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I added the full thing

What is the full question?  

x= 1 , x= 3/2. 

Utiliser la formule quadratique

Solve for x:
2 x^2 + x - 6 = 4 x^2 - 4 x - 3

Subtract 4 x^2 - 4 x - 3 from both sides:
-2 x^2 + 5 x - 3 = 0

The left hand side factors into a product with three terms:
-(x - 1) (2 x - 3) = 0

Multiply both sides by -1:
(x - 1) (2 x - 3) = 0

Split into two equations:
x - 1 = 0 or 2 x - 3 = 0

Add 1 to both sides:
x = 1 or 2 x - 3 = 0

Add 3 to both sides:
x = 1 or 2 x = 3

Divide both sides by 2:
Answer: | x = 1                   or                    x = 3/2

The translation is a bit off with the google translate. Im a russian native and what hes asking is "what kind of formula or steps can you prove this function exist when graphed. I think hes asking about a identity.

Bratan, u tebya yest screenshot? Mozesh tut dat screenshot ha eto website?  a potom mozeme sprashiovat tvoy vapros. Ya ne dumayu tvoyu vapros kak i ne vizu nomeri.

That is not very helpful orestis :/

Это не очень полезно orestis

Την επόμενη φορά προσπαθήστε να τα γράφετε στα αγγλικά και όχι στα ρώσικα

Help me please.

Check whether the function z = z (x, y) satisfies the differential equation. How can this be done?

Mmm 

What differential equation?

It may be my own fault but I am not understanding your question.  Sorry.

Какое дифференциальное уравнение?
Возможно, это моя вина, но я не понимаю ваш вопрос. Сожалею.

Here you go.

You can do the addition

Nevermind i figured it out.

where is the fraction button?

\(Give\ [LaTeX] \)

\( Delete\)   x = {-b \pm \sqrt{b^2-4ac} \over 2a}

\(operate \ [\ \underline{Math}\ ]\)

\(operate \ [\ \Box/\Box\ ]\)

\(Give \ numerator \ and \ denominator \ in\) \\(frac\) {1}{3}

\(Give \ OK\)

\(\frac{1}{3}\)

  !

Chris, why are you doing all of Failurewithasmile's homework for her ?

I have never even seen her say please or thank you.... (not to you or to MJ or to anywone else) !

More importantly l have never seen her query an answer which suggests rather plainly that she is not learning anything at all, she is just cheating.

Failurewithasmile be warned, if you continue to spam the forum like this the Ghost of Web2 is likely to retaliate and your questions may all disappear.

deleted - added on a new question.

We have that

P(LC l S )  =  P( LC and S) / P(S)

.205   =  P(LC and S) / .39        multiply both sides by 39

(.39) (.205)  = P(LC and S)   =  .07995 ≈ .08  

Pay for "A"   - Pay for "B"   =

16 *400  - [ 1300  + 1300(1.10) + 1300(1.10)^2 + 1300(1.10)^3]  =

$366.70  .......just as MJ  found......!!!!!

Let A  = P(hot dog)  = .59

Let B  = P(soda pop)   = .88

And let P(A and B)  = .48

And we want to find this

P(B l A)  =   P(A and B) / P(A)   =   .48 / .59   =  .2832  ≈  .28

Asinus....you have given the odds that   "x"  is no  more than 1 unit away from "A,"  not the probability

To see the difference .....look at the following image :

The probability that we choose the green area in the above circle  =

[ area that is green] / [ area that is green + area that is not green ]  =

[1/4 of circle] / [ 1/4 of circle + 3/4 of circle] =

[1/4 of circle] / [ whole circle]   =    1/4

But the odds that we choose the green area  =

[area that is green] / [area that is not green]  =

[1/4 of circle] / [ 3/4 of the circle] =    1/3   =  1 : 3

                    Truck                 Car              Total

Female           5                      35                40

Male              27                     21                48

Total              32                     56                88

P(buying a car given a male )  =  21/88 = .238 ≈ 24%

Yep...geno is correct

Check :

-24(1/3)^(n - 1)    for terms 1, 2, 3  = -24, -8, -8/3 = -2. 6....

P ( 2 defective cans from 50)  =

14% defective rate means that 7 cans out of 50 should be defective

So.....we have a 7/50 chance of getting a defective can on the first draw   and a 6/49 chance of getting a defective can on the second draw.....so

7/50  * 6/49  =

7/49 * 6/50  =

1/7 *  3/25  =

3/175

 Team        Jersey         No Team         JerseyTotal

Female           9                17                       26

Male             36                  8                       44

Total             45                 25                      70

Probabilty that  a person is male given that they are wearing a team jersey  =

36/45  =  4/5  = .80   = 80%

To test whether this is independent

Let A  = probability of being a male  =  44/70

Let B = probability of wearing a team jersey  =  45/70

P(A ∩ B)   = P(A) x P(B)  ???

.80  = (44/70) * (45/70)   ??

.80  =  .404       not independent

Woops! You're right... I don't know how I misread that! It should be 3a, not 6a....

And I figure the new shape is a rectangle, and the old shape is a square.

Please explain: How can a SQUARE have a side of 6cm and 9cm? Also, how can the AREA increase from a to 6a, when the area is supposed to "triple"?? Thanks.

We have that

-351  =  a₁ (r)^(4 - 1)

-351  = a₁ (r)^3           divide both sides by r^(3)

-351/  r^3   =  a₁         (1)

And

85293  = a1 (r)^(9 -1)

85293  = a1 (r)^8             sub (1)  into this

85293 = [ -351/r^3] r^8     

85293 = -351 r^5             divide both sides by -351

-243  = r^5                     take the 5th root of both sides

(-243)^(1/5)  = r  =  - 3

Using (1)

-351 / (-3)^3   = -351 / -27 =  a₁  = 13