BuilderBoi

There are \({8 \choose 4} = 70\) ways to choose dogs for the group of 5. 

There are \({4 \choose 3} = 4\) ways to choose the dogs for the group of 3. 

Then there is \({1 \choose 1} = 1\) ways to choose the dogs for the group of 1. 

So, there are \(70 \times 4 \times 1 = \color{brown}\boxed{280}\)

I entered \color{brown}\boxed{20\text{%}}

To box it, just do \boxed{}

\({14 \over 70} = {2 \over 10} = \color{brown}\boxed{20\text{%}}\)

b) 

0, 0, 0, 0, 4

3, 1, 0, 0, 0

2, 1, 1, 0, 0

1, 1, 1, 1, 0

For the first case, there are \({5 \choose 1} = 5\) cases

For the second one, there are \(5 \times {4 \choose 1}= 20\) cases

For the third one, there are \(5 \times {4 \choose 2} = 30\) cases

For the final one, there are 5 cases. 

So, there are a total of \(5 + 20 + 30 + 5 = \color{brown}\boxed{60}\) cases.

a)

1, 1, 2

4, 0, 0

3, 1, 0

\(\color{brown}\boxed{3}\) cases

Here's another way: 

The equation simplifies to \(x^2 + 9x + c = 0\)

The equation has at least 1 real root when the discriminant \(b^2 - 4ac \geq 0\)

Substituting in what we have gives us \(81 - 4c \geq 0 \), meaning \(c \leq 20.25\), so the maximum value of c is \(20.25 = \color{brown}\boxed{81\over 4}\)

I don't see a grid either.

What image?

The product of the roots is \({c \over a} = {n \over 2}\)

Can you take it from here?

https://web2.0calc.com/questions/graphing-question_107#r1