at his OKB-1 facilities. At that time, the Americans announced their ambitious plans for the Project Gemini flight schedule. These plans included major advancements in spacecraft capabilities, including a two-person spacecraft, the ability to change orbits, the capacity to perform an extravehicular activity (EVA), and the goal of docking with another spacecraft. These represented major advances over the previous Mercury or Vostok capsules, and Korolev felt the need to try to beat the Americans to many of these innovations. Korolev already had begun designing the Vostok's replacement, the next-generation Soyuz spacecraft, a multi-cosmonaut spacecraft that had at least the same capabilities as the Gemini spacecraft. Soyuz would not be available for at least three years, and it could not be called upon to deal with this new American challenge in 1964 or 1965. Political pressure in early 1964–which some sources claim was from Khrushchev while other sources claim was from other Communist Party officials—pushed him to modify his four remaining Vostoks to beat the Americans to new space firsts in the size of flight crews, and the duration of missions.
The greater advances of the Soviet space program at the time allowed their space program to achieve other significant firsts, including the first EVA "spacewalk" and the first mission performed by a crew in shirt-sleeves. Gemini took a year longer than planned to accomplish its first flight, allowing the Soviets to achieve another first, launching Voskhod 1 on October 12, 1964, the first spacecraft with a three-cosmonaut crew. The USSR touted another technological achievement during this mission: it was the first space flight during which cosmonauts performed in a shirt-sleeve-environment. However, flying without spacesuits was not due to safety improvements in the Soviet spacecraft's environmental systems; rather this innovation was accomplished because the craft's limited cabin space did not allow for spacesuits. Flying without spacesuits exposed the cosmonauts to significant risk in the event of potentially fatal cabin depressurization. This feat was not repeated until the US Apollo Command Module flew in 1968; this later mission was designed from the outset to safely transport three astronauts in a shirt-sleeve environment while in space.
Between October 14–16, 1964, Leonid Brezhnev and a small cadre of high-ranking Communist Party officials deposed Khrushchev as Soviet government leader a day after Voskhod 1 landed, in what was called the "Wednesday conspiracy". The new political leaders, along with Korolev, ended the technologically troublesome Voskhod program, cancelling Voskhod 3 and 4, which were in the planning stages, and started concentrating on the race to the Moon. Voskhod 2 ended up being Korolev's final achievement before his death on January 14, 1966, as it became the last of the many space firsts that demonstrated the USSR's domination in spacecraft technology during the early 1960s. According to historian Asif Siddiqi, Korolev's accomplishments marked "the absolute zenith of the Soviet space program, one never, ever attained since." There was a two-year pause in Soviet piloted space flights while Voskhod's replacement, the Soyuz spacecraft, was designed and developed.
On March 18, 1965, about a week before the first American piloted Project Gemini space flight, the USSR accelerated the competition, by launching the two-cosmonaut Voskhod 2 mission with Pavel Belyayev and Alexei Leonov. Voskhod 2's design modifications included the addition of an inflatable airlock to allow for extravehicular activity (EVA), also known as a spacewalk, while keeping the cabin pressurized so that the capsule's electronics would not overheat. Leonov performed the first-ever EVA as part of the mission. A fatality was narrowly avoided when Leonov's spacesuit expanded in the vacuum of space, preventing him from re-entering the airlock. In order to overcome this, he had to partially depressurize his spacesuit to a potentially dangerous level. He succeeded in safely re-entering the ship, but he and Belyayev faced further challenges when the spacecraft's atmospheric controls flooded the cabin with 45% pure oxygen, which had to be lowered to acceptable levels before re-entry. The reentry involved two more challenges: an improperly timed retrorocket firing caused the Voskhod 2 to land 386 kilometers (240 mi) off its designated target area, the town of Perm; and the instrument compartment's failure to detach from the descent apparatus caused the spacecraft to become unstable during reentry.
Though delayed a year to reach its first flight, Gemini was able to take advantage of the USSR's two-year hiatus after Voskhod, which enabled the US to catch up and surpass the previous Soviet lead in piloted spaceflight. Gemini achieved several significant firsts during the course of ten piloted missions:
- On Gemini 3 (March 1965), astronauts Virgil "Gus" Grissom and John W. Young became the first to demonstrate their ability to change their craft's orbit.
- On Gemini 5 (August 1965), astronauts L. Gordon Cooper and Charles "Pete" Conrad set a record of almost eight days in space, long enough for a piloted lunar mission.
- On Gemini 6A (December 1965), Command Pilot Wally Schirra achieved the first space rendezvous with Gemini 7, accurately matching his orbit to that of the other craft, station-keeping for three consecutive orbits at distances as close as 1 foot (0.30 m).
- Gemini 7 also set a human spaceflight endurance record of fourteen days for Frank Borman and James A. Lovell, which stood until both nations started launching space laboratories in the early 1970s.
- On Gemini 8 (March 1966), Command Pilot Neil Armstrong achieved the first docking between two spacecraft, his Gemini craft and an Agena target vehicle.
- Gemini 11 (September 1966), commanded by Conrad, achieved the first direct-ascent rendezvous with its Agena target on the first orbit, and used the Agena's rocket to achieve an apogee of 742 nautical miles (1,374 km), the crewed Earth orbit record still current as of 2015.
- On Gemini 12 (November 1966), Edwin E. "Buzz" Aldrin spent over five hours working comfortably during three (EVA) sessions, finally proving that humans could perform productive tasks outside their spacecraft. This proved to be the most difficult goal to achieve.
Most of the novice pilots on the early missions would command the later missions. In this way, Project Gemini built up spaceflight experience for the pool of astronauts for the Apollo lunar missions.
Soviet crewed Moon programs
Korolev's design bureau produced two prospectuses for circumlunar spaceflight (March 1962 and May 1963), the main spacecraft for which were early versions of his Soyuz design. Soviet Communist Party Central Committee Command 655-268 officially established two secret, competing crewed programs for circumlunar flights and lunar landings, on August 3, 1964. The circumlunar flights were planned to occur in 1967, and the landings to start in 1968.
The circumlunar program (Zond), created by Vladimir Chelomey's design bureau OKB-52, was to fly two cosmonauts in a stripped-down Soyuz 7K-L1, launched by Chelomey's Proton UR-500 rocket. The Zond sacrificed habitable cabin volume for equipment, by omitting the Soyuz orbital module. Chelomey gained favor with Khruschev by employing members of his family.
Korolev's lunar landing program was designated N1/L3, for its N1 super rocket and a more advanced Soyuz 7K-L3 spacecraft, also known as the lunar orbital module ("Lunniy Orbitalny Korabl", LOK), with a crew of two. A separate lunar lander ("Lunniy Korabl", LK), would carry a single cosmonaut to the lunar surface.
The N1/L3 launch vehicle had three stages to Earth orbit, a fourth stage for Earth departure, and a fifth stage for lunar landing assist. The combined space vehicle was roughly the same height and takeoff mass as the three-stage US Apollo/ Saturn V and exceeded its takeoff thrust by 28%, but had only roughly half the translunar injection payload capability.
Following Khruschev's ouster from power, Chelomey's Zond program was merged into the N1/L3 program.
Outer space treaty
The US and USSR began discussions on the peaceful uses of space as early as 1958, presenting issues for debate to the United Nations, which created a Committee on the Peaceful Uses of Outer Space in 1959.
On May 10, 1962, Vice President Johnson addressed the Second National Conference on the Peaceful Uses of Space revealing that the United States and the USSR both supported a resolution passed by the Political Committee of the UN General Assembly on December 1962, which not only urged member nations to "extend the rules of international law to outer space," but to also cooperate in its exploration. Following the passing of this resolution, Kennedy commenced his communications proposing a cooperative American/Soviet space program.
The UN ultimately created a Treaty on Principles Governing the Activities of States in the Exploration and Use of Outer Space, including the Moon and Other Celestial Bodies, which was signed by the United States, USSR, and the United Kingdom on January 27, 1967, and went into force the following October 10.
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- bars party States from placing weapons of mass destruction in Earth orbit, on the Moon, or any other celestial body;
- exclusively limits the use of the Moon and other celestial bodies to peaceful purposes, and expressly prohibits their use for testing weapons of any kind, conducting military maneuvers, or establishing military bases, installations, and fortifications;
- declares that the exploration of outer space shall be done to benefit all countries and shall be free for exploration and use by all the States;
- explicitly forbids any government from claiming a celestial resource such as the Moon or a planet, claiming that they are the common heritage of mankind, "not subject to national appropriation by claim of sovereignty, by means of use or occupation, or by any other means". However, the State that launches a space object retains jurisdiction and control over that object;
- holds any State liable for damages caused by their space object;
- declares that "the activities of non-governmental entities in outer space, including the Moon and other celestial bodies, shall require authorization and continuing supervision by the appropriate State Party to the Treaty", and "States Parties shall bear international responsibility for national space activities whether carried out by governmental or non-governmental entities"; and
- "A State Party to the Treaty which has reason to believe that an activity or experiment planned by another State Party in outer space, including the Moon and other celestial bodies, would cause potentially harmful interference with activities in the peaceful exploration and use of outer space, including the Moon and other celestial bodies, may request consultation concerning the activity or experiment."
The treaty remains in force, signed by 107 member states. – As of July 2017[update]
Disaster strikes both sides
In 1967, both nations faced serious challenges that brought their programs to temporary halts. Both had been rushing at full-speed toward the first piloted flights of Apollo and Soyuz, without paying due diligence to growing design and manufacturing problems. The results proved fatal to both pioneering crews.
On January 27, 1967, the same day the US and USSR signed the Outer Space Treaty, the crew of the first crewed Apollo mission, Command Pilot Virgil "Gus" Grissom, Senior Pilot Ed White, and Pilot Roger Chaffee, were killed in a fire that swept through their spacecraft cabin during a ground test, less than a month before the planned February 21 launch. An investigative board determined the fire was probably caused by an electrical spark, and quickly grew out of control, fed by the spacecraft's pure oxygen atmosphere. Crew escape was made impossible by inability to open the plug door hatch cover against the greater-than-atmospheric internal pressure. The board also found design and construction flaws in the spacecraft, and procedural failings, including failure to appreciate the hazard of the pure-oxygen atmosphere, as well as inadequate safety procedures. All these flaws had to be corrected over the next twenty-two months until the first piloted flight could be made. Mercury and Gemini veteran Grissom had been a favored choice of Deke Slayton, NASA's Director of Flight Crew Operations, to make the first piloted landing.
On April 24, 1967, the single pilot of Soyuz 1, Vladimir Komarov, became the first in-flight spaceflight fatality. The mission was planned to be a three-day test, to include the first Soviet docking with an unpiloted Soyuz 2, but the mission was plagued with problems. Early on, Komarov's craft lacked sufficient electrical power because only one of two solar panels had deployed. Then the automatic attitude control system began malfunctioning and eventually failed completely, resulting in the craft spinning wildly. Komarov was able to stop the spin with the manual system, which was only partially effective. The flight controllers aborted his mission after only one day. During the emergency re-entry, a fault in the landing parachute system caused the primary chute to fail, and the reserve chute became tangled with the drogue chute, causing descent speed to reach as high as 40 m/s (140 km/h; 89 mph). Shortly thereafter, Soyuz 1 impacted the ground 3 km (1.9 mi) west of Karabutak, exploding into a ball of flames. The official autopsy states Komarov died of blunt force trauma on impact, and that the subsequent heat mutilation of his corpse was a result of the explosive impact. Fixing the spacecraft's faults caused an eighteen-month delay before piloted Soyuz flights could resume.
Onward to the Moon
The United States recovered from the Apollo 1 fire, fixing the fatal flaws in an improved version of the Block II command module. The US proceeded with unpiloted test launches of the Saturn V launch vehicle (Apollo 4 and Apollo 6) and the Lunar Module (Apollo 5) during the latter half of 1967 and early 1968. Apollo 1's mission to check out the Apollo Command/Service Module in Earth orbit was accomplished by Grissom's backup crew commanded by Walter Schirra on Apollo 7, launched on October 11, 1968. The eleven-day mission was a total success, as the spacecraft performed a virtually flawless mission, paving the way for the United States to continue with its lunar mission schedule.
The Soviet Union also fixed the parachute and control problems with Soyuz, and the next piloted mission Soyuz 3 was launched on October 26, 1968. The goal was to complete Komarov's rendezvous and docking mission with the un-piloted Soyuz 2. Ground controllers brought the two craft to within 200 meters (660 ft) of each other, then cosmonaut Georgy Beregovoy took control. He got within 40 meters (130 ft) of his target, but was unable to dock before expending 90 percent of his maneuvering fuel, due to a piloting error that put his spacecraft into the wrong orientation and forced Soyuz 2 to automatically turn away from his approaching craft. The first docking of Soviet spacecraft was finally realized in January 1969 by the Soyuz 4 and Soyuz 5 missions. It was the first-ever docking of two crewed spacecraft, and the first transfer of crew from one space vehicle to another.
The Soviet Zond spacecraft was not yet ready for piloted circumlunar missions in 1968, after five[verification needed] unsuccessful and partially successful automated test launches: Cosmos 146 on March 10, 1967; Cosmos 154 on April 8, 1967; Zond 1967A September 27, 1967; Zond 1967B on November 22, 1967. Zond 4 was launched on March 2, 1968, and successfully made a circumlunar flight. After its successful flight around the Moon, Zond 4 encountered problems with its Earth reentry on March 9, and was ordered destroyed by an explosive charge 15,000 meters (49,000 ft) over the Gulf of Guinea. The Soviet official announcement said that Zond 4 was an automated test flight which ended with its intentional destruction, due to its recovery trajectory positioning it over the Atlantic Ocean instead of over the USSR.
During the summer of 1968, the Apollo program hit another snag: the first pilot-rated Lunar Module (LM) was not ready for orbital tests in time for a December 1968 launch. NASA planners overcame this challenge by changing the mission flight order, delaying the first LM flight until March 1969, and sending Apollo 8 into lunar orbit without the LM in December. This mission was in part motivated by intelligence rumors the Soviet Union might be ready for a piloted Zond flight during late 1968. In September 1968, Zond 5 made a circumlunar flight with tortoises on board and returned safely to Earth, accomplishing the first successful water landing of the Soviet space program in the Indian Ocean. It also scared NASA planners, as it took them several days to figure out that it was only an automated flight, not piloted, because voice recordings were transmitted from the craft en route to the Moon. On November 10, 1968, another automated test flight, Zond 6, was launched. It encountered difficulties in Earth reentry, and depressurized and deployed its parachute too early, causing it to crash-land only 16 kilometers (9.9 mi) from where it had been launched six days earlier. It turned out there was no chance of a piloted Soviet circumlunar flight during 1968, due to the unreliability of the Zonds.
On December 21, 1968, Frank Borman, James Lovell, and William Anders became the first humans to ride the Saturn V rocket into space, on Apollo 8. They also became the first to leave low-Earth orbit and go to another celestial body, entering lunar orbit on December 24. They made ten orbits in twenty hours, and transmitted one of the most watched TV broadcasts in history, with their Christmas Eve program from lunar orbit, which concluded with a reading from the biblical Book of Genesis. Two and a half hours after the broadcast, they fired their engine to perform the first trans-Earth injection to leave lunar orbit and return to the Earth. Apollo 8 safely landed in the Pacific Ocean on December 27, in NASA's first dawn splashdown and recovery.
The American Lunar Module was finally ready for a successful piloted test flight in low Earth orbit on Apollo 9 in March 1969. The next mission, Apollo 10, conducted a "dress rehearsal" for the first landing in May 1969, flying the LM in lunar orbit as close as 47,400 feet (14.4 km) above the surface, the point where the powered descent to the surface would begin. With the LM proven to work well, the next step was to attempt the landing.
Unknown to the Americans, the Soviet Moon program was in deep trouble. After two successive launch failures of the N1 rocket in 1969, Soviet plans for a piloted landing suffered delay. The launch pad explosion of the N-1 on July 3, 1969, was a significant setback. The rocket hit the pad after an engine shutdown, destroying itself and the launch facility. Without the N-1 rocket, the USSR could not send a large enough payload to the Moon to land a human and return him safely.
Apollo 11 was prepared with the goal of a July landing in the Sea of Tranquility. The crew, selected in January 1969, consisted of commander (CDR) Neil Armstrong, Command Module Pilot (CMP) Michael Collins, and Lunar Module Pilot (LMP) Edwin "Buzz" Aldrin. They trained for the mission until just before the launch day. On July 16, 1969, at exactly 9:32 am EDT, the Saturn V rocket, AS-506, lifted off from Kennedy Space Center Launch Complex 39 in Florida.
The trip to the Moon took just over three days. After achieving orbit, Armstrong and Aldrin transferred into the Lunar Module, named Eagle, and after a landing gear inspection by Collins remaining in the Command/Service Module Columbia, began their descent. After overcoming several computer overload alarms caused by an antenna switch left in the wrong position, and a slight downrange error, Armstrong took over manual flight control at about 180 meters (590 ft), and guided the Lunar Module to a safe landing spot at 20:18:04 UTC, July 20, 1969 (3:17:04 pm CDT). The first humans on the Moon waited six hours before they left their craft. At 02:56 UTC, July 21 (9:56 pm CDT July 20), Armstrong became the first human to set foot on the Moon.
The first step was witnessed by at least one-fifth of the population of Earth, or about 723 million people. His first words when he stepped off the LM's landing footpad were, "That's one small step for [a] man, one giant leap for mankind." Aldrin joined him on the surface almost 20 minutes later. Altogether, they spent just under two and one-quarter hours outside their craft. The next day, they performed the first launch from another celestial body, and rendezvoused back with Columbia.
Apollo 11 left lunar orbit and returned to Earth, landing safely in the Pacific Ocean on July 24, 1969. When the spacecraft splashed down, 2,982 days had passed since Kennedy's commitment to landing a man on the Moon and returning him safely to the Earth before the end of the decade; the mission was completed with 161 days to spare. With the safe completion of the Apollo 11 mission, the Americans won the race to the Moon.
The Race winds down
NASA had ambitious follow-on human spaceflight plans as it reached its lunar goal, but soon discovered it had expended most of its political capital to do so.
The first landing was followed by another, precision landing on Apollo 12 in November 1969. NASA had achieved its first landing goal with enough Apollo spacecraft and Saturn V launchers left for eight follow-on lunar landings through Apollo 20, conducting extended-endurance missions and transporting the landing crews in Lunar Roving Vehicles on the last five. They also planned an Apollo Applications Program to develop a longer-duration Earth orbital workshop (later named Skylab) to be constructed in orbit from a spent S-IVB upper stage, using several launches of the smaller Saturn IB launch vehicle. But planners soon decided this could be done more efficiently by using the two live stages of a Saturn V to launch the workshop pre-fabricated from an S-IVB (which was also the Saturn V third stage), which immediately removed Apollo 20. Belt-tightening budget cuts soon led NASA to cut Apollo 18 and 19 as well, but keep three extended/Lunar Rover missions. Apollo 13 encountered an in-flight spacecraft failure and had to abort its lunar landing in April 1970, returning its crew safely but temporarily grounding the program again. It resumed with four successful landings on Apollo 14 (February 1971), Apollo 15 (July 1971), Apollo 16 (April 1972), and Apollo 17 (December 1972).
In February 1969, President Richard M. Nixon convened a Space Task Group to set recommendations for the future US civilian space program, headed by his Vice President Spiro T. Agnew. Agnew was an enthusiastic proponent of NASA's follow-on plans, and the STG recommended plans to develop a reusable Space Transportation System including a Space Shuttle, which would facilitate development of permanent space stations in Earth and lunar orbit, perhaps a base on the lunar surface, and the first human flight to Mars as early as 1986 or as late as 2000. Nixon had a better sense of the declining political support in Congress for a new Apollo-style program, which had disappeared with the achievement of the landing, and he intended to pursue detente with the USSR and China, which he hoped might ease Cold War tensions. He cut the spending proposal he sent to Congress to include funding for only the Space Shuttle, with perhaps an option to pursue the Earth orbital space station for the foreseeable future.
The USSR continued trying to perfect their N1 rocket, finally canceling it in 1976, after two more launch failures in 1971 and 1972.
Salyuts and Skylab
Having lost the race to the Moon, the USSR decided to concentrate on orbital space stations. During 1969 and 1970, they launched six more Soyuz flights after Soyuz 3, then launched the first space station, the Salyut 1 laboratory designed by Kerim Kerimov, on April 19, 1971. Three days later, the Soyuz 10 crew attempted to dock with it, but failed to achieve a secure enough connection to safely enter the station. The Soyuz 11 crew of Vladislav Volkov, Georgi Dobrovolski and Viktor Patsayev successfully docked on June 7, and completed a record 22-day stay. The crew became the second in-flight space fatality during their reentry on June 30. They were asphyxiated when their spacecraft's cabin lost all pressure, shortly after undocking. The disaster was blamed on a faulty cabin pressure valve, that allowed all the air to vent into space. The crew was not wearing pressure suits and had no chance of survival once the leak occurred.
Salyut 1's orbit was increased to prevent premature reentry, but further piloted flights were delayed while the Soyuz was redesigned to fix the new safety problem. The station re-entered the Earth's atmosphere on October 11, after 175 days in orbit. The USSR attempted to launch a second Salyut-class station designated Durable Orbital Station-2 (DOS-2) on July 29, 1972, but a rocket failure caused it to fail to achieve orbit. After the DOS-2 failure, the USSR attempted to launch four more Salyut-class stations up to 1975, with another failure due to an explosion of the final rocket stage, which punctured the station with shrapnel so that it would not hold pressure. All of the Salyuts were presented to the public as non-military scientific laboratories, but some of them were covers for the military Almaz reconnaissance stations.
The United States launched the orbital workstation Skylab 1 on May 14, 1973. It weighed 169,950 pounds (77,090 kg), was 58 feet (18 m) long by 21.7 feet (6.6 m) in diameter, with a habitable volume of 10,000 cubic feet (280 m3). Skylab was damaged during the ascent to orbit, losing one of its solar panels and a meteoroid thermal shield. Subsequent crewed missions repaired the station, and the final mission's crew, Skylab 4, set the Space Race endurance record with 84 days in orbit when the mission ended on February 8, 1974. Skylab stayed in orbit another five years before reentering the Earth's atmosphere over the Indian Ocean and Western Australia on July 11, 1979.
Apollo–Soyuz Test Project
In May 1972, President Richard M. Nixon and Soviet Premier Leonid Brezhnev negotiated an easing of relations known as detente, creating a temporary "thaw" in the Cold War. In the spirit of good sportsmanship, the time seemed right for cooperation rather than competition, and the notion of a continuing "race" began to subside.
The two nations planned a joint mission to dock the last US Apollo craft with a Soyuz, known as the Apollo-Soyuz Test Project (ASTP). To prepare, the US designed a docking module for the Apollo that was compatible with the Soviet docking system, which allowed any of their craft to dock with any other (e.g. Soyuz/Soyuz as well as Soyuz/Salyut). The module was also necessary as an airlock to allow the men to visit each other's craft, which had incompatible cabin atmospheres. The USSR used the Soyuz 16 mission in December 1974 to prepare for ASTP.
The joint mission began when Soyuz 19 was first launched on July 15, 1975, at 12:20 UTC, and the Apollo craft was launched with the docking module six and a half hours later. The two craft rendezvoused and docked on July 17 at 16:19 UTC. The three astronauts conducted joint experiments with the two cosmonauts, and the crew shook hands, exchanged gifts, and visited each other's craft.
Human spaceflight after Apollo
In the 1970s, the United States began developing a new generation of reusable orbital spacecraft known as the Space Shuttle, and launched a range of uncrewed probes. The USSR continued to develop space station technology with the Salyut program and Mir ('Peace' or 'World', depending on the context) space station, supported by Soyuz spacecraft. They developed their own large space shuttle under the Buran program. The USSR dissolved in 1991 and the remains of its space program mainly passed to Russia. The United States and Russia have worked together in space with the Shuttle–Mir Program, and again with the International Space Station.
The Russian R-7 rocket family, which launched the first Sputnik at the beginning of the Space Race, is still in use today. It services the International Space Station (ISS) as the launcher for both the Soyuz and Progress spacecraft. It also ferries both Russian and American crews to and from the station.
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Home > Probability > Probability: AND & OR rule
## Probability: AND & OR rule
### The ‘OR’ rule
The ‘OR’ rule states $P(A\text{ or } B)=P(A)+P(B)$
This makes sense since the number of successes(NoS) = (NOS of A) + (NoS of B)
The number of outcomes has not changed.
So $P(A\mbox{ or } B)=\frac{\mbox{(NOS of A) + (NoS of B)}}{\mbox{possible outcomes}}=\frac{\mbox{(NOS of A)}}{\mbox{possible outcomes}}+\frac{\mbox{(NOS of A)}}{\mbox{possible outcomes}}$
$\implies P(A\mbox{ or } B)=P(A)+P(B)$
Examples
A bag contains a red ball, 5 blue ball and 4 green balls.
1) P(red)
2) P(blue)
3) P(green)
4) P(red or green)
5) P(red or green or blue)
1) $P(red)=\frac{1}{10}$
2) $P(blue)=\frac{5}{10}$
3) $P(green)=\frac{4}{10}$
4) $P(\mbox{red or green})=P(red)+P(green)=\frac{1}{10}+\frac{4}{10}=\frac{5}{10}$
5) $P(\mbox{red or green or blue})=P(red)+P(green)+(blue)=\frac{1}{10}+\frac{4}{10}+\frac{5}{10}=1$
### The ‘AND’ rule
Here we are considering the probability of something happening and then something else happening.
Let’s consider a die rolled twice.
What is the P(two sixes)?
Well we will get a ‘6’ $\frac{1}{6}$ of the time. Of these times we will get another ‘6’ $\frac{1}{6}$ of the time.
So $P(\mbox{two sixes})=\frac{1}{6}\text { of }\frac{1}{6}=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$
So $P(\mbox{A and B})=P(A)\times P(B)$
Examples
A bag contains 4 red(R) balls and 3 black(B) balls. I take out a ball, look at it and then put it back. I then take out another ball and record its colour. What is the:
1. P(R,R) $\rightarrow$ means getting two reds.
2. P(R then B)
3. P(R,B) $\rightarrow$ means getting a red and blue in any order. i.e. red then blue or blue then red
$P(R)=\frac{4}{7}$
$P(B)=\frac{3}{7}$
So
$P(R,R)=\frac{4}{7} \times\frac{4}{7} =\frac{16}{49}$
$P(\mbox{R then B})=\frac{4}{7}\times \frac{3}{7}=\frac{12}{49}$
$P(\mbox{R, B})=P(\mbox{R then B}) + P(\mbox{B then R})$
$P(\mbox{B then R})=\frac{3}{7}\times \frac{4}{7}=\frac{12}{49}$
$P(R,B)=\frac{12}{49}+\frac{12}{49}=\frac{24}{49}$<|endoftext|>
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How German Is American? Building Communities
How German Is American?
Soon after their arrival, German-speaking immigrants began organizing institutions around which community life revolved. Although many of these local groups, such as clubs and religious congregations, were ethnically fairly homogeneous, the new arrivals, having made the difficult decision to migrate, saw themselves as very much a part of their adopted community. Despite place names like New Berlin, New Glarus, and New Holstein, they did not, for the most part, strive to create “little Germanies” on the American landscape. A look at both secular and religious community institutions illustrates nicely the synthesis of Old and New World influences in the (post-)immigration context.
In American communities as far-flung as New York, Cincinnati, La Bahia (Texas), Plymouth (Wisconsin), Lawrence (Kansas), and San Diego, one can find meeting halls and theaters bearing the name “Turner” or “Turn Verein.” The Turner movement, founded in Berlin in 1811 by Friedrich Ludwig Jahn, had an enormous impact on the development of American gymnastics, both as a sport and as a formalized program of instruction within the public schools. The first Turner societies in the United States were organized in 1848 by German immigrants and exiles fleeing their country after the failed democratic revolutions of 1848/49. These “Forty-Eighters” created athletic, cultural, and social organizations throughout the country in the tradition of the German Turnvereine, which in today’s Germany are only one of many types of “Sportvereine.” The Turner motto, “Sound Mind in a Sound Body,” expresses their vision for realizing human potential through the integration of intellectual and physical development.
Some of the more radical Forty-Eighters and Turners were also Freethinkers. Freethinkers promoted an attitude of liberalism and rationality unencumbered by religious dogma, and many supported progressive ideas such as public education reform, improved working conditions, voting rights for women, and the abolition of slavery. These issues were often raised among the Turners as well, and may explain in part the fact that large numbers of Turners enthusiastically responded to Lincoln’s call for volunteers in the Union army.
Photo courtesy of the Milwaukee Turners.
Many Turners, Forty-Eighters, and Freethinkers were motivated by decidedly secular ideals, and admittedly religion was not the primary factor in most immigrants’ decisions to leave their German homelands. Nevertheless, religion was profoundly important to the majority of the German-speaking immigrants, as it is today among their descendants, and the churches they built in America became important reflections of their origins and traditions. Traveling through rural Dodge County in southeastern Wisconsin, for example, one might come upon a little church with an unusual name: “Zum Kripplein Christi,” translated by the congregation as “To the little manger of Christ.” Established in 1849, this Evangelical Lutheran church is an example of the many houses of worship built by German immigrants. Today, Zum Kripplein Christi shares a pastor with nearby St. John’s Church and maintains an elementary school of the same name serving ten students.
Unlike other immigrant groups, German-speakers did not comprise a single, homogeneous religious group, and in America they were represented in numerous denominations. Thus, as early as the 1860s German-speakers in southeastern Wisconsin identified themselves as Catholics, Lutherans, Reformed, Methodists, Baptists, Congregationalists, Episcopalians, Presbyterians, Quakers, Jews, or organized Freethinkers. Embracing the American model of individual religious freedom, German immigrants became more flexible in their choice of church, and individual congregations had a greater degree of autonomy than they would have had in Europe. Nevertheless, their desire to belong to a German community church frequently trumped their religious heritage. When German Americans belonged to historically Anglo-American denominations, they were often at odds with certain Yankee social mores. Especially with regard to alcohol and festival culture, German-American Protestants had more in common with their Catholic countrymen than with other American Protestants.
