Rom

\(slope=\dfrac{y_1-y_0}{x_1-x_0}\\ slope=\dfrac{-8-5}{6-(-2))}=\dfrac{-13}{8}\)

\(midpt=\left(\dfrac{x_0+x_1}{2},\dfrac{y_0+y_1}{2}\right)\\ midpt=\left(\dfrac{-2+6}{2},\dfrac{5+-8}{2}\right)=(2,-3/2)\)

\(dist = \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2}\\ dist= \sqrt{(6--2)^2 + (-8-5)^2}\\ dist=\sqrt{8^2+(-13)^2}=\sqrt{64+169}=\sqrt{233}\)

to elaborate on CPhil's answer.  Consider the sum of the first N odd numbers

\(s=\displaystyle{\sum_{k=1}^N}(2k-1) \\ s = 2 \displaystyle{\sum_{k=1}^N}k -\displaystyle{\sum_{k=1}^N}1\\ s=2\dfrac{N(N+1)}{2}-N\\ s=N^2+N-N=N^2\)

there are two ways to do this

20% of 2400 is

\(r=\dfrac{20}{100} 2400 = \dfrac{2400}{5}=480\)

we are taking this off of 2400 so

\(remaining = 2400-r = 2400-480 = 1920\)

or we could do it by noting that if we remove 20% we are leaving (100-20)=80%

\(remaining = 2400 \dfrac {100-20}{100}=2400 \dfrac{80}{100}=24\times 80=1920\)

109 is the correct number for the sum.

\(109 \bmod{10} = 9\\ 10-9 = 1\\ D[13] = 7 \neq 1\\ \mbox{So this is not a valid ISBN}\)

\(2x-5y=-12\\ x-3y=-7\)

using the 2nd equation

\(x = 3y - 7\)

plugging this into the first equation

\(2(3y-7) - 5y = -12\\ 6y - 14 -5y = -12 \\ y -14 = -12 \\ y=2\)

plugging this into our equation for x we get

\(x = 3(2)-7=6-7=-1\)

As a check

 \(2(-1)-5(2)=-2-10 = -12 \\ (-1)-3(2) = -1 - 6 = -7\)

no

9+10=19

\(12.08 = 12 + \dfrac {8}{100} \\ \mbox{so } 12.08 \Rightarrow 12\dfrac {8}{100}= 12\dfrac {2}{25}\)

density = mass/volume

\(Density = \dfrac{454 g}{2270 cm^3}=0.2 \dfrac {g}{cm^3}\)

\(a = 0.5 b h\)

divide both sides by \(0.5 h\)

\(\dfrac {a}{0.5 h} = b\\ b=\dfrac {2a}{h}\)

you just grind the numbers through and at the end you have a number that is in units km/hr.

\(\dfrac {22560000000 \times 0.00000012km}{ 24.5 hr}=110.498~ km/hr\)