Now, keep in mind that, we'll be using 25mile intervals, notice, that at a cost of 200 bucks, it went for 50 miles, now the charges are just for every 25miles only, so, if you drive 30 miles, you pay for 25miles and if you drive for 49 miles, you still only pay for 25 miles, but because 50/25 is 2 intervals, then if you drive for 50 miles, you're paying for 2 intervals of 25miles, not just one.

now, also let's keep in mind that, when the car was driven for 200 miles, that's 200/25 or 8 25mile intervals, so the 25mile charge kicked in 8 times.

$\bf \stackrel{cost}{y}=\stackrel{fixed~charge}{F}+\stackrel{25mile~intervals}{m}\quad \stackrel{25mile~charge}{x}\implies y=F+mx\\\\
-------------------------------\\\\
\begin{cases}
200=F+2x\implies 200-2x=\boxed{F}\\
245=F+8x\\
----------\\
245=\boxed{200-2x}+8x
\end{cases}
\\\\\\
45=6x\implies \cfrac{45}{6}=x\implies \cfrac{15}{2}=x\impliedby 
\begin{array}{llll}
\textit{so the 25mile charge is}\\
\textit{7 bucks and 50cents}
\end{array}
\\\\\\
$

$\bf 200=F+2\left( \frac{15}{2} \right)\implies 200=F+15\implies 185=F
\\\\\\
\textit{and the \underline{fixed charge} is 185 bucks then, let's check for 450}
\\\\\\
\boxed{y=185+\frac{15}{2}x}\qquad 
\begin{cases}
450~miles\\\\
\cfrac{450}{25}\implies 18~\textit{\underline{25mile intervals}}\\\\
x=18
\end{cases}
\\\\\\
y=185+\cfrac{15}{2}\cdot 18\implies y=185+135\implies y=320$

7 0

2 years ago

The math equation: p(x)=x^2-10x+16 what is the answer?

Gemiola [76]

(x+2)(x+8)

so x=-8 or x=-2

6 0

3 years ago