In the nineteenth century, churches were the centers of German-American religious, social, and cultural activity, especially in rural areas; German-language services, parochial school events, and celebrations of religious holidays were important events in community life. Until the beginning of the twentieth century, German was the language used in services and Sunday school in most of these churches. In the more autonomous Protestant churches, German often lasted several decades longer. Zum Kripplein Christi, for example, offered a Sunday service in German as recently as the 1990s. Despite the shift to English, particularly the Lutheran congregations have not forgotten their German roots. Heritage tours, student exchange programs, and mutual visits by choirs are examples of the enduring ties between Germans and Americans on the local level.
The diversity of religious expression among German-speaking immigrants was paralleled by a high degree of heterogeneity stemming from differences in regional and linguistic origins. This situation differed from that of other nineteenth-century immigrant groups, notably the Irish, but also Italians and people of other European backgrounds. The resulting lack of a unified and clearly definable German-American community explains in part why only few Americans, including those of German descent, have any idea when Steuben Day or German-American Day falls, whereas the Irish St. Patrick’s Day is one of America’s most popular celebrations, and Columbus Day, named after the Italian explorer, is a federal holiday.
This historic heterogeneity was and to some extent still is reflected in the plethora of clubs and societies linked to German ethnicity. These “Vereine” (clubs, societies, associations) allowed members of the growing middle class to associate publicly with one another and became an important social expression of the changes brought on by industrialization in Germany during the nineteenth century. German-speaking immigrants brought the “Vereinswesen” (club culture) with them to America, where it represented not only an example of direct cultural transfer, but also a means through which the transition from the Old Country to the New could be eased.
Many of these clubs did not last beyond the first generation, especially with the rise of mass and consumer culture during the twentieth century, which weakened older social divisions along ethnic lines. But some do still exist today, including the Plattdeutscher Verein (Low German Club) of Watertown, Wisconsin. The Verein was founded in 1882 with a twofold mission: “fraternalism and the perpetuation of the German language, especially the Plattdeutscher tongue.” Low German (Plattdeutsch) dialects derive from the “lower” (flatter) regions of the north, from which many immigrants to Wisconsin hailed. The dialects of this area are so different from those of the “higher” south, notably Switzerland, as to be mutually unintelligible. Though many immigrants had knowledge of the written standard dialect known as High German (so-called for its origins in the south), their identities were rooted in linguistically and culturally distinct particular regions, rather than a single “Germany,” and have endured to the present in American communities such as Watertown.
Most German-American community groups, religious and secular, were founded at least in part to preserve the German language, as is exemplified by the mission statement of Watertown’s Plattdeutscher Verein. Language maintenance was also a matter of concern among a group of Americans of German descent who have historically had little contact with other German-speaking communities in the U.S., namely the Americans known as the Pennsylvania Dutch.
Pennsylvania Dutch is an American language that developed in rural areas of southeastern and central Pennsylvania during the eighteenth century. Because most German-speaking emigrants to colonial Pennsylvania were from the cultural region of Central Europe known as the Palatinate (Pfalz), Pennsylvania Dutch resembles most strongly the German dialects of this area. Nevertheless, approximately 10% to 15% of Pennsylvania Dutch vocabulary is derived from English. Although scholars and some language advocates prefer the term“Pennsylvania German,” the use of “Dutch” here does not reflect a (mis)translation of “Deutsch” or “Deitsch.” The English word “Dutch” was used in earlier times to describe people of both German and Netherlandic origins, often with a “folksy” connotation.
Observers, including many Europeans, frequently assume, incorrectly, that the term “Pennsylvania Dutch” is synonymous with“Amish.” In fact, of the approximately 81,000 German-speaking immigrants who came to Pennsylvania during the eighteenth century, only a few hundred were members of the small, but very visible, Anabaptist sect known today as the Old Order Amish. Until the early part of the twentieth century, most speakers of Pennsylvania Dutch were of either Lutheran or German Reformed (“nonsectarian”) background who, unlike the Amish and other “sectarians,” did not separate themselves for spiritual reasons from the social mainstream. Although the sectarian and nonsectarian Pennsylvania Dutch lived in close proximity to one another in the colonial period, during the nineteenth century the two groups moved into different regions, including outside of Pennsylvania. Today, despite their common language, sectarians and nonsectarians represent two very distinct Pennsylvania Dutch-speaking groups.
With the dramatic demographic changes of the twentieth century, which led especially to greater mobility and the loss of rural isolation across America, maintenance of Pennsylvania Dutch among nonsectarians declined sharply; only members of the conservative Anabaptist sects have resisted these changes and continue actively to speak the language and transmit it to their children. Some nonsectarian Pennsylvania Dutch have attempted to counteract the shift to English monolingualism by creating institutions to promote their language. The most prominent of these are the Grundsow (Groundhog) Lodges, the first of which was founded in Allentown, Pennsylvania, in 1933. Annual lodge meetings coincide with Groundhog Day (February 2), a New World expression of the traditional European mid-winter holiday of Candlemas. The program cover pictured here reads: “The Third Annual Meeting of the Groundhog Lodge Number One on the Lehigh (River). Monday evening after Groundhog Day, at 6:30 p.m., the 3rd of February, 1936.” Most speakers of Pennsylvania Dutch are literate in English only; the result is that English spelling conventions are usually observed when the language is written down.
Despite the virtual disappearance of Pennsylvania Dutch in the everyday lives of nonsectarians, the Grundsow Lodges remain active, and Groundhog Day has become an increasingly popular local holiday. In a uniquely American move, lodge members have recently begun a campaign to have the Pennsylvania Department of Transportation approve their design for a “special organization” license plate.
In addition to specific clubs and religious groups, certain aspects of German culture have become a part of largely deethnicized regional American identities. Examples of this can be found in the “Dutch Country” of southeastern Pennsylvania, which most visitors do not associate with Europe but with early America. Far from becoming submerged, many cultural expressions with clear antecedents in German-speaking Europe, from the forebay bank barn to hard pretzels, not only have survived in Pennsylvania but have spread across America.
In the Milwaukee postcard from around 1900 shown here, the central figure bears an unmistakable resemblance to stereotypical representations of ethnic Germans that were common at the time. The stout, good-natured, and quite evidently beer-loving Dutchman rides in a fanciful beer-barrel automobile through the city. Outfitted with overflowing steins for reflective headlights, the vehicle has compartments for limburger cheese and frankfurters, while a dachshund chases along after a sausage link. In the background one sees a cheese factory, pretzel factory, malt house, and brewery—all the comforts of a Dutchman’s adopted “Heimat.” While the references to Milwaukee’s brewing industry are historically correct, those to cheese and pretzels are not. Wisconsinites are known today as“cheeseheads,” to be sure, but the state’s cheese industry owes more to Yankee immigrants than to Germans.
The emphasis in the image on alcohol reflects an early division between people of German heritage and Yankees over the cultural and political issue of temperance, often arising from the fondness of German Americans for drinking on Sundays, especially in connection with their family-oriented tavern culture. Similar images of and perceptions about Germans, centering on the food and drink of cheerfully hefty “Dutch” men and women, also flourished in such American communities as Chicago, Cincinnati, and St. Louis, where beer, sausages, and pretzels have become standard features not only of local cuisine but also in such mainstream American settings as the ballpark, where fans chomp and swallow while cheering on the Cubs, Reds, and Cardinals.
Not only at sports events are considerable amounts of beer and hot dogs consumed; an increasing number of American communities, many with no German heritage to speak of, now sponsor Oktoberfests. One such community is Oak Park, Illinois, located ten miles west of Chicago. This area’s rapid growth in the nineteenth century coincided with the acceleration of German emigration to the United States, and by the end of the century Germans, with 25% of the population, constituted the largest ethnic group. As in Milwaukee, German Americans were active in business, churches, clubs, theaters, and political and cultural arenas. Despite divisions within their ranks resulting from their different regional and social origins, they presented a more or less unified ethnic group in beer gardens, at fairs, and in parades through neighborhood streets. In the twentieth century, however, Germans moved away from public displays of ethnic pride, as ethnicity gave way to more complex identities formed around class, race, and American popular culture. Midwestern cities, as elsewhere in the U.S., were changed after World War II by newcomers, including large numbers of African Americans, some of whom settled in neighborhoods where German-speaking immigrants had lived.
It thus is a curious phenomenon that the Oktoberfest has become a signature fall event in about 200 communities across the U.S. and Canada. Awash in beer, pretzels, the chicken dance, and the Schnitzelbank song, the typical American Oktoberfest today is less a celebration of German heritage—real or imagined— than it is the expression of a dynamic and culturally diverse local identity. In Oak Park’s festival, this cultural diversity is represented by musical groups as different as Jimmy’s Bavarians, Bumble Bee Bob and the Stingers, and Koko Taylor and Her Blues Machine. Even the more specifically German-themed Oktoberfests nationwide reflect an American phenomenon that is striking to European Germans, namely the predominance of symbols specific to traditional Bavaria, which strive, misleadingly, to evoke a single “German” culture.
The commodification of ethnic culture, as reflected in the explosion of Oktoberfests over the last few decades, is part of a larger trend of American communities to promote economic growth through tourism. In 2003 two New Glarus, Wisconsin, policemen, in uniform and with guns at their side, posed for a Swiss photographer in front of the town’s most prominent sign: a depiction of Switzerland’s coat of arms and its national hero, William Tell. To a foreign visitor this is a quintessentially American picture, confirming every stereotype fostered abroad by cop shows on American television. To an American observer, however, who is drawn more to the “Old World” sign depicting a historic heroic act performed with an ancient weapon, the image speaks to the town’s unique identity rooted in its ethnic heritage.
New Glarus was settled in 1845 by a group of Swiss German immigrants from Canton Glarus. For decades, this rural community looked much like any other Midwest pioneer settlement. Only toward the end of the nineteenth century, as European Americans nationwide began to celebrate their ethnic and national backgrounds publicly, did New Glarus “rediscover” its Swiss heritage. By staging festivals and pageants celebrating Swiss Independence Day, the arrival of the original immigrants, and—beginning in the 1930s—the William Tell story, New Glarus brought together not only members of its local community, but also Swiss Americans from across the country. Seeing the economic potential of tourism, the town eventually decided to remake itself into “America’s Little Switzerland.” Based less on traditions handed down directly from the original settlers to their descendants, and more on a contemporary American image of things stereotypically “Swiss,” buildings were constructed in the chalet style, restaurants adopted Swiss menus, and folk musicians from Switzerland were invited to perform and to teach members of the community.
Today many residents of New Glarus are not of Swiss descent, but the townspeople still perform William Tell every year in both English and German, thereby creating a sense of local identity and culture. At the same time, this unique American community attracts visitors from around the world, including Switzerland. Keen on promoting Switzerland’s image abroad, the Swiss government now has plans to build a cultural center in New Glarus.
Next: Growing into the Nation
[This booklet is available in PDF format]<|endoftext|>
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Quick Method to Solve Mensuration Problems | | GovernmentAdda
Tuesday , August 21 2018
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Home / Quantitative Aptitude Tricks / Quick Method to Solve Mensuration Problems
# Quick Method to Solve Mensuration Problems
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This quick trick to solve Mensuration problems is totally based on your thinking. It involves next to no calculations. This trick will save you precious time in your IBPS Clerk or SBI PO exams.
The trick will help you get answers in just a few seconds. Solve the questions in detail first to understand this method more clearly. Once you get acquainted with the mechanisms of this trick, you will find it extremely helpful.
Below are given types of questions where this trick can be applied.
## How to Use this Trick
Check whether π=22/7 has been used in the formula for finding out the Particular Area, Curved Surface Area, Total Area, Volume, etc. If it is so, then
Here are examples to explain the chart given above.
Example 1:
Find the surface area of a sphere whose volume is 4851 cubic meters.
a) 1380 m2
b) 1360 m2
c) 1368 m2
d) 1386 m2
Using the Trick:
We know that surface area of a sphere = 4πr2
It means ‘π’ has been used in finding out the surface area of the sphere.
We can easily see that only ‘1386’ from the given options is divisible by ‘11’
Hence, surface area of the sphere = 1386 m2
Example 2:
The radius and height of a right circular cylinder are 14 cm & 21 cm respectively. Find its volume.
a) 12836 cm3
b) 12736 cm3
c) 12936 cm3
d) 12837 cm3
Using the Trick:
The know that volume of Cylinder = πr2h
We must check divisibility by ‘11’. Here, both ‘12936’ and ‘12837’ are divisible by 11. But you also notice that radius (14 cm) & height (21 cm) are both multiples of 7. So the option divisible by ‘7’ is your answer.
Hence, volume of a right circular cylinder = 12936 cm3 (since this is the only option divisible by 7)
We can test this as follows:
Volume of the given cylinder
= (22/7) × 14 × 14 × 14 × 21 cm3
= 22 × 14 × 14 × 3 cm3
⇒ Volume must be divisible by ‘7’.
Example 3:
The radius and height of a right circular cone are 7 cm & 18 cm respectively. Find its volume.
a) 814 cm3
b) 624 cm3
c) 825 cm3
d) 924 cm3
Using the Trick:
The option should be divisible by ‘11’ because ‘π’ has been used in finding its volume. One of the parameters is a multiple of 7 without being a higher power. So we must go through fundamentals.
Now, volume of a right circular cone = (1/3)πr2h
= (1/3) × (22/7) × 7 × 7 × 18 cm3
= 22 × 7 × 6
Clearly, we need an answer that is a multiple of 11, 7 as well as 3.
Among the given options, 814, 825 and 924 are all multiples of 11. However, we see that only one option is divisible by 7. So this is the correct answer.
Hence, volume of the given cone = 924 cm3
Example 4:
Find the circumference of a circle whose radius is 49 cm.
a) 208 cm
b) 288 cm
c) 308 cm
d) 407 cm
Using the Trick:
The option should be divisible by ‘11’ because ‘π’ has been used in finding its circumference. One of the parameters is a higher power of 7. Thus, we need to find the only option that is a multiple of 7. If, however, we find more than one option that is a multiple of 7, we need to go through fundamentals.
Among the options, 308 and 407 are both multiples of 11. However, only 308 is a multiple of 7. So circumference of the circle = 308 cm.
We can test this as follows:
Circumference of circle = 2πr cm
= 2 × (22/7) × 7 × 49 cm
= 2 × 22 × 7 cm
Remember: If there is only one parameter equal to ‘7’ or multiple of ‘7’ and this parameter is not in a higher power in the formula, the answer will not be divisible by ‘7’
Example 5:
Find the curved surface area of a right circular cylinder whose radius & height are 14 cm & 50 cm respectively.
a) 3300 cm2
b) 3420 cm2
c) 4440 cm2
d) 4400 cm2
Solution:
Curved surface area of a right circular cylinder = 2πrh
Here only one parameter (r = 14 cm) is a multiple of 7 (without being a higher power of 7). This parameter is used only as ‘r’ and not in its higher powers. So we see that our answer will not be a multiple of 7. However, the presence of π means that it will still be a multiple of 11.
Among the given options, 3300 and 4400 are both multiples of 11. We also see that both are not multiples of 7 either. However, we can see that none of our parameters are multiples of 3, so curved surface area cannot be a multiple of 3 either. So our answer cannot be 3300.
Therefore curved surface area of right circular cylinder must be 4400 cm2.
We can test this as follows:
Curved surface area of a right circular cylinder = 2πrh
=2 × (22/7) × 14 × 50
=2 × 22 × 2 × 50
= 4400 cm3
NOTE:
1. The whole trick is based on the multiplication & divisibility by 11 & 7. So you are advised to use this trick very carefully using your quick mental mathematics. Please learn divisibility rules for both 11 and 7 for this purpose – they are both pretty simple! Please do not waste time solving these types of questions.
2. If there is a ‘None of these’ among the given options, then don’t use this trick.
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Decompose the Vector 6 ^ I − 3 ^ J − 6 ^ K into Vectors Which Are Parallel and Perpendicular to the Vector ^ I + ^ J + ^ K . - CBSE (Arts) Class 12 - Mathematics
ConceptMultiplication of a Vector by a Scalar
Question
Decompose the vector $6 \hat{i} - 3 \hat{j} - 6 \hat{k}$ into vectors which are parallel and perpendicular to the vector $\hat{i} + \hat{j} + \hat{k} .$
Solution
$\text{ Let } \vec{a} =6 \hat{i} - 3 \hat{j} - 6 \hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} + \hat{k}$
$\text{ and } \vec{x} \text{ and } \vec{y} \text{ be such that }$
$\vec{a} = \vec{x} + \vec{y}$
$\Rightarrow \vec{y} = \vec{a} - \vec{x} . . . \left( 1 \right)$
$\text{ Since } \vec{x} \text{ is parallel to } \vec{b} ,$
$\vec{x} = t \vec{b}$
$\Rightarrow \vec{x} = t \left( \hat{i} + \hat{j} + \hat{k} \right) = t \hat{i} + t \hat{j} + t \hat{k} ...(2)$
$\text{ Substituting the values of } \vec{x} \text{ and } \vec{a} \text{ in } (1), \text{ we get }$
$\vec{y} = 6 \hat{i} - 3 \hat{j} - 6 \hat{k} - \left( t \hat{i} + t \hat{j} + t \hat{k} \right) = \left( 6 - t \right) \hat{i} + \left( - 3 - t \right) \hat{j} + \left( - 6 - t \right) \hat{k} . . . \left( 3 \right)$
$\text{ Since } \vec{y} \text{ is perpendicular to } \vec{b} ,$
$\vec{y} . \vec{b} = 0$
$\Rightarrow \left[ \left( 6 - t \right) \hat{i} + \left( - 3 - t \right) \hat{j} + \left( - 6 - t \right) \hat{k} \right] . \left( \hat{i} + \hat{j} + \hat{k}\right) = 0$
$\Rightarrow 1 \left( 6 - t \right) + 1\left( - 3 - t \right) + 1 \left( - 6 - t \right) = 0$
$\Rightarrow - 3 - 3t = 0$
$\Rightarrow t = - 1$
$\text{ From } (2) \text{ and } (3), \text{we get}$
$\vec{x} = - \hat{i} - \hat{j} - \hat{k}$
$\vec{y} = 7 \hat{i} - 2 \hat{j} - 5 \hat{k}$
$\text{ So },$
$\vec{a} = \vec{x} + \vec{y} = \left( - \hat{i} - \hat{j} - \hat{k} \right) + \left( 7 \hat{i} - 2 \hat{j} - 5 \hat{k} \right)$
Is there an error in this question or solution?
Video TutorialsVIEW ALL [1]
Solution Decompose the Vector 6 ^ I − 3 ^ J − 6 ^ K into Vectors Which Are Parallel and Perpendicular to the Vector ^ I + ^ J + ^ K . Concept: Multiplication of a Vector by a Scalar.
S<|endoftext|>
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Relative aperture, the measure of the light-gathering power of an optical system. It is expressed in different ways according to the instrument involved. The relative aperture for a microscope is called the numerical aperture (NA) and is equal to the sine of half the angle subtended by the aperture at an object point times the index of refraction of the medium between the object and the objective lens. For binoculars, telescopes, and photographic lenses in which the object may be distant, the relative aperture is taken as the ratio of focal length of the objective to the diameter of the entrance pupil. The relative aperture of a camera lens is sometimes expressed as a simple ratio—e.g., 1:4.5—or more commonly as its f-number, f/4.5. In either case, a lens of 180-millimetre focal length set at this relative aperture would have a pupil diameter (effectively, the lens diaphragm opening) of 40 mm.<|endoftext|>
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# Subtract the First Rational Number from the Second in Each of the Following: − 8 33 , − 7 22 - Mathematics
Sum
Subtract the first rational number from the second in each of the following:
$\frac{- 8}{33}, \frac{- 7}{22}$
#### Solution
$\frac{- 7}{22} - \frac{- 8}{33} = \frac{- 21}{66} - \frac{- 16}{66} = \frac{( - 21) - ( - 16)}{66} = \frac{- 21 + 16}{66} = \frac{- 5}{66}$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 8 Maths
Chapter 1 Rational Numbers
Exercise 1.3 | Q 1.8 | Page 18<|endoftext|>
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Common Core Standards:
Construct an argument that plants and animals have internal and external structures that function to support survival, growth, behavior, and reproduction. [Clarification Statement: Examples of structures could include thorns, stems, roots, colored petals, heart, stomach, lung, brain, and skin.] [Assessment Boundary: Assessment is limited to macroscopic structures within plant and animal systems.]
Discuss how compost can be made using worms.
Have students learn about the anatomy, diet, and habitat of worms.
Key words: Vocab Tree
Vermicomposting; Anatomy; Anterior; Posterior; Cilia; Prostomium; Clitellum; Red Wigglers; Esophagus; Gizzard; Intestines
(Using Key words: Students can create a glossary, in books or on wall in classroom. Students are encouraged to practice using vocab in written or verbal sentences – perhaps writing example sentences and displaying them. Students could earn points for using the vocab in novel sentences each week)
- Worm anatomy blank picture
- Worm anatomy key
- Colored pencils or crayons
- Powerpoint: Worms: What’s Crawling Through Our Dirt?
In order to give students a better idea of how this method of composting works we will introduce them to the anatomy of the worm. This activity can be altered for the age and level of the students by excluding or including terms as you see fit.
Each student will receive the blank image of the worm.
As a class, go through each blank area of the worm diagram and discuss which anatomic part it is and what purpose it serves (it is assumed most students will not have previous knowledge of these parts)
For younger students focus on the brain, mouth, anus, and heart.
For older students, include other terms such as the intestines (The intestines are a long, continuous tube running from the stomach to the anus. Most absorption of nutrients and water happen in the intestines. The intestines include the small intestine, large intestine, and rectum), esophagus (tube in throat through which food travels to the stomach), and gizzard (specialized stomach constructed of thick, muscular walls is used for grinding up food, often aided by particles of stone or grit).
Point out common anatomic parts that the worm does not have such as ears, eyes, legs, or arms.
After the parts have been labeled allow the students to color in their pictures.
Ask students to name their worms and come up with a creative backstory about them answering these questions:
Where is your worm from? What does he/she like to eat? What does he/she not like to eat? Where do they live now? Do they live with lots of friends and family or alone?
The anatomy of the worm is much different from that of a human or other mammal. They are blind creatures with a very simple cardiac and gastrointestinal system, but the job they do as decomposers is incredibly important. Without decomposers like them the food cycle would be broken, nutrient rich soil would no longer be regenerated, and plants would no longer grow. For this reason, understanding more about the worm and this process is pivotal for students to understand where their own food comes from.
Further Activities/ Homework:
Have students go out into their neighborhood and find a worm. Have them note the location of the worm, the length, what is surrounding it, what it was doing when found, and how far underground it was.<|endoftext|>
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Science is a subject that describes the world around us. From the billions of stars in the night sky to the microscopic world of microorganisms, there are a lot of things to learn about and ponder about. However, some of us might find science less than interesting, and in an academic environment, it might sound down right tedious. Therefore, here are a few tips to make learning science engaging and interesting.
Develop an Interest
Studying is hard when you don’t get good grades or lack interest in the subject. But regardless, you need to score good marks to affect your overall result. Therefore, try to develop interest for science in your own way. This could be done by studying with a close friend, or studying while listening to music. Just ensure that the learning experience is pleasant.
Image Source: Pixabay
One of the most effective methods to recall what you have read is by making notes during lectures or sessions. This helps you reinforce what you studied and also, is an excellent tool to use for exams. You would be able to understand your handwritten answers better than any other textbook or class notes. And not just science, you can implement this strategy in all subjects including maths NCERT class 8.
Use Charts and Diagrams
When you read a big block of text on any concept or topic, it can be very monotonous and dry. That might make it difficult for you to memorize and recall. Therefore, use charts and diagram to represent information. This can be even more effective when you want to revise and recall a topic. If you have enough time to incorporate the same into your answer script, then you probably have chances of scoring more. In conclusion, these are the tips and tricks you can implement if you want to score better in science.
Preparing for a science exam? Logon to BYJU’s and use our ncert science class 9 to jumpstart your learning. Alternatively, you can subscribe to our YouTube channel to watch related videos.<|endoftext|>
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It’s been well over a year since NASA’s incredibly trusty Cassini probe performed its final act, plunging into Saturn’s atmosphere where it was obliterated by the intense friction. It was a bittersweet day for the astronomy community, but the wealth of information Cassini gathered during its over 13 years orbiting Saturn is still yielding new discoveries.
In a new paper published in Nature Astronomy, researchers on the Cassini project reveal that the spacecraft’s bounty of radar data now shows that Saturn’s moon Titan are even more special than scientists already realized. The moon’s lakes, which were observed by Cassini during its final pass in 2017, are far deeper than anyone thought.
Earth has a lot of liquid on its surface, but the same can’t be said for many other worlds in our solar system, Titan is the exception, with vast lakes visible from space. The lakes aren’t filled with liquid water, however, as Titan is far too cold for that to be possible. Instead, Titan’s lakes are filled with methane, chilled to the point where it becomes a liquid rather than a gas as we think of it on Earth.
Scientists have known these lakes exist for some time, but Cassini revealed how deep they really are. In the new study, the researcher team reveals that the lakes are over 300 feet deep, and they’re replenished by similar mechanisms that we see on Earth with water in the form of liquid, vapor, and rain.
“Every time we make discoveries on Titan, Titan becomes more and more mysterious,” lead author Marco Mastrogiuseppe of Caltech said in a statement. “But these new measurements help give an answer to a few key questions. We can actually now better understand the hydrology of Titan.”
The discovery is also a great reminder that missions like Cassini can yield new developments even years after they wrap, and the efficiency with which NASA’s spacecraft and rovers gather data rapidly outpaces the ability of human scientists to sift through it.<|endoftext|>
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Emmeline Pankhurst addressing a crowd in Trafalgar Square in 1908 / Public Domain
History is messy. And so are politics.
By Dr. John Buescher
History is messy. And so are politics. A good historian resists the urge to reduce the many causes or meanings of an event to a single one. One of the most persistent urges of students of American history is to try to decide whether the Civil War was “really” about slavery or about states’ rights. Another contender for the “real” cause of the war has been the regional tensions between an agrarian and an industrial economy, and another contender, the unequal unfolding in various segments of society of the universal implications of the Enlightenment’s principle of individual freedom.
The “real” cause was all of these and more. Those on each side of the conflict acted with a variety of goals, and individuals were commonly motivated by more than one reason.
The 13th Amendment Abolishes Slavery
The complexity of interests, goals, and motivations continued throughout the Reconstruction period after the war. The radical Republicans, who dominated Congress, were determined to complete the task of eliminating slavery. But this meant more than simply abolishing slavery itself, which occurred through the adoption of the 13th Amendment at the end of 1865. (The 13th Amendment wrote the abolition of slavery into the deepest level of American law, making it permanent. Northern abolitionists had worried that the Emancipation Proclamation of 1863 would be attacked after the end of the war as merely a temporary emergency measure.)
The passage of the 13th Amendment did not end the problem, however, because the freed slaves’ legal status was undefined and unclear. From the freed slaves’ point of view, this left them without legal protection against attempts in the South to coerce them into a permanent underclass status.
The 14th Amendment Makes Ex-Slaves Citizens
The problem was constitutionally complicated because the pre-war Supreme Court Dred Scott decision had declared black slaves to be non-persons. A 14th Amendment was necessary, therefore, to explicitly establish the status of blacks as persons and citizens through a natural right, inhering simply in having been born in the country and in recognizing their allegiance to it.
This was a philosophical expansion of who was included in the “We the People” phrase in the preamble to the Constitution, but the plight of the still-disenfranchised freed slaves in the South increased the urgency of passing the Amendment. Because the southern states were still occupied federal territory, the freed slaves—for the time being—could be given direct federal protection. However, the states were agitating for readmission to the Union, and their legislative representation had to be calculated. The Constitution had calculated it by counting slaves as three-fifths of a person. That language obviously now had to be amended. In addition, it was urgent that blacks be given full legislative representation to thwart Southern efforts to turn them into a permanent underclass without the full rights of citizens.
Problematic Language in the 14th Amendment
For the advocates of women’s rights, this is where it got messy, and where some of the various motivations and goals of those who had previously been working together began to unravel. The radical Republicans who drafted the language of the 14th Amendment realized that by making a “natural rights” case for including blacks as full citizens, with all the rights and obligations, they would be making the same case for women. Had the amendment contained only the language of Section 1, women’s rights advocates would have been thrilled because it would have strengthened their argument for female suffrage, even though it had to do with establishing citizenship rather than the right to vote per se:
“Section 1. All persons born or naturalized in the United States, and subject to the jurisdiction thereof, are citizens of the United States and of the State wherein they reside. No State shall make or enforce any law which shall abridge the privileges or immunities of citizens of the United States; nor shall any State deprive any person of life, liberty, or property, without due process of law; nor deny to any person within its jurisdiction the equal protection of the laws.”
But this wording would have made the amendment impossible to pass. There was wide political support for protecting the freed slaves, but not for giving women the right to vote.
This political dilemma was “solved” through the language of Section 2, which was needed to specify how the inhabitants of states would be counted for the purpose of legislative representation. It amended the Constitution’s “three-fifths” clause.
And a penalty would be exacted from a recalcitrant state for any effort to deny blacks their votes. For each black denied the vote, the state’s basis for representation would be reduced by one:
“Section 2. Representatives shall be apportioned among the several States according to their respective numbers, counting the whole number of persons in each State, excluding Indians not taxed. But when the right to vote… is denied to any of the male inhabitants of such State, being twenty-one years of age, and citizens of the United States … the basis of representation therein shall be reduced in the proportion which the number of such male citizens shall bear to the whole number of male citizens twenty-one years of age in such State.”
Before the war, a slave state was able to count each slave as three-fifths of a person for the purposes of its apportion representation in Congress. Now, for each black who was denied the vote, the state would be forced to deduct a whole person from its basis for apportioned representation. This would apply strong pressure on the state not to disenfranchise blacks. Simultaneously, however, the language of this section of the amendment, in precisely specifying the calculation to be used, qualified the words “inhabitants” and “citizens” with the word “male.”
