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Help With Math 1050 Rational Functions

How do you find all rational zeros for f(x) =x^3-2x^2-5x+6?

As the coefficient of x^3 is 1, rational zeros are integers. The numerical term has only eight divisors, 1, -1, 6, -6, 2, -2, 3 and -3. So, if there are any rational roots, one of them must be one. Try them. 1^3 - 2(1^2) - 5(-1) + 6 = 0.Yes, you have found a zero straightaway.Now factorise: (x-1)(x^2 - Ax - 6) = x^3 - (A+1)x^2 + (A-6)x + 6 so A must be 1.Therefore x = 1 or x^2 - x - 6 = 0. The latter is a quadratic with discriminant 1+24 = 25. As this is a perfect square there must be two integer solutions, so factorise it. You should find (x - 3)(x + 2) = 0 so solutions are 1, 3 and -2.They are all rational.

Math help! Graphing Rational Functions?

Math help: Be sure to indicate (if any) the zeroes, vertical asymptotes, end behaviour written as limits, holes, y-intercept and behaviour near vertical asymptotes written as limits.


(6x^2-24x-30)/(x^2-x-56)

I got it factored to
6x(x-5)+6(x-5) / x(x-8)+7(x-8)

So vertical Asymptotes are: -7 and 8
Zeroes: 5 and -1

I can't figure out the end behaviors for the asymptotes, or how to graph it.

How do you write a pow function?

The easiest thing to do is to create a pow function for positive powers with a loop:Function pow(number, power){
var outcome = 1
for(i=0;i outcome = outcome*number
}
return outcome
}
This will simply do all the multiplications till the outcome is reached.Now a little more advanced, because we also want negative powers to work. Luckily, numbers to a negative power are just one divided by the same number to a positive power. So, by a little expansion, we can solve this:Function pow(number, power){
var outcome = 1
if(i<0){
return 1/pow(number, -power)
}
else{
for(i=0;i outcome = outcome*number
}
return outcome
}
}
This function will, when given a negative number, run the same function with a positive power, and then divide one by the outcome, and return that.note: this was written in JS, but will probably work in most languages.

Determine if the function is a polynomial function: f(x)=1050?

A: This is not a polynomial. Horizontal lines can not be polynomials as there are no turning points.

B:This is not a polynomial. Only vertical lines of the form x = b can be classified as a polynomial.

C: This is a polynomial function of degree zero with a leading coefficient of 1,050.

D:This is not a polynomial because there are no variables.

My final Algebra 2 Qustions!?

1. The time required to finish a test in normally distributed with a mean of 70 minutes and a standard deviation of 10 minutes. What is the probability that a student chosen at random will finish the test in less than 60 minutes?

2. The time required to finish a test in normally distributed with a mean of 70 minutes and a standard deviation of 10 minutes. What is the probability that a student chosen at random will finish the test in less than 90 minutes?

3. Suppose you have a normally distributed set of data pertaining to a standardized test. The mean score is 1050 and the standard deviation is 200. What is the z-score of 1000 point score?

4. Suppose you have a mean standardized score of 1100 points with a standard deviation of 100 points. This data is normally distributed. What is the z-score of 1125 points?

5. Suppose you did a survey of male shoes sizes and the mean of that survey was size 10.5 with a standard deviation of 1.5. What is the z-score of a size of 11?

6. Suppose you have a normally distributed set of data pertaining to a standardized test. The mean score is 1050 and the standard deviation is 200. What is the z-score of 1400 point score?

Any help is much appreciated and thank you so much in advance!

What does the Laplacian of a given function imply, in general?

Do you mean taking the Laplacian of a vector field, or the Laplace Transform of a function?The Laplacian is the divergence of the gradient, and a measurement of how much the average value centered around the point changes when you move away. It's the core part of the Laplace, Heat, Wave, and Helmholtz equations.The Laplace Transform on the other hand is a particular type of Integral Transform, where our kernal is composed of exponential functions. The Laplace Transform is very helpful for differential equations as they have the nice property of mapping derivatives from the domain to multiplication in the destination space. So you can work with just algebra, and then map it back. Initial value problems in particular have the solution baked right in!As to the meaning behind an integral transform. You can think of it as creating a new function where it has pulled out all of the interesting bits and putting them together in a new space. Fourier transforms break up functions into their frequencies, and laplace transforms break up functions into their 'moments'.

