## algebra precalculus – How to compare which interest rate is better compounded annual vs compounded 3 times a year

Having a little trouble with getting the answer to this question.

: How much better is the return on a 4% yearly interest rate investment that is compounded 3 times per year as opposed to compounded yearly?

I tried to set up the equation as :

10000(1.04)^n = 10000(1+.04/3)^3n n=1

then to compare the them :

10400/ 1.040535704 = 0.999485

I’m guessing I am not setting things up right…

I am suppose to get answer 1% – 1.5% better.

## algebra precalculus – Prove that the square root of distinct prime numbers is irrational

Prove that if $$p_1,…,p_k$$ are distinct prime numbers, then $$sqrt{p_1p_2…p_k}$$ is irrational.

I do not usually prove theorems, so any hint is appreciated. I have taken a look at this and tried to repeat that argument over and over, but I messed up. Perhaps there is an easier to do it. Thanks in advance.

## algebra precalculus – Unable to Solve a quiz question asked in mathematics exam ( Quantitative Aptitude)

I am self studying for an exam and I am unable to solve this quiz question.

I tried by finding numbers in the sentences but couldn’t find and I think that’s a wrong approach.

Can anyone please tell how to solve this question.

## algebra precalculus – How should I write the factors in polynomial division if I know the polynomials roots?

If a polynomial P(x) has a root at x=0.5 and I do polynomial division on P(x), would I divide it by (x-0.5) or (2x-1)? Are both equally valid? Furthermore, would it be okay to write the polynomial in factored form as P(x) = (x-0.5) * Q(x), or should I have integer coefficients?

## algebra precalculus – Actuarial theory of interest question with effective discount

I have spent several hours trying to solve this problem. $$A$$ and $$B$$ both open up new bank accounts at time $$0$$. The principle for $$A$$ (the amount deposited at $$t=0$$) is $$100$$. The principle of $$B$$ is $$50$$ (the amount deposited at $$t=0$$). Each account earns an annual discount rate of $$d$$. The amount of interest earned in $$A$$ during the 11th period is equal to $$X$$. The amount of interest earned in $$B$$ during the 17th period is equal to $$X$$. Calculate X.

Given that we are dealing with effective discount rate, for $$A$$ and $$B$$ we have $$a(t)^{-1} = (1-d)^t$$ clearly the amount of interest earned is $$(1-d)^t$$ for both $$A$$ and $$B$$. This means that $$(1-d)^{11}=(1-d)^{17}$$
Is this the correct set up? if not why? Given that the principles are $$100, 50$$ we have:

$$frac{100}{(1-d)^{11}}=frac{50}{(1-d)^{17}}$$ that is if we are setting the amount functions equal to each other during the $$17$$ and $$11$$ period. I thought this would translate to $$50(1-d)^{11} = 100(1-d)^{17}$$ then substituting $$X$$ for $$(1-d)^{17}$$ and $$(1-d)^{11}$$ we have something that makes no sense $$50X=100X$$ which gives $$X=0$$ after solving. If I don’t substitute for $$X$$ I have $$50(1-d)^{11}=100(1-d)^{17} to 50=100(1-d)^6$$ which translates to $$frac{1}{2}=(1-d)^{6}$$ which gives a decimal less than one. The answer is $$38.88$$ I have no idea where I am going wrong. What am I missing? Am I even any where close on my thought processes?

## Algebra precalculus – polynomial transformations and Vietas formulas

To let $$f (x)$$ be a monic, cubic polynomial with $$f (0) = – 2$$ and $$f (1) = – 5$$. Is the sum of all solutions too $$f (x + 1) = 0$$ and to $$f big ( frac1x big) = 0$$ are equal to what is $$f (2)$$?

From $$f (0)$$ I got that $$f (x) = x ^ 3 + ax ^ 2 + bx-2$$ and from $$f (-1) = – 5$$ The $$a + b = -4$$ However, I'm not sure how to use the transform information to find it $$f (2).$$ It appears that $$(x + 1)$$ is a root for $$f (x + 1)$$ and the same logic applies to $$f big ( frac1x big)$$?

Should I use Vieta's here or what is the right way?

## Algebra precalculus – Prove that \$ X ^ n + aX ^ {n-1} + … + aX-1 \$ in \$ mathbb {Z} cannot be reduced.[X]\$.

Prove that $$X ^ n + aX ^ {n-1} + … + aX-1$$ is irreducible in $$mathbb {Z} (X)$$, Where $$n ge 2$$ and $$a in mathbb {Z} ^ {*}$$.
To let $$f = X ^ n + aX ^ {n-1} + … + aX-1$$. I was able to show this through direct calculations $$f = (X-1 + a) (X ^ {n-1} + X ^ {n-2} + … + X + 1) -a$$ and then I tried to assume that it is not irreducible, i.e. $$exists g, h in mathbb {Z} (X)$$ so that $$f = g cdot h$$ and $$deg g, deg h .
That was not useful because I could only get that $$g (0) = 1$$ and $$h (0) = – 1$$ or $$g (0) = – 1$$ and $$h (0) = 1$$ (which could be obtained from $$f$$Initial form). I also tried to take advantage of the fact that $$f (1-a) = – a$$, but without success.

## Algebra precalculus – Find the equation of the parabola that passes through point (2, -1), whose vertex is at (-7, 5), and that opens to the right.

I've seen several questions like this, but I want to know how to write the equation $$(y-k) ^ 2 = 4p (x-h)$$ form instead of vertex form.

The question is this: Find the equation of the parabola that passes through point (2, -1), whose apex is at (-7, 5), and that opens to the right.

1. When I insert the corner point, the equation is $$(y-5) ^ 2 = 4p (x + 7)$$,
2. I understand that if the parabola opens to the right, $$p> 0$$,

How would i find $$p$$ but for this equation?

## Algebra precalculus – solving after the closed form of the repetition relation, given by \$ x_ {n + 1} = (x_ {n-1}) ^ 2 + n (x_ {n}) \$

Problem: Get closed form of $$x_n$$ using the following: $$x_ {n + 1} = (x_ {n-1}) ^ 2 + n (x_ {n})$$ ; $$x_0 = 3, x_1 = 4$$

Try: I understand $$x_n = n + 3$$ by observation after inserting values ​​of $$x_0$$ and $$x_1$$ to get $$x_2 = 5$$ and then $$x_3 = 6$$,

To prove this, I have proven this by induction $$x_n = n + 3$$ is a valid solution.

Provided $$x_ {k-1} = k + 2$$ and $$x_ {k} = k + 3$$ for an integer $$k$$,
Clear $$x_0$$ and $$x_14$$ to meet the conditions.
Now, $$x_ {k + 1} = (k + 2) ^ 2-k (k + 3) = k + 4$$ as required; This completes the induction.

So we get $$x_n = n + 3$$ for all $$n$$ in whole numbers.

Doubt: How do I show that this is the only solution?

Is there also an alternative method to derive this result? If so, please share, thanks.

## Algebra precalculus – inequality in relation to non-negative numbers

I would like to demonstrate the following inequality:

$$text {max} {s_n ^ 2, s_ {n-1} ^ 2 } + 4 sum_ {i = 1} ^ ns_ {i-1} (s_i-s_ {i-1}) leq 4s_n ^ 2$$

where I follow from $$n + 1$$ real numbers, ALL of which are non-negative, and that $$s_0 leq s_1 leq … leq s_ {n-1}$$ ($$s_n$$ can be any non-negative real number).