This section, therefore, both enraged women’s rights advocates while also allowing the amendment’s proponents enough cover to find the votes for passage because it appeared to limit its effects to expanding the male population eligible to vote. The words “male” and “female” had not appeared in the Constitution before this. And women had been making their strongest Constitutional argument for the right to vote based on the “natural rights” reasoning upon which the Constitution relied. They argued that women already had the right to vote (and had always had it), at least implicitly, in the Constitution, but that mere outmoded convention had prevented that right from being recognized. They had been arguing for woman suffrage, in other words, based on the universal human rights they saw as affirmed by implication in the Constitution.
The Reformers’ Coalition Unravels
Most of those who had argued for women’s rights before and during the war had also allied themselves strongly with the movement to abolish slavery, linking the two causes on the basis of natural rights. But now, by the insertion of the word “male” into the amendment, the Constitution would no longer be technically gender-blind, but would actively “disfranchise” women. Women’s rights advocates were particularly stung by the fact that the amendment was written and was being pushed by the very same reformers, such as Senator Charles Sumner, with whom they had stood shoulder to shoulder in the agitation against slavery.
As Elizabeth Cady Stanton remarked on the Republican Congress’s determination to extend voting rights to blacks: “to demand his enfranchisement on the broad principle of natural rights, was hedged about with difficulties, as the logical result of such action must be the enfranchisement of all ostracized classes; not only the white women of the entire country, but the slave women of the South … the only way they could open the constitutional door just wide enough to let the black man pass in, was to introduce the word ‘male’ into the national Constitution.”
Wendell Phillips, in 1865, as the new head of the American Anti-Slavery Society, turned the society’s sights on ensuring black Americans’ civil and political rights, especially suffrage. The old-line anti-slavery agitators understood that trying to extend suffrage to African-Americans would require a huge political battle. Trying to extend suffrage to women, too, at the same time, would be impossible. So now he told the society’s annual convention, “I hope in time to be as bold as [British reformer John] Stuart Mill and add to that last clause ‘sex’!! But this hour belongs to the negro. As Abraham Lincoln said, ‘One War at a time’; so I say, One question at a time. This hour belongs to the negro.” Elizabeth Cady Stanton and Susan B. Anthony both understood immediately that this meant that their erstwhile supporters among the abolitionists—many of whom were now in the councils of legislative influence in the Republican Party—were putting the “woman’s cause … in deep water.”
Congress proposed the 14th Amendment on June 13, 1866. It was ratified and became law on July 9, 1868. Its adoption caused a deep rift among those who, until then, had made common cause. Many of the supporters of the amendment hoped that the issues of black suffrage and woman suffrage could be separated out and treated sequentially, one after the other. And many of them were acting on the pressing need to deal with the issue of black citizenship and suffrage separate from the issue of woman suffrage out of the necessity to cope with the unfolding events in the aftermath of the war.
Nevertheless, many women’s rights activists felt that their cause had been betrayed by their former friends in reform, and that the cause of blacks and women had not just been separated, out of a temporary necessity, but that the cause of women had been set back. Historian Ellen DuBois has noted that this was a watershed event in that women’s rights activists, after this, began focusing their organizing efforts specifically on gaining for women the right to vote, rather than relying on broader reforms. They organized both the National Woman’s Suffrage Association and the American Woman Suffrage Association in 1869, and began petitioning for a constitutional amendment that would guarantee women the right to vote.
The 15th Amendment Makes Ex-Slaves Voters
As events unfolded in the South, blacks were often excluded from voting by local restrictions of one kind or another, and Congress recognized that constitutionally defining blacks as citizens, through the 14th Amendment, did not absolutely guarantee their right to vote. Consequently, Congress proposed the 15th Amendment on February 26, 1869. It was ratified and became law on February 3, 1870:
“Section 1. The right of citizens of the United States to vote shall not be denied or abridged by the United States or by any State on account of race, color, or previous condition of servitude.”
For women’s rights advocates, this amendment added nothing new to their struggle for suffrage. Especially frustrating for them was the fact that antebellum reformers had often railed against legal limits to freedom based on “race, color, or sex,” and the language of this new amendment seemed to them to be a kind of parody of that, in which “sex” was deliberately replaced by “previous condition of servitude,” that is, slavery.
It was a painful irony for many women’s rights activists, therefore, that they found themselves actively opposing the passage of the amendment (as some of them had opposed the 14th Amendment). The amendment that would guarantee them the right to vote—the 19th—would not become law until 1920.
Elizabeth Cady Stanton, Susan B. Anthony, Matilda Joslyn Gage, eds., History of Woman Suffrage, Volume 2: 1861-1876. Rochester, NY: Privately Printed, 1881, pp. 90-106, 333-362, 407-416.
Ellen Carol DuBois, Feminism and Suffrage: The Emergence of an Independent Women’s Movement in America, 1848-1869. Ithaca: Cornell University Press, 1978, pp. 53-72.
Eleanor Flexner and Ellen Fitzpatrick, Century of Struggle: The Woman’s Rights Movement in the United States, rev. ed. Cambridge, MA: Harvard University Press, 1996, pp. 136-148.
Originally published by Teaching History under a Creative Commons Attribution Non-Commercial Share Alike 3.0 license.<|endoftext|>
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Subtraction Breaking Apart Year 1 Addition and Subtraction Learning Video Clip | Classroom Secrets
MathsYear 1Autumn Block 2 (Addition and Subtraction)13 Subtraction Breaking Apart › Subtraction Breaking Apart Year 1 Addition and Subtraction Learning Video Clip
# Subtraction Breaking Apart Year 1 Addition and Subtraction Learning Video Clip
## Step 13: Subtraction Breaking Apart Year 1 Addition and Subtraction Learning Video Clip
Dot and Stan are having a fantastic day playing in the sunshine! Use your knowledge of subtraction (breaking apart) to help them work out some tricky problems.
More resources for Autumn Block 2 Step 13.
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Discussion points for teachers
1. Write a calculation in words to show this.
Discuss how many grey pterodactyls there were. Discuss how many brown pterodactyls there were. Discuss which pterodactyls have flown away. Discuss how many pterodactyls are left. Discuss how we would write this into a calculation in words.
nine – six = three
2. Help Dot and Stan to work out how many ants are left by completing the calculation in words.
Discuss how many ants there were altogether. Discuss how many ants were big. Discuss how many ants were small. Discuss how many ants walk away first. Discuss how many more ants walk away. Discuss how we would show this in a calculation. Discuss how we would write the calculation in words.
seven – four – one = two
3. Write a calculation to show how many leaves Dot has eaten. Discuss how many leaves there were to begin with.
Discuss how many leaves Dot has eaten. Discuss how many leaves are left. Discuss how we would show this in a calculation.
10 – 5 = 5
4. Write a new calculation to show how many ferns Dot and Stan have eaten altogether.
Discuss how many ferns Stan has eaten. Discuss how how many ferns Dot ate. Discuss how we would show these numbers in a calculation. Discuss what the new answer would be.
10 – 5 – 0 = 5
5. Write a subtraction calculation to show which groups of stones they could have taken away.
Discuss what the different groups of stones could be (smooth, spiky, big, small, small spiky, big, spiky, small smooth, big smooth). Discuss how we could take one of these groups away from the total number of stones. Discuss how we would show this in a calculation. Discuss how we could take two groups away from the total. Discuss how we would show this in a calculation. This question is open-ended for the children to explore.
Various answers, for example: all of the stones – all of the smooth stones (10 – 5 = 5); all of the stones – all of the spiky stones ( 10 – 5 = 5)
National Curriculum Objectives
Mathematics Year 1: (1C1) Represent and use number bonds and related subtraction facts within 20
This resource is available to play with a Premium subscription.<|endoftext|>
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Very High Frequency :
ITU designationfor the range of radio frequency electromagnetic waves (radio waves) from 30 to 300 megahertz (MHz), with corresponding wavelengths of ten to one meter. Frequencies immediately below VHF are denoted high frequency (HF), and the next higher frequencies are known as ultra high frequency (UHF).
Common uses for VHF are FM radio broadcasting, television broadcasting, two way land mobile radio systems (emergency, business, private use and military), long range data communication up to several tens of kilometers with radio modems, amateur radio, and marine communications. Air traffic control communications and air navigation systems (e.g. VOR & ILS) work at distances of 100 kilometers or more to aircraft at cruising altitude.<|endoftext|>
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An allergy is a condition where an individual suffers a mild to a severe reaction to something they have come in contact with – this can be chemical, edible, or environmental in nature. The body’s immune system over-reacts to the substance, causing an allergic response. The result can manifest in many forms, including hives, rash, or swelling of a localized area.
The general causes of allergies can be traced to food, drinks, insect bites/stings, synthetic or natural materials, plants, or medications. The top ten allergens are listed below.
Tree nuts include Brazil nuts, almonds, cashews, hazelnuts, pecans, and any other type of nut that grows on trees. Rooted nuts like peanuts are also of concern. Milk and dairy (including cheese) is common, as well as seafood, shellfish (mollusks, crab and lobster, shrimp, etc.) and fish. Some grains like soy, wheat, cocoa, and even certain food additives can be identified as ‘allergens’ by the body.
The Range of Symptoms that can Appear
In cases here in Ottawa where contact is made with any of the aforementioned items, the symptomatic reactions will vary depending on the individual’s immune system. In extreme cases we have what is called ‘anaphylaxis’, which is a severe life-threatening reaction – this can manifest in difficulty breathing, a drop in blood pressure or even shock – leading to unconsciousness and sometimes death.
The following conditions are common symptoms of an allergic reaction:
- Flushed face, hives or rash, red and itchy skin
- Swelling of the eyes, face, lips, throat and tongue
- Trouble breathing, speaking or swallowing
- Anxiety, distress, faintness, paleness, sense of doom, weakness
- Cramps, diarrhea, vomiting
- A drop in blood pressure, rapid heartbeat, loss of consciousness
Here are the Top Ten Food Allergens:
- Root nuts (Peanuts)
- Seafood (Fish, Crustaceans and Shellfish)
- Tree Nuts<|endoftext|>
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The abolitionists were divided over strategy and tactics, but they were very active and very visible. Many of them were part of the organized Underground Railroad that flourished between 1830 and 1861. Not all abolitionists favored aiding fugitive slaves, and some believed that money and energy should go to political action. Even those who were not abolitionists might be willing to help when they encountered a fugitive, or they might not. It was very difficult for fugitives to know who could be trusted.
Southerners were outraged that escaping slaves received assistance from so many sources and that they lived and worked in the North and Canada. As a part of the Compromise of 1850, a new Fugitive Slave Act was passed that made it both possible and profitable to hire slave catchers to find and arrest runaways. This was a disaster for the free black communities of the North, especially since the slave catchers often kidnapped legally-free blacks as well as fugitives. But these seizures and kidnappings brought the brutality of slavery into the North and persuaded many more people to assist fugitives. Vigilance Committees acted as contact points for runaways and watched out vigilantly for the rights of northern free blacks. They worked together with local abolition societies, African American churches and a variety of individuals to help fugitives move further on or to find them homes and work. Those who went to Canada in the mid-nineteenth century went primarily to what was then called Canada West, now Ontario.
Last Modified: EST<|endoftext|>
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# BODMAS Rules - Involving Integers
We will obey the rules for simplifying an expression using BODMAS rules - involving integers for solving order of operations.
Follow the order of operation as:
1. Bracket Solve inside the Brackets before Of, Multiply, Divide, Add or Subtract.
For example:
7 × (15 + 5)
= 7 × 20
= 140
2. Of Then, solve Of part (Powers, Roots, etc.,) before Multiply, Divide, Add or Subtract.
For example:
6 + 3 of 7 - 5
= 6 + 3 × 7 - 5
= 6 + 21 - 5
= 27 - 5
= 22
3. Division/Multiplication Then, calculate Multiply or Divide before Add or Subtract start from left to right.
For example:
20 + 21 ÷ 3 × 2
= 20 + 7 × 2
= 20 + 14
= 34
4. Addition/Subtraction At last Add or Subtract start from left to right.
17 + (8 - 5) × 5
= 17 + 3 × 5
= 17 + 15
= 32
Worked-out problems for solving BODMAS rules - involving integers:
Simplify using BODMAS rule:
(a) 25 - 48 ÷ 6 + 12 × 2
Solution:
25 - 48 ÷ 6 + 12 × 2
= 25 - 8 + 12 × 2, (Simplifying ‘division’ 48 ÷ 6 = 8)
= 25 - 8 + 24, (Simplifying ‘multiplication’ 12 × 2 = 24)
= 17 + 24, (Simplifying ‘subtraction’ 25 - 8 = 17)
= 41, (Simplifying ‘addition’ 17 + 24 = 41)
(b) 78 - [5 + 3 of (25 - 2 × 10)]
Solution:
78 - [5 + 3 of (25 - 2 × 10)]
= 78 - [5 + 3 of (25 - 20)], (Simplifying ‘multiplication’ 2 × 10 = 20)
= 78 - [5 + 3 of 5], (Simplifying ‘subtraction’ 25 - 20 = 5)
= 78 - [5 + 3 × 5], (Simplifying ‘of’)
= 78 - [5 + 15], (Simplifying ‘multiplication’ 3 × 5 = 15)
= 78 - 20, (Simplifying ‘addition’ 5 + 15 = 20)
= 58, (Simplifying ‘subtraction’ 78 - 20 = 58)
(c) 52 - 4 of (17 - 12) + 4 × 7
Solution:
52 - 4 of (17 - 12) + 4 × 7
= 52 - 4 of 5 + 4 × 7, (Simplifying ‘parenthesis’ 17 - 12 = 5)
= 52 - 4 × 5 + 4 × 7, (Simplifying ‘of’)
= 52 - 20 + 4 × 7, (Simplifying ‘multiplication’ 4 × 5 = 20)
= 52 - 20 + 28, (Simplifying ‘multiplication’ 4 × 7 = 28)
= 32 + 28, (Simplifying ‘subtraction’ 52 - 20 = 32)
= 60, (Simplifying ‘addition’ 32 + 28 = 60)
Related Concept<|endoftext|>
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Graphical Solution - Page 17d
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GRAPHICAL SOLUTION A fourth method of solving a quadratic equation is by means of graphing. In graphing linear equations using both axes as reference, we recall that an independent variable, x, and a dependent variable, y, were needed. The coordinates of points on the graph of the equation were designated (x, y). Since the quadratics we are considering contain only one variable, as in the equation x2 - 8x + 12 = 0 we cannot plot values for the equations in the present form using both x and y axes. A dependent variable, y, is necessary. If we think of the expression x2 - 8x + 12 as a function, then this function can be considered to have many possible numerical values, depending on what value we assign to x. The particular value or values of x which cause the value of the function to be 0 are solutions for the equation x2 - 8x + 12 = 0 For convenience, we may choose to let y represent the function X2 - 8x + 12 If numerical values are now assigned to x, the corresponding values of y may be calculated. When these pairs of corresponding values of x and y are tabulated, the resulting table provides the information necessary for plotting a graph of the function. EXAMPLE: Graph the equation x2 + 2x - 8 = 0 and from the graph write the roots of the equation. SOLUTION: 1. Let y = x2 + 2x - 8. 2. Make a table of the y values corresponding to the value assigned x, as shown in table 16-1. Table 16-1.-Tabulation of x and y values for the function y = x2 + 2x - 8. 3. Plot the pairs of x and y values that appear in the table as coordinates of points on a rectangular coordinate system as in figure 16-1 (A). 4. Draw a smooth curve through these points, as shown in figure 16-1 (B). Notice that this curve crosses the X axis in two places. We also recall that, for any point on the X axis, the y coordinate is zero. Thus, in the figure we see that when y is zero, x is -4 or +2. When y is zero, furthermore, we have the original equation, Figure 16-1.-Graph of the equation y = x2 + 2x - 8. (A) Points plotted; (B) curve drawn through plotted points. x2 + 2x - 8 = 0 Thus, the values of x at these points where the graph of the equation crosses the X axis (x = -4 or +2) are solutions to the original equation. We may check these results by solving the equation algebraically. Thus, The curve in figure 16-1 (B) is called a PARABOLA. Every quadratic of the form ax2 + bx + c = y will have a graph of this general shape. The curve will open downward if a is negative, and upward if a is positive. Graphing provides a fourth method of finding the roots of a quadratic in one variable. When the equation is graphed, the roots will be the X intercepts (those values of x where the curve crosses the X axis). The X intercepts are the points at which y is 0. Practice problems. Graph the following quadratic equations and read the roots of each equation from its graph 1. x2 - 4x - 8 = 0 2. 6x - 5 - x2 = 0 Answers: 1. See figure 16-2. x = 5.5; x = -1.5 2. See figure 16-3. x = 1; x = 5<|endoftext|>
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*****THIS UNIT IS SHORTER THAN OUR TYPICAL UNITS! We designed this unit as an alternative to traditional Valentine’s Day activities. We are always looking for ways to connect beloved thematic topics to global themes. We made this shorter because we believe there are other topics we want to spend time teaching in February, but we wanted to provide enough activities for one day.
The topics that we cover include:
: Did you know that February is Heart Health Awareness month? This is an important time for everyone to raise awareness about doing everything we can to have a healthy heart. Since Valentine’s Day is an exciting time and hearts are everywhere, let’s talk about our actual hearts that we should all take care of! We also wanted to expose students to ways to raise awareness (about any topic, really) and how to do that well.
Valentine’s day usually means FLOWERS! Lots and lots of flowers. Where do they come from? There are many developing countries that produce a good amount of the flowers that we will use for Valentine’s day. The unfortunate reality is that many famers and workers on these farms are producing all the flowers, but are not reaping the benefits. Many are paid poorly, have little to no vacation, and are often working in dangerous conditions. People that buy flowers usually experience and intense rise in costs. While there area few different reasons for this, a lot of this can be traced back to unfair policies that prevent the farmers from getting paid more.
For history, we will take a look at how conversation hearts became the most popular candy for Valentine’s Day. Conversation hearts are over 100 years old. The sayings used to be longer because the candy used to be bigger and shaped like a seashell. Over the years, it became shaped like a heart and the phrases were shortened. The inventors eventually started to ask the public for ideas for new phrases. They rely on pop culture for a lot of ideas- which means that the phrases can become outdated. Your students will always remember this when they see a box of hearts!
Make SOCIAL STUDIES important again!
LaNesha Tabb & Naomi O’Brien<|endoftext|>
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Inspiratory and expiratory ratio is a figure that shows the relationship between a person's inspiratory time and his expiratory time. Typically, the expiratory time will be greater than the inspiratory time because it takes a person longer to exhale than inhale. If this isn't, the inspiratory time is longer than the expiratory time, and the person may be experiencing gas trapping, which could lead to barotrauma or even cardiac arrest in extreme cases.
Lay down or sit in a comfortable position and set your stopwatch to 30 seconds.
Inhale and at the same time start your stop watch. Count how many times you take a breath within 30 seconds. As an example, you inhale 10 times in 30 seconds.
Sciencing Video Vault
Multiply the number of times you inhale in 30 seconds by 2 to determine your inhalation rate in one minute. In the example, multiply 10 by 2. You inhale 20 times in 1 minute.
Subtract the inspiratory rate from 60 to determine the expiratory rate. In the example, subtract 20 from 60. You have an expiratory rate of 40 seconds in 1 minute.
Write the inspiratory and expiratory rate as a ratio. List the inspiratory rate first. In the example, the ratio would look like this: 20:40.
Reduce the ratio by dividing each number by the greatest common factor shared by the two numbers. In the example, the shared greatest common factor is twenty. Dividing 20 by 20 gives you 1. Dividing 40 by 20 gives you 2. The inspiratory and expiratory ratio is 1:2.<|endoftext|>
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3.5: Subtract Integers (Part 1) - Mathematics
Skills to Develop
• Model subtraction of integers
• Simplify expressions with integers
• Evaluate variable expressions with integers
• Translate words phrases to algebraic expressions
• Subtract integers in applications
be prepared!
Before you get started, take this readiness quiz.
1. Simplify: (12 − (8−1)). If you missed this problem, review Example 2.1.8.
2. Translate the difference of (20) and (−15) into an algebraic expression. If you missed this problem, review Example 1.3.11.
3. Add: (−18 + 7). If you missed this problem, review Example 3.2.6.
Model Subtraction of Integers
Remember the story in the last section about the toddler and the cookies? Children learn how to subtract numbers through their everyday experiences. Real-life experiences serve as models for subtracting positive numbers, and in some cases, such as temperature, for adding negative as well as positive numbers. But it is difficult to relate subtracting negative numbers to common life experiences. Most people do not have an intuitive understanding of subtraction when negative numbers are involved. Math teachers use several different models to explain subtracting negative numbers.
We will continue to use counters to model subtraction. Remember, the blue counters represent positive numbers and the red counters represent negative numbers.
Perhaps when you were younger, you read (5 − 3) as five take away three. When we use counters, we can think of subtraction the same way.
We will model four subtraction facts using the numbers (5) and (3).
[5 − 3 qquad −5 − (−3) qquad −5 − 3 qquad 5 − (−3) onumber ]
Example (PageIndex{1}): model
Model: (5 − 3).
Solution
The difference between (5) and (3) is (2).
Exercise (PageIndex{1})
Model the expression: (6 − 4)
(2)
Exercise (PageIndex{2})
Model the expression: (7 − 4)
(3)
Example (PageIndex{2}): model
Model: (−5 − (−3)).
Solution
The difference between (−5) and (−3) is (−2) .
Exercise (PageIndex{3})
Model the expression: (−6 − (−4))
(-2)
Exercise (PageIndex{4})
Model the expression: (−7 − (−4))
(-3)
Notice that Example (PageIndex{1}) and Example (PageIndex{2}) are very much alike.
• First, we subtracted (3) positives from (5) positives to get (2) positives.
• Then we subtracted (3) negatives from (5) negatives to get (2) negatives.
Each example used counters of only one color, and the “take away” model of subtraction was easy to apply.
Figure (PageIndex{1})
Now let’s see what happens when we subtract one positive and one negative number. We will need to use both positive and negative counters and sometimes some neutral pairs, too. Adding a neutral pair does not change the value.
Example (PageIndex{3}): model
Model: (−5 − 3).
Solution
Interpret the expression. −5 − 3 means −5 take away 3. Model the first number. Start with 5 negatives. Take away the second number. So we need to take away 3 positives. But there are no positives to take away. Add neutral pairs until you have 3 positives. Now take away 3 positives. Count the number of counters that are left.
The difference of (−5) and (3) is (−8).
Exercise (PageIndex{5})
Model the expression: (−6 − 4)
(-10)
Exercise (PageIndex{6})
Model the expression: (−7 − 4)
(-11)
Example (PageIndex{4}): model
Model: (5 − (−3)).
Solution
Interpret the expression. 5 − (−3) means 5 take away −3. Model the first number. Start with 5 positives. Take away the second number, so take away 3 negatives. But there are no negatives to take away. Add neutral pairs until you have 3 negatives. Then take away 3 negatives. Count the number of counters that are left.
The difference of (5) and (−3) is (8).
Exercise (PageIndex{7})
Model the expression: (6 − (−4))
(10)
Exercise (PageIndex{8})
Model the expression: (7 − (−4))
(11)
Example (PageIndex{5}): model
Model each subtraction.
1. (8 − 2)
2. (−5 − 4)
3. (6 − (−6))
4. (−8 − (−3))
Solution
1. (8 − 2): This means (8) take away (2).
1. (−5 − 4): This means (−5) take away (4).
1. (6 − (−6)): This means (6) take away (−6).
1. (−8 − (−3)): This means (−8) take away (−3).
Exercise (PageIndex{9})
Model each subtraction.
1. (7 - (-8))
2. (-7 - (-2))
3. (4 - 1)
4. (-6 - 8)
Exercise (PageIndex{10})
Model each subtraction.
1. (4 - (-6))
2. (-8 - (-1))
3. (7 - 3)
4. (-4 - 2)
Example (PageIndex{6}): model
Model each subtraction expression:
1. (2 − 8)
2. (−3 − (−8))
Solution
(2 − 8 = −6)
(−3 − (−8) = 5)
Exercise (PageIndex{11})
Model each subtraction expression.
1. (7 − 9)
2. (−5 − (−9))
(-2)
(4)
Exercise (PageIndex{12})
Model each subtraction expression.
1. (4 − 7)
2. (−7 − (−10))
(-3)
(3)
Simplify Expressions with Integers
Do you see a pattern? Are you ready to subtract integers without counters? Let’s do two more subtractions. We’ll think about how we would model these with counters, but we won’t actually use the counters.
• Subtract (−23 − 7). Think: We start with (23) negative counters. We have to subtract (7) positives, but there are no positives to take away. So we add (7) neutral pairs to get the (7) positives. Now we take away the (7) positives. So what’s left? We have the original (23) negatives plus (7) more negatives from the neutral pair. The result is (30) negatives. (−23 − 7 = −30) Notice, that to subtract (7), we added (7) negatives.
• Subtract (30 − (−12)). Think: We start with (30) positives. We have to subtract (12) negatives, but there are no negatives to take away. So we add (12) neutral pairs to the (30) positives. Now we take away the (12) negatives. What’s left? We have the original (30) positives plus (12) more positives from the neutral pairs. The result is (42) positives. (30 − (−12) = 42) Notice that to subtract (−12), we added (12).
While we may not always use the counters, especially when we work with large numbers, practicing with them first gave us a concrete way to apply the concept, so that we can visualize and remember how to do the subtraction without the counters.
Have you noticed that subtraction of signed numbers can be done by adding the opposite? You will often see the idea, the Subtraction Property, written as follows:
Definition: Subtraction Property
[a − b = a + (−b)]
Look at these two examples.
Figure (PageIndex{2})
We see that (6 − 4) gives the same answer as (6 + (−4)).
Of course, when we have a subtraction problem that has only positive numbers, like the first example, we just do the subtraction. We already knew how to subtract (6 − 4) long ago. But knowing that (6 − 4) gives the same answer as (6 + (−4)) helps when we are subtracting negative numbers.
Example (PageIndex{7}): simplify
Simplify:
1. (13 − 8) and (13 + (−8))
2. (−17 − 9) and (−17 + (−9))
Solution
Subtract to simplify. 13 − 8 = 5 Add to simplify. 13 + (−8) = 5 Subtracting 8 from 13 is the same as adding −8 to 13.
Subtract to simplify. −17 − 9 = −26 Add to simplify. −17 + (−9) = −26 Subtracting 9 from −17 is the same as adding −9 to −17.
Exercise (PageIndex{13})
Simplify each expression:
1. (21 − 13) and (21 + (−13))
2. (−11 − 7) and (−11 + (−7))
(8), (8)
(-18), (-18)
Exercise (PageIndex{14})
Simplify each expression:
1. (15 − 7) and (15 + (−7))
2. (−14 − 8) and (−14 + (−8))
(8), (8)
(-22), (-22)
Now look what happens when we subtract a negative.
Figure (PageIndex{3})
We see that (8 − (−5)) gives the same result as (8 + 5). Subtracting a negative number is like adding a positive.
Example (PageIndex{8}): simplify
Simplify:
1. (9 − (−15)) and (9 + 15)
2. (−7 − (−4)) and (−7 + 4)
Solution
1. (9 − (−15)) and (9 + 15)
Subtract to simplify. 9 − (−15) = 24 Add to simplify. 9 + 15 = 24
Subtracting (−15) from (9) is the same as adding (15) to (9).
1. (−7 − (−4)) and (−7 + 4)
Subtract to simplify. −7 − (−4) = −3 Add to simplify. −7 + 4 = −3
Subtracting (−4) from (−7) is the same as adding (4) to (−7).
Exercise (PageIndex{15})
Simplify each expression:
1. (6 − (−13)) and (6 + 13)
2. (−5 − (−1)) and (−5 + 1)
(19), (19)
(-4), (-4)
Exercise (PageIndex{16})
Simplify each expression:
1. (4 − (−19)) and (4 + 19)
2. (−4 − (−7)) and (−4 + 7)
(23), (23)
(3), (3)
Look again at the results of Example (PageIndex{1}) - Example (PageIndex{4}).
Table (PageIndex{1}): Subtraction of Integers
5 – 3–5 – (–3)
2–2
2 positives2 negatives
When there would be enough counters of the color to take away, subtract.
–5 – 35 – (–3)
–88
5 negatives, want to subtract 3 positives5 positives, want to subtract 3 negatives
need neutral pairsneed neutral pairs
When there would not be enough of the counters to take away, add neutral pairs.<|endoftext|>
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ZingPath: Counting Principles
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Counting Principles
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Lesson Focus
Fundamental Counting Principle
Geometry
The number of possible outcomes for a compound event using a tree diagram or the fundamental counting principle is determined.
Now You Know
After completing this tutorial, you will be able to complete the following:
• Determine the number of possible outcomes for a compound event by using a tree diagram.
• Determine the number of possible outcomes for a compound event by using the Fundamental Counting Principle.
Everything You'll Have Covered
A tree diagram is a visual display of all possible outcomes in a compound event.
The following is an example of a tree diagram for the possible outcomes of flipping a coin twice.
Possible outcomes would be as follows {heads, heads}, {heads, tails}, {tails, heads}, {tails, tails} which equal to 4 possible outcomes for the event of flipping a coin twice.
A tree diagram is an excellent way to visually introduce possible outcomes to students; students can see how each combination is found by following the "branches" of the tree diagram. However, as the outcomes become more complex another way to find a solution is needed.
Using the Fundamental Counting Principle is the ideal method to use when finding possible outcomes when there are a large amount of combinations.
The Fundamental Counting Principle states that if an event has m possible outcomes and another independent event has n possible outcomes, then there are m × n possible outcomes for the two events together.
An example of using the Fundamental Counting Principle is provided below.
Event:
Sandy is deciding what uniform to wear tomorrow. She has a blue, pink, and white shirt. She has blue and beige pants and black or brown sneakers. What are all the possible combinations?
Because she has 3 shirts, 2 pants, and 2 sneakers we can multiply these numbers together to find the amount of possible outcomes for this event.
3 × 2 × 2 = 12 possible outcome.
The following key vocabulary terms will be used throughout this Activity Object:
• Fundamental Counting Principle ¬- if an event has m possible outcomes and another independent event has n possible outcomes, then there are m × n possible outcomes for the two events together
• multiplication - a mathematical operation where a number is added to itself for a determined amount of times (i.e. 4 + 4 + 4 + 4 + 4 = 4 × 5 = 20)
• outcomes - the result of an event or experiment.
