## Precalculus of algebra: how to calculate the area of ​​the triangle with perpendicular lines

1. In the following figure, the lines have slopes of 3 and 5. The lines intersect at (10,15). How far is it between the x-intercepts of the lines?

The equation of the line with slope of 3 is $$y = 3x + b$$ and with the point $$(10,15)$$ is $$15 = 3 (10) + b$$ then the equation is $$y = 3x-15$$

Similarly, the slope is 5. Then the equation is $$y = 5x-35$$

The x interceptions are $$x = 5$$ for $$y = 3x-15$$ Y $$x = 7$$ for $$y = 5x-35$$. The distance between the two 2. Is this correct?

1. In the following figure, the two lines are perpendicular, and intersect in $$(6.8)$$. The intersections in and of the lines have a sum of zero. Find the area of ​​the shaded region.

the equations of the lines are $$y = mx + b$$ Y $$y = – ( frac {1} {m}) x-b$$ since the intersections in and are opposite each other. If I reconnect the point, I have:

$$8 = 6m + b$$ Y $$8 = – ( frac {1} {m}) (6) -b$$

If I put these two equations together, I would have

$$6m + b = – ( frac {1} {m}) (6) -b$$

$$6m = – frac {6} {m}$$

$$6m ^ 2 = -6$$

$$6m ^ 2 + 6 = 0$$

$$6 (m ^ 2 + 1) = 0$$

Here I am stuck because I have a negative root, so I'm pretty sure I tried this problem badly.

## Precalculus of algebra: interrelated constraints through linear combinations

Dice $$x$$ Y $$and$$ Real variables are such that:

$$left | x right | le alpha, left | and right | le alpha,$$

where $$alpha$$ It is a positive constant.
I want to determine the limits of $$u, v$$ where $$u, v$$ are determined by a combination of $$x, y$$ as follows:
begin {align} & u text {} = text {} x text {} – text {} y, \ & v text {} = text {} y. \ end {align}
My approach is:
begin {align} & -2 alpha le u = x text {} – text {} and le 2 alpha, \ & – alpha le v text {} = text {} and le alpha. \ end {align}
Therefore, the limits of $$u, v$$ are :
$$left | u right | le 2 , left | v right | le alpha,$$

I tried to verify my result by transforming the original equation into:
begin {align} & x text {} = text {} u text {} + text {} v, \ & y text {} = text {} v. \ end {align}

Then using the derived restrictions, we can conclude
$$left | x right | le 3 , left | and right | le alpha,$$

which is different from the original statement.

## Precalculus algebra – Order to prove inequalities

at some point when I'm trying inequalities
Example
For positive a, b, c, d and $$abcd = 16$$
Show that $$( frac {a} {b}) ^ 3 + ( frac {b} {c}) ^ {3} + ( frac {c} {d}) ^ 3 + ( frac {d} {a }) ^ 3 + 4 geq a + b + c + d$$

My test But I can not finish …

First, L.H.S we use the power, the inequalities that we will obtain.

$$LHS geq ( frac { frac {a} {b} + frac {b} {c} + frac {c} {d} + frac {d} {a}} {16}) ^ { 3} +4$$
And R.H.S. I multiply with $$frac {2} { sqrt[4]{abcd}}$$ Now the inequality is homogeneous, I will ignore the condition abcd = 16

We have to show that $$( frac { frac {a} {b} + frac {b} {c} + frac {c} {d} + frac {d} {a}} {16}) ^ {3} + 4 geq frac {2 (a + b + c + d)} { sqrt[4]{abcd}}$$

From A.M.-G.M; $$sum_ {cyc} frac {a} {b} geq 4$$…. so at the end

I have $$4 sqrt[4]{abcd} geq a + b + c + d$$ But following with wrong inverse sign ….

So, is there a trick about what I should do first to prove the inequality? A.M-G.M or Cauchy ….. that is
Thank you !

## Precalculus of algebra – Is there a mathematical function for x + x ^ 2 + x ^ 3 + x ^ 4 + x ^ 5 …. etc?

I am a UI programmer and I am trying to calculate an effect related to mouse movement. The behavior that I'm noticing is equivalent to `x + x ^ 2 + x ^ 3 + x ^ 4 + x ^ 5 ...` and so on.

For the purposes of my application, I just have to go to Like x ^ 3 before the result is close enough, but I wonder if there is a mathematical function or operator, ideally a version in common programming languages, designed for this type Of what has more precision.

Thank you!

