## algebra precalculus – What will be the solution set of the equation \$ frac {1} {x} times0 = 0?

What will be the solution set of the equation?
$$frac {1} {x} times0 = 0$$

The problem is, I can't decide if we will take $$0$$ or not because if I take $$0$$ then it becomes undefined by zero, which I don't know what it is equal to. Which is correct $$(undefined) times0 = 0$$ $$O$$
$$(undefined) times0 = undefined$$

So I had these following two questions that are synonyms in the given context (here):

$$1.$$ What is it $$(undefined) times0$$ ?

$$2.$$ What will be the solution set of the equation? $$frac {1} {x} times0 = 0$$ ?

## Algebra Precalculus – Polynomial with Grade 3

Can anyone help me with this problem? Thank you!

Leave $$f$$ a polynomial function such that $$deg (f) = 3$$ Y $$alpha < beta$$, $$alpha, beta in mathbb {R}$$ such that the equations $$f (x) = alpha$$ Y $$f (x) = beta$$ have all different and real zeros.
Show the previous image of the interval $$(?,?)$$ through $$f$$ it's meeting of $$3$$ disjoint intervals, one of them with the length equal to the sum of the length of the others.

$$text {My approach:}$$

used to $$f = aX ^ 3 + bX ^ 2 + cX + d$$ and I tried to get some iphotesys but I don't understand …

## Algebra precalculus: How do I find the minimum force that should be applied to a block that slides on a slope as it moves with constant speed?

The problem is the following:

The figure shows a block on a slope. Find the minimum force that should be applied to the block so that the body of mass $$m = 2 , kg$$ as that body moves with constant speed up the slope. It is known that the coefficient of friction between surfaces is $$mu = 0.3$$ and the angle of inclination is $$alpha = 30 ^ { circ}$$. The alternatives given in my book are the following:

$$begin {array} {ll} 1. & 21 , N \ 2. & 23 , N \ 3. & 18 , N \ 4. and 20 , N \ 5. and 2.2 , N \ end {array}$$

I really need help with this problem. Initially I thought I should break down strength and weight. Which I assumed that the figure of the force is parallel to the floor, which is the base of the inclination.

By doing this and considering the coefficient of friction (which I assumed to be static), this would translate as follows:

$$F cos alpha – mu N = 0$$

The normal or the tilt reaction I found using this logic:

$$N- mg cos alpha – F sin alpha = 0$$

$$N = mg cos alpha + F sin alpha$$

Inserting this in the previous equation:

$$F cos alpha – mu left (mg cos alpha + F sin alpha right) = 0$$

$$F cos alpha – mu mg cos alpha – mu F sin alpha = 0$$

$$F left ( cos alpha – mu sin alpha right) = mu mg cos alpha$$

$$F = frac { mu mg cos alpha} { cos alpha – mu sin alpha}$$

Therefore, inserting there the given information would become:

$$F = frac { frac {3} {10} (2 times 10) cos 30 ^ { circ}} { cos 30 ^ { circ} – frac {3} {10} sen 30 ^ { circ}}$$

$$F = frac { frac {6 sqrt {3}} {2}} { cos 30 ^ { circ} – frac {3} {10} without 30 ^ { circ}}$$

$$F = frac { frac {6 sqrt {3}} {2}} { frac { sqrt {3}} {2} – frac {3} {10} times frac {1} { 2}}$$

This is where simplification becomes ugly:

$$F = frac {3 sqrt {3}} { frac {10 sqrt {3} -3} {20}}$$

$$F = frac {60 sqrt {3}} {10 sqrt {3} -3} approx 27.25$$

Therefore, in the end I get that value for strength. But it is not close to the answers. Can anyone help me with this? What could I have done wrong? How could I simplify this? Can anyone offer an FBD help for this problem?

## (Precalculus) How would you graph \$ 𝑦 = (𝑥 ^ 2 – 1) (𝑥 – 2) ^ 2 \$ by hand?

Hello mathematical exchange of stack.

I have been doing some summer practice tasks for my next calculation class, and I have been assigned the task of graphing $$𝑦 = (𝑥 ^ 2 – 1) (𝑥 – 2) ^ 2$$ by hand.
At first I started to make a table of values, but I quickly realized that both $$x = 1$$ Y $$x = 2$$ to have $$and$$ values ​​of $$0,$$ with a relative maximum in the middle. I am struggling to discover how to find the relative minimum / maximum without the use of my calculator and without the use of differential calculation (thanks google) since I have not yet learned this. If you were in my place, how would you solve this? (Without the use of a calculator).

## precalculus algebra – Factoring in an induction test

I have to try the following:

$$1 + 3 ^ 3 + … + (2n + 1) ^ 3 = (n + 1) ^ 2 (2n ^ 2 + 4n + 1)$$ Inductively.

My intent:

Base case, $$n = 1$$:

$$1 + 3 ^ 3 = (2) ^ 2 (2 cdot1 ^ 2 + 4 cdot1 + 1)$$, which is true.

