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What is the Constant in 3:4

What is the constant in 3:4

Research, Knowledge and Information :


Mathematical constant - Wikipedia


Mathematical constant A mathematical constant is a special number, usually a real number, that is "significantly interesting in some way". Constants arise ...
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Constant | Define Constant at Dictionary.com


Constant definition, not changing or varying; uniform; regular; invariable: All conditions during the three experiments were constant. See more. Dictionary.com;
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How do I use the binomial theorem to find the constant term ...


How do I use the binomial theorem to find the constant term? Precalculus The Binomial Theorem The Binomial Theorem. 1 Answer ? 53 Sayani R ...
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Constant term - Wikipedia


... and powers of variables, the notion of constant term cannot be used in this sense, since that would lead to calling "4" the constant term of ...
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Identifying the Constant of Proportionality - Video & Lesson ...


Identifying the Constant of Proportionality. One Saturday morning, you find yourself at the local grocery store helping out with a little shopping for your family.
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What is the constant of variation for the relationship shown ...


What is the constant of variation for the relationship shown in the table? x 1 2 3 4 y 4 8 12 16 2. Ask for details ; Follow; Report; by west3160 02/15/2017. Log in ...
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Which term in the expansion of '((1/(2x^3))-x^5)^8' is a ...


Which term in the expansion of "((1/(2x^3))-x^5)^8" is a constant? ... ^2 = 4x^4 + 12x^3 + 9x^2 Doing this for ((1/ ... and this will be your constant term.
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Constant Contact - Official Site


With Constant Contact, you can create effective email marketing and other online marketing campaigns to meet your business goals. Start your FREE trial today!
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Polynomial functions - Topics in precalculus


POLYNOMIAL FUNCTIONS. Terms and factors. Variables versus constants. Definition of a monomial in x. ... Leading coefficient: 1 2 · 1 3 · 2 4 = 16. Constant term: ...
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Mathwords: Constant Term


Constant Term. The term in a simplified algebraic expression or equation which contains no variable(s). If there is no such term, the constant term is ...
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Suggested Questions And Answer :


Differentiation of terminal velocity final equation

When the drag force and force due to gravity are equal the terminal velocity is reached. Various factors are involved: 0.5dv^2AC=mg where d=air density, v=terminal velocity, A=surface area presented to the air, C=drag coefficient, assumed constant. The air density changes gradually with altitude, being densest near the ground. Although g does change with altitude, the change is very small compared to other factors so it can be regarded as constant at about 9.8m/s^2. A is constant as long as the falling object is not rotating. The mass, m, of the object is constant for a coherent object. From this v=√(2mg/dAC). DERIVATION USING CALCULUS We have to start by assuming the drag force at some velocity v is 0.5dv^2AC. So if x is the altitude where x=0 is ground level, then gravity acts in the negative direction and drag in the positive direction. F=ma is Newton's Law and we can replace F with 0.5dv^2AC-mg, the net result of drag and gravitation. Initial velocity at time t=0 is zero so the only force is gravity -mg. As the object accelerates under gravity v increases. The acceleration a can be written dv/dt or v'. So we have mv'=0.5dv^2AC-mg or v'=0.5dv^2AC/m-g=g(0.5dv^2AC/mg-1). Note that v is the vertical velocity upwards. We can write this in the form of integrands, but before we do, let's simplify this a bit. We know which of these are constants so we can combine them into just one called k, where k^2=0.5dAC/mg. v'=g(k^2v^2-1). ∫(dv/(k^2v^2-1))=g∫dt.  Integrating by parts we have: ∫(dv/(2(kv-1)))-∫(dv/(2(kv+1)))=gt. (We'll insert the constant of integration later.) ln(kv-1)/2k-ln(kv+1)/2k=gt; ln((kv-1)/(kv+1))=2kgt+C, where C is the constant of integration. However, when v=0, t=0 so that would give us ln(-1)=C which has no meaning. So let's rephrase the original integrand: ∫(dv/(1-k^2v^2))=-g∫dt. Now we have ∫(dv/(2(1-kv))+∫(dv/(2(1+kv))=-gt and -ln(1-kv)+ln(1+kv)=-2kgt+C. This time when v=t=0, C=0, so ln((1+kv)/(1-kv))=-2kgt. So (1+kv)/(1-kv)=e^(-2kgt). After some time t=T the object reaches terminal velocity, but by definition the velocity stays constant and the object keeps falling at the same constant rate. If we imagine the object falls from a very great height so as never to reach the ground, t can be made infinite for all intents and purposes, and the right-hand side approaches zero. Therefore 1+kv=0 and v=-1/k=-√(2mg/dAC). The negative sign simply means the velocity is in the downward direction. A truer interpretation is that the falling object never reaches a terminal velocity no matter how far it falls, so the terminal velocity is in fact an asymptote. v=-(1/k)(1-e^(-2kgt))/(1+e^(-2kgt)) is the actual equation for v. For practical purposes, though, given sufficient altitude, over the range for which the quantities assumed to be constant are approximately constant, the terminal velocity has meaning.    
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how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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How do special products help us factor polynomials? Give examples.

