Melody

a has to be 2, 4, or 6

and b has to be 2,4, or 6

so yes,  0.5*0.5 = 0.25

LLNN plus Q

26*1*10*10*10*2 = 52000 - 10*10*10 = 51000

That is assuming the letters have to come first.

If you want me to explain why I subtracted 1000 then ask me and I will explain

I originally have 5x   Blue and  4x red.

I add 4 of each colour

Now I have   5x+4   blue and   4x+4 rad    and the ratio is  7:5

so

\(\frac{5x+4}{4x+4}=\frac{7}{5}\\ so\\ 25x+20=28x+28\\ -8=3x\\ x=-\frac{3}{8}\\ \)

This means that the original number of marbels was not an integer so I think there is an error in your question.

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

You can ask whatever academic thing you want.   It won't necessarily be answered.

I would just suggest that for some things google would quicker and maybe more reliable.

I assume you are cutting the rectangular prism into smaller triangular prisms.

the block of chese does not have a height of half an inch so what do you mean when you say BOTH prisms have a height of 0.5 inches.

If just the triangular prismas have a height of 0.5inch then you can cut 

3*2*6 *2  wedges = 72 wedges.

1 / 72

Edit:

Thus is the ratio of 1 piece to the original block.     It is not actually a weight.

ok, we wouldn't want the Marians to get you.              (that is you disguised as  CPhills )

You are not given enough information to answer this question.  

You need to know something more about the siecond sequence.

Breakstick added the assumption that the Difference for the second sequence was 0.  If this was the case than their answer would be accurate.

Ok That is more your style Ginger.

Why don't you just log on when you post then there would be no (or less) confusion?