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See my answer at your original post https://web2.0calc.com/questions/hey-guys-really-need-help-hints-would-even-help-really-stuck#r10

Let ABCD be a trapezoid with bases AB and CD.  Let AD = 5 and BC = 7, and let P be a point on side CD such that CP/PD = 7/5.  If Q is a point on AB such that PQ is parallel to AD, then find AQ/QB.

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https://web2.0calc.com/questions/trapezoid_11

Everything depends on this given assumption
\[
    2f(a^2+b^2) = [f(a)]^2 + [f(b)]^2\qquad\qquad(*)
\]
for all nonnegative integers $a$ and $b$.

Letting $b=0$ in $(*)$ we have another general formula
\[
    2f(a^2) = f(a)^2+f(0)^2\qquad\qquad\qquad(**)
\]
for all nonnegative integer $a$.

Letting $a=b=0$ in $(*)$ we get 
\[
    2f(0) = 2f(0)^2.
\]
Solving for $f(0)$ we get $f(0) = 0, 1$.

Letting $a=1$ in $(**)$ we get 
\[
    2f(1)=f(1)^2+f(0)^2.
\]
When $f(0)=0$, we solve for $f(1)$ to get $f(1) = 0, 2$;
when $f(0)=1$, we solve for $f(1)$ to get $f(1)=1$.

The current situation is like this

   | f(0) | 0 | 0 | 1 |
   | f(1) | 0 | 2 | 1 |

Setting $a = b = 1$ in $(*)$ we get 
\[
    f(2) = f(1)^2
\]
This implies that $f(2) = 0$ when $f(1) = 0$;
$f(2) = 4$ when $f(1) = 2$;
and $f(2) = 1$ when $f(1) = 1$;

The current situation is like this

 
   | f(0) | 0 | 0 | 1 |
   | f(1) | 0 | 2 | 1 |
   | f(2) | 0 | 4 | 1 |
 

Setting $a=2$ and $b=1$ in $(*)$ we have
\[
    2f(5) = f(2)^2 + f(1)^2
\]

When $f(1)=f(2)=0$, we solve for $f(5)$ to get $f(5) = 0$.
When $f(1)=2$ and $f(2)=4$, we solve for $f(5)$ to get $f(5)=10$.
When $f(1)=f(2)=1$, we solve for $f(5)$ to get $f(5)= 1$.

The current situation is like this

   | f(0) | 0 |   0 | 1 |
   | f(1) | 0 |   2 | 1 |
   | f(2) | 0 |   4 | 1 |
   | f(5) | 0 | 10 | 1 |
 

Letting $a=5$ in $(**)$ we get
\[
    2f(25) = f(5)^2 + f(0)^2
\]

When $f(0)=f(5)=0$, we solve for $f(25)$ to get $f(25) = 0$.
When $f(0)=0$ and $f(5)=10$, we solve for $f(25)$ to get $f(25)=50$.
When $f(0)=f(5)=1$, we solve for $f(25)$ to get $f(25)= 1$.

The final situation is like this

   | f(0)  | 0 |   0 |  1 |
   | f(1)  | 0 |   2 |  1 |
   | f(2)  | 0 |   4 |  1 |
   | f(5)  | 0 | 10 |  1 |
   | f(25) | 0 | 50 | 1 |

So $n = 3$ and $s=0+50+1=51$, and $n\times s = 3\times 51 = 153$.
 

4X=35N+1   where N is an integer and x is between 0 and 34

4*9=35+1     (by inspection)

so  \(4^{-1}=9 \mod{35}\)

I don't know if this is the only answer  

Thanks Jugoslav :)

here is the original thread.  NaturalDisaster failed to say so but this is a repost.

I am disappointed to see that you have reposted this without referencing this original post.

If you repost you must give the address of the original and request that people answer on the original thread.

Thank you.

Also, if you believe an answer is wrong please give your reason.  (Even if It is just 'I put it in the answer box and was told it was wrong')

Repost:

Thank you so much! :DDD

No it is fine.  They are exactly the same.

nC1 always equals n

Think about it, there are n ways to choose an item from n.