By inductive hypothesis, suppose $$n = k$$:

$$1 + 3 ^ 3 + … + (2k + 1) ^ 3 = (k + 1) ^ 2 (2k ^ 2 + 4k + 1)$$

by $$n = k + 1$$

$$1 + 3 ^ 3 + … + (2k + 1) ^ 3 + (2k + 3) ^ 3 = (k + 2) ^ 2 (2 (k + 1) ^ 2 + 4 (k + 1) + 1)$$

Using the inductive hypothesis we need to prove that

$$(k + 1) ^ 2 (2k ^ 2 + 4k + 1) + (2k + 3) ^ 3 = (k + 2) ^ 2 (2 (k + 1) ^ 2 + 4 (k + 1) + 1)$$

And here I have the problem, because I do not know how to manipulate any of the sides of the equation to prove this. I tried it on both sides but I can not find the way. One of my last attempts ended here:

$$(k + 1) ^ 2 (2k ^ 2 + 4k + 1) + (2k + 3) ^ 3 = (k + 2) ^ 2 (2 (k + 1) (k + 3) + 1)$$

Which is true for alpha wolfram.

PS: I am aware that there are many documents and information about this test, but I am not looking for another form or something, because I saw many publications on how to prove this claim through induction, but all were used. the last term is (2n-1) ^ 3, and I need to prove it when it is (2n + 1) ^ 3, and the final expression is a bit different. I just need help factoring my last step.

Corrections in the inductive steps that I followed are also appreciated.

## Precalculus of algebra – Polynomial expansion (trick plus / minus in statistics)

Let's suppose the random variables $$X_1, …, X_n$$ They are independent and identically distributed. Leave $$mu_x$$ denotes the expected value of $$X$$. I often see the following plus / minus trick used in statistics, that is,

begin {align *} frac {1} {n} sum_ {i = 1} ^ n (X_i – bar {X}) ^ 2 & = frac {1} {n} sum_ {i = 1 } ^ n (X_i – mu_X + mu_X – bar {X}) ^ 2 \ & = frac {1} {n} sum_ {i = 1} ^ n (X_i – mu_x) ^ 2 – ( bar {X} – mu_x) ^ 2 end {align *}

Can someone formulate this in an identity for me? I mean, something like

$$(a – b) ^ 2 = (a – c + c – b) ^ 2 = (a – c) ^ 2 + (c – b) ^ 2 + text {some crossed term}?$$

What exactly is that term crossed?

In a similar token, I've also seen
begin {align *} frac {1} {n} sum_ {i = 1} ^ n (X_i – bar {X}) (Y_i – bar {Y}) & = frac {1} {n } sum_ {i = 1} ^ n (X_i – mu_X) (Y_i – mu_Y) – ( bar {X} – mu_x) ( bar {Y} – mu_Y) end {align *}

What identity of polynomial expansion is at stake here?

## Precalculus of algebra – Is there any way to maximize the given function?

While working on a question, I came across the following equation

$$c = frac {(2-a_0) + sqrt {-a_0 (3a_0-4)}} {2}$$

We are obliged to maximize the function. $$c$$ and find its corresponding value

I drew his graph in Desmos and found that his maximum was 1/3, however I have no idea how to get to this result.

Nor are we allowed to use calculus (at least it discourages you)
How are we supposed to proceed with this?

## Precalculus of algebra: why is not there a solution set for \$ | x-7 |

They ask me to find the set of solutions for $$| x-7 | <-4$$?

I arrived at $$(- infty, 3) cup (11, infty)$$

by $$x – 7> 0$$:

$$x-7 <-4$$ = $$x <3$$

by $$x-7 <0$$ = $$– (x-7) <- 4$$ = $$-x + 7 <-4$$ = $$-x <-11$$ = $$x> 11$$

Then, I come to a solution of:

$$(- infty, 3) cup (11, infty)$$

However, my textbook says "there is no solution". Why is there no solution?

## precalculus algebra: solve the equation on reals: \$ sqrt {5x ^ 2 + 27x + 25} – 5 sqrt {x + 1} = sqrt {x ^ 2 – 4} \$.

Solve the equation over reais: $$sqrt {5x ^ 2 + 27x + 25} – 5 sqrt {x + 1} = sqrt {x ^ 2 – 4}$$.

This problem is an adaptation of a recent competition. And I can not solve it.

The solutions are $$sqrt {5} + 1$$ Y $$dfrac {13 + sqrt {65}} {8}$$, as Wolfram Alpha says.

## precalculus algebra – Solve fr x y \$ x ^ 2 + y ^ 2 = 1 \$ y \$ x pm y = frac pi4 \$

Solve fr x y $$x ^ 2 + y ^ 2 = 1$$ Y $$x pm y = frac pi4$$

I tried to solve this by a substitute method. And using the quadratic formula, but that creates many cases.
The original problem was to solve. $$arcsin x + arcsin y = frac pi2$$ Y $$without 2x = cos2y$$.