The special products that students are usually asked to identify and remember to help in factorisation are the squares and difference of squares: (a+b)^2=a^2+2ab+b^2; (a-b)^2=(b-a)^2=a^2-2ab+b^2; (a-b)(a+b)=a^2-b^2. a and b can be composite quantities like 3x, xy, 2xyz, etc. Examples: (3xy-z)^2=9x^2y^2-6xyz+z^2; (1/x+1/y)^2=1/x^2+2/xy+1/y^2; (5-t)(5+t)=25-t^2. I include some other types of products you might find useful: The constant term in a polynomial is a strong clue to how it factors, assuming that it is meant do so. If the constant is a prime number p, then the factors will consist of p and at least one 1, because p*1*...*1= p. The degree of the polynomial tells you how many factors there are: a degree 2 (quadratic) has two, for example. Examples: x^3-7x^2-x+7=(x-1)(x+1)(x-7); x^2-12x-13=(x-13)(x+1). If the constant is a composite number (i.e., not a prime number), then you need to factorise it and group the factors according to the degree. For example, if the constant is 15 and the degree is 4, then the factors arranged in fours are (1,1,1,15), (1,1,3,5). Example: x^4+2x^3-16x^2-2x+15=(x-1)(x+1)(x-3)(x+5). If the degree is n and the constant term is p^n, where p is a prime number, then the polynomial may consist of factors +p and -p. Examples: x^3-5x^2-25x+125=(x-5)^2(x+5). But x^2-26x+25=(x-25)(x-1). The sign of the constant is significant. If the degree n is even, and the sign is plus, then the zeroes of the polynomial will consist of an even number of pluses and minuses. If the sign is minus, there is an odd number of pluses and an odd number of minuses. If the degree is odd, and the sign is plus, the factors contain an even number of minuses, or none at all, and the rest are plus. If the sign is minus the factors contain an odd number of minuses, and the rest are pluses, or there are no other factors.
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Looking for an answer to the question below

Looking for an answer to the question below If a sprinter runs with constant acceleration for the first 25 yd. of a 100 yd-dash and then runs the rest of the race at constant velocity, what would his constant acceleration  have to be if he is to run the race in 10 sec?   Distance left to  traveld = 100 - 25 = 75 yards   Distance = initail Velocity x time +  ½ Accelaration x Time ² 75 = 0 + ½ Accelarrtion (10)² (75 x 2) 100 = Accelaration   Accelaration = 1.5  yard/sec² ◄ Ans  
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how can i tell what number to add or subtact from when solving varibals

Go to each side of the equal sign and combine your x's (or variables of like kind) then your numbers (constants) without x's (or other variables of like kind) and use the sign in front of the term, and when there is no sign, it may be positive or + in most cases! How does this answer your question? If you have two terms one has an "x" and the other just a number, alone those are not "like kind" (one is a constant(for example "5") and the other a variable(Ffor example "x")) and should not be combined because one is part of the variables and the other is part of the constants. When I say "combine" this is where you "add or subtract" those variables or like terms. In the example you provided the variable with the equal sign, erasing the rest of the problem reads "2x+3x=" Notice how they are both on the same side of the equal sign, in this case you just do as it reads "2x+3x=" or "5x=" do not forget to keep that equal sign so you can determine where the rest of the terms go. Sometimes you will encounter variables or other terms on the other side of the equal sign, in a case like this simply combine like terms on one side of the equal sign and then  do the same to the other, no particular order but some say it is easier to use a method or pattern. EXAMPLE 2x+5x+5+2= -5x-13x+32. Do not get scared it only looks more challenging then it is. IF you picture your variables with everything else other than the equal sign erased you will have "2x+5x=-5x-13x" I know this will not work out as a problem, just use it as a visual. and combine. You should get "7x=-18x". You could go even further and move your x's to one side or just wait until that step. In the example I provided it requires an extra step that some call reversing/negating the terms and others in your book call it something similiar. Just a thought!! Lastly, move your variables to one side of the equal sign and your constants to the other side of the equal sign. It really does not matter what side unless you see an easy way to solve, but variables must always be on that one side of the equal sign you pick and constants on the other side of the equal sign. Usually 25x=25 or x=1, and 25=x or 1=x no matter what order if you do not mind the multiplication rule of multiplying negatives and positives. that is where you create you easy method.
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What are the terms, variables, coefficients, and constants of y=3x+7?

