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Hey, there's no need to be so rude. If I were doing your homework for you, I definitely would've written a problem for you to solve. But that's just because I'm a really nice and generous person. You don't always have to ask for help, you know, sometimes I just like to volunteer it.

No one asked you hw cheater!

You are so shameless! This is obviously an a o p s question! The picture is from an ao ps link! But of course this question will be answered since this corrupt site has the same regard for academic honesty as Brainly. I mean, at least Brainly pretends they care! We don't!

nvm 51000 was the anwser 

I got 26000 but they say its wrong

cause 

The first digit guaranteed to be = 1 

The next letter can be all possible = 26

Then all the next have to be either ( 0 ,1 , 2 , 3 , 4 , 5 , 6 ,7 , 8 , 9)

So : 26 x 10 x 10 x 10 = 26000

a) Let's evaluate 2x^3 + 5y^2 and x^3 + 5xy^2 for x = 7 and y = 11.

For x = 7, ad y = 11:

2(7)^3 + 5(11)^2 = 2(343) + 5(121) = 686 + 605 = 1291

for the second expression:

(7)^3 + 5(7)(11)^2 = 343 + 5(7)(121) = 343 + 5(847) = 343 + 4235 = 4578

So the first expresion is equal to 1291 and the second one is equal to 4578.

b) Now, let's find the values of x and y for which 2x^3 + 5y^2 is equal to x^3 + 5xy^2 and also the values for which they are not equal.

To find the values for which they are equal, we set the two expressions equal to each other:

2x^3 + 5y^2 = x^3 + 5xy^2

Simplifying the equatoin:

2x^3 - x^3 = 5y^2 - 5y^2

x^3 = 5y^2(x - 1)

Now, we can see two scenes where the expressions are equal:

When x = 0, the equaton becomes 0 = 0, which is true for all y (all real numbers).

When x = -y, the equation becomes (-y)^3 = 5y^2(-y - 1).

Simplifying further:

-y^3 = -5y^3 - 5y^2

4y^3 + 5y^2 = 0

This equation is true when y = 0 or y = -5/4.

So, the values for which the expressions are equal are: (x = 0, y is any real number) and (x = -y, y = 0 or y = -5/4).

On the other hand the values for which the expressoins are not equal, we will consider any non-zero value of x and y that do not satisfy the condtions we found above So if x and y are both nonzero numbers and they are not negotiations of each other (x is not eqaul to -y) then the expressions are equal.

Now that I think about it I guess talking to trolls wasn't the best idea (but again I had nothing to do). Anyways, I wasn't given any prior notice.

TB

When we open 1 side of the equation  (x-4)^3 we get :

x^3 - 12x^2 + 48x - 64 =8

Now we simply 

x^3 - 12x^2 + 48x = 72 

Now we move everything to 1 side to isolate then set = 0 , then factor 

thus x=6 

>>> maybe I'll have to start adding 'TB' to the end of my guest posts <<<   

If you really are the banned poster managing to gain access through a guest account, you'd be well-advised to squelch any urges to resume the type of activity that got you banned in the first place.  The administrators can identify, and block, the IPA you post from.  If you are really the TB you imply that you are, then from now on. please make only "some actual contributions to this forum" and don't engage in feuds and bickering.  For what it's worth, my opinion is that you weren't the worst offender ... I would have issued an official warning to you, first.  Or maybe they did, and you didn't heed it.  

Well thanks a lot!

I apologize if calling out blatant ChatGPT answers and talking to trolls out of boredom is not allowed. Throughout, I have remained civil and not resorted to vulgar name-calling or creating alts. I was messaged vulgar hand gestures by an alt of the troll. I have made some actual contributions to this forum also. But if admin thinks this is too much, so be it, it's his choice. But consider this as an official reinstatement request. If this doesn't work, maybe I'll have to start adding 'TB' to the end of my guest posts :).

TB

Let the 3 consecutive co-efficients be 

     \(\binom{n}{r-1},\qquad \binom{n}{r},\qquad \binom{n}{r+1}\\ \text{And these are in the ratio }\qquad 1:7:35\\~\\ so\\ \binom{n}{r-1}=\frac{1}{7}\cdot \binom{n}{r}\qquad \color{red}{(1)} \color{black}{\quad and } \qquad \binom{n}{r+1}=5\cdot \binom{n}{r}\quad \color{red}{(2)}\\~\\ \)

\(\binom{n}{r-1}=\frac{1}{7}\cdot \binom{n}{r}\qquad \color{red}{(1)} \\~\\ LHS=\frac{n!}{(r-1)!(n-r+1)!}\\~\\ LHS=\frac{n!}{\frac{r!}{(r)}(n-r)!(n-r+1)}\\~\\ LHS=\frac{n!\qquad (r)}{r!(n-r)!\quad (n-r+1)}\\~\\ LHS=\binom{n}{r}\frac{ (r)}{ (n-r+1)}\\~\\ so\\ \frac{1}{7}=\frac{r}{n-r+1}\\ n-r+1=7r\\ n=8r-1 \)

Now simplify equation 2 and then solve simulataneously to find n and r.     Then check your answer

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LaTex:

\binom{n}{r-1},\qquad \binom{n}{r},\qquad \binom{n}{r+1}\\
\text{And these are in the ratio  }\qquad 1:7:35\\~\\
so\\
\binom{n}{r-1}=\frac{1}{7}\cdot \binom{n}{r}\qquad \color{red}{(1)} \color{black}{\quad and } \qquad 
 \binom{n}{r+1}=5\cdot \binom{n}{r}\quad \color{red}{(2)}\\~\\

\binom{n}{r-1}=\frac{1}{7}\cdot \binom{n}{r}\qquad \color{red}{(1)} \\~\\
LHS=\frac{n!}{(r-1)!(n-r+1)!}\\~\\
LHS=\frac{n!}{\frac{r!}{(r)}(n-r)!(n-r+1)}\\~\\
LHS=\frac{n!\qquad (r)}{r!(n-r)!\quad (n-r+1)}\\~\\
LHS=\binom{n}{r}\frac{ (r)}{ (n-r+1)}\\~\\
so\\
\frac{1}{7}=\frac{r}{n-r+1}\\
n-r+1=7r\\
n=8r-1