• tree diagram - a visual map that lists all possible outcomes or choices.
Tutorial Details
Approximate Time 20 Minutes Pre-requisite Concepts Students should know how to multiply whole numbers, and be able to recognize a tree diagram. Course Geometry Type of Tutorial Concept Development Key Vocabulary counting outcomes, counting principles, probability<|endoftext|>
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Arsenic has a bad reputation-for good reason. In part, it’s a potent poison, as everyone familiar with murder mysteries knows. But while it’s rarely used as a weapon in the real world, it does turn up as a contaminant in the environment-and mere traces of the stuff ingested from, say, water, can eventually lead to cancer. Arsenic taints more than 700 hazardous-waste sites across the nation, including lakes and rivers in or near many major cities. Typically the toxin comes from waste dumped decades ago by manufacturers of insecticides and other chemicals.
The question becomes what to do with sites contaminated with arsenic-a metal that can combine with oxygen, carbon, and other elements into various compounds, some that are more toxic and that travel more easily through the environment than others and hence are more likely to be ingested. Ideally, health officials would like to remove arsenic from water and soil. At the least, they’d like to minimize its exposure to humans. But neither can be adequately accomplished until researchers learn the basics of how arsenic moves in the environment and how compounds containing it change over time.
Harry Hemond, a professor of civil and environmental engineering at MIT, has been examining such matters for almost 10 years. Hemond, who also directs the Ralph M. Parsons Laboratory, which conducts studies in water resources and environmental engineering at MIT, has been researching arsenic in the Aberjona Watershed, a 25-square-mile polluted stretch of water north of Boston.
In this aquatic ecosystem Hemond finds that arsenic often cycles between two compounds. One, arsenite, is a highly toxic substance that dissolves and moves easily in water and that humans can thus ingest. The other, arsenate, is a less soluble and less toxic variety that iron-rich particles tend to absorb before settling deep in lake sediments.
Factors such as temperature, dissolved oxygen, organic matter, and microbe populations determine the relative proportions of arsenate and arsenite in water, Hemond says. He and his students have therefore been sampling lakes in the watershed under many conditions.
The group is constantly developing new techniques to obtain results that are as accurate as possible. For instance, David B. Senn, an environmental-engineering doctoral student, has designed a pump to collect and filter water samples from murky, oxygen-free lake bottoms without introducing any oxygen. Because the highly toxic arsenite differs chiefly from its chemical cousin arsenate in containing less oxygen, any addition of that element could lead to unrealistically high readings of the safer compound. Typically, water-filtering devices allow some air-and hence oxygen-to enter the samples. But Senn’s apparatus starts out filled with nitrogen. That physically prevents any arsenic-oxygen reactions when researchers introduce lake water into the instrument. The fact that the MIT device filters below the water surface rather than in a laboratory, as is the norm, also helps, Hemond explains.
Through its years of sampling work Hemond’s group has discovered that arsenic levels in bottom waters can change dramatically. In 1992 and 1994, the team found roughly the same amount of arsenic along the bottom of a lake that receives Aberjona drainage. But sediment levels skyrocketed in 1993. Thus, says Hemond, health researchers can’t “just grab one sample” to analyze an area’s level of contamination. The fluctuations could reflect the activity of organisms that absorb and release the metal, he suggests.
Indeed, Hemond’s group has discovered that bacteria play a major role in converting arsenate to arsenite. In 1994, then-graduate student Dianne Ahmann-now an assistant professor of microbial ecology at Duke University-and Franois Morel, a professor of geochemistry at Princeton University, found in sediment samples a bacterium that takes up arsenate, uses its oxygen for energy, and releases arsenite. Ahmann’s lab is studying the potential of the microbe, named MIT-13, for remediating arsenic-laden sites so that technicians do not have to excavate large quantities of contaminated soil. Engineers might be able to treat small, isolated areas with MIT-13 to mobilize the arsenic. Although that would result in a more toxic compound, engineers could then pump it out and purify it for reuse in chemicals employed in industries that today are subject to tight environmental regulations. Or engineers could pump out, concentrate, and dispose of the arsenic in designated landfills.
Meanwhile, Hemond’s lab, in its continuing quest to catalog environmental factors that influence arsenic, has started investigating other life forms and physical processes that may affect the pollutant. For example, Laurel Schaider, an undergraduate student in environmental-engineering science at MIT, is working to cull from sediment a bacterium that is apparently converting the more toxic arsenite to arsenate. Hemond says that beyond identifying the species, determining how quickly it oxidizes arsenite and how it can grow are necessary steps for indicating whether it could be a catalyst useful in a “low-cost arsenic removal system for small villages” where drinking water contains the metal.<|endoftext|>
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Czechoslovakia Table of Contents
Czechoslovakia's centralized political structure might have been well suited to a single nation-state, but it proved inadequate for a multinational state. Constitutional protection of minority languages and culture notwithstanding, the major nonCzech nationalities demanded broader political autonomy. Political autonomy was a particularly grave issue for the Czechs' partners, the Slovaks. In 1918 Masaryk signed an agreement with American Slovaks in Pittsburgh, promising Slovak autonomy. The provisional National Assembly, however, agreed on the temporary need for centralized government to secure the stability of the new state. The Hlasists, centered on the journal Hlas, continued to favor the drawing together of Czechs and Slovaks. Although the Hlasists did not form a separate political party, they dominated Slovak politics in the early stages of the republic. The Hlasists' support of Prague's centralization policy was bitterly challenged by the Slovak Populist Party. The party had been founded by a Catholic priest, Andrej Hlinka, in December 1918. Hlinka argued for Slovak autonomy both in the National Assembly and at the Paris Peace Conference. He made Slovak autonomy the cornerstone of his policy until his death in August 1938.
The Slovak Populist Party was Catholic in orientation and found its support among Slovak Catholics, many of whom objected to the secularist tendencies of the Czechs. Religious differences compounded secular problems. The Slovak peasantry had suffered hardships during the period of economic readjustment after the disintegration of the Hapsburg Empire. Moreover, the apparent lack of qualified Slovaks had led to the importation of Czechs into Slovakia to fill jobs (formerly held by Hungarians) in administration, education, and the judiciary. Nevertheless, at the height of its popularity in 1925, the Slovak Populist Party polled only 32 percent of the Slovak vote, although Catholics constituted approximately 80 percent of the population. Then, in 1927, a modest concession by Prague granted Slovakia the status of a separate province, and Slovak Populists joined the central government. Monsignor Jozef Tiso and Marko Gazlik from Slovakia were appointed to the cabinet.
Although Hlinka's objective was Slovak autonomy within a democratic Czechoslovak state, his party contained a more radical wing, led by Vojtech Tuka. From the early 1920s, Tuka maintained secret contacts with Austria, Hungary, and Hitler's National Socialists (Nazis). He set up the Rodobrana (semimilitary units) and published subversive literature. Tuka gained the support of the younger members of the Slovak Populist Party, who called themselves Nastupists, after the journal Nastup.
Tuka's arrest and trial in 1929 precipitated the reorientation of Hlinka's party in a totalitarian direction. The Nastupists gained control of the party; Slovak Populists resigned from the government. In subsequent years the party's popularity dropped slightly. In 1935 it polled 30 percent of the vote and again refused to join the government. In 1936 Slovak Populists demanded a Czechoslovak alliance with Hitler's Germany and Mussolini's Italy. In September 1938, the Slovak Populist Party received instructions from Hitler to press its demands for Slovak autonomy.
Data as of August 1987<|endoftext|>
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By Nadia Cassim
Research Contributor: Sumaya Seedat
As Elisabeth Siddiqui so aptly states, “Art is the mirror of a culture and its world view” She goes on to state that in the Islamic world, this is truer than ever as Islamic art reflects the different lands, culture and people which Muslims have ruled over. Furthermore, the spiritual ethos of Islam is central to the art form. As Siddiqui explains, “for the Muslim, reality begins with and centres around God (“Allah” in Arabic), the One, the Unique, the Sovereign, the Holy, the Almighty, the All-Knowing, the Loving, the Most Merciful. All existence is subject to His will and His laws. He is the centre of conscious Muslims’ worship and aspirations, the focus of their lives”¹
This ethos finds its way into the physical expression of art in Islam in the form of painting, calligraphy, ceramics, architecture, carpets and rugs, glass work and metal work². Islamic art is more about beautification and ornamentation, then “art for art’s sake”¹. Produced from the 7th century onwards, Islamic art styles have been adapted from early Christian art, Byzantine and Roman styles as well as pre-Islamic Persian influences (Sassanian)². Thus, Islamic art is not necessarily religious art, but depictive of the time and place that Islam ruled.
A method employed in classifying Islamic art is according to the dynasty or influential power that ruled at the time the art piece was produced, which in turn had an influence on the art style. The various dynasties began with the Umayyad dynasty and the Abbasid dynasty that governed a vast Islamic state, concluding with the more powerful dynasties such as the Safavids, Ottomans and Mughals².
For instance, during the reign of Suleyman “the magnificent” or “the law maker”, the influence of Islamic art spread throughout what was then known as the Ottoman Empire (as far as Vienna, Hungary, Iraq, and important North African Ports). Suleyman’s reign thus covered large areas of Europe, Asia and Africa, thus dubbing it the Golden Age as it encapsulated an age of cultural activity, trade and economic expansion³. An example of the influence of his reign on significant Islamic architecture can be seen by the Tile Revetment in the Dome of The Rock, Jerusalem.
Arabic calligraphy is one of the most common expressions of Islamic art throughout the ages. It is not unique in terms of its employment as a method of cultural expression- others include “Chinese and Japanese calligraphy and illuminated bibles from north-west Europe including the famous Book of Kells”⁴– but it has transcended the ordinary by being more creative and versatile. It has been able to convey the deeper meaning of a script (often Quranic), with adaptability and balance. Arabic calligraphy is often drawn on mediums such as glass, ceramics, sculptures and paintings, wood and textiles.⁴
Largely influenced by the Roman and Sassanian glass-making techniques, Islamic glass found its recognition in the late 8th and early 9th century. Glass was used for purposes such as inkwells, perfume sprayers, mosque lamps and beads for jewellery. Glass was also used for medicinal purposes such as in the production of test tubes and cuppers.⁵
Islamic-influenced rugs and carpets have become famous throughout the world. They vary in size and shape and range from cushions to rugs and even bags. The Islamic prayer mat also boasts patterns of Islamic art, used in homes and mosques alike, hung on walls and draped over furniture². Under the Ottoman, Safavid and Mughal dynasties the making of carpets and rugs became more than just a domestic practice passed down from generation to generation, but rather a worldwide trade and industry. Under Shah Abbas (1587-1629) carpets were traded throughout Europe, in particular, England, France and Spain. In Ottoman Turkey, after the conquest of Egypt and Persia, carpet weaving techniques and patterns changed to accommodate new influences. Since most carpets are not dated their depiction in Flemish paintings are often used as a method of dating their presence and influence. Common patterns and colours on most Islamic rugs and carpets are floral designs, geometric shapes in hues of blue and red⁶. The significance of rugs and carpets in Islam is further emphasized by the Quran’s mention of them in Surah 88, Verses 8-20
“In the Islamic period, conquests and trade within the region and beyond resulted in technological innovations such as metallic glazing, a wide colour palette, and the imitation and adaptation of Chinese production techniques long before these innovations reached the West”⁷Ceramics from the Muslim world not only shed light on the culture and nature of society at the time, but it also gives insight into non-religious or nonspiritual art in Islam. Whilst the depiction of figures are frowned upon in Islam for fear of idol worship emerging as a result, there exist ceramics from the Muslim world with figurines alongside calligraphy and geometric designs⁷
Modern and contemporary artwork has reached new heights and expression. Whilst designs dating back from the 17th Century are being translated into modern works, many of the patterns are being reproduced using machines. There are still a lot of controversial artworks being produced by local artists worldwide, depicting the human figure or eyes in portraits². With regards to architecture, the continuous revamping on the Prophet’s Mosque in Medina and the Holy Mosque in Mecca is testimony of the ever-evolving nature of Islamic art.
⁶Marka S, 2011, Department of Islamic Art The Metropolitan Museum of Art, http://islamic-arts.org/2011/carpets-from-the-islamic-world/<|endoftext|>
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# Basic Math Facts – Exponents
Exponents comprise a juicy tidbit of basic-math-facts material. Exponents allow us to raise numbers, variables, and already expressions to powers, consequently achieving repeated multiplication. The ever present exponent in all kinds of mathematical problems requires that the student be thoroughly conversant with its features and similarities. Here we look at the laws, the knowledge of which, will allow any student to master this topic.
In the expression 3^2, which is read “3 squared,” or “3 to the second strength,” 3 is the base and 2 is the strength or exponent. The exponent tells us how many times to use the base as a factor. The same applies to variables and variable expressions. In x^3, this average x~x~x. In (x + 1)^2, this method (x + 1)*(x + 1). Exponents are omnipresent in algebra and indeed all of mathematics, and understanding their similarities and how to work with them is extremely important. Mastering exponents requires that the student be familiar with some basic laws and similarities.
Product Law
When multiplying expressions involving the same base to different or equal powers, simply write the base to the sum of the powers. For example, (x^3)(x^2) is the same as x^(3 + 2) = x^5. To see why this is so, think of the exponential expression as pearls on a string. In x^3 = x~x~x, you have three x’s (pearls) on the string. In x^2, you have two pearls. consequently in the product you have five pearls, or x^5.
Quotient Law
When dividing expressions involving the same base, you simply subtract the powers. consequently in (x^4)/(x^2) = x^(4-2) = x^2. Why this is so depends on the cancellation character of the real numbers. This character says that when the same number or variable appears in both the numerator and denominator of a fraction, then this term can be canceled. Let us look at a numerical example to make this completely clear. Take (5~4)/4. Since 4 appears in both the top and bottom of this expression, we can kill it—well not kill, we don’t want to get violent, but you know what I average—to get 5. Now let’s multiply and divide to see if this agrees with our answer: (5~4)/4 = 20/4 = 5. Check. consequently this cancellation character holds. In an expression such as (y^5)/(y^3), this is (y~y~y~y~y)/(y~y~y), if we expand. Since we have 3 y’s in the denominator, we can use those to cancel 3 y’s in the numerator to get y^2. This agrees with y^(5-3) = y^2.
strength of a strength Law
In an expression such as (x^4)^3, we have what is known as a strength to a strength. The strength of a strength law states that we simplify by multiplying the powers together. consequently (x^4)^3 = x^(4~3) = x^12. If you think about why this is so, notice that the base in this expression is x^4. The exponent 3 tells us to use this base 3 times. consequently we would acquire (x^4)*(x^4)*(x^4). Now we see this as a product of the same base to the same strength and can consequently use our first character to get x^(4 + 4+ 4) = x^12.
Distributive character
This character tells us how to simplify an expression such as (x^3~y^2)^3. To simplify this, we spread the strength 3 outside parentheses inside, multiplying each strength to get x^(3~3)*y^(2~3) = x^9~y^6. To understand why this is so, notice that the base in the original expression is x^3~y^2. The 3 outside parentheses tells us to multiply this base by itself 3 times. When you do that and then rearrange the expression using both the associative and commutative similarities of multiplication, you can then apply the first character to get the answer.
Zero Exponent character
Any number or variable—except 0—to the 0 strength is always 1. consequently 2^0 = 1; x^0 = 1; (x + 1)^0 = 1. To see why this is so, let us consider the expression (x^3)/(x^3). This is clearly equal to 1, since any number (except 0) or expression over itself yields this consequence. Using our quotient character, we see this is equal to x^(3 – 3) = x^0. Since both expressions must provide the same consequence, we get that x^0 = 1.
Negative Exponent character
When we raise a number or variable to a negative integer, we end up with the reciprocal. That is 3^(-2) = 1/(3^2). To see why this is so, let us consider the expression (3^2)/(3^4). If we expand this, we acquire (3~3)/(3~3~3~3). Using the cancellation character, we end up with 1/(3~3) = 1/(3^2). Using the quotient character we that (3^2)/(3^4) = 3^(2 – 4) = 3^(-2). Since both of these expressions must be equal, we have that 3^(-2) = 1/(3^2).
Understanding these six similarities of exponents will give students the substantial foundation they need to tackle all kinds of pre-algebra, algebra, and already calculus problems. Often times, a student’s stumbling blocks can be removed with the bulldozer of foundational concepts. Study these similarities and learn them. You will then be on the road to mathematical expert.<|endoftext|>
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This timeline shows some pivotal events in the history of the Universe which is assumed to have begun with the Big Bang some 13.7 billion years ago. The "Redshift Desert" is a region where the light from galaxies has been redshifted (stretched by the expansion of the Universe) into a region of the spectrum where a natural glow in the Earth's atmosphere interferes with key spectroscopic features of many of these galaxies. This interference is especially problematic when trying to study dimmer galaxies in the early Universe. Using a sophisticated observing technique that overcomes this problem, the Gemini Deep Deep Survey revealed that a large number of galaxies from this period of cosmic history were fully formed and more massive than the widely accepted Hierarchical Model of galaxy formation predicts. This timeline also illustrates the concept of "look-back time," which is what happens when astronomers look at more and more distant objects in space. Because light travels at a finite speed (about 300,000 km/s or 186,000 miles/s), it takes time for the light to reach our telescopes to be studied. This results in a cosmic "time-machine" because the light that we see from distant galaxies has traveled for billions of years. Thus, we see the galaxies as they were long ago when that light began its journey to our telescopes. Full Resolution TIF (8.5 MB)Medium Resolution JPG (248 KB)Gemini Observatory Image/Video Usage Policy
Photoshop version with text layers available upon request.<|endoftext|>
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Grades K-5 | Technology Skills | App & Robot
Teaching standard curriculum using robotics
Wednesday, July 13, 2016
Robotics can be used to help students learn important core topics such as English Language Arts, Science, Social Studies, and Mathematics. As personable characters that ignite curiosity and confidence, Dash & Dot are especially useful in projectstyle lessons that are crosscurricular and exploratory. Students may need to program Dash to complete a science experiment, using mathematics to calculate distances and angles, and then write a report on their findings to present to the class. Robotics is a great way to keep kids engaged in topics that they might not find interesting on their own.
Here are some example lessons:
Mathematics: Students can compare fractions on a number line by programming one Dash to move some distance to model a fraction and another Dash to model the other fraction. Which is larger? After recording a number of fractions with a common numerator, how can you tell which fraction is larger just by looking at it?
English Language Arts: Students can write a story starring the robots. Then students can sketch storyboards for their story and film it, programming the robots to act out the story and using other fun props. Students can be as creative as they like, but once the play has been programmed, it can’t be changed on the fly.
Science: Students can study magnetism by attaching magnets to two Dashes. Students will program the Dashes to move toward each other. They can see that when opposite poles are lined up, the Dashes will snap together, and when similar poles are lined up, the magnets repel each other. Students can design and conduct experiments to test the strength of the magnets and their attractive/repulsive properties on different objects. They can then write a report on their use of the scientific method to test their hypotheses.<|endoftext|>
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Biology and ecology: Feral and domestic cats are descended from wildcats
(Felis silvestris) native to Eurasia and Africa. Cats share a long history of association with humans that has helped their spread across the globe. Archaeological evidence of cat taming dates back approximately 9500 years ago to Crete. Today, cats can be found on every continent except the North and south poles.
Feral cats have the body shape, acute senses and fine coordination of the ultimate predator.
Feral cats weigh about on average 3–4 kg, with body lengths ranging from 40–60 cm. Males are usually larger than females. Most feral cats are short haired and not showy breeds like Siamese or Persian. Coat colours range between ginger, tabby, tortoiseshell, grey and black. In Australia, tabby and ginger cats are the most abundant, this is all due to breeding in the wild where stealth and camouflage is of great importance in the hunting of prey species.
Breeding: Female cats can reproduce at between 10–12 months of age, with males reaching maturity at about one year. Cats generally do not breed during winter where cold weather can reduce the survival of kittens. They produce up to three litters a year (65 days gestation) averaging four kittens per litter. Kittens are weak hunters and can take up to six months to learn all the necessary skills to become independent hunters. Kittens and juveniles are often killed by foxes, dingoes, large reptiles like goanna’s and wedge-tailed eagles. Female feral cats are likely to reproduce for all of their adult lives. This high reproductive ability keeps populations growing, despite the high death rates of young. Feral cat populations do not need a supply of new domestic or stray cats to maintain their numbers.
Habitat: Feral cats live in a diverse range of habitats including deserts, forests, woodlands and grasslands. They usually reach their highest densities on small islands or in human-modified habitats such as farms and rubbish tips due to the large amounts of food available. However, most of the time they are found in low numbers with relatively large home ranges (may exceed 10 km2). The distance travelled by ranging cats depends on the availability of prey, breeding season of the cat and habitat. Males tend to roam over larger range sizes than female cats. Feral cats are generally nocturnal and will rest during the day in den sites such as hollow logs, piles of debris, rabbit warrens or dense scrub.
Diet: Feral cats are carnivores and can survive with limited access to water, as they use the moisture from their prey. They generally eat small mammals, but also catch birds, reptiles, amphibians, fish and insects - taking prey up to the size of a brush-tail possum. In pastoral regions, they feed largely on young rabbits, but in other areas feral cats prey mainly on native animals. In a 2012 report, the AWC estimates that each feral cat kills between five and 30 animals a day. It says taking the lower figure in that range and multiplying it by a "conservative population estimate" of 15 million feral cats gives a minimum estimate of 75 million native animals killed daily by feral cats.
Current Australian distribution: There is on average one feral cat for every one to two-kilometres square but this may be larger if food supplies are scarce. Cats now occupy 99% of Australia, including many offshore islands.
Economic impacts: The cost of feral cat’s due to management and research has been estimated at $2 million per year. The loss inflicted by feral and domestic cats, based on bird predation alone, has been estimated at $144 million annually.
Environmental impacts: Feral cats are exceptional hunters and pose a significant threat to the survival of many native species including small mammals, birds and reptiles. Feral cats have been implicated in extinctions of Australian native animals and have added to the failure of endangered species reintroduction programs (eg numbat and bilby). About 80 endangered and threatened species are at risk from feral cat predation in Australia according to Australia’s Environment Protection & Biodiversity Conservation Act (1999)
Social impacts: Feral cats pose a serious health risk to humans, livestock and native animals as carriers of diseases such as toxoplasmosis and sarcosporidiosis. Cat-related toxoplasmosis can cause wide ranging health problems from debilitation, miscarriage to congenital birth defects in humans and other animals. Feral cats also represent a high-risk reservoir for exotic diseases such as rabies if an outbreak were to occur in Australia.
Strategies for control
Baiting - In order to kill a cat using poison baits, cats must first find and then ingest the bait, unfortunately cats hunt mainly using sight and sound so finding an inert sausage may be hard for a cat to locate. Many cats fail to find a poison bait before it breaks down and are no longer toxic. Even when cats do find baits, up to 80% of encounters do not lead to bait ingestion, the cats often ignoring, sniffing or avoiding baits when detected. This is because cats prefer to catch their own live prey and will only ingest a bait when very hungry.
Successful baiting relies on using large densities of baits in areas with low food availability at the right time of year when cats are hungriest this being in late autumn to late winter before prey becomes abundant in spring.
Grooming trap- A grooming trap squirts a poisonous paste onto the fur of the cat as it walks past a trap station, which it then ingests through compulsive grooming.
Cage and leghold traps- These techniques are very useful in urban and rural situation; the cage trap is the preferred method for dealing with feral cats as they are quite vicious and lively when caught and this will reduce the likelihood of the animal hurting the operator or itself in the process. This also helps for euthanasia if they are caught in an urban area. Padded-jaw, leg-hold traps should only be used at sites where the animal can be killed by shooting whilst still held in the trap. Leg-hold traps may be more effective than cage traps for hard-to-catch cats that have had minimal exposure to humans.
Ground shooting – Shooting is one of the main methods of controlling feral cats, usually done at night from a vehicle fitted with spotlights.
This method is fairly labour intensive but when done by a skilled operator it can quickly reduce the local numbers of feral cats very quickly. It is most probably the most humane way of control where the right calibre and bullet placement is paramount.
Integrated pest management –
With all the techniques explained above we can see that with an integrated approach the feral cat numbers could be reduces quite quickly if all steps are taken. Traps can be set, baiting programs be used and a night time shooting schedule can be devised, with all these running at the same time as a singular program they will be much more affective together magnifying the effectiveness of each other.<|endoftext|>
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Geology deals with vastness of time that is far outside human experience. Our common experience of time, and our ancestors' experiences that shaped our brain's evolution, span intervals from a moment, to a day, to a season, to a year, to a lifetime. Geologists, in contrast, deal with thousands to millions of years, up to the 4.5 billion year age of the Earth. This vastness of time is central to geological thought, because it allows time for imperceptibly slow processes to effect monumental changes. Given enough time, erosion levels mountains. Given enough time, natural selection produces new species. Frodeman (2003) argues that understanding the brevity of human existence relative to the span of Earth history requires an "innovation in our sense of reality."
In addition, time is a clue to causality in geology. If it can be shown that A happened before B, then A can have caused or influenced B, but B cannot have caused or influenced A. And finally, rate is important because it is a clue to the energy budget of the causative process; a process that moved a cubic kilometer of material in an hour is a profoundly different phenomenon than a process that moved a cubic kilometer of material in a million years.
We wish to know:
- How do humans build on our experiential concept of time to deal with temporal scales of thousands to billions of years? What distortions are introduced into a learner's view of earth history and process if the immenseness of geological time is not grasped?
- What is the role of narrative, of analogy (Frodeman, 1995), and of representation in helping people understand unfamiliar times and time spans?
- How do learners make the conceptual leap from observable processes (e.g. convection of cream in a cup of coffee at a rate of cm/s) to processes that are too slow to observe (e.g. convection of the Earth's mantle at a rate of mm/yr)?
- By what sequence of understandings does a learner come to accept that a landscape that looks static according to the standards of our life experiences is actually dynamic, undergoing constant change, but at a rate too slow for us to see?
- What is necessary for a learner to accept as "known" an event or sequence of events that happened when no humans were there to observe it?
- What is the relationship between a learner's understanding of time and causality? What is the relationship between a learner's understanding of the rate of earth processes and the energy of the formative processes?
Researchers have begun to develop instruments and methodologies to probe learner's knowledge of, interest in, and ability to reason about geological time. For example, Trend (1998, 2000, 2001, 2002) used card sorting, questionnaires, concept mapping, group discussions, and responding to objects, to explore what students and teachers know about the sequence and timing of events in geological history. He found that children cluster geologic events into just two categories ("extremely ancient" and "less ancient"). Their assignment of geo-events to these categories is generally accurate, but there is confusion over the relative timing within each time cluster (Trend, 1998). Dodick and Orion (2003a, 2003b) have developed an instrument, the Geological Time Aptitude Test, to probe students' ability to employ diachronic thinking in a geological context. "Diachronic" refers to the study of the development of something through time (Montagnero, 1996), for example, the ability to determine the sequential order of stages, or to identify precursor/successor linkages between events. They found that many students found such tasks difficult, that diachronic thinking continues to improve with age in the absence of instruction among 11 th and 12th graders, and that study of geology accelerates the acquisition of diachronic thinking skills even in a non-geological scenario.Browse our Growing Reference Collection addressing Geologic Time in Geoscience Learning
Dodick, J. and N. Orion (2003a). Measuring student understanding of geological time. Science Education, 87, 708-731.
Dodick, J. and N. Orion (2003b). Cognitive factors affecting student understand of geological time. Journal of Research in Science Teaching, 40, 415-442.
Frodeman, R. (1995). Geological reasoning: Geology as an interpretive and historical science. Geological Society of America Bulletin, 107, 960-968.
Frodeman, R. (2003). Geo-Logic: Breaking Ground Between Philosophy and the Earth Sciences, State University of New York Press, 159 p.
Montangero, J. (1996). Understanding Changes in Time. Taylor and Francis, London.
Trend, R. D. (1998). An investigation in into understanding of geological time among 10- and 11-year-old children. International Journal of Science Education, 20, 973-988.
Trend, R. D. (2000). Conceptions of geological time among primary teacher trainees, with reference to their engagement with geosciences, history and science. International Journal of Science Education, 22, 539-555.
Trend, R. D. (2001). Deep time framework: A preliminary study of U.K. primary teachers' conceptions of geological time and perceptions of geoscience. Journal of Research in Science Teaching, 38, 191-221.
Trend, R. D. (2002). How Important is Deep Time? Unpublished m manuscript. Downloaded May 16, 2006, from www.cgu.org.tw/2004jga/dach/paper/07/07-O-01.doc<|endoftext|>
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Judenräte, (German: Jewish Councils) Jewish councils established in German-occupied Poland and eastern Europe during World War II to implement German policies and maintain order in the ghettos to which the Nazis confined the country’s Jewish population. Reinhard Heydrich, chief of Nazi Germany’s Gestapo, established the Judenräte (singular: Judenrat) by decree on September 21, 1939, three weeks after the German invasion of Poland. No aspect of Jewish behaviour during the Holocaust was more controversial than the conduct of the Judenräte.
The Judenräte were composed of up to 24 Jewish men, chosen from “remaining authoritative personalities and rabbis.” When the Judenräte were first established, the Jews did not know the ultimate intensions of the Germans toward them nor, according to most scholars, were the intentions of the Germans yet clear. Jewish leaders assumed that their responsibility was to provide for the needs of Jews, who they assumed would remain in the ghetto indefinitely. The Judenräte became a municipal authority providing sanitation, education, commerce, and food for their increasingly beleaguered community. With meager resources at their disposal, they struggled to meet the basic needs of starving ghetto residents and to make life bearable. Their German oppressors provided the basis of their power. At first unaware of their people’s fate, in time they understood their role in maintaining communities destined for annihilation.
The Judenräte relied on forms of taxation to support their activities. Jewish police forces were established to enforce Judenräte decrees and provide order in the ghetto. The individual Judenräte used different models of governance. In Warsaw, the largest of the ghettos, laissez-faire capitalism was the rule under Judenrat chairman Adam Czerniaków. Private enterprise continued for as long as possible. In Łódź, under the chairmanship of Mordecai Chaim Rumkowski, authority was more centralized. Commerce, trade, and all municipal services, including the distribution of food and housing, were tightly controlled.
