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} $.

Problem Sketch

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 $ ๐‘ฆ = (๐‘ฅ ^ 2 – 1) (๐‘ฅ – 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 $ ๐‘ฆ = (๐‘ฅ ^ 2 – 1) (๐‘ฅ – 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 โ€‹โ€‹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 $.