How about proofreading your question!

"Note: All solutions should be expressed in the form , where  and  are real numbers."

Could you perhaps fill in the blanks for me? Thank you! =^-^=

Thank you for helping!

A question though: would it also work if you split up the 6C3 into, say

6C1, 5C1, 4C1?

So the result would be:

4*4*6C1*5C1*4C1

or

4*4*6*5*4?

I'm not sure if this is an one in a million accidentally correct answer.

Thank you for your help once more =^-^=

Answer: 

44/3

Solution:

Hi Guest! Sorry for the late help 

Let's get  \(\mathrm{Line_1}\) into standard form first, yeah?

\(\mathrm{Line_1}\) = 

3x+4y=-14

      4y=-3x-14

       y=-3x/4 -7/2

So now that \(\mathrm{Line_1}\) is now happily in standard form, let's start looking for \(\mathrm{Line_2}\)'s equation.

First, let's find the slope.

To find a perpendicular line, you have to take the opposite reciprocal of \(\mathrm{Line_1}\)'s slope, which is -3/4.

So \(\mathrm{Line_2}\)'s new slope is 4/3. 

What we have so far for \(\mathrm{Line_2}\): y=4/3 x + b

Now we need to find b.

We can do that by plugging in (-5,7), a point on the line, into x and y in the partially done equation for  \(\mathrm{Line_2}\).

y = 4/3 x + b <-- before plugging in

7 = 4/3 * -5 + b <-- after plugging in

7 = -20/3 + b <-- simplifying

41/3 = b <-- even more simplifying

b = 41/3 <-- just flipping it nothing here to see

So we have the full equation for \(\mathrm{Line_2}\), or, y = 4/3 x + 41/3.

From here, we look at the question again. What is it asking for?

"find m+b"

m = 4/3, and b= 41/3. 

Hence, the answer is 44/3.

=^-^=

If you find a mistake, please let me know in the replies!

o e o e o e o e o  

For it to alternate it must start and finnish with an odd    

5 odd places and 4 even places

5!*4! = 2880 

Just as TheOddOne already told you.   

Ok I will try and explain.

Your answer assumes that the 3 objects are identical.  But they are not, they are different

So if you took your answer of  4*4*6C3  and then ordered the last 3 you would have to multiply by 3! (or 6)

4*4*6C3*6 = 1920 

Now the two answers are the same  

Final Answer: 

2880

Reasoning:

Odd numbers: 1,3,5,7,9

Even numbers: 2,4,6,8

So there are 5 odd numbers and 4 even numbers.

^ The number possibilities

For the first box, there are five possibilities, because there are five odd numbers.

For the second slot, there are four because there are four even numbers.

For the third, there are four possibilities, because there are four odd numbers left, since you can not use whatever was the first option.

For the fourth slot, there are three possibilities, because there are three even numbers remaining.

The pattern continues, until we are down to the ninth box, the unit digit, where there is only one possibility remaining.

Multiply what you get from the slots:

5*4*4*3*3*2*2*1*1 

= 2880

Please reply if you find a mistake lurking!

=^-^=

Solve
\(xy + x + y = 23 \\ yz + y + z = 31 \\ zx + z + x = 47\)
in real numbers.

\(\begin{array}{|rcll|} \hline xy + x + y &=& 23 \\ (x+1)(y+1) -1 &=& 23 \\ \mathbf{(x+1)(y+1)} &=& \mathbf{24} \qquad (1) \\ \hline yz + y + z &=& 31 \\ (y+1)(z+1) -1 &=& 31 \\ \mathbf{(y+1)(z+1)} &=& \mathbf{32} \qquad (2) \\ \hline zx + z + x &=& 47 \\ (z+1)(x+1) -1 &=& 47 \\ \mathbf{(z+1)(x+1)} &=& \mathbf{48} \qquad (3) \\ \hline \end{array}\)

\(\text{Let $a=x+1,~b=y+1,~c=z+1$}\)