Help with math problem?

a).
speed = distance / time
Therefore
time x speed = distance
time = distance / speed
t = d/s

b).
speed = 360 mph and distance = 700 miles
time = 700 miles / 360 mph
= 1.944444 hours
= 1 hour 56 minutes 40 seconds.

(If the speed was meant to be 350mph not 360 then this would obviously be 2 hours!)

c). speed = (360 + x)mph
with a tail wind, the speed of the plane will be faster. Therefore you need to add "x" to the normal speed of 360 mph.

d). The time of the outward trip = 700/360 hours
The time for the homeward flight = 700 / (360 + x) hours
Therefore as the total time taken was 3.5 hours

3.5 = 700/360 + 700 /(360 +x)

multiply both sides of the equation by 360(360 + x) to get rid of the fractions
3.5*360(360+x) = 700(360 +x) + 700*360
1260(360 +x) = 252000 +700x + 252000
453600 + 1260x = 700x +504000
Simplify by subtracting 453600 and 700x from both sides
560x = 50400
x = 50400/560 = 90

The tailwind is 90mph.
Therefore on the return journey the speed would be 360+90 = 450mph

Check
700 miles at 360 mph = 1.94444 hours
700 miles at 450 mph = 1.55555 hours
Total time = 3.5 hours therefore correct.

Again are you sure it was not a typo and the speed was meant to be 350 mph ? The maths would have been a lot easier.

What is the value of [math]\frac{x^5-1}{x-1}[/math] at [math]x=1[/math] ?

This is an interesting question.Let me show you something from a mathematics paper first. The screenshot below is taken from the 2002 paper Some results on graph spectra by Elias Hagos (Linear Algebra and its Applications, 356, 1–3, 103–111):Here, [math]P_{G_j}(x)[/math] and [math]P^\prime_G(x)[/math] are polynomials. It so happens that, sometimes, [math]\lambda_i[/math] is a root of both [math]P_{G_j}(x)[/math] and [math]P^\prime_G(x)[/math], in which case the expression[math]\dfrac{P_{G_j}(\lambda_i)}{P^\prime_G(\lambda_i)}[/math]reduces to [math]\frac{0}{0}[/math], which is undefined. Thus, the author of the paper is explicitly stating to first cancel the common terms of these two polynomials and then substitute [math]x=\lambda_i[/math]. He is denoting this by[math]\left.\dfrac{P_{G_j}(x)}{P^\prime_G(x)}\right|_{x=\lambda_i}.[/math]In the same way, the expression[math]\dfrac{x^5-1}{x-1}[/math]reduces to [math]\frac{0}{0}[/math] when [math]x=1[/math], which is undefined, but the expression[math]\left.\dfrac{x^5-1}{x-1}\right|_{x=1}[/math]is equal to [math]5[/math], because [math]\frac{x^5-1}{x-1}[/math] simplifies to [math]x^4+x^3+x^2+x+1[/math].Such ‘fractional polynomials’ are technically called rational functions[1] . Rational functions sometimes have a so-called removable singularity[2] , as in this question’s case, where [math]x=1[/math] is a removable singularity. This is also the case for the above expression in Theorem 4.1 of the 2002 paper by Hagos, where [math]\lambda_i[/math] is, at worst, a removable singularity.However, sometimes singularities are not removable; for example, the rational function[math]\dfrac{1}{x}[/math]has a singularity at [math]x=0[/math], which is not removable. Thus, in the case of [math]\frac{1}{x}[/math], we cannot assign any value to it when [math]x=0[/math], but in the case of [math]\frac{x^5-1}{x-1}[/math], we may assign the value [math]5[/math] to it when [math]x=1[/math], though only after removing the singularity.(To be honest, the above explanation is perhaps too simplistic. Technically, the topic of singularities in rational functions is studied in complex analysis[3] , which I have not delved into here at all.)Footnotes[1] Rational Function[2] Removable Singularity[3] Complex Analysis

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