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Thanks Ginger :)     And thanks ilorty too   

Why ?

You do not need to understand latex.

The latex coding that i have posted is for my benefit, and for the possible benefit of someone who specifically wants to learn LaTex, it is not a part of the answer.

The answer is written in normal maths symbols.

(it is rendered LaTex, but it is not LaTex coding)

Just telling you, guest did not go into your computer and steal all those files.

What he/she did was long press (if on mobile) the image, or left click (if on PC or MAC) and select "copy image". This will give the link to the image, which gives the picture.

I think the reason why guest's pictures were not blocked (or sent for moderation) was because they were the web2.0calc links, which the system probably recognizes as 'innocent'. Maybe.

Have a good day Gwen!

:)

18 + 108 + 1008 =1,134

This can be done algebraicly, and without what seems to look like a not understandable copy and pasted answer.

Using some factoring:

6xyz + 30xy + 21xz + 2yz + 105x + 10y + 7z = 812

3x(2yz + 10y + 7z + 35) + (2yz + 10y + 7z) = 812      <------ factor out 3x, and "join like terms"

We see that the difference between the first and the second is "35", so we add 35 to both sides:

3x(2yz + 10y + 7z + 35) + (2yz + 10y + 7z + 35) = 847 

We can factor out a 2y and a 7 from the left:

3x(2y + 7)(z + 5) + 1(2y + 7)(z +5) = 847

Now, we can factor out (2y + 7)(z + 5).

(3x + 1)(2y + 7)(z + 5) = 847

The prime factorization of 847 is 7 * 11 * 11, which can be applied to our factored form (they are both factors....get it?):

So: 

3x + 1 = 7: x = 2

2y + 7 = 11: y = 2

z + 5 = 11: z = 6

Therefore, x + y + z = 10

:)

How was the guest able to post them though?

Hi thank you for the reply. It took me a bit to understand how to do it but I think I got it! So does this mean that the minimum is always zero, seeing as multiplying by zero will always result in that same number?

OK. So what is the question?

The smallest multiple of 11 that is > 1000 is:

x = 1001

The greatest multiple of 11 <11^2 is:

y = 110

x - y =1001 - 110 =891

x - y = 37.

Since angles that are across from each other (formed by two 180 degree lines: anyone know what the term is? I forgot...), we can say that:

6x - 3 = 2x + 13

6x = 2x + 16

4x = 16

x = 4

Therefore, x = 4.

:)

x=0;p=0; y=0;z=0;n=6*x*y*z+30*x*y+21*x*z+2*y*z+105*x+10*y+7*z;if(n==812, goto loop, goto next);loop:p=p+1;printp," =",x,y,z;next:x++;if(x<200, goto4,0);x=0;y++;if(y<200, goto4, 0);x=0;y=0;z++;if(z<200, goto4,0)

OUTPUT:  All the following 5 values will balance the equation, but I think only the 2nd one meets your condition of "positive integers".

       x      y        z

1  = 0     57      2
2  = 2     2       6
3  = 0     35     6
4  = 0     2     72
5  = 0     0     116
 

Wow thanks so much!!

I need to calculate the point in which the graph of f(x) = (1/3)x^3 - 4x + 2 declines the most using the derivative. Could anyone help me? 

The derivative is df(x) = x^2 -4

To find the minimum of df(x) we differentiate again to find ddf(x) = 2x

To find where this is a minimum we set it equal to zero.  Clearly it is zero when x = 0, so this is the point at which f(x) has the steepest negative slope.

Let the 3 numbers be:
n,  n + 2,  n + 4
5n + 2[n + 4] =6[n + 2] + 33, solve for n
5n + 2n + 8 = 6n + 12 + 33
5n + 2n - 6n = 12 + 33 - 8
n = 37 - the smallest number
37 + 2 = 39 - the middle number
37 + 4 = 41 - the larger number.

