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why isn't it three

See https://web2.0calc.com/questions/counting-problem_49

area of shaded region is called x.

Total area minus blank region= x

lets call area of blank region y

lets call total area z

\(z-y=x\)

\(z = (12\times9)/2\)

\(z=54\)

\(y=\frac{12\times5}{2}\)

\(y=30\)

Plug them in to get

\(54-30=x\)

\(x=24\)

1/2*4*13=26

504/131

Factor b:

\(b(\frac{1}{7}+\frac{b}{7^2}+\frac{b^2}{7^3}+...)=1\)

Now, what is inside the bracket is a geometric series with a common ratio, \(r=\frac{b}{7}\)

The formula is:

\(Sum=\frac{1}{1-r}=\frac{1}{1-\frac{b}{7}}=\frac{7}{7-b}\)   (But, r has to be less than 1).

So:

\(b(\frac{7}{7-b})=1\), can you solve it from here?

is it -12

https://web2.0calc.com/questions/1-what-is-the-smallest-distance-between-the-origin 

LOOK

P = (4,-7).

try 3 if it not it then idk

it is 16. ty

That is what I thought before as well. 10 is incorrect

Not 14 either. Tried it before 13

there are 10 numbers

wait is it 16

Thats. what I got but my answer is incorrect.

the answer is 540

1,1,2          13

1,2,2

2,2,1

2,1,2

1,1,1,1,1

1,1,1,2

3,1,1

1,1,3

1,3,1

3,2

2,3

4,1

1,4

 so 13