The level and tenor of interaction between the Judenräte and the Germans differed ghetto by ghetto, leader by leader, and meeting by meeting. Some meetings with Nazi officials were courteous and might even appear friendly, others were harsh and threatening. Generally, the Germans would make demands of the Judenräte, who, in return, would beg for supplies and relief on behalf of their beleaguered populations.
Among the ghetto residents, the Judenräte often drew anger. Many viewed their role in enforcing German decrees and conditions as indistinguishable from the role of the Germans who had ordered them. This anger grew when conditions in the ghettos deteriorated under an intensified German campaign of deprivation.
Perhaps the defining test of the courage and the character of Judenrat leaders occurred when the Germans ordered lists drawn up indicating those to be protected by work permits and those to be deported to concentration camps. Judenrat members knew that deportation meant near-certain death. Thus, while the Judenräte used tactics such as bribery, postponement, importuning, and appeasement to secure work permits for as many residents as possible, only a specified number of work permits were available and decisions were required. This became especially wrenching when it came to children and the elderly, who were incapable of working.
In Łódź, Rumkowski cooperated with the deportations. He argued, “I must cut off the limbs to save the body itself. I must take the children because if not, others will be taken as well. The part that can be saved is much larger than the part that must be given away.” Similar decisions were made by Judenrat leaders in Vilna (now Vilnius, Lithuania) and Sosnowiec.
In Warsaw, Czerniaków committed suicide rather than participate in the deportation of children and the liquidation of the entire ghetto. “They have asked me to kill the children with my own hands,” he said in despair. To some Jews, Czerniaków’s suicide was an act of integrity. Others saw it as a sign of weakness and condemned his failure to call for resistance.
Leaders who openly refused to cooperate in delivering their own people to concentration camps soon paid with their lives. Dr. Joseph Parnas, first Judenrat leader of Lwów (now Lviv, Ukraine), refused an order to deport thousands of Jews and was shot, as were several other Judenrat leaders. Megalif, the leader of the Judenrat at Nieśvież (now Nesvizh, Belarus), marched to his death rather than participate in the deportation.
When the Germans ordered the final liquidation of the ghetto, there could be little pretense that many Jews could be saved. The Jewish resistance in several ghettos began to take control. While some Judenrat leaders, such as Dr. Elchanan Elkes of Kovno (now Kaunas, Lithuania) and his counterpart in Minsk (now in Belarus), Eliyahu Mushkin, cooperated with the underground and the resistance, most Judenrat leaders considered the resistance a threat to their efforts to maintain order and sustain the ghettos. As a consequence, Judenrat leaders and Jewish police were often the first to be assassinated by the Jewish resistance, even before direct battle with the Germans.
At the end of the war, virtually all Judenrat leaders, regardless of their level of accommodation with the Germans, were dead. Rumkowski, who perhaps tried the hardest to cooperate with the Germans to save “the body” of his ghetto, met the same fate as that body—death at an extermination camp.
In her book Eichmann in Jerusalem (1963), Hannah Arendt revived the controversy over the role of the Judenräte by implying that their complicity actually increased the Holocaust’s death toll. She wrote, “The whole truth was that if the Jewish people had really been unorganized and leaderless, there would have been chaos and plenty of misery but the total number of victims would hardly have been between four and a half and six million people.” Her work triggered a storm of controversy but also provoked research that yielded a more subtle understanding of the impossible task these leaders faced in confronting the Nazis’ overwhelming power and fervent, disciplined commitment to annihilate the Jewish people.
Learn More in these related Britannica articles:
Holocaust: German expansion and the formation of ghettos… ordered the establishment of the Judenräte (“Jewish Councils”), comprising up to 24 men—rabbis and Jewish leaders. Heydrich’s order made these councils personally responsible in “the literal sense of the term” for carrying out German orders. When the Nazis sealed the Warsaw Ghetto, the largest of German-occupied Poland’s 400 ghettos, in…
Ghetto, formerly a street, or quarter, of a city set apart as a legally enforced residence area for Jews. One of the earliest forced segregations of Jews was in Muslim Morocco when, in 1280, they were transferred to segregated quarters called millahs. In some Muslim countries, rigid ghetto systems were…
Reinhard Heydrich, Nazi German official who was Heinrich Himmler’s chief lieutenant in the Schutzstaffel (“Protective Echelon”), the paramilitary corps…
Nazi Party, political party of the mass movement known as National Socialism. Under the leadership of Adolf Hitler, the party came to power in Germany in 1933 and governed by totalitarian methods until 1945.…
Gestapo, the political police of Nazi Germany. The Gestapo ruthlessly eliminated opposition to the Nazis within Germany and its occupied territories and, in partnership with the Sicherheitsdienst (SD: “Security Service”), was responsible for the roundup of Jews throughout Europe for deportation to…
More About Judenräte1 reference found in Britannica articles
- establishment by Heydrich<|endoftext|>
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# Business Statistics I
Chapter 6
1-1
Calendar
10/18 10/23 10/25 10/30 Chapter 6 Finish Chapter 6 Homework Chapter 5 - DELAYED Homework Chapter 6 Review Chapter 4-6 Mid-Term 2 on Chapters 4-6
1-2
Continuous Variables
Chapter 5 introduced Discrete variables: a finite number of values or an infinite sequence of values
## Chapter 6 covers Continuous variables: any number in an interval or collection of intervals
1-3
Probability Distributions
Discrete Distributions provide the probability for any particular value Probability Density Functions, also denoted by f(x), do not directly provide the probability at that point, but the area under its curve provides the probability
1-4
## Uniform Probability Distribution
Similar to the discrete uniform probability distribution function: =
1
= 0
1-5
## Area as a Measure of Probability
Ex: Uniform flight times from 120 to 140 minutes - Fig 6.1 (p 235)
For the probability that the flight is between 120 and 130 minutes, thats half of the rectangle: HxW = (1/20) x (130-120) = 0.50
1-6
## Area as a Measure of Probability
Ex: Same flight time distribution Q: What is the probability that the flight takes 128 to 136 minutes?
1-7
## Area as a Measure of Probability
Just like the discrete probability distributions, two rules:
1-8
## Normal Probability Distribution
The normal curve, or bell-shaped curve, describes many naturally occurring events Notice that its symmetric
1-9
## Normal Probability Distribution
Formula is on p 239 Two parameters describe a normal distribution: = mean (see bottom p 239) = standard deviation (see p 240)
1-10
1-11
## Standard Normal Probability Distribution
=0 =1
1-12
Example 1
P 242: What is the probability that z 1.00 ? See the problem as a picture See the solution from the table See Table 1, pp 978-979
1-13
Example 2
P 243: What is the probability that -0.50 z 1.25 ? See the problem as a picture Find the solution from the tables f(-0.50) = 0.3085 f(1.25) = 0.8944
1-14
## Transformation of Data into a Standard Normal Curve
z =
= +
Then use Table 1, pp 978-979
1-15
Example 3
P 246: What is the probability that tires will last 40,000 miles? P(x 40,000) = ? Process: See the problem as a picture Transform to a standard normal dist Find the solution from the tables Solution: =
= 0.70
## Table p 979: P(0.70) = 0.7580 P(x 40,000) = 1 0.7580 = 0.2420
1-16
Example 4
P 247: How many miles represent 10% or less? P(x) 10% ? Process: See the problem as a picture Find the solution from the tables Transform to a standard normal dist Solution: Table p 978: P(x) 0.1000; x = -1.28 = + = 5,000 1.28 + 36,500 = 30,100
1-17
Chapter 6 Exclusions
1-18
Questions?
1-19<|endoftext|>
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# Activities to Teach Students to Multiply Three or More Mixed Numbers, Fractions, and Whole Numbers
Multiplication is an essential mathematical concept that students must master in order to progress to advanced math courses. Multiplying three or more mixed numbers, fractions, and whole numbers can be a daunting task for some students. However, classroom activities can provide hands-on experiences that help students learn how to multiply mixed numbers, fractions, and whole numbers efficiently.
Here are some effective activities to teach students to multiply three or more mixed numbers, fractions, and whole numbers:
1. The Pizza Party Game:
This game is an ideal way for students to learn about multiplication of fractions and mixed numbers. Draw a large pizza on a piece of paper and divide it into several equal parts. Give each student a fraction or a mixed number that they must multiply with their partner’s fraction or mixed number. Then, they can add up all the products to determine how many slices of pizza they’ve created. The team with the most slices wins the game.
2. The Arrays Race:
This activity requires students to create arrays to solve multiplication problems that involve fractions and whole numbers. Divide students into small groups and provide them with a set of multiplication questions, and ask them to create arrays with tiles or objects to solve the problems. The first group to complete the arrays wins the race.
3. The Multiplication Hopscotch:
This activity is a fun way to teach students multiplication of mixed numbers and whole numbers. Draw a hopscotch grid on the floor with various multiplication problems in each of the squares. Students must toss a beanbag onto the hopscotch grid, then hop on the numbered square where the beanbag landed and solve the multiplication problem in that square. They then have to hop on the correct answers to complete the hopscotch.
4. The Shopping Game:
This game is perfect for teaching students multiplication of whole numbers. Ask students to pretend they have a certain amount of money to go shopping. Provide a list of items with prices, and ask them to choose what they want to buy and calculate the total cost of their shopping. It will help them understand how multiplication applies to real-world situations.
5. The Group Round-Robin:
This activity is a perfect way to engage students in multiplication practice. Break students into small groups and ask them to solve a multiplication problem in their notebook. After a set amount of time, every group passes their notebook to the next group, and then repeats the process with the new problem. This encourages students to work together to solve problems quickly.
In conclusion, classroom activities can help students learn how to multiply mixed numbers, fractions, and whole numbers. Teachers can explore various activities to engage students in math practice that provide hands-on experiences and help them understand the concept of multiplication in a more tangible way. These activities encourage students to build confidence, discover the steps involved in the multiplication process, and engage with their peers while mastering mathematical concepts.<|endoftext|>
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# How do you find the arc length of the curve sqrt(4-x^2) from [-2,2]?
Jul 28, 2015
The answer turned out to be $2 \pi$ units.
The arc length is essentially the usage of the distance formula with a small, independent change in x and a small change in y, where y changes according to $f \left(x\right)$. Thus:
$D = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}}$
$= \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} / {\left(\Delta x\right)}^{2} \cdot {\left(\Delta x\right)}^{2}}$
$= \sqrt{1 + {\left(\Delta y\right)}^{2} / {\left(\Delta x\right)}^{2}} \Delta x$
Make it a sum, and you've got the arc length over an interval $\left[a , b\right]$.
$s = {\sum}_{a}^{b} \sqrt{1 + {\left(\Delta y\right)}^{2} / {\left(\Delta x\right)}^{2}} \Delta x$
Turn it into an integral expression to get:
$s = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$
So, we could take the derivative first to get:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{4 - {x}^{2}}} \cdot - 2 x = \frac{- x}{\sqrt{4 - {x}^{2}}}$
Then, let's square it and plug it in.
${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = {x}^{2} / \left(4 - {x}^{2}\right)$
$\to s = {\int}_{a}^{b} \sqrt{1 + {x}^{2} / \left(4 - {x}^{2}\right)} \mathrm{dx}$
Cross-multiply:
$= {\int}_{a}^{b} \sqrt{\frac{4 - {x}^{2} + {x}^{2}}{4 - {x}^{2}}} \mathrm{dx}$
$= {\int}_{a}^{b} \sqrt{\frac{4}{4 - {x}^{2}}} \mathrm{dx}$
$= 2 {\int}_{a}^{b} \frac{1}{\sqrt{4 - {x}^{2}}} \mathrm{dx}$
And we've got ourselves a u-substitution, if you know that $\frac{d}{\mathrm{dx}} \left[\arcsin \left(x\right)\right] = \frac{1}{\sqrt{1 - {x}^{2}}}$.
$= 2 {\int}_{a}^{b} \frac{1}{\sqrt{4}} \cdot \frac{1}{\sqrt{1 - {x}^{2} / 4}} \mathrm{dx}$
$= {\int}_{a}^{b} \frac{1}{\sqrt{1 - {\left(\frac{x}{2}\right)}^{2}}} \mathrm{dx}$
Let $u = \frac{x}{2}$ and you get:
$\mathrm{dx} = 2 \mathrm{du}$
$= 2 {\int}_{a}^{b} \frac{1}{\sqrt{1 - {u}^{2}}} \mathrm{du}$
$= \left[2 \arcsin \left(\frac{x}{2}\right)\right] {|}_{- 2}^{2}$
$= 2 \arcsin \left(\frac{2}{2}\right) - 2 \arcsin \left(\frac{- 2}{2}\right)$
$= 2 \arcsin \left(1\right) - 2 \arcsin \left(- 1\right)$
Let:
$y = \arcsin \left(1\right) \to \sin y = 1 \to y = \frac{\pi}{2}$
$y = \arcsin \left(- 1\right) \to \sin y = - 1 \to y = \frac{3 \pi}{2}$
$\to 2 \left(\frac{\pi}{2}\right) - 2 \left(\frac{3 \pi}{2}\right) \to 2 \left(\frac{\pi}{2}\right) - 2 \left(\frac{- \pi}{2}\right)$
(so that our answer is positive)
$= \pi + \pi = \textcolor{b l u e}{2 \pi \text{ u}}$<|endoftext|>
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# Test of Mathematics Solution Subjective 65 – Minimum Value of Quadratic
This is a Test of Mathematics Solution Subjective 65 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
## Problem
Show that for all real x, the expression ${ax^2}$ + bx + C ( where a, b, c are real constants with a > 0), has the minimum value ${\frac{4ac - b^2}{4a}}$ . Also find the value of x for which this minimum value is attained.
## solution:
f (x) ${ax^2}$ + bx + c
Now minimum derivative = 0 & 2nd order derivative > 0.
${\frac{df(x)}{dx}}$ = 2ax + b
Or ${\frac{d^2f(x)}{dx^2}}$ =2a
Now 2a> so 2nd order derivative > 0 so ${\frac{d^2f(x)}{dx^2}}$ = 2.
So minimum occurs when
${\frac{df(x)}{d(x)}}$ = 0 or 2ax + b = 0
or 2ax = -b
or x = ${\frac{-b}{2a}}$ (ans)
At x = ${\frac{-b}{2a}}$
${ax^2}$ + bx + c
= ${a\times {\frac{b^4}{4a^2}}}$ + ${b\times {\frac{-b}{2a}}}$ + c
= ${\frac{4ac-b^2}{4a}}$ (proved)<|endoftext|>
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### Partial Orderings
```Partial Orderings:
Selected Exercises
Partial Order
• Let R be a relation on A.
• R is a partial order when it is:
– Reflexive
– Antisymmetric
– Transitive.
2
Exercise 10
a
b
c
d
Is this directed graph a partial order?
3
Exercise 10 Solution
a
b
c
d
Is this directed graph a partial order?
Is it reflexive?
Is it antisymmetric?
Is it transitive?
4
Exercise 20
Draw the Hasse diagram for the “≥” relation on
{ 0, 1, 2, 3, 4, 5 }.
5
Exercise 20 Solution
Draw the Hasse diagram for the “≥” relation on
{ 0, 1, 2, 3, 4, 5 }.
In a Hasse diagram:
1.
1
Direction is implied (up), hence omitted
I.e., we use edges instead of arcs.
2.
0
Edges implied by transitivity are omitted
2
3
4
5
6
Exercise 40
a) Show that there is exactly 1 greatest element of a poset, if
such an element exists.
7
Exercise 40
a) There is exactly 1 greatest element of a poset, if such an
element exists.
Proof:
a) By contradiction: Assume x & y are distinct greatest elements.
b) x y
(Step a: y is a greatest element)
c) y x
(Step a: x is a greatest element)
d) x = y.
(Step b & c & antisymmetry)
8
Exercise 40 continued
b) Show that there is exactly 1 least element, if such an
element exists.
Proof: Similar to part a)
9
•
Let S be a set with n elements.
•
Consider the poset ( P( S ), ).
•
What does the Hasse diagram look like when:
1. Let |S| = 0
2. Let |S| = 1
3. Let |S| = 2
4. Let |S| = 3
5. Let |S| = 4
6. Let |S| = n
10
|S| = 0; | P( S ) | = 20
Hasse diagram: a 0-cube: Just a single point.
Ø
11
|S| = 1; | P( S ) | = 21
Represent each subset by a 1-bit string:
0 represents the empty set
1 represents the set with 1 element.
Hasse diagram: a 1-cube: Just a single edge.
1
0
12
|S| = 2; | P( S ) | = 22
Represent each subset by a 2-bit string: b1 b2
Hasse diagram: a 2-cube: Just a square.
11
10
01
00
13
|S| = 3; | P( S ) | = 23
Represent each subset by a 3-bit string: b1 b2 b3
Hasse diagram: a 3-cube.
111
110
101
011
100
010
001
000
14
|S| = 4; | P( S ) | = 24
Represent each subset by a 4-bit string: b1 b2 b3 b4
Hasse diagram: a 4-cube.
15
1111
1101
1110
1100
1000
1010
1011
1001
0111
0101
0110
0100
0011
0001
0010
0000Cappello 2011
16
1111
1101
1110
1100
1000
1010
1011
1001
0101
0110
0100
Sub-diagram
For elements 1, 2, 3
0111
0011
0001
0010
0000Cappello 2011
17
1111
1101
1110
1100
1000
1010
1011
1001
0101
0110
0100
Sub-diagram
For elements 2, 3, 4
0111
0011
0001
0010
0000Cappello 2011
18
1111
1101
1110
1100
1000
1010
1011
1001
0101
0110
0100
Sub-diagram
For elements 1, 2, 4
0111
0011
0001
0010
0000Cappello 2011
19
In the Connection Machine, 216 processors were
connected as a 16-cube.
20
Topological Sorting
• Total ordering T is compatible with partial
ordering P when a, b ( a ≤P b a ≤T b ).
• Element a is minimal when there is no
element b with b ≤ a.
21
Topological Sorting
• Problem (Topological Sort)
– Input: A finite partial ordering ( S, ≤ ).
– Output: A compatible total ordering.
– Algorithm:
While ( S ≠ )
output ( S.removeAMinimalElement() );
• What are good data structures for
finding a minimal element?
22
End 8.6
23
Exercise 30
Let ( S, ) be a poset, and let x, y S.
Notation: x < y means x y and x ≠ y.
Definitions:
• y covers x if x < y and z S ( x < z < y ).
• The covering relation of (S, ) = { ( x, y ) | y covers x }.
Show:
( x, y ) is in the covering relation of finite poset ( S, )
x is lower than y and an edge joins x & y in the Hasse diagram.
A poset’s covering relation defines the edge set of its Hasse diagram.
24
Exercise 30 Solution
x is lower than y and an edge joins x & y in the Hasse diagram
(x, y) is in the covering relation of finite poset (S, ).
Proof:
1.
Assume x is lower than y and an edge joins x & y in the Hasse
diagram.
2.
x < y.
(Defn. of Hasse diagrams)
3.
(An edge joins x to y) z S ( x < z < y ).
(Defn. of Hasse
diagrams)
4.
An edge joins x to y.
(Step 1)
5.
z S ( x < z < y ).
(Step 3 & 4 & modus ponens)
6.
Therefore, x is covered by y. (Step 2 & 5, defn. of covers)
25
Exercise 30 Solution
( x, y ) is in the covering relation of finite poset ( S, )
x is lower than y and an edge joins x & y in the Hasse diagram.
Proof:
1.
Assume ( x, y ) is in the covering relation of finite poset ( S, ).
2.
x<y
3.
x is lower than y in diagram. (Step 2 & Defn. of Hasse diagram)
4.
z ( x < z < y ).
5.
An edge joins x to y.
(Defn of y covers x)
(Defn. of y covers x)
(Step 2 & 4 & Defn. of Hasse diagram)
26
50
Defn. If (S, ) is a poset & every 2 elements are
comparable, S is totally ordered.
Defn. x is the least upper bound of A if x is an upper bound
that is less than every other upper bound of A.
Defn. x is the greatest lower bound of A if x is a lower bound
that is greater than every other lower bound of A.
Defn. A poset in which every 2 elements have a least upper
bound & a greatest lower bound is a lattice.
Show that every totally ordered set is a lattice.
27
50 continued
Prove S is totally ordered S is a lattice.
Proof
1. Assume S is totally ordered.
2. a, b (a b b a).
(Defn. of total order)
3. Select 2 arbitrary elements a, b S.
4. Assume without loss of generality a b.
5. a is the greatest lower bound of {a, b}.
(Step 3)
6. b is the least upper bound of {a, b}.
(Step 3)
7. S is a lattice.
(Step 4 & 5, Defn. of lattice)
28
60
Defn. a is maximal in poset (S, ) if b S ( a < b ).
Show:
Poset (S, ) is finite & nonempty a S, a is maximal.
Proof:
1. Assume poset (S, ) is finite & nonempty.
2. Let a S.
(Step 1: S )
3. for ( max := a; S ; S := S – {b} )
1. Let b S.
2. If max < b, max := b.
4. max is maximal.
5. Step 3 terminates. (S is finite; smaller each iteration)<|endoftext|>
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# Mathematics Paper: Simultaneous Equation Study Guide
If you haven’t read our detailed article about the tips for Mathematics paper, visit that page here first before diving into the specific topics! The article talks about specific tips to tackle specific papers of the Mathematics examinations. In addition to that, it also includes a downloadable summary cheat sheet for your revision!
Now, this article will serve as the ultimate guide to help you master the simultaneous equations for your mathematics examinations.
## Methods for simultaneous equations
### 1 – Elimination method
Watch a youtube video to help you better understand the method better!
### 2 – Substitution method
Watch a youtube video on the method of substitution here!
Here’s Mr Eddie Woo, solving simultaneous equations for his students.
## When do I use the elimination or substitution method in mathematics paper?
The answer to this question is – if the coefficient of any variable is 1, which means you can easily solve for it in terms of the other variable, then substitution is a very good bet.
If all the coefficients are anything other than 1, then you can use elimination, but only if the equations can be added together to make one of the variables disappear. You can make use of common multiples to get rid of one of the variables to solve for the other.
## Additional practice for simultaneous equations
Remember to pace yourself because knowing how to manage your time is as crucial as getting these answers correct.
Additionally, you can also use this website to help you find the answers by keying in the equations.
## Tips for doing simultaneous equations in mathematics paper
1 – label your equations when using the substitution method
2 – for the elimination method, it might be easier to know the common multiple of the two equations
3 – always be careful and pay extra attention to the operations, and make sure your pluses become minuses when you bring them over from one side of the equation to the other
In a nutshell, simultaneous equations in mathematics paper are easy to get a hang of when you keep practising. As you continue to practice more, you will build muscle memory around it and these methods will come to you naturally.<|endoftext|>
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# WORD PROBLEMS USING LAW OF SINES AND COSINES
## About "Word Problems Using Law of Sines and Cosines"
Word Problems Using Law of Sines and Cosines :
Here we are going to see some practical problems using the concept of law of sines and cosines.
## Word Problems Using Law of Sines and Cosines - Examples
Question 1 :
A researcher wants to determine the width of a pond from east to west, which cannot be done by actual measurement. From a point P, he finds the distance to the eastern-most point of the pond to be 8 km, while the distance to the western most point from P to be 6 km. If the angle between the two lines of sight is 60°, find the width of the pond.
Solution :
By finding the missing side, we may find the width of pond.
AP = a, PB = b and AB = c
cos C = (a2 + b2 - c2) / 2ab
cos 60 = (62 + 82 - c2) / 2(6)(8)
1/2 = (36 + 64- c2) / 96
48 = 100 - c2
c2 = 100 - 48
c2 = 52
c = √52
= 2 √13 km
Hence the width of pond is √13 km.
Question 2 :
Two Navy helicopters A and B are flying over the Bay of Bengal at same altitude from the sea level to search a missing boat. Pilots of both the helicopters sight the boat at the same time while they are apart 10 km from each other. If the distance of the boat from A is 6 km and if the line segment AB subtends 60° at the boat, find the distance of the boat from B.
Solution :
To find the missing side, we have to use the cosine formula.
cos C = (a2 + b2 - c2)/2ab
Here we use the formula for cos C, because we know the length of b and c.
cos 60 = (a2 + 62 - 102)/2(6)(10)
1/2 = (a2 -64)/2a(6)
1 = a2 - 64/6a
a2 - 64 = 6a
a2 - 6a - 64 = 0
= (-b ± √b2 - 4ac)/2a
= (6 ± √(36 - 4(1)(-64))/2(1)
= (6 ± √292)/2
= (6 ± 2√73)/2
= 3 ± √73
Hence the distance from the helicopter B to boat is 3 + √73 km.
Question 3 :
A straight tunnel is to be made through a mountain. A surveyor observes the two extremities A and B of the tunnel to be built from a point P in front of the mountain. If AP = 3km, BP = 5 km and ∠APB = 120, then find the length of the tunnel to be built
Solution :
Given that :
AP = 3 km = b, BP = 5 km = a and ∠APB = 120
cos C = (a2 + b2 - c2)/2ab
cos 120 = (52 + 32 - c2)/2(5)(3)
cos 120 = cos (90+30) = sin 30 = -1/2
-1/2 = (25 + 9 - c2)/30
-15 = 34 - c2
c = 49, c = 7
After having gone through the stuff given above, we hope that the students would have understood, "Word Problems Using Law of Sines and Cosines"
Apart from the stuff given in "Word Problems Using Law of Sines and Cosines", if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
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Remainder when 2 power 256 is divided by 17
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Sum of all three digit numbers divisible by 6
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Sum of all three digit numbers formed using 1, 3, 4
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Monday, May 4, 2015
Circumcenter and 3 circumcircles
Given an acute triangle $ABC$ and its circumcircle is centered at $O$ and has radius $R$.
$AO$ meets the circumcircle of $\triangle OBC$ at $X$, $BO$ meets the circumcircle of $\triangle OAC$ at $Y$, and $CO$ meets the circumcircle of $\triangle OAB$ at $Z$.
Show that $$AX + BY + CZ \geq 9R$$
Determine when equality occurs.
Solution
Let $AX$ meets $BC$ at $D$. And let $OO'$ be the diameter of the circumcircle of $\triangle OBC$, and let $M$ be the intersection of $OO'$ and $BC$.
$$\frac{OX}{OO'} = \frac{OM}{OD}$$ $$OX.OD = OO'.OM = OC^2 = R^2$$ $$AX = AO + OX = R + \frac{R^2}{OD} = \frac{R(R + OD)}{OD} = R\frac{AD}{OD}$$
If we define $E$ as the intersection of $BY$ and $AC$, and $F$ as the intersection of $CZ$ and $AB$, then similarly: $$BY = R \frac{BE}{OE}$$ $$CZ = R \frac{CF}{OF}$$
So: $$AX + BY + CZ = R(\frac{AD}{OD} + \frac{BE}{OE} + \frac{CF}{OF})$$ $$\geq \frac{9R}{\frac{OD}{AD} + \frac{OE}{BE} + \frac{OF}{CF}}$$ But $OD / AD = A_{\triangle OBC} / A_{\triangle ABC}$ and so on, so: $$\frac{OD}{AD} + \frac{OE}{BE} + \frac{OF}{CF} = 1$$ $$AX + BY + CZ \geq 9R$$
and equality happens when the area of $\triangle OBC$ is exactly $\triangle ABC / 3$. And this happens only when $O$ is the centroid of $\triangle ABC$, which means $ABC$ is an equilateral.<|endoftext|>
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Breaking News
15 Percent Of 2500
15 Percent Of 2500. Replacing the given values in formula (a) we have: 15 / 100 = part / 2500 cross multiply: What is 15 percent of 500? Amount saved = original price x discount in percent / 100.
1) what is 15% of 2500? 100 it's 100% (one hundred percent). How to calculate 15 percent of 2500?
How to calculate 15 percent of 2500?
15 / 100 = part / 2500 cross multiply: 15 percent of 500 is 75. Now we can solve our fraction by writing it as an equation:
If The Number Refers To A Unit Of.
15 / 100 = part / 2500 cross multiply: 15 x 2500 = 100 x part, or. Amount saved = original price x discount in percent / 100. ( x y ) × 100 copy x refers to the small number whose value we want to calculate from the large number.
15% Of 2500 Is 375.
15 percent of 2500 is 375.
Kesimpulan dari 15 Percent Of 2500.
Now we can solve our fraction by writing it as an equation: So, amount saved = 2500 x 15 / 100. ( x 100 ) × y copy.<|endoftext|>
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# Addition Using Parallelogram Law Of Vectors
This post categorized under Vector and posted on July 10th, 2019.
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Statement of Parallelogram Law If two vectors acting simultaneously at a point can be represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram pvectoring through that point.1. Triangle Law and Parallelogram Law yield the same resultant. 2. Each method is suited for different situations more of which you will learn in future chapters. 3. But at any time both are equally valid methods to find the sum of two vectors.Vector addition is commutative in nature i.e. if C A B then C B A. Similarly if we have to subtract both the vectors using the triangle law then we simply reverse the direction of any vector and add it to other one as shown. Now we can mathematically represent this as C A B. Parallelogram Law of Vector Addition This law is also very similar to triangle law of vector addition. Consider the two
Triangle Law of Vector Addition. A vector ( vecAB ) in simple words means the displacement from point A to point B. Now imagine a scenario where a boy moves from point A Parallelogram law of vector addition Questions and Answers . Note vectors are shown in bold. scalars are shown in normal type. The diagram above shows two vectors 31.03.2016 Addition of vectors Triangle and Parallelogram law of vectors As vectors have both magnitude and direction they cannot be added by the method of ordinary algebra. Vectors can be added graphically or geometrically.
Addition of Vectors basically found their origin from the Triangular Law of Vector Addition. The triangle law of vectors basically is a process that allows one to take two vectors draw them proportional to each other connect them head to tail then draw the resultant vector as a result of the third side that is missing.<|endoftext|>
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# How do you sketch the graph of y=x^2+0.5 and describe the transformation?
Apr 27, 2017
Pick some easy points for $x$.