[700 / 250] x 5.90 =$16.52 cost of 700 grams.

OR:

$5.90 / 250 =0.0236 cents per gram

0.0236 x 700g =$16.52 cost of 700 grams.

[60 / 150] x 2.5 = 1 hour.

Let his speed downstream =S
His speed upstream =S - 3
[S - 3] * 0.75 =S * 0.5, solve for S
0.75S - 2.25 =0.5S
- 2.25 =0.5S - 0.75S
- 2.25 = - 0.25S
S = - 2.25 / - 0.25
S = 9 mph - his speed downstream
9 - 3 = 6 mph - his speed upstream.

The price of 1 gram is 5.90/250 = 0.02
0.02 * 700 = $14 

Not sure if this is the way to do it but if you divide 150 by 2.5 you get 60. So maybe just divide 2.5 hours by 2.5 = 1 hour?

Tá céad fáilte romhat 

Thanks very much guest.        

Thanks Loki and Heureka,

I went with a more straight forward (but not shorter) approach, 

I just divided by (x+1) three times to find the coefficients.

$f(x)=ax^5+bx^4+cx^3\\ f(x)+2=ax^5+bx^4+cx^3+2\\ [f(x)+2]\div(x+1)=[ax^5+bx^4+cx^3+2]\div(x+1)\\ \text{I did this by algebraic division and got a remainer of }-c+b-a+2\\ \text{The remainder must be 0 so}\\ -c+b-a+2=0\\ c=b-a+2$

So now I have

$f(x)+2=ax^5+bx^4+(b-a+2)x^3+2\\ [f(x)+2]\div(x+1)\\\qquad=[ax^5+bx^4+(b-a+2)x^3+2]\div(x+1)\\ \qquad =ax^4+(b-a)x^3+2x^2-2x+2\\~\\ (ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\ \\\text{I did this by algebraic division and got a remainder of }6-b+2a\\ \text{The remainder must be 0 so}\\ 6-b+2a=0\\ b=6+2a\\~\\ (ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\ =(ax^4+(6+a)x^3+2x^2-2x+2)\div(x+1)\\ =ax^3+6x^2-4x+2$

$(ax^3+6x^2-4x+2)\div(x+1)\\ \\\text{I did this by algebraic division and got a remainder of }12-a\\ \text{The remainder must be 0 so}\\ a=12\\~\\ so\\ a=12\\ b=6+2a=30\\ c=b-a+2=30-12+2=20\\ so\\ f(x)=12x^5+30x^4+20x^3$

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LaTex:

f(x)=ax^5+bx^4+cx^3\\
f(x)+2=ax^5+bx^4+cx^3+2\\
[f(x)+2]\div(x+1)=[ax^5+bx^4+cx^3+2]\div(x+1)\\
\text{I did this by algebraic division and got a remainer of }-c+b-a+2\\
\text{The remainder must be 0 so}\\
-c+b-a+2=0\\
c=b-a+2

f(x)+2=ax^5+bx^4+(b-a+2)x^3+2\\
[f(x)+2]\div(x+1)\\\qquad=[ax^5+bx^4+(b-a+2)x^3+2]\div(x+1)\\
\qquad =ax^4+(b-a)x^3+2x^2-2x+2\\~\\
(ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\
\\\text{I did this by algebraic division and got a remainder of }6-b+2a\\
\text{The remainder must be 0 so}\\
6-b+2a=0\\
b=6+2a\\~\\
(ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\
=(ax^4+(6+a)x^3+2x^2-2x+2)\div(x+1)\\
=ax^3+6x^2-4x+2

(ax^3+6x^2-4x+2)\div(x+1)\\
\\\text{I did this by algebraic division and got a remainder of }12-a\\
\text{The remainder must be 0 so}\\
a=12\\~\\
so\\
a=12\\
b=6+2a=30\\
c=b-a+2=30-12+2=20\\
so\\
f(x)=12x^5+30x^4+20x^3