Algebra Word Problem Solutions

A polynomial whose roots form an arithmetic progression
Let f be a fourth degree polynomial whose roots form an arithmetic progression. Prove that f′'s roots also form an arithmetic progression.
I didn' t make much progress, I just wrote $f(x)=a(x-<!--\; -\; -->b-<!--\; -\; -->r)(x-<!--\; -\; -->b-<!--\; -\; -->2r)(x-<!--\; -\; -->b-<!--\; -\; -->3r)(x-<!--\; -\; -->b-<!--\; -\; -->4r)$ and I tried to differentiate, which obviously doesn't help too much.

Find a1 given an = 24, r = 2/3, n = 4

A Mighty Kids meal includes 3.2 ounces of applesauce. If a child eats two frozen meals each week, how many ounces of applesauce will they consume in one year?

Find the integers if the sum of three consecutive odd integers is 177

How to compute $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\displaystyle \frac{\ln <!--\; \; -->(1+e^{\mathrm{\alpha <!--\; \alpha \; -->}x})}{\ln <!--\; \; -->(1+e^{\mathrm{\beta <!--\; \beta \; -->}x})}}$?
Where $\alpha >0$ and $\beta >0$

I'm interested in solving a heat equation with a discontinuous source term in one dimension on the real line:
$\frac{\mathrm{\partial <!--\; \partial \; -->}u}{\mathrm{\partial <!--\; \partial \; -->}t}(x,t)=-<!--\; -\; -->x\mathrm{\theta <!--\; \theta \; -->}(x)u(x,t)-<!--\; -\; -->x\mathrm{\theta <!--\; \theta \; -->}(-<!--\; -\; -->x)u(-<!--\; -\; -->x,t)+D\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}u}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}(x,t)$
where $\mathrm{\theta <!--\; \theta \; -->}(\cdot <!--\; \cdot \; -->)$ is the Heaviside function. The boundary conditions are that $\underset{|x|\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}u(x,t)=0$ The initial condition is arbitrary, but even for a symmetric Gaussian or a symmetric exponential an analytical solution would be interesting.
Because of the discontinuity in the right-hand side induced by $\mathrm{\theta <!--\; \theta \; -->}(\cdot <!--\; \cdot \; -->)$, my intuition would be to solve in the two domains x>0 and x<0 and patch them together at x=0. However, the source terms have the effect of causing exponential decay with rate x at the point x for x>0 and corresponding growth at the point −x. This introduces an asymmetry, which should cause a diffusive flux back to the side x>0. Solving for the solution in the two domains does not seem to capture this feedback, and I'm not sure how to solve globally. What are some avenues to proceed?

$a:[0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathbb{R}$ is a continous and bounded and
$x^{\prime }(t)=\left(\begin{array}{cc}0& 1\\ -<!--\; -\; -->a(t)& 0\end{array}\right)x(t)$
has a non-zero solution like $y(t)$ such that $\underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}y(t)=0$.
Show that this equation has an unbounded solution on $[0,\mathrm{\infty <!--\; \infty \; -->})$.

Is the following a geometric sequence 1.5, 2.25, 3.375, 5.0625,…

Calculus - limit of a function with logarithms
Compute
$\underset{x\to <!--\; \to \; -->0_{+}}{lim}\ln <!--\; \; -->(x\ln <!--\; \; -->a)\ln <!--\; \; -->\left(\frac{\ln <!--\; \; -->(ax)}{\ln <!--\; \; -->(\frac{x}{a})}\right)$
where $a>1$
I am trying to get to a result withou using any advanced methods or even things such as l'Hospital's rule etc..
I got to a phase where I took the limit of the first logarithm which we can see tends to 0 from rewriting it as $\ln <!--\; \; -->a^x$. Then I wanted to make some adjustments to the second part of the expression and I got to the stage where I have the limit of
$(\ln <!--\; \; -->a^2)\left(\frac{x-<!--\; -\; -->a}{a}\right)$
That wont give me exact result but I should be able to justify that it the expression is defined and by multiplying it with the first limit which is $0$, the result should also be $0$
Can somebody please tell me how correct or wrong I am? Thanks.

If a polynomial is evaluated over an interval (by the use of interval arithmetic and Horner's method), what does the result say about the polynomial's roots?
I mean, if the resulting interval contains zero, does it mean there must be one more more roots inside the original interval? Could it happen that there might be no zero?
If the resulting interval contains no zero (i.e. both of its end-points have the same sign), does it mean there is no polynomial root inside the original interval? What if one of the end-points of the result is zero?