\(\begin{array}{|rcll|} \hline ab &=& 24 \qquad (4) \\ bc &=& 32 \qquad (5) \\ ca &=& 48 \qquad (6) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(4)*(6)}{(5)}: & \dfrac{ab*ca}{bc} &=& \dfrac{24*48}{32} \\ & a^2 &=& 36 \\ & \mathbf{a} &=& \pm \mathbf{6} \\ & x+1 &=& \pm 6 \\ & x_1 &=& 5 \\ & x_2 &=& -7 \\ \hline & \mathbf{(x+1)(y+1)} &=& \mathbf{24} \\ & (\pm6)(y+1) &=& 24 \\ & 6(y_1+1) &=& 24 \\ & y_1 +1 &=& 4 \\ & y_1 &=& 3 \\\\ & (-6)(y_2+1) &=& 24 \\ & y_2 +1 &=& -4 \\ & y_2 &=& -5 \\ \hline &\mathbf{(z+1)(x+1)} &=& \mathbf{48} \\ & (\pm6)(z+1) &=& 48 \\ & 6(z_1+1) &=& 48 \\ & z_1 +1 &=& 8 \\ & z_1 &=& 7 \\\\ & (-6)(z_2+1) &=& 48 \\ & z_2 +1 &=& -8 \\ & z_2 &=& -9 \\ \hline \end{array}\)

Yes I know you were not trying to be mean. I am well aware of that.

But you ignored the questions that I asked you and just reposted,  That was downright rude.

I accuse you of rudeness, not meaness.  I see no evidence that you are seriously trying to learn. 

If you were trying to learn you would have responded to my questions.

This is my response that you chose to ignore.

If you did not understand my response then you cannot do the question.

Reposting was downright rude.

You have not shown us anything at all that you have done. 

You have presented 2 numbers that could have come from anywhere.

$N$ divided by 3 leaves a remainder of 2 means $N = 3k+2$ for some integer $k$.  $N$ divided by 4 leaves a remainder of 2 means $N = 4\ell+2$ for some integer $\ell$.   It follows that $3k+2=4\ell+2$, so that $3k=4\ell$.  This tells us that $\ell = 3m$ for some integer $m$.   Replacing $\ell$ by $3m$ in the equation $N = 4\ell+2$ should give you the answer.

By Vieta's Theorem the equation $3x^2+5x=7=0$ gives you the values of $u+v$ and $uv$.   Noticing that $u^2+v^2 = (u+v)^2 - 2uv$ should tell you what to do.

i got 18 guys. 3*3*2, because if u use a b c d as the people, d is youngest so 12*2-6=24-6=18

I'm not Phill or Melody, but I think that I can help you. 

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First, I'm gonna correct your question:

"The total length of the lines she draws is 180 cm."     (NOT 18 cm)

On the second piece of paper, Pat drew only 4 lines, and each line was 18 cm long. 

From the assumptions it follows that 2, 3, and 5 are all divisors of $n$.   Also $n$ should not be divisible by any other primes because if it were then dividing $n$ by the product of all these other primes would result in still another integer satisfying all the given assumptions.

Thus we conclude that $n= 2^a\cdot 3^b\cdot 5^c$ for some nonnegative integer $a$, $b$, and $c$.

The assumption $2n$ being a perfect square then implies $2 \mid a+1$, $2 \mid b$, and $2 \mid c$.

The assumption $3n$ being a perfect cube then implies $3 \mid a$, $3 \mid b+1$, and $3 \mid c$.

The assumption $5n$ being a perfect fifth power then implies $5 \mid a$, $5 \mid b$, and $5 \mid c+1$.

So $a$ has to be a multiple of ${\rm lcm}(3, 5)$ such that $a+1$ is even,
$b$ has to be a multiple of ${\rm lcm}(2, 5)$ such that $b+1$ is divisible by 3,

and $c$ has to be a multiple of ${\rm lcm}(2, 3)$ such that $c+1$ is divisible by 5.

The least numbers satisfying these conditions are $a=15$, $b=20$, and $c=24$.  

Thus $n=2^{15}\cdot 3^{20}\cdot 5^{24}$.

Continuing the pattern from the 3 by 3 piece of paper, you get that the total length is 36 for a 6 by 6 piece of paper.