What is the question??

Note that in standard form (Ax + By = C), the slope of the equation is just \(-{A \over B}\).

So, we have \(-{A \over B} = -{1 \over 3}\), meaning \(3A = B\)

Now, substitute the point \((-7, 2)\) into the equation. This gives us \(-7A + 2B = 3\)

Now, just solve the system for A and B, then do B - A

:)

https://web2.0calc.com/questions/need-help-plz_17

answered on original :)

We know there must be five 1's, two 2's, and four 3's. 

There are \({11! \over 2! \times 5! \times 4!} = \color{brown}\boxed{6930}\) numbers that work. 

How many two-digit numbers are there?

How many numbers can you list out that have at least 1 7?

If you are still confused, feel free to ask for further help 

Thanks Jasonkiln,

You are right of course, I cannot just divide by 2 at the end

.I can tie my two Cs together though. becasue any answer with 2 Cs tied together will not be included and I will have to subtract.

So I have

aauryC    and I will add the other c  later

6!/2! = 360

Now where can the last c go

  |a|a|u|r|y|C

It can go where any of those bars are.   It can't go after the C becasue that would be the same as before the C     Cc = cC

There are 6 bars.

360*6=2160

There are 2160 permutations where 2 or more Cs are together

3360-2160 = 1200 

There are 1200 permutations where no 2 Cs are together.

This is now the same as Ginger's answer,    Builderboi agrees too.  I think his method is similar to mine.

Ginger's method is a little more simple.

need some help

1 x 60 x 60 x 24 x 365 x 15 = 473,040,000

1,2,2,3,6,6,6 ==7 digits

7! / [3!*2!] == 420 permutations

Each of the above 420 permutations begins with one the 7 digits this many ways:

420 / 7 ==60 permutations begin with 1, 2, 1 each

60  x  3 ==180 permutations that are less than 3,000,000

THerefore: 420 - 180 ==240 permutations that are > 3,000,0000

As stated in the question,  "How many different combinations of 3 blocks can be selected from that set so that no two are in the same row or column?"

Order does matter

Wait, I see my mistake... I subtracted 300 instead of 360. 

So the correct answer becomes \(3360- 360-600-1200 = \color{brown}\boxed{1200}\).

Now, GingerAle and I are in agreement. 

just chilling bro, what about you?

The smaller triangle is the medial triangle of the other, so M:T=4

OR

Use similar triangles: The length's ratio is 2, so the area's ratio is \(2^2=4\)

Use Vieta's Formulas:

\(r^2+s^2=(r+s)^2-2rs=(-\frac{4}{3})^2+8=\boxed{\frac{88}{9}}\)

There are 25 ways to choose the first square. To select second square there are 16 ways. To select the third square there are 9 ways. So, number of ways to choose 3 square blocks is 25×16×9=3600.

I assume that the question states that the order doesn't matter, so then each collection of three blocks may be chosen in any of six different orders, which leaves the final answer as 3600/3! = 600

If order does matter the answer would be 3600 instead since no overcounting has been done.

Solution: \(\dfrac{5!}{2!} * \dbinom {6}{3}\)

Description of process:

Removing the ‘C’s from the set gives (5!/2! =60) distinguishable arraignments for the remaining five (5) letters. The three ‘C’s can then occupy three (3) of the six (6) positions between the letters and at either end of the distinguishable arraignments

Graphic demonstrating positions:  _A_U_R_A_Y_   

Select the three (3) positions as \(\dbinom {6}{3} = 20\)

Total number of arrangements where no ‘C’s are adjacent to another ‘C’ \(= (60 * 20) = 1200 \)

GA

--. .- 

In how many ways can the letters of the word ACCURACY be rearranged so that no two C's appear as consecutive letters in the rearrangement?
The word accuracy can be arranged 3360 times

Restatement of question:

How many arrangements of the letters in the word ACCURACY are there, such that no ‘C’s are adjacent to another ‘C’?

--. .- 

Multiply the ratio by 3 to get 15:24.

I assume you can take it from here...

The only constraint is that the leading digit must be a 3 or greater. 

Can you take it from here?