## Precalculus of algebra: how can I prove that some functions are injective?

Definition of an injective function.

$$f (x_1) = f (x_2) => x_1 = x_2$$

Well I have two functions that I can not prove …

$$f (x) = x ^ 3 + x$$
$$f (x) = frac {x} {1-log (x)}$$
log (x) denotes the base 10 logarithm function, well, it does not matter for this problem, but anyway.

the first function
$$x_1 ^ 3 + x_1 = x_2 ^ 3 + x_2$$
What can I do after this? The same with the second function, how can I manipulate to achieve $$x_1 = x_2$$

## Precalculus algebra – When to apply PEDMAS in reverse?

They told me that when undoing operations in an equation, you should start by following the rules of PEDMAS, but vice versa. For example:

8x + 16 / x = 4x + 16

According to the council, you should start this problem by subtracting 16 / x, since the addition is the first operation performed if you perform the following PEDMAS backwards. However, the online solvers of this problem perform multiplication first (multiplying everything on both sides by x) to undo the division. I have seen similar problems resolved in the same way, with the cancellation of the division as the first operation. My question is when are you supposed to apply the inverse PEDMAS rule? Or is it that the rule really matters?

## Precalculus of algebra: how to simplify \$ frac {3+ log_ {p} 3} {5 ^ 2} = frac {-5+ log_ {p} 5} {3 ^ 2} \$?

I have found this problem in my algebra book and I still can not find the right way to solve it.

The problem is the following:

Find $$log_ {p ^ 5} left (3 ^ 5 times 5 ^ 3 right)$$ since:

$$frac {3+ log_ {p} 3} {5 ^ 2} = frac {-5+ log_ {p} 5} {3 ^ 2} = frac {10} {3 ^ 3 + 5 ^ 3}$$

The alternatives given are:

$$begin {array} {ll} 1. & 2 \ twenty-one \ 3. & frac {1} {2} \ 4. and 3 \ 5. & frac {1} {3} \ end {array}$$

So far, what I tried is the following, but I must say that this process was tediously slow and in the end I could not get the value of $$p$$.

What I started was to obtain a relationship between the functions:

$$log_ {p} 3 = frac {5 ^ 2 left (10 right)} {3 ^ 3 + 5 ^ 3} -3$$

$$log_ {p} 5 = frac {3 ^ 2 left (10 right)} {3 ^ 3 + 5 ^ 3} + 5$$

Then I went to the question, so what is asked is:

$$log_ {p ^ 5} left (3 ^ 5 times 5 ^ 3 right)$$

This is reduced to:

$$log_ {p ^ 5} left (3 ^ 5 times 5 ^ 3 right) = frac {5} {5} log_ {p} 3+ frac {3} {5} log_ {p } 5$$

So all that remained was to replace the first two expressions in the last one:

$$left ( frac {1} {5} right) left ( left[5right] left ( frac {5 ^ 2 left (10 right)} {3 ^ 3 + 5 ^ 3} -3 right) + left[3right] left ( frac {3 ^ 2 left (10 right)} {3 ^ 3 + 5 ^ 3} +5 right) right)$$

$$frac {5} {5} log_ {p} 3+ frac {3} {5} log_ {p} 5 = frac {1} {5} left ( frac {5 ^ 3 left (10 right) -15 left (3 ^ 3 + 5 ^ 3 right) + 3 ^ 3 left (10 right) +15 left (3 ^ 3 + 5 ^ 3 right)} {3 ^ 3 + 5 ^ 3} right)$$

$$frac {5} {5} log_ {p} 3+ frac {3} {5} log_ {p} 5 = frac {1} {5} left ( frac {10 left (3 ^ 3 + 5 ^ 3 right) -15 left (3 ^ 3 + 5 ^ 3 right) +15 left (3 ^ 3 + 5 ^ 3 right)} {3 ^ 3 + 5 ^ 3} right)$$

$$frac {5} {5} log_ {p} 3+ frac {3} {5} log_ {p} 5 = frac {1} {5} left ( frac {10 left (3 ^ 3 + 5 ^ 3 right)} {3 ^ 3 + 5 ^ 3} right)$$

$$frac {5} {5} log_ {p} 3+ frac {3} {5} log_ {p} 5 = frac {1} {5} left (10 right)$$

$$frac {5} {5} log_ {p} 3+ frac {3} {5} log_ {p} 5 = 2$$

Therefore the answer would be $$2$$. However, as mentioned using this procedure it is not necessary to find $$p$$ However, I wonder if I try to find $$p$$ Would you give a whole number or would it be a correct way? The method I used took a lot of time. Is there a solution to be easier or faster? I would appreciate someone helping me with this.

