Guide :

# 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 ...

### 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;

### 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 ...

### 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 ...

### 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.

### 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 ...

### 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.

### 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!

### 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: ...

### 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 ...

## 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.

### 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.

### 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

### 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.

### 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.

### 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

### 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

### 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.