What is the remainder on division of $z^{400}+z^{303}+1$ by $z^4-<!--\; -\; -->1$?
I am asked to determine the remainder when the polynomial $f(z)=z^{400}+z^{303}+1$ is divided by the polynomial $g(z)=z^4-<!--\; -\; -->1$.
I expressed f as $f(z)=h(z)(z^4-<!--\; -\; -->1)+r(z)$ where r(z) is a polynomial
Realising that $z^4-<!--\; -\; -->1=(z^2-<!--\; -\; -->1)(z^2+1)=(z-<!--\; -\; -->1)(z+1)(z^2+1)$, I expressed f as
$f(z)=h(z)(z-<!--\; -\; -->1)(z+1)(z^2+1)+r(z)$
I also wrote f as $f(z)=q(z)(z-<!--\; -\; -->1)+3$, since $f(1)=3$, and q is some polynomial.
So I had $f(z)=q(z)(z-<!--\; -\; -->1)+3=h(z)(z-<!--\; -\; -->1)(z+1)(z^2+1)+r(z)$
Hence, for some constant k, $q(z)=h(z)(z+1)(z^2+1)+k$
We know $f(-<!--\; -\; -->1)=1$
From $f(z)=q(z)(z-<!--\; -\; -->1)+3$, we have $f(-<!--\; -\; -->1)=q(-<!--\; -\; -->1)(-<!--\; -\; -->2)+3=1$, which implies $q(-<!--\; -\; -->1)=1$.
Then $q(-<!--\; -\; -->1)=h(-<!--\; -\; -->1)(0)+k=1$. Thus, $k=1$
We then have $q(z)=h(z)(z+1)(z^2+1)+1$, and from this, we have $f(z)=q(z)(z-<!--\; -\; -->1)+3=(h(z)(z+1)(z^2+1)+1)(z-<!--\; -\; -->1)+3$. Which simplifies to $f(z)=h(z)(z^4-<!--\; -\; -->1)+z+2$
And gives $r(z)=z+2$.
But I am given the answer is $r(z)=z^3+2$. What is going wrong here? Can I use modular arithmetic instead? How?

What is 2.8 as a fraction?

Missy used a coupon 20% off the price of her meal at a buffet restoraunt and received $3 discount. What was the original price of Missy's mea; before the discount?

Find the slope of the line perpendicular to ${\displaystyle y=\frac{8}{5}x-3}$

Find a standard form equation for the line with A(2, −4) and is parallel to the line 5x-2y=4

partial fraction decomposition of $\frac{(2l+1)(l+1)}{K+(l+1)(l+2)(l-<!--\; -\; -->1)}$
I am struggling with the partial fraction decomposition of
$\frac{(2l+1)(l+1)}{K+(l+1)(l+2)(l-<!--\; -\; -->1)},$
where $K$ is a positive real number.
I would like to decompose it in a sum of simpler polynomial fractions.
I tried the Mathematica's function "Apart" but it is returning the fraction so I am thinking there is no simple partial fraction decomposition.
Could you help me?

Simplify. Assume that variable a is nonzero.
$(2a^{-<!--\; -\; -->4}{)}^{-<!--\; -\; -->4}$

Tealeaf, a company that offers management solution to companies that sell online, announced in a 2008 consumer survey that nearly 9 out of 10 customers reported problems with transactions online. The survey sampled 2010 adults in the United States, 18 years and older, who had conducted an online transaction in the past year. Estimate the number of adults who reported problems with transactions online.

Can all programs be modeled as operations of elementary arithmetic operations on inputs?
In mathematics and computabiltiy theory, we treat all inputs and intermediate results and final outputs as natural number. While algorithms/programs themselves are considered natural numbers, here we treat these programs/functions/algorithms as just computable functions.
The question is, when the function operates on an input to produce an output, can we consider the operation of function as using only a number of arithmetic operations (addition, subtraction, multiplication and division) on an input? Or does the use of if/else make the aforementioned not true?
If this is true, is the number of arithmetic operations polynomially proportional to the lowest time complexity bound possible for solving a problem? (That is, if the lowest time complexity is O(whatever), then the number of arithmetic operations is ${\text{O(whatever}}^k)$ where k is some rational number.)

Evaluation of polynomial modulo in GF(2)
I'm new to this forum and I have a question about about modular arithmetics. My background is in computational chemistry and I work now as a software developer. For my work I need to learn about cryptography and I'm reading the book "Cryptography Made Simple" by Nigel P. Smart.
In paragraph 1.2 on Finite Fields, the author gives an example of polynomial modulo in ${\mathbb{F}}_2(x)/f(x){\mathbb{F}}_2(x)$ as such:
$(1+x+x^2)(x+x^3)(\mathrm{mod}{x}^4+1)=x^2+1$
However, when I do the evaluation myself I get a different result, which is confirmed by using the PolynomialMod function in Mathematica. I will show the steps I took below. First of all I expand the multiplication and evaluate modulo 2:
$x^5+x^4+2x^3+x^2+x(\mathrm{mod}2)=x^5+x^4+x^2+x$
Then:
$x^5+x^4+x^2+x(\mathrm{mod}{x}^4+1)$
$x(x^4+1)+x^4+x^2(\mathrm{mod}{x}^4+1)$
$x(x^4+1)+(x^4+1)+x^2-<!--\; -\; -->1(\mathrm{mod}{x}^4+1)$
Which finally results in:
$(x+1)(x^4+1)+x^2-<!--\; -\; -->1(\mathrm{mod}{x}^4+1)=x^2-<!--\; -\; -->1$
I put in attachment a screenshot 1 from Mathematica to confirm it matches my result. Initially I thought it was an error in the book, but in a following section the author gives another example:
$(x^3+1)(x^4+1)(\mathrm{mod}{x}^7+x+1)=x^4+x^3+x$
Again, evaluating the expression manually I obtain:
$(x^3+1)(x^4+1)(\mathrm{mod}{x}^7+x+1)=x^4+x^3-<!--\; -\; -->x$
Which is confirmed once more by Mathematica, see screenshot 2.
Does anyone know if there is a reason behind the difference in the results? Any help would be much appreciated.