## Precalculus of algebra: find the equation of the line in standard form with slope \$ m = -3 \$ and go through the point \$ (1, frac {1} {3}) \$

They give me the question "Find the equation of the line in standard form with slope $$m = -3$$ and going through the point $$(1, frac {1} {3})$$"

The solution is provided as $$x + 3y = 2$$

I arrived at $$3y + 3x = 4$$

Here is my work:

Point slope formula:
$$y – y1 = m (x – x1)$$

My given slope m is $$– frac {1} {3}$$ and the point is $$(1, frac {1} {3})$$

So:

$$y – frac {1} {3} = – frac {1} {3} (x – 1)$$

Then I multiplied both sides by 3 to get rid of the fractions:

$$3 (y – frac {1} {3}) = 3 (- frac {1} {3} (x – 1))$$

$$3y – 1 = -3 (x – 1)$$

$$3y – 1 = -3x + 3$$

$$3y = -3x + 4$$

$$3y + 3x = 4$$

Where did I go wrong and how can I get to $$x + 3y = 2$$?

## precalculus algebra: find the solution set of a complicated system of conditions (complex numbers)

Leave $$x, and in mathbb {C ^ *}$$ have the following properties:

$$| x | ^ 2 + | and | ^ 2 = 1$$, $$x ^ n + y ^ m> 0$$ ($$x, y$$ they are complex numbers but this sum is strictly positive real) where $$gcd (n, m) = 1$$

Leave $$Delta theta in mathbb {R}$$, $$Delta varphi in mathbb {R}$$ Y $$Delta t in mathbb {R}$$ be such that

$$e ^ {i ( Delta theta + Delta t / n)} = 1$$ (*)

$$e ^ {i ( Delta varphi + Delta t / m)} = 1$$ (**)

Y

$$(xe ^ {i Delta theta}) ^ n + (ye ^ {i Delta varphi}) ^ m> 0$$ (***)

Is it true that, under these conditions, the only solutions are $$Delta t = 2k pi$$, $$Delta theta equiv- frac { Delta t} {n} pmod {2 pi m}$$ Y $$Delta varphi equiv – frac { Delta t} {m} pmod {2 pi n}$$ for $$k en mathbb {Z}$$? I know these are solutions, but are they the only ones?

I've been trying to get this result in vain. I'm not sure of the condition $$| x | ^ 2 + | and | ^ 2 = 1$$ It is useful.

I know that the conditions (*) and (**) imply:

$$Delta theta + frac { Delta t} {n} equiv0 pmod {2 pi}$$

$$Delta varphi + frac { Delta t} {m} equiv0 pmod {2 pi}$$

which is close to what I want but not exactly.

It is much less clear to me what I can get from (***) knowing that $$x ^ n + y ^ m> 0$$. A geometric interpretation helps if $$x$$ Y $$and$$ They are conjugated, but otherwise it is more difficult. As $$x$$ Y $$and$$ are complex numbers, the algebra quickly becomes complicated … I think it might be possible to obtain from this a relationship between $$Delta theta$$ Y $$Delta varphi$$, and maybe conclude that at least $$Delta t$$ is a multiple of $$2 pi$$, but I have not managed to do it in my scribbles.

Any attempt to solve this led me to drown in several equations … Is there a trick I can not see? That's wrong? This comes from an elaborate problem that would take a long time to explain, so maybe it is not true and I made a mistake in my ~ 15 page manuscript. An answer to this question would allow me to conclude (if it is true) some tedious test of a result unrelated to complex algebra.

Any ideas or suggestions?

## precalculus algebra: show the inequality \$ sum x + 6 ge 2 ( sum sqrt {xy}) \$

Leave $$x; Y; z in R ^ +$$ such that $$x + y + z + 2 = xyz$$. Such that $$x + y + z + 6 ge 2 ( sqrt {xy} + sqrt {yz} + sqrt {xz})$$

This inequality is not homogeneous and look at the condition that I thought
I would replace the variables $$x; Y; z$$ such that

+)$$x ^ 2 + y ^ 2 + z ^ 2 + 2xyz = 1$$. Leave $$x = frac {2a} { sqrt { left (a + b right) left (a + c right)}}$$

+)$$xy + yz + xz + xyz = 4$$. Leave $$a = frac {2 sqrt {xy}} { sqrt { left (y + z right) left (x + z right)}}$$.Leave $$x = frac {2a} {b + c}$$

but failed. Please explain to me how I can obtain this substitution (if I have a solution by substitution)

I also tried to solve it by $$u, v, w$$.Leave $$sum_ {cyc} x = 3u; sum_ {cyc} xy; Pi_ {cyc} a = w ^ 3 (3u + 2 = w ^ 3; u, v, w> 0)$$ so $$u le w ^ 3-3u$$ or $$4u le w ^ 3$$ but stagnant (I'm very bad at $$uvw$$)