#### Explanation:
Let's choose $x = - 3 , - 2 , - 1 , 0 , 1 , 2 , 3$
With these points we get the following
$y \left(- 3\right) = {\left(- 3\right)}^{2} + 0.5 = 9 + 0.5 = 9.5$
$y \left(- 2\right) = {\left(- 2\right)}^{2} + 0.5 = 4 + 0.5 = 4.5$
$y \left(- 1\right) = {\left(- 1\right)}^{2} + 0.5 = 1 + 0.5 = 1.5$
$y \left(0\right) = {\left(0\right)}^{2} + 0.5 = 0 + 0.5 = 0.5$
$y \left(1\right) = {\left(1\right)}^{2} + 0.5 = 1 + 0.5 = 1.5$
$y \left(2\right) = {\left(2\right)}^{2} + 0.5 = 4 + 0.5 = 4.5$
$y \left(3\right) = {\left(3\right)}^{2} + 0.5 = 9 + 0.5 = 9.5$
We can plot the following points to get
graph{x^2+0.5 [-11.21, 11.29, -0.75, 10.5]}
Since 0.5 is added after $x$ is "modified," it will move the function $y = {x}^{2}$ up by half a unit.<|endoftext|>
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A team of scientists have managed to capture some powerful images that depict aurorae occurring on a brown dwarf.
A display of natural lights, which also occur on Earth known as the northern or southern lights, were spotted in the atmosphere of a small stellar object that is located outside of our solar system. The authors, led by Gregg Hallinan of the California Institute of Technology, speculate it is likely a brown dwarf, which has been dubbed LSR J1835.
Aurorae occur on all magnetised planets within our solar system and are powered by strong currents in the magnetic field that surrounds a planet – otherwise known as the magnetosphere – which allow precipitation of energetic electrons in regions of the upper atmosphere, the report in Nature explains.
However, with massive stars, such as the sun, the phenomenon are powered by the same occurrence but in the lower atmosphere.
With the LSR J1835, though, the scientists found the aurorae was emanating from its upper atmosphere, with the findings suggesting such natural lights may be a regular occurrence of large-scale magnetospheres that can produce much stronger luminosities than those spotted in the solar system we reside in.<|endoftext|>
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The intensity of land-use influences the speed of regeneration in tropical rainforests, says new research. Tropical rainforests are a priority for biodiversity conservation; they are hotspots of endemism but also some of the most threatened global habitats. The Atlantic Forest stands out among tropical rainforests, hosting an estimated 8,000 species of endemic plants and more than 650 endemic vertebrates. However, only around 11 percent of these forests now remain. The quality of what remains is also a concern: 32 to 40 percent of remnants are small areas of secondary forest. Although the restoration of these secondary forests would go a long way toward mitigating the loss of forest cover and biodiversity elsewhere, it is not always possible to recover richness, diversity and floristic composition. Land-use history can make these changes irreversible.
Published in mongabay.com’s open access journal Tropical Conversation Science, the study surveyed the effects of different land-use histories on the restoration process of Atlantic rainforest areas, in the Michelin Ecological Reserve, Brazil.
“Disturbance history influenced the richness, diversity and the floristic composition of the disturbed forest studied,” wrote Rocha-Santos and Talora, lead authors of the study.
The Atlantic Forest outside of Rio de Janeiro. Photo by: Rhett A. Butler.
The highest values of richness and diversity were found in areas with a history of low-impact logging. This area recorded 93 tree species from 35 families. Areas that suffered slash-and-burn agriculture returned 77 species from 38 families, while those with a history of high-impact logging had the lowest richness and diversity amongst the three land-use histories: its tree community structure comprised 75 species from 32 families.
“The area in recovery after low-impact logging showed the highest richness and diversity, indicating that this was the disturbance that allowed the fastest recover…on the other hand, the area subjected to high-impact logging showed the lowest richness and diversity, suggesting that the higher the intensity of exploitation, the longer the time required for the recovery of this forest’s characteristics,” said the authors.
While, the diversity of the slash-and-burn areas were similar to high-impact logging sites, previous research has shown that slash-and-burn areas only present low richness in the early years (5-20 years), and that after 40 to 60 years of recovery, these forests are similar in composition and diversity to those of less disturbed tropical forests. Studies in tropical forests also indicate that the greatest species dominance (where few species account for most of the coverage) is found in areas of an early successional stage. Areas subjected to selective logging, regardless of the intensity, appear to be in a more advanced stage of recovery than areas subjected to slash-and-burn. This was also recorded in African tropical forests. Enrichment actions may therefore be necessary to accelerate forest recovery in such forest remnants.
“Such actions may assist in the recovery of ecological processes such as nutrient cycling, biotic pollination and dispersal, which may have been corrupted by previous land-use”, concluded the authors.
The fact that only minor intervention is necessary in areas where forest cover is already established significantly reduces restoration costs. Political will aside, we have no excuses when it comes to the recovery of Atlantic rainforests.
CITATION: Rocha-Santos, L. and Talora, D. C. 2012. Recovery of Atlantic Rainforest areas altered by distinct land-use histories in northeastern Brazil. Tropical Conservation Science Vol. 5(4):475-494 .
(11/26/2012) On paper, the northern muriquis (Brachyteles hypoxanthus) look like a conservation comeback story. Three decades ago, only 60 of the gentle, tree-dwelling primates lived in a fragment of the Atlantic Forest along the eastern coast of Brazil. Now there are more than 300. But numbers don’t tell the whole story, according to anthropologist Karen Strier and theoretical ecologist Anthony Ives of the University of Wisconsin, Madison. The pair analyzed 28 years of data on the demographics of the muriquis, one of the longest studies of its kind. They found surprising patterns about birth and death rates, sex ratios, and even how often the monkeys venture out of their trees. These findings raise questions about the muriquis’ long-term survival and how best to protect them, the scientists wrote in the Sept 17 issue of PLoS ONE.
(10/31/2012) If you suffer from acute arachnophobia, this is the perfect Halloween discovery for you: a spider expert has discovered nine new species of arboreal (tree-dwelling) tarantulas in the Brazil. Although tarantula diversity is highest in the Amazon rainforest, the new species are all found in lesser-known Brazilian ecosystems like the Atlantic Forest, of which less than 7 percent remains, and the cerrado, a massive savannah that is being rapidly lost to agriculture and cattle ranching.
(08/15/2012) When the Portuguese first arrived on the shores of what is now Brazil, a massive forest waited for them. Not the Amazon, but the Atlantic Forest, stretching for over 1.2 million kilometers. Here jaguars, the continent’s apex predator, stalked peccaries, while tapirs waded in rivers and giant anteaters unearthed termites mounds. Here, also, the Tupi people numbered around a million people. Now, almost all of this gone: 93 percent of the Atlantic Forest has been converted to agriculture, pasture, and cities, the bulk of it lost since the 1940s. The Tupi people are largely vanished due to slavery and disease, and, according to a new study in the open access journal PLoS ONE, so are many of the forest’s megafauna, from jaguars to giant anteaters.
(06/06/2012) Deforestation of Brazil’s Mata Atlântica — a forest ecosystem more threatened than the Amazon rainforest — fell to 133 square kilometers between 2010 and 2011, down about 14.7 percent from the annual average between 2008 and 2010, reports Brazil’s National Institute for Space Research (INPE) and Fundação SOS Mata Atlântica.<|endoftext|>
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### Home > PC3 > Chapter 9 > Lesson 9.1.3 > Problem9-54
9-54.
Perform each of the following matrix operations without using a calculator. If the operation is impossible, explain why.
$\begin{bmatrix} {1} & { -4 } & { 0} \\ { -2 } & { 5 } & { 3 } \end{bmatrix}$ $\begin{bmatrix} {-1} & { 0 } \\ { 2 } & { 4 } \\ { -5 } & { -3 }\end{bmatrix}$ $=\begin{bmatrix} {1(-1)+(-4)(2)+0(-5)} & { 1(0)+(-4)(4)+0(-3) } \\ { (-2)(-1)+5(2)+3(-5) } & { (-2)(0)+5(4)+3(-3) } \end{bmatrix}$ $=\begin{bmatrix} {-9} & { -16 } \\ { -3 } & { 11 } \end{bmatrix}$ $2 \times 3$ $3 \times 2$ $2 \times 2$
1. $\left[ \begin{array} { l l l } { 4 } & { 9 } & { 2 } \\ { 6 } & { 0 } & { 5 } \end{array} \right] \left[ \begin{array} { l } { a } \\ { b } \\ { c } \end{array} \right]$
Multiplying a $2 \times 3$ matrix by a $3 \times 1$ matrix results in a $2 \times 1$ matrix.
1. $2 \left[ \begin{array} { r r r } { 4 } & { 9 } & { 2 } \\ { 6 } & { 0 } & { - 5 } \end{array} \right] + \left[ \begin{array} { r r r } { 1 } & { 0 } & { - 3 } \\ { 0 } & { 4 } & { 1 } \end{array} \right]$
$\begin{bmatrix}8&18&4\\12&0&-10\end{bmatrix}+\begin{bmatrix}1&0&-3\\0&4&1\end{bmatrix}$
1. $\left[ \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right] \left[ \begin{array} { l l } { e } & { f } \end{array} \right]$
Is it possible to multiply a $2 \times 2$ matrix by a $1 \times 2$ matrix?
1. $\left[ \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right] + \left[ \begin{array} { l l } { e } & { f } \end{array} \right]$
Is it possible to add two matrices with different dimensions?
1. $\begin{vmatrix} 10 & 1 \\ -9 & 2 \\ \notag \end{vmatrix}$
$\begin{vmatrix} a & b \\ c & d \\ \notag \end{vmatrix} = ad-cb$
1. $\begin{vmatrix} 4 & 2 & 5 \\ 10 & 3 & -6 \\ \notag \end{vmatrix}$
See the hint in part (e). Is this a square matrix?<|endoftext|>
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What is bullying?
Bullying is defined in the Anti-Bullying Procedures for Primary and Post-Primary Schools as "unwanted negative behaviour, verbal, psychological or physical conducted by an individual or group against another person (or persons) and which is repeated over time". More information on the definition of bullying, types of bullying and the impact and indicators of bullying is available in Sections 2 and 3 of the procedures for schools.
Who is responsible for dealing with bullying in schools?
Responsibility for tackling bullying in schools falls to the level of the individual school.
New Anti-Bullying Procedures for Primary and Post-Primary Schools were published in September 2013. The New Procedures and the associated Department Circular 0045/2013 apply to all recognised primary and post-primary schools and to centres for education (as defined in the Education Act 1998) which are attended by pupils under the age of 18 years.
School authorities and school personnel are required to adhere to these procedures in dealing with allegations and incidents of bullying.
The purpose of these procedures is to give direction and guidance to school authorities and school personnel in preventing and tackling school-based bullying behaviour amongst its pupils and in dealing with any negative impact within school of bullying behaviour that occurs elsewhere.
The new procedures require all schools to formally adopt and implement an anti-bullying policy that fully complies with the requirements of the procedures.
All Boards of Management are now required to immediately commence the necessary arrangements for developing and formally adopting an anti-bullying policy which fully complies with the requirements of these procedures.
It is expected that this will be completed by each school as early as possible in the 2013/14 school year but in any event by no later than the end of the second term of the 2013/14 school year.
The school's anti-bullying policy must then be made available to school personnel, published on the school website (or where none exists, be otherwise readily accessible to parents and pupils on request) and provided to the Parents' Association (where one exists).
The Board of Management of each school in developing its anti-bullying policy must formulate the policy in co-operation with both teaching and non-teaching school staff under the leadership of the Principal and in consultation with parents and pupils.
Word versions of the templates which must be used by all schools are provided via the following links:
Where can I get more information about bullying?
The Anti-Bullying Procedures set out the requirements on schools in relation to preventing and dealing with bullying behaviour. There are also a number of bodies/groups that provide information and assistance on the topic of bullying including the Anti-Bullying Centre at Dublin City University and the National Parents Council.
Action Plan on Bullying
The Action Plan on Bullying, which was published in January 2013, sets out the Department of Education and Skills’ approach to tackling bullying and promoting an anti-bullying culture in schools. The twelve actions in the plan focus on support for schools, teacher training, research and awareness raising and aim to ensure that all forms of bullying are addressed. Implementation of the actions in the Plan is ongoing and good progress has been made in all areas:
- New Anti-Bullying Procedures, which were published in September 2013, are currently being implemented by all 4,000 primary and post primary schools in the country. Training materials for parents, teachers and Boards of Management are being developed and rolled out. More information about free anti-bullying training for parents can be at www.npc.ie - National Parents Council Primary.
- The Department supports a number of anti-bullying awareness raising initiatives including the Europe-wide Safer-Internet day and the Stand Up! Against Homophobic and Transphobic bullying campaign, organised by BeLonG To Youth Services. Funding is also provided to the internet safety initiative, Webwise, which has launched a number of resources aimed at tackling cyberbullying.
- "Being LGBT in School" A Resource for Post-Primary Schools to Prevent homophobic and Transphobic Bullying and Support LGBT Students was developed by the Gay and Lesbian EqualityNetwork (GLEN) as part of the implementation of the Action Plan on Bullying. It will support schools in the implementation of the Department’s Anti-Bullying Procedures.
- A new national anti-bullying website, www.tacklebullying.ie, was launched in November 2015. The website will provide a single point of information and support for young people, parents and teachers affected by bullying.
- In addition, two pieces of research suggested in the Plan around children with special needs and social media have been published.<|endoftext|>
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# What is the n th term for an AP with a common difference of -4.5?
Oct 30, 2016
The sequence will be: ${T}_{n} = a - 4.5 n + 4.5$
#### Explanation:
We know the general equation for nth term of an A.P.series is
${T}_{n} = a + \left(n - 1\right) d$
where $a$ is the first term of the series and $d$ is the common difference.
Now for the common difference ($d = - 4.5$) the sequence will be:
${T}_{n} = a + \left(n - 1\right) \left(- 4.5\right)$
Simplifying gives: ${T}_{n} = a - 4.5 n + 4.5$
The summation of all the terms will be:
${S}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$ where $d$ is the common difference.
In your question $d = - 4.5$.<|endoftext|>
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- Diagnostic Testingdropdown
- Anatomy of the Eye
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- Dry Eye
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- Ocular Migraines
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- Variations in Visiondropdown
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WHAT IS DRY EYE?
When you blink, a film of tears spreads over the eye, making the surface of the eye smooth and clear. Without this tear film, good vision would not be possible.
Sometimes people don't produce enough tears or the right quality of tears to keep their eyes healthy and comfortable. This condition is known as dry eye.
The tear film consists of three layers:
- An oily layer;
- A watery layer;
- A layer of mucus.
Each layer has its own purpose. The oily layer, produced by the meibomian glands, forms the outermost surface of the tear film. Its main purpose is to smooth the tear surface and reduce evaporation of tears.
The middle watery layer makes up most of what we ordinarily think of as tears. This layer, produced by the lacrimal glands in the eyelids, cleanses the eye and washes away foreign particles or irritants.
The inner layer consists of mucus produced by the conjunctiva. Mucus allows the watery layer to spread evenly over the surface of the eye and helps the eye remain moist. Without mucus, tears would not stick to the eye. Normally, the eye constantly bathes itself in tears. By producing tears at a slow and steady rate, the eye stays moist and comfortable.
The eye uses two different methods to produce tears. It can make tears at a slow, steady rate to maintain normal eye lubrication. It can also produce a lot of tears in response to eye irritation or emotion. When a foreign body or dryness irritates the eye, or when a person cries, excessive tearing occurs.
It may not sound logical that dry eye would cause excess tearing, but think of it as the eye's response to discomfort. If the tears responsible for maintaining lubrication do not keep the eye wet enough, the eye becomes irritated. Eye irritation prompts the gland that makes tears (called the lacrimal gland) to release a large volume of tears, overwhelming the tear drainage system. These excess tears then overflow from your eye.<|endoftext|>
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Circle A has a radius of 2 and a center at (5 ,2 ). Circle B has a radius of 5 and a center at (3 ,4 ). If circle B is translated by <-2 ,1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?
Nov 5, 2016
circles overlap.
Explanation:
What we have to do here is $\textcolor{b l u e}{\text{compare}}$ the distance ( d) between the centres of the circles to the $\textcolor{b l u e}{\text{sum of the radii}}$
• If sum of radii > d , then circles overlap
• If sum of radii < d , then no overlap
Before calculating d, we must find the ' new' centre of circle B under the given translation, which does not change the shape of the circle only it's position.
Under a translation $\left(\begin{matrix}- 2 \\ 1\end{matrix}\right)$
$\left(3 , 4\right) \to \left(3 - 2 , 4 + 1\right) \to \left(1 , 5\right) \leftarrow \text{ new centre of B}$
To calculate d, use the $\textcolor{b l u e}{\text{distance formula}}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$
The 2 points here are (5 ,2) and (1 ,5)
let $\left({x}_{1} , {y}_{1}\right) = \left(5 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 5\right)$
$d = \sqrt{{\left(1 - 5\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$<|endoftext|>
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Pre-Algebra, Algebra 1, and Algebra 2 all require you to master new math skills. Do you find solving equations and word problems difficult in Algebra class? Are the exponents, proportions, and variables of Algebra keeping you up at night? Intercepts, functions, and expressions can be confusing to most Algebra students, but a qualified tutor can clear it all up! Our Algebra tutors are experts in math and specialize in helping students like you understand Algebra. If you are worried about an upcoming Algebra test or fear not passing your Algebra class for the term, getting an Algebra tutor will make all the difference.
Pre-algebra - The goal of Pre-algebra is to develop fluency with rational numbers and proportional relationships. Students will: extend their elementary skills and begin to learn algebra concepts that serve as a transition into formal Algebra and Geometry; learn to think flexibly about relationships among fractions, decimals, and percents; learn to recognize and generate equivalent expressions and solve single-variable equations and inequalities; investigate and explore mathematical ideas and develop multiple strategies for analyzing complex situations; analyze situations verbally, numerically, graphically, and symbolically; and apply mathematical skills and make meaningful connections to life's experiences.
Algebra I - The main goal of Algebra is to develop fluency in working with linear equations. Students will: extend their experiences with tables, graphs, and equations and solve linear equations and inequalities and systems of linear equations and inequalities; extend their knowledge of the number system to include irrational numbers; generate equivalent expressions and use formulas; simplify polynomials and begin to study quadratic relationships; and use technology and models to investigate and explore mathematical ideas and relationships and develop multiple strategies for analyzing complex situations.
Algebra II - A primary goal of Algebra II is for students to conceptualize, analyze, and identify relationships among functions. Students will: develop proficiency in analyzing and solving quadratic functions using complex numbers; investigate and make conjectures about absolute value, radical, exponential, logarithmic and sine and cosine functions algebraically, numerically, and graphically, with and without technology; extend their algebraic skills to compute with rational expressions and rational exponents; work with and build an understanding of complex numbers and systems of equations and inequalities; analyze statistical data and apply concepts of probability using permutations and combinations; and use technology such as graphing calculators.
College Algebra – Topics for this course include basic concepts of algebra; linear, quadratic, rational, radical, logarithmic, exponential, and absolute value equations; equations reducible to quadratic form; linear, polynomial, rational, and absolute value inequalities, and complex number system; graphs of linear, polynomial, exponential, logarithmic, rational, and absolute value functions; conic sections; inverse functions; operations and compositions of functions; systems of equations; sequences and series; and the binomial theorem.
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Because Jacksonville is the third most populous city on the East Coast, after New York City and Philadelphia, we have come to appreciate the needs of all the students within Duval County. Here, we provide not only our services to private individuals, but we work hand-in-hand with the local school board to provide free services to those children most in need. Jacksonville’s diverse population of graduate students, high-tech and industrial workers, retirees, and of course an abundant number of teachers provides us with an almost unlimited number of highly qualified and experienced tutors to meet the needs of all learners. We have been providing tutoring services to Jacksonville students for many, many successful years. Our reputation as a premium service is evident in the hundreds of testimonials we have received from parents, students, and schools in Jacksonville and the surrounding communities.
Our Tutoring Service
We offer our clients only the very best selection of tutors. When you request a tutor for a certain subject, you get what you ask for. Our tutors are expertly matched to your individual needs based on the criteria you provide to us. We will provide you with the degrees, credentials, and certifications that each selected tutor holds. Equally important is the peace of mind we offer you knowing that each of our tutors has been cleared by a nation-wide criminal background check, a sexual predator check, and social security verification. We want you to have the same confidence in our tutors as we do.<|endoftext|>
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 4/12 = 1/3 ≅ 0.3333333
Spelled result in words is one third.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• In the cafeteria
There are 18 students in Jacob’s homeroom. Six students bring their lunch to school. The rest eat lunch in the cafeteria. In simplest form, what fraction of students eat lunch in the cafeteria?
• In fractions
An ant climbs 2/5 of the pole on the first hour and climbs 1/4 of the pole on the next hour. What part of the pole does the ant climb in two hours?
• Fraction to decimal
Write the fraction 3/22 as a decimal.
• Denominator 2
Denominator of a fraction is 5 and numerator is 7. Write the fraction .
• Pipe
Steel pipe has a length 2.5 meters. About how many decimetres is 1/3 less than 4/8 of this steel pipe?
• Teacher
Teacher Rem bought 360 pieces of cupcakes for the outreach program of their school. 5/9 of the cupcakes were chocolate flavor and 1/4 wete pandan flavor and the rest were a vanilla flavor. How much more pandan flavor cupcakes than vanilla flavor?
• Larry 2
Larry spends half of his workday teaching piano lessons. If he sees 6 students, each for the same amount of time, what fraction of his workday is spent with each student?
• Spending
Peter spends 1/5 of his earnings on his rent, and he saves 2/7. What fraction of his earnings is left?
• Kelly
Kelly and Dan are collecting clothes for a clothing drive. Dan collected 1/2 as many clothes as Kelly did. If Kelly collected 3 bags of clothes, how many bags of clothes did Dan collect?
Rhea answered 5/11 in the questions correctly and Precious answered 7/11 of it correctly. If each problem is worth the same amount, who got the higher score?
• Paper clips
Mrs. Bright is organizing her office supplies. There are 5 open boxes of paper clips in her desk drawer. Each box has 1/2 of the paper clips remaining. How many boxes of paper clips are left?
• Scouts 4
4/7 of the students in a school are boys. If 3/8 of the boys are scouts, how many scouts are there in a school of 1878 students?
• Eq-frac
Solve the following equation with fractions: h + 1/3 =5/3<|endoftext|>
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ELA I and II
Ohio’s Learning Standards for English Language Arts
The Standards describe what students should understand and be able to do. Standards do not dictate curriculum or teaching methods.
Appendix A: Research Supporting Key Elements of the Standards
Appendix A contains supplementary material on reading, writing, speaking and listening, and language that helps educators understand the details of each strand. It also includes information describing the parts and importance of text complexity, along with annotated text excerpts that explain how grade-level texts represent these parts of text complexity.
Appendix B: Text Exemplars and Sample Performance Tasks
Appendix B offers a guide for educators reviewing samples of high quality text that meet grade-level standards. Districts and educators should not use this as a reading list; rather, they should use it as a guide to the quality of texts they should choose at each grade level.
Appendix C: Samples of Student Writing
Appendix C offers a guide for educators examining student responses that meets the grade level standards for Writing.
The model curriculum provides clarity to the standards, the foundation for aligned assessments, and guidelines to assist educators in implementing the standards. It also provides instructional strategies and resources.
- Model Curriculum
- Reading Literature - pp. 1-7
- Reading Informational - pp. 11-16
- Writing - pp. 20-21, 23-27
- Language - pp. 34-40
Students may log in as a guest to access any of the practice tests. Half-length practice tests are available for each of the high school end of course exams in English language arts. Practice test items allow students to become familiar with the online test environment by showcasing the different item types, features, and functionality available to students during online testing.
Two writing rubrics are available for grades 6 through high school: informative/explanatory and argumentation. Each rubric describes the score point characteristics across three domains. Text in red in these documents denotes changes that will apply beginning with the spring 2017 administration.
Test blueprints serve as a guide for test construction and provide an outline of the content and skills to be measured on the test. They contain information about individual tests, including the number of test items and the number of points on the test, and they show how the learning standards are grouped in order to report the test results
The test blueprint for ELA I and II is grouped together; the same standards will be assessed at both levels, but the complexity and the texts will be higher on the ELA II assessment.
Performance Level Descriptors
The Performance Level Descriptors (PLDs) are the link between Ohio’s Learning Standards and performance standards. They were developed by Ohio educators and other content experts to illustrate the typical demonstration of the learning for each of the five performance levels: Limited, Basic, Proficient, Accelerated, and Advanced. Teachers can review the PLDs for each course to match each descriptive statement with its course standards. Teachers can then review local curriculum materials to see if students have opportunities to demonstrate deeper conceptual
Below are other states with released online test items aligned to standards similar to Ohio’s Learning Standards for English language arts. With the exception of PARCC, all these states use the vendor AIR. Teachers can use these tests for additional practice with the standards and for students to become familiar with the online test environment by showcasing the different item types, features, and functionality available to students during online testing. Since educators outside of Ohio approved these released test items, the items may not reflect the level of depth, conceptual understanding, or rigor of Ohio’s State Tests.
PearlTrees is a collaborative website to store thoughts, links, videos, and notes. Click through these pearls to find more information and resources on ELA assessments.
A guide used to assist curriculum developers and teacher leaders with choosing texts that are aligned to Ohio’s Learning Standards.
Last Modified: 5/21/2019 1:24:19 PM<|endoftext|>
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Safety-certified tools Tools for Automotive Applications C-STAT Static analysis C-RUN Runtime analysis Debugging and trace probes
Anyone who has ever read a piece of C source code has seen them—the preprocessor directives. For example, you can find include directives (#include) at the beginning of most source files. The preprocessor is a system that rewrites the source before the actual compiler sees it. Clearly, this is a very powerful tool—the downside is that you could accidentally shoot yourself in the foot.
In this article we will introduce the preprocessor and covered the basics, including object- and function-like macros, the include directive, conditional compilation, and end with the two special directives #error and #pragma.
The most straight-forward preprocessor directive is #include. When the preprocessor finds this directive it simply opens the file specified and inserts the content of it, as though the content of the file would have been written at the location of the directive.
It can take two forms, for example:
The first is used to include standard headers like stdio.h. The latter is for your own application-specific headers.
One of most useful features of the preprocessor is to allow the user to define macros, which simply is an identifier that is mapped to a piece of source code. Whenever the preprocessor finds the macro in the application source code it replaces the macro with the definition.
Basically, there are two types of macros, object-like macros and function-like macros, the difference is that function-like macros have parameters.
By convention, macro names are written using upper-case only. The only exception is when a macro is used to replace something that should have been a function but is implemented using a macro for the sake of efficiency.
The directive #define can be used to define a macro. In the following example, we define NUMBER_OF_PLAYERS to the constant "2". This is an object-like macro.
#define NUMBER_OF_PLAYERS 2
Here, the function-like macro PRINT_PLAYER is mapped a more complex piece of source code.
#define PRINT_PLAYER(no) printf("Player %d is named %s", no, names[no])
The definition of a macro should, technically, be specified on a single source line. Fortunately, the C standard allows you to end a line with a backslash and continue on the next physical line. For example:
#define A_MACRO_THAT_DOES_SOMETHING_TEN_TIMES \
for (i = 0; i < 10; ++i)\
In addition to the macros that a user can define, the C standard specifies a number of predefined and library-provided macros that could be used. For example, the macro __FILE__ contains the name of the current source file, as a string.
If you ever would like to make a macro name undefined you can use the #undef directive.
Object-like macros can be used to replace an identifier in the source code with some kind of replacement source code.
Typically, macros can be used to declare a constant that could be configured in one location. Also, constants could be used to make the source code more readable, even if the value is not intended to change. For example:
#define SQUARE_ROOT_OF_TWO 1.4142135623730950488016887
double convert_side_to_diagonal(double x)
return x * SQUARE_ROOT_OF_TWO;
double convert_diagonal_to_side(double x)
return x / SQUARE_ROOT_OF_TWO;
Preprocessor macros could be used for very weird things, since all that the preprocessor does is to replace an identifier with an arbitrary piece of source code. For example, the following is legal code (although, you will probably have to answer to your boss if you ever try to write something like this):
#define BLA_BLA );
int test(int x)
printf("%d", x BLA_BLA
Function-like macros are macros that take parameters. When you use them they look like a function call. A function-like macro could look like the following:
#define SEND(x) output_array[output_index++] = x
When the preprocessor finds a function-like macro in your application source code it will replace it with the definition. The parameters of the macro will be inserted into the resulting source code at the location of the formal parameter variables.
So, if you write the following:
Then the compiler will see:
output_array[output_index++] = 10
In the second part of this article we will revisit function-like macros and discuss some of the traps that are easy to fall into.
One of the most powerful features of the preprocessor is the so-called conditional compilation—this means that portions of the code could be excluded in the actual compilation under the certain conditions.
This means that your source could contain special code for, say, the ARM processor. Using conditional compilation, this code could be ignored when compiling for all other processors.
The preprocessor directives #ifdef, #ifndef, #if, #elif, and #else are used to control the source code. The#ifdef (#ifndef) directive includes a section if a preprocessor symbol is defined (undefined). For example:
The #if directive can handle any type of integer and logical expression, for example:
#if (NUMBER_OF_PROCESSES > 1) && (LOCKING == TRUE)
The #elif directive works like a combined #else and #if.
#if and #elif can use the special operator defined to check if a symbol is defined. This is useful in combination with complex tests, for example:
#if defined(VERSION) && (VERSION > 2)
In part two of this article we will revisit conditional compilation and discuss whether you should prefer #if:s or #ifdef:s in your application.
One typical use for conditional compilation is to ensure that the content of include files are only seen once. This will not only speed up the compilation, but also ensures that the compiler will not issue an error (e.g. for redeclaration of a struct) if the header file is included twice.
An include guard typically looks like the following:
/* The content of the header file goes here. */
Clearly, the first time the header file is included the symbol MYHEADER_H is not defined and the content is included in the compilation. The second time the header file is read the symbol is defined and the content is excluded.
