0 is an integer and thus it is a rational number.
\(0=\dfrac 0 1\)
is the ratio of two integers
for a postive integer x, x! is the product
x(x-1)(x-2)...(3)(2)(1)
for example
1! = 1
2! = 2
3! = 6
...
6! = 6 x 5 x 4 x 3 x 2 = 720
0! = 1 is also used on occasion
\(a=\dfrac {v_f - v_i}{\Delta t}\)
\(a= \dfrac{15 - 0}{10}=\dfrac {15}{10}=\dfrac 3 2\dfrac {m}{s^2}\)
\(-5(3k+1)=-14-5k\\ -15k-5=-14 -5k \\ 9=10k \\ k = \dfrac {9}{10}\)
\(\dfrac{5(4x-2)}{8(2x-1)}=\dfrac{10(2x-1)}{8(2x-1)}=\dfrac{10}{8}=\dfrac 5 4\)
Not really. However you slice it you're going to end up with
\(6n = 14\\ n= \dfrac {14}{6} = \dfrac 7 3\)
\(\dfrac 1 3 \times -\dfrac 3 4=-\dfrac{1 \times 3}{3 \times 4}=-\dfrac {3}{12}=-\dfrac 1 4\)
\(2|x-1|=-10 \\ |x-1|=-5 \\ \mbox{ and this has no solution since } |x|\geq 0,~\forall x\)
6x-3
you have two coefficients, 6 and 3.
the greatest common factor of 6 and 3 is 3
so you can rewrite
(6x-3) as 3(2x-1)
\(boxers = 4\times cspaniels\\ boxers+cspaniels=30\\ 4\times cspaniels + cspaniels = 30\\ 5\times cspaniels = 30\\ cspaniels = \dfrac {30}{5}=6 \\ boxers = 4\times 6 = 24\)
+6251 