Problem: What are the terms, variables, coefficients, and constants of y=3x+7? Hello, I am working on a question that requires me to find the terms, variables, coefficients, and constants of y=3x+7. There are three terms: y, 3x and 7. The variables are y and x. The coefficients are 1 (the coefficient of y) and 3 (the coefficient of x). The constant is 7.
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help me!! :(

Let's look at each answer in turn. A. It can't be the answer. The reason is that the constant amount increases the wage year on year, so the wage is getting bigger but the increment is becoming a smaller fraction, or percentage, of the wage as time goes on. B. This can't be the answer even though the increment itself is increasing. The percentage increase isn't in step with this. 0.50/13.50=3.7% approx. The wage is $14/hr. Next year the percentage is 0.75/14=5.4%, so the percent increase is not constant. C. This is the correct answer because the wage rises by $2, which is 2/20=10% increase. Next year the increase percent is 2.20/22=10%. That's a constant percent increase. D. This can't be correct if C is correct. Let's see what we've got. 0.75/15=5%; 1.75/17=10.3%. Not a constant increase.
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calculus work problem. Leaky bucket.

Question: A leaky 17 kg bucket is lifted from the ground to a height of 62 m at a constant speed with a rope that weighs 0.2 kg/m. Initially the bucket contains 32 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 62-m level. How much work is done? Loss of weight of water from bucket is 32 kg over a distance of 62 m  = 32/62 kb/m = 16/31 kg/m. Weight of bucket, including water, at distance x from ground is Wb = 17 + 32 – (16/31).x kg Weight of rope diminishes by 0.2 kg for every metre the bucket is drawn upwards. Initial length of rope is 62 m. Initial weight of rope, being pulled up is, 62*0.2 = 12.4 kg Length of rope when at a distance x m from the ground is 62 – x m. Weight of rope when at a distance x m from the ground is Wr = (62 – x)*0.2 = 12.4 – 0.2.x kg. Total weight being lifted at distance of x m from ground is W = Wb + Wr W = (49 – (16/31).x) + (12.4 – 0.2.x) = 61.4 – (111/155).x Work done over a small distance δx δW=F∙δx Where the lifting force, F, is just equal to the current weight, Wg. Total work done WD=∫dW=∫_0^x F dx=∫_0^62 Wg dx WD=g∫_0^62 61.4 – (111/155).x dx WD=g(61.4x - (111/155).x^2/2)   [62 ..0] WD=9.81(3,806.8 - 1376 2/5) WD=23,842 J WD=24KJ A quicker way to do this problem is to notice that the rates (of loss of weight) are linear relationships. E.g. 0.2.x kg/m rather than 0.2.x^2 kg/m. Since the relationships are linear, then just take average values and work from that. Weight of rope = 62*0.2 = 12.4 kg. Average weight of rope = 6.2 kg. Average weight of water = 32/2 = 16 kg. Weight of bucket (constant) = 17 kg. Average weight being lifted = 17 + 16 + 6.2 = 39.2 kg Distance over which the average weight is being lifted = 62 m. Work done = Fx = Wg.x = 39.2*g*62 = 23,842.224 J = 24 KJ  
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constant acceleration?

constant acceleration? If an object starts from rest, what constant acceleration will allow it to travel 500m in 10s?   Distance = intial Velocity x time + ½ accelaration in Time² initial Velocity = 0 inital time = 0 distances = 500 m Time = 10 s   500 = 0 + ½ (Accelaration) 10² 500 = 100/2 Acceleration Accelaraion = 500/50   Accelaration = 100  m/s²  ◄ Ans
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Graph the circle x^2+y^2-8x+4y+16=0

Complete the squares: x^2-8x+16+y^2+4y+4-4=0⇒(x-4)^2+(y+2)^2=4. This form of the equation tells us where the centre of the circle is and what the radius is. The centre is where x-4=0, so x=4, and where y+2=0, so y=-2. The constant on the right is the square of the radius. This constant was the result of completing the squares by providing the right constant, and adjusting the original constant, 16, so 16 was changed to 16-16-4=-4, which became 4 when moved to the right hand side. The centre of the circle is at (4,-2) and the radius is sqrt(4)=2. Because we know where the centre is we know that if we move a radius to the right and left we get the diameter. So the diameter goes from (2,-2) to (6,-2) horizontally; and from (4,-4) to (4,0) vertically. These parameters help to draw the circle, perhaps with the help of a pair of compasses. 
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