The #error directive can be used to generate a compiler error message. This is useful when performing consistency checks, for example:
#if USE_COLORS && !HAVE_DISPLAY
#error "You cannot use colors unless you have a display"
Another preprocessor directive is #pragma. This directive allows the programmer to control the behavior of the compiler and gives compiler vendors the opportunity to implement extensions to the C language.
The #pragma directice is not covered further in this article since it has little to do with the main task of the preprocessor.
This article is written by Anders Lindgren, Development Engineer at IAR Systems.<|endoftext|>
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Taeniasis is an infection caused by the tapeworm, a type of parasite. Parasites are small organisms that attach themselves to other living things in order to survive. The living things the parasites attach to are called hosts.
Parasites can be found in contaminated food and water. If you consume contaminated food or drinks, you may contract a parasite that can live and sometimes grow and reproduce inside your body.
- Taenia saginata (beef tapeworm)
- Taenia solium (pork tapeworm)
Most people who have taeniasis don’t have any symptoms. If signs and symptoms are present, they may include:
Some people with taeniasis may also experience irritation in the perianal area, which is the area around the anus. Worm segments or eggs being expelled in the stool cause this irritation.
People often become aware that they have a tapeworm when they see worm segments or eggs in their stool.
Infections can take between 8 and 14 weeks to develop.
You can develop taeniasis by eating raw or undercooked beef or pork. Contaminated food can contain tapeworm eggs or larvae that grow in your intestines when eaten.
Fully cooking beef or pork will destroy the larvae so that they can’t live in your body.
The tapeworm can grow up to 12 feet in length. It can live in the intestines for years without being discovered. Tapeworms have segments along their bodies. Each of these segments can produce eggs. As the tapeworm matures, these eggs will be passed out of the body in the stool.
Poor hygiene can also cause the spread of taeniasis. Once tapeworm larvae are in human stool, they can be spread through contact with the stool. You should wash your hands properly to help prevent the spread of the infection.
- Eastern Europe and Russia
- East Africa
- sub-Saharan Africa
- Latin America
- parts of Asia, including China, Indonesia, and South Korea
According to the
Taeniasis is more likely to develop in people who have weakened immune systems and aren’t able to fight off infections. Your immune system can weaken due to:
See your doctor if you see worm segments or eggs in your stool. Your doctor will ask you about your health history and recent travel outside of the United States. Doctors will often be able to make a diagnosis of taeniasis based on the symptoms.
Taeniasis is typically treated with medications prescribed by your doctor. Medications for the treatment of taeniasis include praziquantel (Biltricide) and albendazole (Albenza).
Both drugs are antihelmintics, which means that they kill parasitic worms and their eggs. In most cases, these medications are provided in a single dose. They can take a few weeks to fully clear an infection. The tapeworm will be excreted as waste.
Most cases of this infection go away with treatment. Medications prescribed for this condition are typically effective and will cure the infection.
In rare cases, serious complications from the infection can occur. Tapeworms may block your intestines. This may require surgery to correct.
In other cases, a pork tapeworm may travel to other parts of your body such as the heart, eye, or brain. This condition is called cysticercosis. Cysticercosis can cause other health problems such as seizures or infections in the nervous system.
The most effective way to prevent taeniasis is to cook food thoroughly. This means cooking meat to a temperature above 140°F (60°F) for five minutes or more. Measure the meat temperature with a cooking thermometer.
After cooking meat, allow it to stand for three minutes before cutting it. This can help destroy any parasites that may be in the meat. Learn more about meat safety.
In the United States, laws requiring the inspection of animals and meat help reduce the chance that tapeworms will be spread.
Proper hand hygiene is also important for preventing the spread of this disease. Always wash your hands after using the bathroom and teach your children to do the same.
Also, drink bottled water if you live in or travel to an area where water must be treated.<|endoftext|>
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Fibroblasts are one of the main components of connective tissue, which is the main reason scientists typically exploit them for experiments. A collaborative team of scientists from the University of Pennsylvania, Boston University, and the New Jersey Institute of Technology have invented a way to reprogram fibroblasts without going through a pluripotent stage.
The senior author of this study, Xiaowei Xu, associate professor of pathology and laboratory medicine at the University of Pennsylvania School of Medicine, said, “Through direct reprogramming, we do not have to go through the pluripotent stem cell stage, but directly convert fibroblasts to melanocytes . So these cells do not have tumorigenicity” (the ability to cause tumors).
Melanocytes are found in the skin and they are responsible for the pigment in our skin. They are in the uppermost layer of the skin, known as the epidermis, and produce melanin, a brown pigment that helps screen against the harmful effects of UV light.
Turning a fibroblast into a melanocytes might seem trivial for a stem cell scientist; just reprogram the fibroblast into an induced pluripotent stem cells and then differentiate it into a melanocytes. However, this procedure utilized direct reprogramming, in which the fibroblast was converted into a melanocytes without traversing through the pluripotent stage. The difficultly with direct reprogramming is finding the right cocktail of genes and/or growth factors that will accomplish the deed.
Xu and his colleagues began their search by examining the genes that are specific to melanocytes. They found 10 different transcription factors that are important for melanocytes development. Next they screened these ten genes for their ability to convert a fibroblast into a melanocytes. They found that of the ten melanocytes-specific genes, three of them, Sox10, MITF, and PAX3 could do the job effectively. They called this gene combination “SMP3.”
When Xu and others tested SMP3 on mouse embryonic fibroblasts, they quickly expressed melanocytes-specific genes. When Xu’s group used SMP3 on human fetal dermal cells, once again, the cells rapidly differentiated into melanocytes. Xu and his team referred to these cells as hi-Mel, which is short for human, induced melanocytes.
When hi-Mel were grown in culture they produced melanin a plenty. When they were implanted into the skin of pigmentless mice, once again they rose to the challenge and made a great deal of pigment. Thus hi-Mel express the same genes as melanocytes and they behave for all intents and purposes as melanocytes.
Xu and his colleagues think that their procedure might be able to treat human patients with a condition called vitiligo in which the skin has patches that are devoid of pigment.
Another potential use of this technology is a way to effectively study melanoma, one of the most dangerous skin cancers known to human medicine. My good friend and SAU colleague died over a year ago from melanoma and having better ways to treat this monster would have been marvelous for Charlie, and his family, who miss him dearly. By generating melanocytes from the fibroblasts of melanoma patients, they can “screen not only to find why these patients easily develop melanoma, but possibly use their cells to screen for small compounds that can prevent melanoma from happening.”.
Also, because so the body contains so many fibroblasts in the first place, this reprogramming technique is well-suited for other cell-based treatments.<|endoftext|>
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# Mensuration - Important Concepts and Notes
By Abhinav Gupta|Updated : August 24th, 2021
Mensuration - Important Concepts and Notes
Important Formulas on Quadrilateral and Circle
Rectangle
A four-sided shape that is made up of two pairs of parallel lines and that has four right angles; especially: a shape in which one pair of lines is longer than the other pair.
The diagonals of a rectangle bisect each other and are equal.
Area of rectangle = length x breadth = l x b
OR Area of rectangle = if one sides (l) and diagonal (d) are given.
OR Area of rectangle = if perimeter (P) and diagonal (d) are given.
Perimeter (P) of rectangle = 2 (length + breadth) = 2 (l + b).
OR Perimeter of rectangle = if one side (l) and diagonal (d) are given.
Square
A four-sided shape that is made up of four straight sides that are the same length and that has four right angles.
The diagonals of a square are equal and bisect each other at 900.
Area (a) of a square
Perimeter (P) of a square
= 4a, i.e. 4 x side
Length (d) of the diagonal of a square
Circle
A circle is a path travelled by a point which moves in such a way that its distance from a fixed point remains constant.
The fixed point is known as the centre and the fixed distance is called the radius.
(a) Circumference or perimeter of circle =
where r is radius and d is the diameter of the circle
(b) Area of circle
is circumference
(c) The radius of circle =
Sector :
A sector is a figure enclosed by two radii and an arc lying between them.
here AOB is a sector
length of arc
Area of Sector
Ring or Circular Path:
area=π(R2-r2)
Perimeter=2π(R+r)
Rhombus
Rhombus is a quadrilateral whose all sides are equal.
The diagonals of a rhombus bisect each other at 900
Area (a) of a rhombus
= a × h, i.e. base × height
Product of its diagonals
since d2
since d2
Perimeter (P) of a rhombus
= 4a, i.e. 4 x side
Where d1 and d2 are two-diagonals.
Side (a) of a rhombus
Parallelogram
A quadrilateral in which opposite sides are equal and parallel is called a parallelogram. The diagonals of a parallelogram bisect each other.
Area (a) of a parallelogram = base × altitude corresponding to the base = b × h
Area (a) of the parallelogram
where a and b are adjacent sides, d is the length of the diagonal connecting the ends of the two sides and
In a parallelogram, the sum of the squares of the diagonals = 2
(the sum of the squares of the two adjacent sides).
i.e.,
Perimeter (P) of a parallelogram
= 2 (a+b),
Where a and b are adjacent sides of the parallelogram.
Trapezium (Trapezoid)
A trapezoid is a 2-dimensional geometric figure with four sides, at least one set of which are parallel. The parallel sides are called the bases, while the other sides are called the legs. The term ‘trapezium,’ from which we got our word trapezoid has been in use in the English language since the 1500s and is from the Latin meaning ‘little table.’
Area (a) of a trapezium
1/2 x (sum of parallel sides) x perpendicular
Distance between the parallel sides
i.e.,
Where, l = b – a if b > a = a – b if a > b
And
Height (h) of the trapezium
Pathways Running across the middle of a rectangle:
X is the width of the path
Area of path= (l+b-x)x
perimeter= 2(l+b-2x)
Outer Pathways:
Area=(l+b+2x)2x
Perimeter=4(l+b+2x)
Inner Pathways:
Area=(l+b-2x)2x
Perimeter=4(l+b-2x)
Some useful Short trick:
• If there is a change of X% in defining dimensions of the 2-d figure then its perimeter will also change by X%
• If all the sides of a quadrilateral are changed by X% then its diagonal will also change by X%.
• The area of the largest triangle that can be inscribed in a semi-circle of radius r is r2.
• The number of revolution made by a circular wheel of radius r in travelling distance d is given by
number of revolution =d/2πr
• If the length and breadth of the rectangle are increased by x% and y% then the area of the rectangle will be increased by.
(x+y+xy/100)%
• If the length and breadth of a rectangle are decreased by x% and y% respectively then the area of the rectangle will decrease by:
(x+y-xy/100)%
• If the length of a rectangle is increased by x%, then its breadth will have to be decreased by (100x/100+x)% in order to maintain the same area of the rectangle.
• If each of the defining dimensions or sides of any 2-D figure is changed by x% its area changes by
x(2+x/100)%
where x=positive if increase and negative if decreases.
Thanks
Prep Smart. Stay Safe. Go BYJU'S Exam Prep.
## Important Mensuration (3D) Formulas
Cube
• s = side
• Volume: V = s^3
• Lateral surface area = 4a2
• Surface Area: S = 6s^2
• Diagonal (d) = s√3
Cuboid
• Volume of cuboid: length x breadth x width
• Total surface area = 2 ( lb + bh + hl)
Right Circular Cylinder
• Volume of Cylinder = π r^2 h
• Lateral Surface Area (LSA or CSA) = 2π r h
• Total Surface Area = TSA = 2 π r (r + h)
Hollow-Cylinder
* Volume of Hollow Cylinder = π(pie) h(r1(Square) - r2(Square))
Right Circular Cone
• l^2 = r^2 + h^2
• Volume of cone = 1/3 π r^2 h
• Curved surface area: CSA= π r l
• Total surface area = TSA = πr(r + l )
Important relation between radius, height and slant height of similar cone.
Frustum of a Cone
• h = height, s = slant height
• Volume: V = π/ 3 (r^2 + rR + R^2)h
• Surface Area: S = πs(R + r) + πr^2 + πR^2
Sphere
• Volume: V = 4/3 πr^3
• Surface Area: S = 4π^2
Hemisphere
• Volume-Hemisphere = 2/3 π r^3
• Curved surface area(CSA) = 2 π r^2
• Total surface area = TSA = 3 π r^2
Quarter-Sphere
Let 'r' is the radius of given diagram. You have to imagine this diagram, this is 1/4th part of Sphere.
Prism
• Volume = Base area x height
• Lateral Surface area = perimeter of the base x height
Pyramid
• Volume of a right pyramid = (1/3) × area of the base × height.
• Area of the lateral faces of a right pyramid = (1/2) × perimeter of the base x slant height.
• Area of whole surface of a right pyramid = area of the lateral faces + area of the base.
Important:
1.From a solid cylinder no. of maximum solid cone of same height and radius as cylinder are 3.
2.From a solid sphere, no. of maximum solid cone having height and radius equal can be made are 4.
3.From a solid hemisphere, no. of maximum solid cone having height and radius equal can be made are 2.
## CLICK HERE FOR PDF OF ALL IMPORTANT FORMULAS TO REMEMBER FOR Reasoning
You can avail of Online Classroom Program for all AE & JE Exams:
## Online Classroom Program for AE & JE Exams (12+ Structured LIVE Courses and 160+ Mock Tests)
You can avail of BYJU'S Exam Prep Test Series specially designed for all AE & JE Exams:
## BYJU'S Exam Prep Test Series AE & JE (160+ Mock Tests)
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Sahi Prep Hai to Life Set Hai !!!<|endoftext|>
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alesterp
2021-08-19
To simplify the expression, $3\left(4x-3\right)=6\left(2x+1\right)-15$.
okomgcae
Calculation:
We have given as equation as $3\left(4x-3\right)=6\left(2x+1\right)-15$
We will use various properties of expression for example distributive property, associative property, commutative property as required.
Distributive property: $a\left(b+c\right)=ab+ac$
Associative property: $a+\left(b+c\right)=\left(a+b\right)+c$ and $a\left(bc\right)=\left(ab\right)c=\left(ac\right)b$
Commutative property:
$a+b=b+a$ and $ab=ba$
Now let us simplify given equation.
$3\left(4x-3\right)=6\left(2x+1\right)-15$
$⇒12x-9=12x+6$
$-15$ [Opening parentheses using distributive property]
$⇒12x-9x=12x-9$ [Simplifying]
Above equation will be true for all real number.
Hence, we can say that solution for given equation will be all real numbers.
Do you have a similar question?<|endoftext|>
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A subtropical estuary* of clear waters, the bay supports rich ecological communities and a diverse variety of fish and wildlife. Specific habitat is provided for each species living in the bay, including species that migrate in and out of the bay daily or seasonally. If their habitat becomes unsuitable or disappears, animals and plants lose their home and people lose the benefits of their presence.
Coastal wetland habitat and benthic** habitat, including seagrasses, are most important to aquatic life. Intertidal habitat and adjacent upland areas, including coastal scrub and tropical hardwood hammock, are also critical to the bay ecosystem and are affected by changes to watershed hydrology and development.
The benthic communities of the bay help maintain water clarity, stabilize sediments, and provide food and shelter for fish, shrimp, lobster, crab and other species — especially in sensitive early juvenile stages. When these communities are damaged or eliminated, they cannot serve these vital roles and the increased turbidity that results can prevent their return. More than 40 percent of the bottom area in north Biscayne Bay has been altered by spoil displacement and dredging; turbidity problems persist in the northern part of the bay partly as a result. Shoreline bulkheads, which deflect waves and boat wakes and cause re-suspension of bottom sediments, also contribute to turbidity problems in the northern bay, as does increasingly heavy boat traffic. Protection and restoration of vital benthic communities are needed.
In the past, freshwater and brackish marsh habitats bordered the bay’s western shoreline and barrier islands. Over the last century, the bay’s western shorelines and eastern barrier islands have undergone extensive loss of marsh and mangroves. The western bay was heavily affected by the construction of 19 water management canals that drained wetlands and now release water in pulses to prevent intermittent coastal flooding and facilitate agriculture. The result is a highly modified delivery of freshwater to the bay, rapidly fluctuating salinity regimes adjacent to canals, and loss of estuarine habitat.
Diversion of freshwater flow away from coastal wetlands degrades their ability to support wildlife and the bay ecosystem by preventing a natural estuarine salinity gradient from wetlands to the bay. Development of coastal wetlands and inappropriate development of adjacent uplands are other major threats to coastal wetlands that prevent restoration of more natural timing and distribution of freshwater flow to the bay. Restoration of coastal wetlands is key to restoring natural habitat and estuarine fish communities and wildlife in Biscayne Bay.
* An estuary is the wide part of a waterway where fresh water and salt water mix.
**Benthic refers to habitat on the bottom of a sea, lake or bay.
To help preserve and restore bay habitats, the Biscayne Bay Partnership Initiative has recommended these actions:
* The Florida legislature should provide funding to support continuation of major habitat restoration projects and development and implementation of a process for comprehensive planning and oversight of bay habitat restoration.
* Management of Biscayne Bay must focus on protecting and enhancing natural areas and open land, particularly in south Miami-Dade, as well as on restoring and improving environmental value and infrastructure in the built areas.
* The Florida legislature should provide adequate funding for acquisition of lands currently identified as important in providing protective buffer or water redistribution functions and should accelerate identification of additional lands needed for these functions. More about land acquisition.
* The Florida legislature should provide funding to mark channels, seagrass beds and coral areas and provide maintenance for markers and signage.
* The Biscayne Bay Aquatic Preserve Act is inviolate and should never be weakened.<|endoftext|>
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Women have been suffering discrimination in societies for years. They are considered inferior to men. In two different societies women’s have different roles, in Buddhist societies women are thought o have evolved socially than in Brahmanic societies . Women lead very different lives in Brahmanic and Buddhist societies. Buddhism was created by Buddha(565–485 BCE). The Buddha wanted equality. Instead of trying to overcome the strict caste system , he created a second world where these social barriers no longer existed. Hinduism believes in the worships of gods and goddesses. The origin of Hinduism is unknown. There are multiple symbolism of women in Hinduism for instance women can symbolize the goddesses. the roles of epic heroines as behavior models for women, and the roles of women worshipers or their virtues. Were women better off in Buddhist societies than in Hindu societies.
Women in Buddhism have a more positive than negative role. Buddha has encouraged Dharma(the teaching of the Buddha) to both men and women and he made sure that everyone understands that the differences between men and women are meaningless. During the first few years of Buddhism women were treated the same as they are in Brahmanic societies. They were dominated by men. Buddha created a new way of living but he was still living in a society that has been treating women as inferior to men. After the passing of Buddha’s father, his stepmother asked him if she would be allowed to join the order and he denied her for reasons unknown. She shaved her head and dressed in a yellow robe and she went to him with 500 or more women who also wanted to join. Ananda, Buddha’s cousin saw them at the gate and after realizing that they wanted to join, he went to Buddha and asked if he denied their access because women could not reach the highest spiritual goal. Buddha assured Ananda that women could reach enlightenment. This was the first time in history that religion believed women could be enlightened. Then Buddha agreed to let his step mother, Maha Pajapati to join with the condition that she had to agree the 8 Garudhammas(heavy rules Buddha added to the bhikkhuni rules). Maha Pajapati was responsible the bhikkhuni sangha(orders of the nun), which allowed women to participate in religious practices and rituals.
Although, women were allowed to join the Buddhist orders, it was quite difficult for them to enter the orders in the early stages of Buddhism. Both men and women had to go through a thorough medical examination but only women had to have their sexual organs examined. There were only a few lower class members and most of the women that entered were well educated. Illiterates were prohibited from joining. Women also were discriminated in these societies as well.
There were three general attitudes toward women in Buddhism. The first was that being reincarnated as a woman is negative karma from your past life. The second is that the Buddha was a male and that means being...<|endoftext|>
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Thread: Integrate cos^2 t using integration by parts
1. Integrate cos^2 t using integration by parts
How do we do this, please? Thanks.
Find $\int cos(t)^2\ dt$.
Let $u=cos(t), du=-sin(t)dt, v=sin(t), dv=cos(t)dt$.
$\int cos^2(t) dt \\=\int u\ dv \\= uv-\int v\ du \\= cos(t)sin(t)-\int -sin(t)^2\ dt \\= cos(t)sin(t)+\int sin(t)^2\ dt.$
Let $u=sin(t), du=cos(t)dt, v=-cos(t), dv=sin(t)dt$.
$cos(t)sin(t)+\int sin(t)^2\ dt \\=cos(t)sin(t)+\int u\ dv \\=cos(t)sin(t) + \left[uv-\int v\ du \right] \\=cos(t)sin(t) + \left[-cos(t)sin(t) + \int cos(t)^2\ dt \right] \\=\int cos(t)^2\ dt.$
Thus proving, once and for all, that
$\int cos(t)^2\ dt=\int cos(t)^2\ dt$
2. Re: Integrate cos^2 t using integration by parts
You can say $\displaystyle \sin^2 t = \frac{1-\cos 2t}{2}$ might make it easier.
3. Re: Integrate cos^2 t using integration by parts
How do we do this, please? Thanks.
Find $\int cos(t)^2\ dt$.
Let $u=cos(t), du=-sin(t)dt, v=sin(t), dv=cos(t)dt$.
$\int cos^2(t) dt \\=\int u\ dv \\= uv-\int v\ du \\= cos(t)sin(t)-\int -sin(t)^2\ dt \\= cos(t)sin(t)+\int sin(t)^2\ dt.$
Let $u=sin(t), du=cos(t)dt, v=-cos(t), dv=sin(t)dt$.
$cos(t)sin(t)+\int sin(t)^2\ dt \\=cos(t)sin(t)+\int u\ dv \\=cos(t)sin(t) + \left[uv-\int v\ du \right] \\=cos(t)sin(t) + \left[-cos(t)sin(t) + \int cos(t)^2\ dt \right] \\=\int cos(t)^2\ dt.$
Thus proving, once and for all, that
$\int cos(t)^2\ dt=\int cos(t)^2\ dt$
If you MUST use integration by parts (which is the most tedious method, when, as Pickslides says, the double angle formula for cosine simplifies the integrand greatly)...
Let \displaystyle \begin{align*} u = \cos{x} \implies du = -\sin{x}\,dx \end{align*} and \displaystyle \begin{align*} dv = \cos{x}\,dx \implies v = \sin{x} \end{align*}, then
\displaystyle \begin{align*} \int{\cos^2{x}\,dx} &= \sin{x}\cos{x} - \int{-\sin^2{x}\,dx} \\ \int{\cos^2{x}\,dx} &= \sin{x}\cos{x} + \int{\sin^2{x}\,dx} \\ \int{\cos^2{x}\,dx} &= \sin{x}\cos{x} + \int{1 - \cos^2{x}\,dx} \\ \int{\cos^2{x}\,dx} &= \sin{x}\cos{x} + \int{1\,dx} - \int{\cos^2{x}\,dx} \\ 2\int{\cos^2{x}\,dx} &= \sin{x}\cos{x} + x \\ \int{\cos^2{x}\,dx} &= \frac{1}{2}\sin{x}\cos{x} + \frac{1}{2}x + C \end{align*}
5. Re: Integrate cos^2 t using integration by parts
Hi (slow response on my part, sorry). Thanks for all your replies.
Yes, we had to use integration by parts. Integral of cos^2 was actually part of a larger trig substitution problem (which I'm not going to type up right now) where the teacher wanted them to "unsubstitute" using a triangle, and we couldn't figure out how to do that using the identity of pickslides response.
In other words, the whole problem was a trig substitution where we had to substitute (something like) x=3tan(t), and use use an identity to simplify sqrt(9-x^2) in the denominator and something else, blah blah. We resolved all the integrals, but were left with integral cos^2(t). At the end, the teacher wants them to unsubstitute by using tan(t)=x/3 and labeling the triangle's sides and figuring out how to "unsubstitute" (return to an equation in variable x). But the first trig identity leaves us with a cos(2t), and we do not see a way "unsubstitute" a trig function(2t) when all we have is tan(t)=x/3. I insisted on integration by parts because I wanted to keep the trig functions in t, rather than 2t.
6. Re: Integrate cos^2 t using integration by parts
Hi (slow response on my part, sorry). Thanks for all your replies.
Yes, we had to use integration by parts. Integral of cos^2 was actually part of a larger trig substitution problem (which I'm not going to type up right now) where the teacher wanted them to "unsubstitute" using a triangle, and we couldn't figure out how to do that using the identity of pickslides response.
In other words, the whole problem was a trig substitution where we had to substitute (something like) x=3tan(t), and use use an identity to simplify sqrt(9-x^2) in the denominator and something else, blah blah. We resolved all the integrals, but were left with integral cos^2(t). At the end, the teacher wants them to unsubstitute by using tan(t)=x/3 and labeling the triangle's sides and figuring out how to "unsubstitute" (return to an equation in variable x). But the first trig identity leaves us with a cos(2t), and we do not see a way "unsubstitute" a trig function(2t) when all we have is tan(t)=x/3. I insisted on integration by parts because I wanted to keep the trig functions in t, rather than 2t.
\displaystyle \begin{align*} \cos{2t} &\equiv 2\cos^2{t} - 1 \\ &\equiv \frac{2}{\sec^2{t}} - 1 \\ &\equiv \frac{2 - \sec^2{t}}{\sec^2{t}} \\ &\equiv \frac{2 - \left(1 + \tan^2{t}\right)}{1 + \tan^2{t}} \\ &\equiv \frac{1 - \tan^2{t}}{1 + \tan^2{t}} \end{align*}
ALL trigonometric functions can be written in terms of every other trigonometric function.
7. Re: Integrate cos^2 t using integration by parts
Originally Posted by Prove It
\displaystyle \begin{align*} \cos{2t} &\equiv 2\cos^2{t} - 1\end{align}
...
Ah ha! I'll bet this was the way his teacher wanted him to answer. Thx
8. Re: Integrate cos^2 t using integration by parts
Ah ha! I'll bet this was the way his teacher wanted him to answer. Thx
What? It shows up perfectly fine for me...
9. Re: Integrate cos^2 t using integration by parts
Originally Posted by Prove It
What? It shows up perfectly fine for me...
We can see it perfectly in your post, it's just quoting that seems to destroy your beautiful masterpiece.
10. Re: Integrate cos^2 t using integration by parts
Originally Posted by Prove It
What? It shows up perfectly fine for me...
Fixed. I only wanted to quote part of your post and I screwed up the editing.
,
,
,
,
,
,
,
,
,
,
,
,
,
,
antiderivative with parts cos ^2X
Click on a term to search for related topics.<|endoftext|>
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This newly released NASA image shows Africa and Europe from a million miles away.
Africa is front and center in this image of Earth taken by a NASA camera on the Deep Space Climate Observatory (DSCOVR) satellite. The image, taken July 6 from a vantage point one million miles from Earth, was one of the first taken by NASA’s Earth Polychromatic Imaging Camera (EPIC).
Central Europe is toward the top of the image with the Sahara Desert to the south, showing the Nile River flowing to the Mediterranean Sea through Egypt. The photographic-quality color image was generated by combining three separate images of the entire Earth taken a few minutes apart. The camera takes a series of 10 images using different narrowband filters — from ultraviolet to near infrared — to produce a variety of science products. The red, green and blue channel images are used in these Earth images.
The DSCOVR mission is a partnership between NASA, the National Oceanic and Atmospheric Administration (NOAA) and the U.S. Air Force, with the primary objective to maintain the nation’s real-time solar wind monitoring capabilities, which are critical to the accuracy and lead time of space weather alerts and forecasts from NOAA.
DSCOVR was launched in February to its planned orbit at the first Lagrange point or L1, about one million miles from Earth toward the sun. It’s from that unique vantage point that the EPIC instrument is acquiring images of the entire sunlit face of Earth. Data from EPIC will be used to measure ozone and aerosol levels in Earth’s atmosphere, cloud height, vegetation properties and a variety of other features.
Image Credit: NASA<|endoftext|>
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# What is the equation of the line with slope m= 14/25 that passes through (23/5, (-23)/10) ?
##### 1 Answer
May 7, 2016
$y = \frac{14 x}{25} + 4 \frac{219}{250}$
This is a somewhat unrealistic question, and becomes an exercise in arithmetic rather than maths.
#### Explanation:
There are 2 methods:
Method 1 . uses the formula $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$
This is great to use if you know the slope (m) and one point, which is exactly what we have here. It involves one step of substitution and a bit of simplifying.
$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$
$\left(y - \left(- \frac{23}{10}\right)\right) = \frac{14}{25} \left(x - \frac{23}{5}\right)$
$y + \frac{23}{10} = \frac{14 x}{25} - \frac{14}{25} \times \frac{23}{5} \text{ } \times 250$
$250 y + 250 \times \frac{23}{10} = 250 \times \frac{14 x}{25} - 250 \times \frac{14}{25} \times \frac{23}{5}$
$250 y + 575 = 140 x - 28 \times 23$
$250 y = 140 x + 1219$
$y = \frac{14 x}{25} + 4 \frac{219}{250}$
Method 2 uses $y = m x + c$
Subst for $m , x \mathmr{and} y$ to find $c$
$\left(- \frac{23}{10}\right) = \frac{14}{25} \times \frac{23}{5} + c \text{ } \times 250$
$250 \times \left(- \frac{23}{10}\right) = 250 \times \frac{14}{25} \times \frac{23}{5} + 250 c$
$- 575 = 644 + 250 c$
$1219 = 250 c$
$c = \frac{1219}{250} = 4 \frac{219}{250}$
This leads to the same equation, using values for m and c.
$y = \frac{14 x}{25} + 4 \frac{219}{250}$.<|endoftext|>
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The Hyper Text Transfer Protocol (HTTP) is the client-server network protocol that has been in use by the World-Wide Web since 1990. Whenever you surf the web, your browser will be sending HTTP request messages for HTML pages, images, scripts and styles sheets. Web servers handle these requests by returning response messages that contain the requested resource.
1.1 HTTP Request Message
The HTTP request message has a simple text based structure. For example, here is the the request message sent by Internet Explorer (IE) for this web page:
GET /httpgallery/introduction/ HTTP/1.1 Accept: */* Accept-Language: en-gb Accept-Encoding: gzip, deflate User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64; Trident/7.0; rv:11.0) like Gecko Host: www.httpwatch.com Connection: Keep-Alive
The first line of the message, known as the request line, contains:
- The HTTP method. (See 6. HTTP Methods for more information)
- The relative URL of the resource or a full URL if you are using an HTTP proxy
- The version of HTTP that is being used. Most modern HTTP clients and servers will use HTTP version 1.1 as defined in RFC 2616.
