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adelina 88 [10]
3 years ago
5

Can anyone help lend me a hand solving these problems please?​

Mathematics
2 answers:
masha68 [24]3 years ago
8 0

Answer:

Timmy's tubes = 11.25$

Fred's floats = 11.375$

Step-by-step explanation:

For Timmy's tubes

We use unitary method to solve

For 2 1/2 hours, it charges $18.75

So, for 1 1/2 hours it charges $11.25

Similarly for Fred's floats

Base fee = $2

Fee per hour = $6.25

So, for 1 1/2 hour fee = 2 + 6.25 + 3.125 = $11.375

So, Terry should choose Timmy's tubes

Assoli18 [71]3 years ago
3 0

well, let's see, what do we know about Timmy's Tubes, is a direct variation equation, since he's charging depending on the hours it's rented, thus

\qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}

\stackrel{\textit{\underline{c}ost varying with \underline{h}ours}}{c = kh}\qquad \textit{we know that } \begin{cases} c=18.75\to \frac{75}{4}\\ h=1\frac{1}{2}\to ~\hfill \frac{5}{2} \end{cases} \\\\\\ \stackrel{cost}{\cfrac{75}{4}}=\stackrel{hours}{\cfrac{5}{2}h}\implies 2\cdot 75=4\cdot 5h\implies 150=20h\implies \cfrac{150}{20}=h\implies \cfrac{15}{2}=h \\\\\\ therefore\qquad c=\cfrac{15}{2}h\impliedby \textit{Timmy's rate}

now, Fred charges $2 flat right off, plus $6.25 for every added hour, so Fred's equation will more or less be

1 hour 2 + 6.25(1)

2 hours 2 + 6.25(2)

3 hours 2 + 6.25(3)

<h3>h hours 2 + 6.25(h)</h3>

1)

how much will Timmy charge for 1 1/2 hour?

c = \cfrac{15}{2}\cdot \stackrel{h}{1\frac{1}{2}}\implies c = \cfrac{15}{2}\cdot \stackrel{h}{\cfrac{3}{2}}\implies c = \cfrac{15}{4}   or just 3.75

how much will Fred charge for the same time?

c=2+6.25\left( \frac{3}{2} \right)\implies c=9.375

2)

using only $12 for each, who is the better bargain?

\stackrel{\textit{Timmy's}}{c = \cfrac{15}{2}h}\implies 12=\cfrac{15h}{2}\implies 24=15h\implies \cfrac{24}{15}=h\implies \boxed{1.6=h} \\\\\\ \stackrel{\textit{Fred's}}{c = 2 + 6.25h}\implies 12=2+6.25h\implies 10=6.25h \\\\\\ \cfrac{10}{6.25}=h\implies \boxed{ 1.6=h}

well, let's notice, Terri is going to get the same amount of hours with either, so, it really doesn't matter, both will only give 1.6 hours, or 1 hour and 36 minutes.

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