Баг Спрашивает - Монополь (или другие знатоки математики) - Отвечают. Без Оффтопа

[BagirkaN]

Мнонополь, вот система уравнений.

6х - 5у =7
4х - 6у = 7

Что нужно делать в первую очередь - то есть КАК узнать, что проще найти первым, х или у.

И вообще, я научилась решать (с горем пополам) системы уравнений, где х или у хотя бы раз даны сами по себе, без цифры.

Вот такое

3х + у = -11
6х - 2у = -2

Здесь я сразу превращаю игрек в -11-3х и переношу во второе уравнение. А с первый системой как поступать??

[BagirkaN]

Нет, правильный ответ 2 в последнем.... то есть, не точка х = -1 Там такой и нет), а "х-value" - =2, - вот чего я не понимаю...

А трайномиял - это состоящий из трех мономиалов.
Там правильный ответ (X^2+5X+25)....
СПАСИБО!
В первой - найти домен фукнкции икса...

[Monopole]

Originally Posted by BagirkaN

Нет, правильный ответ 2 в последнем.... то есть, не точка х = -1 Там такой и нет), а "х-value" - =2, - вот чего я не понимаю...

A funkciya kakaya ? Ta zhe, chto i v pervom primere ?

А трайномиял - это состоящий из трех мономиалов.
Там правильный ответ (X^2+5X+25)....
СПАСИБО!
В первой - найти домен фукнкции икса...

Nu v pervom ya pravilno napisal, vse krome x=0, vtoroj tepr tozhe yasen, trinomial - eto trinom, ya prosto sperva podumal, chto eto prilagatelnoe. A s trejim chto-to ne yasno......

A...tak tam grafik, a ne funkciya iz pervogo primera !! Togda delaesh vot chto, nahodish na osi y tochku "1" provodish in nee liniyu, parallelnuyu osi x, nagodish tochku, v kotoroj eta liniya persekaet grafik, opuskaesh ottuda perpendikulyarna os x, eto i budet zanchenie x.

[BagirkaN]

Класссссссссссссс!!

[ttt]

#2
As Monopole factored it out: x^3-125=x^3-5^3=(x-5)(x^2+5x+25).
The trinomal factor is the factor that is a trinomial - meaning has three terms i.e. it's x^2+5x+25

#3
As I see this problem is different from the first one (f(x)= 8/x+9), right ?

As I undesrtand you are given a line that goes through the points (1,0) and (0,-1), i.e. it intersects the x-axis at at 1 and y-axis at -1. If this is the problem you are given, then:

1. You need to find the equation of the line going through the points (1,0) and (0,-1). One way (not the only way) is to find the slope
m= (y_2-y_1)/(x_2-x_1),
where your points (1,0) = (x_2,y_2) and (0,-1) = (x_1,y_1).

So m=(0-(-1))/(1-0) = 1
and thus the equation of your line is y=mx+b ,
where m is the slope defined above and y is the y-intercepts, i.e. the point where the line crosses the y-axes,
so you have
f(x)=1*x-1 = x-1
then you need to find x such that f(x)=1, so plug the enumbers in to get f(x)=x-1=1 which gives you x=2

#1
Do you mean here f(x)= 8/x+9 or f(x)= 8/(x+9)?
In first case, as Monopole mentioned above you domain is all the real numbers except 0.
In the second case you want to exclude numbers when x+9=0, so x=-9, so your answer will be all real numbers but -9

[BagirkaN]

ttt jan, in my first post today here I gave 3 (three) different problems, but, as I see, Monopole was little bit messed up in them...

[ttt]

Originally Posted by BagirkaN

ttt jan, in my first post today here I gave 3 (three) different problems, but, as I see, Monopole was little bit messed up in them...

Are you OK now?

[BagirkaN]

Originally Posted by ttt

#1
Do you mean here f(x)= 8/x+9 or f(x)= 8/(x+9)?
In first case, as Monopole mentioned above you domain is all the real numbers except 0.
In the second case you want to exclude numbers when x+9=0, so x=-9, so your answer will be all real numbers but -9

I mean 8 divided by x+9

[ttt]

Originally Posted by BagirkaN

I mean 8 divided by x+9

Then see my answer above.

[BagirkaN]

Originally Posted by ttt

Are you OK now?

With # 3 - totally great! Thank you!

(and trying to get the rest part!)

[ttt]

Originally Posted by BagirkaN

With # 3 - totally great! Thank you!

What about others? I explained them as well.

[BagirkaN]

Getting in, thanks...

Also, I would appreciate your explanation on simplifying expressions like 5x-15/x (as a fraction) multiplied by x^3/x-3

Please...

[BagirkaN]

Or on this one: m^2 - 7m/m-2 (fraction) + another fraction 10/m-2

Thank you!

[ttt]

Originally Posted by BagirkaN

Getting in, thanks...

Also, I would appreciate your explanation on simplifying expressions like 5x-15/x (as a fraction) multiplied by x^3/x-3

Please...

I assume you mean
(5x-15)/x * x^3/(x-3) = 5(x-3)/x * x^3/(x-3) =
= ( 5(x-3)* x^3 ) / ( x*(x-3) )
= 5 * x^2

[ttt]

Originally Posted by BagirkaN

Or on this one: m^2 - 7m/m-2 (fraction) + another fraction 10/m-2

Thank you!

Please use parentheses.
I assume you mean ( m^2 - 7m ) / ( m-2 ) + 10/(m-2) =
= ( m^2 - 7m + 10) / ( m-2 ) = (m-2)*(m-5) / (m-2) = m-5

[BagirkaN]

Oh, I'm really impressed by your ability to solve these problems (), but would you please use words to explain what the heck I have to do to solve them!