The rest of the message consists of a set of name/value pairs, known as headers (See 2. HTTP Headers). HTTP clients use header values to control how the request is processed by the server. For example, the Accept-Encoding header indicates that the browser can handle content compressed using the gzip or deflate algorithms (see 8. HTTP Compression).
1.2 HTTP Response Message
The web server's response message has a similar structure, but is followed by the contents of the HTML page:
HTTP/1.1 200 OK Server: Microsoft-IIS/8.0 Date: Mon, 04 Jan 2015 12:04:43 GMT X-Powered-By: ASP.NET X-AspNet-Version: 4.0.30319 Cache-Control: no-cache, no-store Expires: -1 Content-Type: text/html; charset=utf-8 Content-Length: 14990 <!DOCTYPE html> <html>...
The first line, or status line, returns a status code from the server that indicates whether the request was successful (see 3. Status codes and errors). The value 200 is returned if the request was processed correctly and content is being returned to the client.
The next eight lines of text contain header values that describe the data and the way in which it is being returned to the client. For example, Content-Type has the value text/html because the page is in HTML format. The response headers are terminated with a double CRLF (carriage return, line feed) and are followed by the contents of the requested resource.
Images are not directly embedded into web pages. Instead, they are specified as separate resources using HTML <img> tags:
<img src="images/logo.gif" width="50" height="50">
Whenever the browser encounters an <img> tag, it checks to see if it has a valid copy of the image either loaded in memory or saved in its cache. If no suitable match is found, it sends out another HTTP request to retrieve it. This means that a web page will usually generate multiple HTTP requests; one for the HTML page and one for each of the images.
Clicking the Refresh button will redisplay this web page by sending HTTP requests to download the HTML of the page and its associated styles sheets and images.
Using HttpWatch with Example 1
- Open HttpWatch by right clicking on the web page and selecting HttpWatch from the context menu
- Click on Record to start logging requests in HttpWatch
- Click on the Refresh HTTP request button above
The HttpWatch window will display about 8 entries. The first is for the page itself; the others are for the style sheets and images used by this page. Select the first entry and then click on the Stream tab to see the HTTP request and response messages described above.<|endoftext|>
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# The Combined Gas Law.
## Presentation on theme: "The Combined Gas Law."— Presentation transcript:
The Combined Gas Law
Manipulating Variables in equations
Often in an equation we want to isolate some variable, usually the unknown From math: what ever you do to one side of an equation you have to do to the other side Doing this keeps both sides the same E.g. x + 5 = 7, what does x equal? We subtract 5 from both sides … x + 5 – 5 = 7 – 5, thus x = 2 Alternatively, we can represent this as 5 moving to the other side of the equals sign … x + 5 = 7 becomes x = 7 – 5 or x = 2 Thus, for addition or subtraction, when you change sides you change signs
Multiplication and division
We can do a similar operation with multiplication and division E.g. 5x = 7, what does x equal? We divide each side by 5 (to isolate x) … 5x/5 = 7/5 … x = 7/5 … x = 1.4 Alternatively, we can represent this as 5 moving to the other side of the equals sign … 5x = 7 becomes x = 7/5 Thus, for multiplication and division, when you change sides you change position (top to bottom, bottom to top)
Multiplication and division
Let’s look at a more complicated example: (x) (y) 5 = 7a b Isolate a in the equation: Move b to the other side (from bottom to top) 5 b (x) (y) = 7a Move 7 to the other side (from top to bottom) (x)(y)(b) 5 = 7a (x)(y)(b) (35) = a (x)(y)(b) (5)(7) = a or
Multiplication and division
This time, isolate b in the equation: (x) (y) 5 = 7a b Move b to the other side (it must be on top) … (x) (y) 5 = 7a b Move everything to the other side of b 35a xy = b (b)(x)(y) 5 = 7a Q - Rearrange the following equation to isolate each variable (you should have 6 equations) P1V1 P2V2 T T2 =
Combined Gas Law Equations
P1 = P2T1V2 T2V1 P2 = P1T2V1 T1V2 T1 = P1T2V1 P2V2 T2 = P2T1V2 P1V1 V1 = P2T1V2 T2P1 V2 = P1T2V1 P2T1
These are all subsets of a more encompassing law: the combined gas law
Combining the gas laws So far we have seen two gas laws: Robert Boyle Jacques Charles Joseph Louis Gay-Lussac V1 T1 = V2 T2 P1 T1 = P2 T2 P1V1 = P2V2 These are all subsets of a more encompassing law: the combined gas law P1V1 P2V2 T T2 = Read pages 437, Do Q 26 – 33 (skip 31)
Q 26 V1 = 50.0 ml, P1 = 101 kPa V2 = 12.5 mL, P2 = ? T1 = T2 P1V1 T1 =
(101 kPa)(50.0 mL) (T1) = (P2)(12.5 mL) (T2) (101 kPa)(50.0 mL)(T2) (T1)(12.5 mL) = 404 kPa = (P2) Notice that T cancels out if T1 = T2
Q 27 V1 = 0.10 L, T1 = 298 K V2 = ?, T2 = 463 P1 = P2 P1V1 T1 = P2V2
(P1)(0.10 L) (298 K) = (P2)(V2) (463) (P1)(0.10 L)(463 K) (P2)(298 K) = 0.16 L = (V2) Notice that P cancels out if P1 = P2
Q 28 P1 = 150 kPa, T1 = 308 K P2 = 250 kPa, T2 = ? V1 = V2 P1V1 T1 =
(150 kPa)(V1) (308 K) = (250 kPa)(V2) (T2) (250 kPa)(V2)(308 K) (150 kPa)(V1) = 513 K = 240 °C = (T2) Notice that V cancels out if V1 = V2
Q 29 P1 = 100 kPa, V1 = 5.00 L, T1 = 293 K P2 = 90 kPa, V2 = ?, T2 = 308 K P1V1 T1 = P2V2 T2 (100 kPa)(5.00 L) (293 K) = (90 kPa)(V2) (308 K) (100 kPa)(5.00 L)(308 K) (90 kPa)(293 K) = 5.84 L = (V2) Note: although kPa is used here, any unit for pressure will work, provided the same units are used throughout. The only unit that MUST be used is K for temperature.
Q 30 P1 = 800 kPa, V1 = 1.0 L, T1 = 303 K P2 = 100 kPa, V2 = ?, T2 = 298 K P1V1 T1 = P2V2 T2 (800 kPa)(1.0 L) (303 K) = (100 kPa)(V2) (298 K) (800 kPa)(1.0 L)(298 K) (100 kPa)(303 K) = 7.9 L = (V2)
For more lessons, visit www.chalkbored.com
Q 32 P1 = 6.5 atm, V1 = 2.0 mL, T1 = 283 K P2 = 0.95 atm, V2 = ?, T2 = 297 K P1V1 T1 = P2V2 T2 (6.5 atm)(2.0 mL) (283 K) = (0.95 atm)(V2) (297 K) (6.5 atm)(2.0 mL)(297 K) (0.95 atm)(283 K) = 14 mL = (V2) 33. The amount of gas (i.e. number of moles of gas) does not change. For more lessons, visit
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January 07, 2019
In optical design, aberrations occur when light from one point of an object doesn’t converge into or diverge from a single point after transmission through the system. The non-linear terms in Snell’s Law cause deviations from perfect imagery: optical systems that form images will produce images that aren’t as sharp. Optical engineers must correct the aberrations to produce the sharpest images possible.
Defocus: The optical system is out of focus, reducing the sharpness of images produced by the system.
Radial distortion: Images with radial distortion usually have symmetric distortion due to the symmetry of a lens. There are three types of radial distortion: barrel, in which image magnification decreases with distance from the optical axis; pincushion distortion, in which image magnification increases with distance from the optical axis; and mustache distortion, which is a blend of the two types. Chromatic aberration is radial distortion that depends on wavelength.
Astigmatism: Rays that propagate in two perpendicular planes have different foci. Where the horizontal and vertical axes cross, the axes will be in sharp focus at two different distances. There are two forms of astigmatism: third-order aberration, which occurs for objects away from the optical access; and when the optical system is not symmetric about the optic axis.
Coma: Defined as a variation in magnification over the entrance pupil, coma is seen often in telescope design, resulting in stars or other objects appearing to have a tail.
Spherical aberration: This occurs when there is increased refraction of light rays striking a lens or when there is a reflection of light rays striking a mirror near the edge versus closer to the center.
Petzval field curvature: Named after physicist Joseph Petzval, one of the founders of geometrical optics, this is an aberration in which a flat object cannot be brought into focus on a flat image plane.
Many optical aberrations can be easily solved with the right optical design software. OpticStudio has many features and functionalities that help correct common aberrations.
For example, OpticStudio offers the Full-field Aberration Analysis feature, which allows optical engineers to improve freeform designs by analyzing how various aberrations degrade the image or beam quality of a system across the full field of view. Freeform optical surfaces are a key element in many modern optical systems, used for example, in virtual and augmented reality designs. This analysis feature displays aberrations and contributions to image degradation across the full XY field of view. Users can analyze freeform systems in the same way they would with any other regular sequential system, identifying the change in aberrations across the field of view and getting an indication of how to correct these.
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Hint: First find the volume of the solid sphere and later find the volume of one ball by using the given radius in the question. To find the number of balls divide the volume of the solid sphere by volume of one ball.
As, it is given that the radius of the solid sphere is R = 8cm.
And we know that the volume of solid sphere is $\dfrac{4}{3}\pi {R^3}$ where R is the radius of solid sphere. So, volume of given Solid sphere will be,
$\Rightarrow$ Volume of Solid sphere= ${\text{ }}\dfrac{4}{3}\pi {R^3} = \dfrac{4}{3}\pi {\left( 8 \right)^3} = \dfrac{{2048\pi }}{3}c{m^3}$
The radius of the ball is given as, r = 1cm.
So, volume of one ball will be,
$\Rightarrow$ Volume of one ball = ${\text{ }}\dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {\left( 1 \right)^3} = \dfrac{{4\pi }}{3}c{m^3}$
So, it is clear that number of balls that can be made is equal to,
$\Rightarrow$ Number of balls= $\dfrac{{{\text{Volume of Solid sphere}}}}{{{\text{Volume of one ball}}}}{\text{ }} = {\text{ }}\dfrac{{\dfrac{{2048\pi }}{3}c{m^3}}}{{\dfrac{{4\pi }}{3}c{m^3}}} = 512{\text{ balls}}{\text{.}}$
So, the number of balls that can be made from a given solid sphere is 512 balls.
NOTE: - Whenever we come up with this type of problem then the best way is to find the total volume/area and then divide it by volume/area of 1 ball, as per required.<|endoftext|>
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The Journey Revealing The Scope Of Earth’s History.
The Scottish highlands, in the region of the Assynt, are perhaps one of the most studied geological sites in the world by early scientist attempting to understand the Earth’s underlying rock formation. They date back over 3000 million years and include some of the most ancient rocks in Europe. The oldest are found at Lewisian Gneiss, with the most extensive cave network in the land in the limestone valleys around Inchnadamph and Elphin.
Assynt is a tiny parish tucked into the North Western corner of Scotland. It is bounded on four corners by the Point of Stoer (Old Man of Stoer), Kylesku, Ledmore and Inverkirkaig. The fishing village of Lochinver is now the principle settlement in Assynt, with the remainder of the population scattered around the coastal communities from Inverkirkaig north to Clachtoll, Stoer and Drumbeg. In earlier times the small inland hamlets of Inchnadamph & Elphin were key settlements – surrounded as they are by fertile limestone soils.
Read Article Here:<|endoftext|>
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# Factors Of 68
Factors of 68 are the numbers which divide the number 68 to give a quotient as a whole number. The factors of a number divide the original number uniformly. When we multiply the factors of 68 in pairs, we get the results as the original number. 68 is a composite number and has more than 2 factors, unlike the prime numbers. Hence, the factors of 68 are 1, 2, 4, 17, 34 and 68.
Multiples are different from the factors, as they give extended times of 68, such as 68, 136, 204, 272, 340, 408, 476, 544, 612, 680 and so on. Learn to find the factors of number 68, along with its prime factorisation, by reading the article completely.
## What are Factors of 68?
Factors of 68 are the numbers that can divide the original number evenly. Factors of 68 are its divisors. There are a total of six factors of 68, they are 1, 2, 4, 17, 34 and 68. Each of these factors can divide 68 in equal number of parts.
Since the number of factors is more than 2. Therefore, 68 is a composite number. Here, the smallest factor of 68 is 1 and the highest factor is 68. Thus, we have simply evaluated the two factors of 68. After finding the factors, we can arrange them in ascending order also. Let us learn to find the other factors in the next section.
## How to Find Factors of 68?
Factors are the real numbers that divide the original number (in this case, 68), evenly or completely. As we learned, 1 is the factor of all the whole numbers. Also, a number is a factor of itself. Now, 68 is an even number. Therefore, it is divisible by 2. Hence, the quotient we get after the division of 64 by 2, is also a factor of 68. Let us find all the factors now.
• 68 ÷ 1 = 68
• 68 ÷ 2 = 34
• 68 ÷ 4 = 17
• 68 ÷ 17 = 4
• 68 ÷ 34 = 2
• 68 ÷ 68 = 1
Thus, the factors of 68 are 1, 2, 4, 17, 34, and 68
## Factor Of 68 in Pairs
We can find the pair factors of 68, by multiplying two numbers in a pair to get the original number, such as;
• 1 × 68 = 68
• 2 × 34 = 68
• 4 × 17 = 68
From the above process, we get the pair factors of the number 68 are (1, 68), (2, 34) and (4, 17).
In the same way, we can write the negative pair factors of 68 since the multiplication of two negative numbers will result in a positive number.
• -1 × -68 = 68
• -2 × -34 = 68
• -4 × -17 = 68
Therefore, negative pair factors are (-1, -68), (-2, -34) and (-4, -17).
### Factors of 68 in Ascending Order
Since we have already obtained all the factors here for the number 68, let us arrange them in ascending order, such as:
1<2<4<17<34<68
You can see from the above arrangement the smallest factor is 1 and the largest factor is 68.
## Prime Factorisation of 68
The number 68 is a composite number. Now let us find its prime factors.
• The first step is to divide the number 68 with the smallest prime factor, i.e. 2 and divide the output again by 2 till you get a fraction or odd number.
68 ÷ 2 = 34
• Now, again divide 34 by 2.
34 ÷ 2 = 17
• Now, we know 17 is a prime number with only two factors; 1 and 17. Therefore, we cannot continue with the division method further.
Therefore, the prime factors of 64 are 2 and 17.
Prime Factorisation of 64 = 2 × 2 × 17 or 22 × 17
## Important Facts on Factors of 68
• Factors of 68: 1, 2, 4, 17, 34, and 68
• Prime Factors of 68: 2 and 17
• Prime Factorisation of 68: 22 × 17
• Pair Factors of 68: (1,68), (2,34), and (4,17)
• Sum of Factors of 68: 126
• Highest factor of 68: 68
• Smallest factor of 68: 1
## Solved Examples on Factors of 68
Q.1: There are 68 students in the class. If there are 17 benches in the class, how many students can sit on each bench?
Solution: Given,
Number of students = 68
Number of benches in classroom = 17
Number of students on each bench = 68/17 = 4
Therefore, on each bench, 4 students can be seated.
Q.2: Find the sum of all the factors of 68.
Solution: The factors of 68 are 1, 7 and 68.
Sum = 1 + 2 + 4 + 17 + 34 + 68 = 126
Therefore, 126 is the required sum.
Q.3: What is the greatest common factor of 60 and 68?
Answer: Let us write the factors of both numbers.
60 → 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
68 → 1, 2, 4, 17, 34, 68
Hence, the greatest common factor is only 4.
## Frequently Asked Questions (FAQs) on Factors of 68
Q1
### What are the factors of 68?
There are a total of six factors of 68. These factors are 1, 2, 4, 17, 34 and 68.
Q2
### How is 68 represented as the product of prime factors?
The prime factorisation of 68 is:
68 = 2 x 2 x 17
Q3
### Is 3 a factor of 68?
No, 3 is not a factor of 68. 3 cannot divide 68 completely.
Q4
### Is 68 a prime number?
No, 68 is a composite number because it has more than 2 factors.
Q5
### Is 68 a perfect square?
No, 68 is not a perfect square because the root of 68 is not a whole number but an irrational number.
√68 = 8.2462112512353
Test your knowledge on Factors of 68<|endoftext|>
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Next: 3.2 Sets Defined by Up: 3. Propositions and Functions Previous: 3. Propositions and Functions Index
# 3.1 Propositions
3.1 Definition (Proposition.) A proposition is a statement that is either true or false. I will sometimes write a proposition inside of quotes ( ''), when I want to emphasize where the proposition begins and ends.
3.2 Examples.
If '', then is a true proposition.
If '', then is a false proposition.
If '', then is a true proposition.
If '', then I will not consider to be a proposition (unless lucky number has been defined.)
3.3 Definition (And, or, not.) Suppose that and are propositions. Then we can form new propositions denoted by and '', or '', and not ''.
and '' is true if and only if both of are true.
or '' is true if and only if at least one of is true.
not '' is true if and only if is false.
Observe that in mathematics, or'' is always assumed to be inclusive or: If '' and '' are both true, then or '' is true.
3.4 Examples.
and '' is false.
or '' is true.
or '' is true.
not(not )'' is true if and only if is true.
For each element of Q let be the proposition
''. Thus = '', so is true, while = '', so is false. Here I consider to be a rule which assigns to each element of Q a proposition
3.5 Definition (Proposition form.) Let be a set. A rule that assigns to each element of a unique proposition is called a proposition form over .
Thus the rule defined in the previous paragraph is a proposition form over Q. Note that a proposition form is neither true nor false, i.e. a proposition form is not a proposition.
3.6 Definition ( , Equivalent propositions.) Let be two propositions. We say that is equivalent to '' if either ( are both true) or ( are both false). Thus every proposition is equivalent either to '' or to '' We write '' as an abbreviation for is equivalent to '' If are propositions, then '' is a proposition, and
'' is true if and only if (( are both true) or ( are both false)).
Ordinarily one would not make a statement like
)''
even though this is a true proposition. One writes '' in an argument, only when the person reading the argument can be expected to see the equivalence of the two statements and .
If and are propositions,then
(3.7)
is an abbreviation for
Thus if we know that (3.7) is true, then we can conclude that is true. The statement '' is sometimes read as if and only if ''.
3.8 Example. Find all real numbers such that
(3.9)
Let be an arbitrary real number. Then
Thus the set of all numbers that satisfy equation (3.9) is {2,3}.
3.10 Definition ( , Implication.) If and are propositions then we say implies '' and write '', if the truth of follows from the truth of . We make the convention that if is false then is true for all propositions , and in fact that
(3.11)
Hence for all propositions and
(3.12)
3.13 Example. For every element in Q
(3.14)
In particular, the following statements are all true.
(3.15) (3.16) (3.17)
In proposition 3.16, is false, is true, and is true.
In proposition 3.17, is false, is false, and is true.
The usual way to prove is to assume that is true and show that then must be true. This is sufficient by our convention in (3.11).
If and are propositions, then '' is also a proposition, and
(3.18)
(the right side of (3.18) is true if and only if are both true or both false.) An alternate way of writing '' is if then ''.
We will not make much use of the idea of two propositions being equal. Roughly, two propositions are equal if and only if they are word for word the same. Thus '' and '' are not equal propositions, although they are equivalent. The only time I will use an '' sign between propositions is in definitions. For example, I might define a proposition form over N by saying
for all '',
or
for all .
The definition we have given for implies'' is a matter of convention, and there is a school of contemporary mathematicians (called constructivists) who define to be true only if a constructive'' argument can be given that the truth of follows from the truth of . For the constructivists, some of the propositions of the sort we use are neither true nor false, and some of the theorems we prove are not provable (or disprovable). A very readable description of the constructivist point of view can be found in the article Schizophrenia in Contemporary Mathematics[10, pages 1-10].
3.19 Exercise.
a) Give examples of propositions such that '' and '' are both true, or else explain why no such examples exist.
b) Give examples of propositions such that '' and '' are both false, or explain why no such examples exist.
c) Give examples of propositions such that '' is true but '' is false, or explain why no such examples exist.
3.20 Exercise. A Let be two propositions. Show that the propositions '' and '' are equivalent. ( '' is called the contrapositive of the statement ''.)
3.21 Exercise. Which of the proposition forms below are true for all real numbers ? If a proposition form is not true for all real numbers , give a number for which it is false.
a)
.
b)
.
c)
.
d)
. (Here assume .)
e)
.
f)
.
g)
.
h)
.
3.22 Exercise. Both of the arguments A and B given below are faulty, although one of them leads to a correct conclusion. Criticize both arguments, and correct one of them.
Problem: Let be the set of all real numbers such that . Describe the set of all elements such that
(3.23)
Note that if then is defined.
ARGUMENT A: Let be an arbitrary element of . Then
Hence the set of all real numbers that satisfy inequality (3.23) is the set of all real numbers such that .
ARGUMENT B: Let be an arbitrary element of . Then
Now
Hence the set of all real numbers that satisfy inequality (3.23) is the set of all such that either or .
Next: 3.2 Sets Defined by Up: 3. Propositions and Functions Previous: 3. Propositions and Functions Index
Ray Mayer 2007-09-07<|endoftext|>
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Writing tools are the things we use to write with. Most writing today is done with a computer system, usually a word processing software. But for about 5,000 years, much simpler tools were used. The tools were different in different parts of the world. The raw materials had to be near where people lived.
The Middle EastEdit
The Sumerian system used clay as the basic material to write on. They developed a type of writing called cuneiform. Triangular marks were pressed into soft clay tablets by a metal or wooden tool. After the clay had dried in the sun the tablets were baked. Then they were carried somewhere else for others to read. We know that its first uses were for trade, accounting and administration.
This is the most famous of the old forms of writing. It was invented at roughly the same time as cuneiform, yet was quite different in style, and used different materials. The writing tools used varied according to the material which was written on. Egyptians ended up with three writing systems for the same language. The tools were:
- 1. Hieroglyphic: the famous pictorial language on stone monuments. The hieroglyphs were carved into stone (hammer & chisel) or painted onto stone surfaces. Many survive, some with the original colours intact.
- 2. The two cursive ('running') scripts, hieratic and demotic, were written with reed pens and carbon inks onto papyrus. If the material was cloth, then the writing was done with a brush.
When writing on papyrus, Egyptians used a reed pen. The was hollow, with a nib-like end structure. This was a good example of how writing tools must be got from materials in the local area.
Although Chinese characters can be written with many materials, for the longest period they were written with a brush. The ink was carbon-based. It was moulded before use into an inkstick. This small solid object could be carried easily. To get the ink the writer had to grind the inkstick against an inkstone with some added water. The earliest Chinese inks can be dated back to 12th century BC, with the use of charred (partly burnt) materials and plant dyes. Mineral inks based on graphite were most common. A bit later, soot was used as the carbon source, and animal glue used to bind the ink to the paper (Egyptian ink used gum arabic for the same reason).
The Romans used lead styli with wax tablets which could be "erased" by rubbing the beeswax surface smooth again. For permanent records they wrote on parchment (treated animal skin) with ink. They invented the idea of a book with sewn-together parchment pages. It was called a codex, a word which means something like 'block of wood'. Before, long documents were kept in a roll.
Dip-in pens were used in Europe for a thousand years or more. At first they were quill pens made of the stem ('rachis') of a pigeon feather. The quill pen was cut by a knife to give the slanting nib at the writing end. They were used from the middle ages until the first part of the 19th century.
Metal pens were tried next, but many metals were not flexible enough. The writing styles which developed with quill pens used thick and thin strokes. People wanted the same flexibility from a metal pen. Between 1839 and 1849 the steel-nibbed pen solved the problem. For the next century and more, industrial production lines turned out hundreds of millions of steel nibs. They had the advantage that different types of nib could be attached to the same holder. This allowed for different styles of writing to be used. Eventually the best nibs were tipped with gold or iridium, a rare metal which arrives on Earth via meteorites. All stages in the making of nibs was done by machine. The great secret of all nibs from the Egyptian reed to the steel pen is the slit down the middle of the tip. This leads the ink from the small blob under the nib to the point where the ink goes onto the paper. Without the slit, a nib does not work well.
- 'Cuneus' being Latin for 'wedge', hence cuneiform = wedge-shaped.
- Boltz, William G. 1986. Early Chinese writing. World Archeeology, 17, (3) Early Writing Systems, pp. 420–36 (436).
- Boltz, William G. 1994; revised 2003. The origin and early development of the Chinese writing system. American Oriental Series, vol. 78. American Oriental Society, New Haven, Connecticut. ISBN 0-940490-18-8
- Qiu Xigui 2000. Chinese writing. Transl. Gilbert L. Mattos and Jerry Norman. Early China Special Monograph Series #4. Berkeley: The Society for the Study of Early China and the Institute of East Asian Studies, University of California, Berkeley. ISBN 1-55729-071-7
- Whalley J.I. 1975. Writing implements and accessories, from the Roman stylus to the typewriter. Newton Abbot: David & Charles.<|endoftext|>
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Common and Scientific NamesCommon names are names given to many compounds, but they may mislead the uninitiated. For instance, the scientific names of alkenes contain the suffix –en(e) as part of their name. But acetylene is not an alkene. It is an alkyne. The scientific names of alkynes contains the suffix –yne. Acetylene is scientifically named ethyne. Yet, it nomenclature is not the only difference between alkanes, alkenes, and alkynes.
AlkanesAlkanes are completely saturated compounds. That means every carbon atom bonds to four other atoms. Ethene has two carbon atoms and six hydrogen atoms. Why not eight hydrogen atoms? Because the two carbon atoms bond to each other, using up one bond of each carbon atom, resulting in 8 – 2 = 6 remaining. Thus ethane’s formula is written:
AlkenesEthene is much like ethane, except instead of a single bond between carbon atoms, it has a double bond. This uses two of the four bonds from each carbon atom, leaving a total of 8 – 4 = 4 bonds to attach to hydrogen atoms. We write ethene’s formula:
AlkynesEthyne has a triple bond between the two carbon atoms, leaving 8 – 6 = 2 bonds to attach to hydrogen atoms. Ethyne is written:
CombinationsWhat about compounds that contain a number of single bonds and one double bond? Although the majority of the compound resembles an alkane, it is classified as an alkene. Thus the following structure is named 1-butene:
Alkanes, Alkenes, and Alkynes – Pecking OrderWhen it comes to naming organic compounds, reference is made to various rings, pendant groups, and bonding combinations. It follows a definite pecking order. For full details, visit the International Union of Pure and Applied Chemistry, or IUPAC.
For only a combination of single, double, and triple carbon-carbon bonded simple alkanes, the order is triple bonds take precedence over double bonds, which, in turn, take precedence over single bonds. Thus the structure:
is hept-3-en-1-yne. The chain is seven carbons long, so the root is “hept”. The double bond lies between carbons 3 and 4. The lower number is used to specify location. The triple bond is between carbons 1 and 2.
Types of BondIt ought to be pointed out that a double bond is not equal to two single bonds. Neither is a triple bond the equivalent of three single bonds. A single bond is inline and is called more formally a sigma bond, whereas a double bond consists of one sigma and one pi bond. A triple bond consists of one sigma (σ) and two pi (π) bonds. The pi bond greatly influences reactivity and reaction behavior.
Note: You might also enjoy Straight Chain Alkanes: Predicting Properties
← Back to Classic Science<|endoftext|>
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# Ex.11.1 Q8 Perimeter and Area - NCERT Maths Class 7
Go back to 'Ex.11.1'
## Question
A door of length $$2\,\rm{}m$$ and breadth $$1\,\rm{}m$$ is fitted in a wall. The length of the wall is $$4.5\,\rm{}m$$ and the breadth is $$3.6\,\rm{}m$$ (See the below Fig). Find the cost of white washing the wall, if the rate of white washing the wall is $$\rm{}Rs\, 20 \,per\, m^2$$
Fig 11.6.
Video Solution
Perimeter And Area
Ex 11.1 | Question 8
## Text Solution
What is known?
A door of length $$2\,\rm{}m$$ and breadth $$1\,\rm{}m$$ is fitted in a wall of length $$4.5\,\rm{}m$$ and the breadth $$3.6\,\rm{}m$$
What is unknown?
The cost of white washing the wall, if the rate of white washing the wall is $$\rm{}Rs\, 20 \,per\, m^2.$$
Reasoning:
Since door will not be whitewashed, we will have to subtract area of the door from the area of wall. After, finding area to be whitewashed, multiply the area with rate of white washing $$\rm{}per\, m^2$$ to get the cost.
Steps:
Given,
Length of wall $$= 4.5\,\rm{}m$$
Breadth of wall $$= 3.6\,\rm{}m$$
\begin{align}\text{Area of wall}&= {\rm{Length}} \times {\rm{Breadth}}\\ &= 4.5 \times 3.6{\rm{ }}\\&= 16.2{\rm{ }}{\rm \,m^2}\end{align}
Given,
Length of door $$= 2\,\rm{}m$$
Breadth of door $$= 1\,\rm{}m$$
So,
\begin{align}\text{Area of door}&= {\rm{Length}} \times {\rm{Breadth}}\\&= 2 \times 1\\&= 2{\rm{ }}{\rm \,m^2}\end{align}
\begin{align}&\text{Area of wall for white wash}\\ &= {\text{Area of wall}} - {\text{Area of door}} \end{align}
\begin{align} &= 16.2{\rm{ }}{\rm \,m^2} - 2{\rm{ }}{\rm \,m^2}\\&= 14.2{\rm{ }}{\rm \,m^2}\end{align}
The rate of white washing the wall
$$= \rm{Rs }\,20\,{\rm{ per\, }}{\rm m^2}$$
Therefore,
\begin{align}{\text{The cost of white washing}}\;1618{\rm{ }}{\,\rm m^2} \end{align}
\begin{align} &= 14.2 \times 20\\ &= {\rm{Rs\, }}284\end{align}
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