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3 years ago

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5. Marvin Melvin purchased a lawnmower for $328.61, a lawn sprinkler for $13.27. and a

1 answer:

iris [78.8K]3 years ago

7 0

Answer:

c. $403.13

Step-by-step explanation:

Cost of lawnmower = $328.61

Cost of lawn sprinkler = $13.27

Cost of hose reel = $27.96

total cost of purchase without tax = $\$328.61+\$13.27+\$27.96=\$369.84$

Taxes applied:

State tax = 5.5%

County tax = 1.5%

City tax = 2%

Total percent of taxes applies = $5.5+1.5+2= 9\%$

Amount added as tax = $\textrm{Total cost of purchase }\times\textrm{ Total percentage of tax applied}$

⇒ $369.84\times 9\%$

⇒ $369.84\times \frac{9}{100}$

⇒ $33.29$

Tax amount = $33.29

∴ Purchase price including tax = Cost price + Tax amount = $\$369.84+\$33.29=\$403.13$ (Answer)

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In which:

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How i am going to solve this problem.

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$V_{n} = {6*10^{-18}, 12*10^{-18},...}$

$a = 6*10^{-18}$

$r = 2$

And the following arithmetic sequence that are the time(in hours).

$T_{n} = {0,0.5,1,...}$

$a = 0$

$d = 0.5$

$T_{n} = 0.5n$

How long will it take for a single bacterium to grow to fill a thimble with volume 1 cm3 ?

I am going to find the value of n for which $V_{n} = 1cm^{3}$ , then i find the value at this position in the arithmetic sequence. So

$1cm^{3} = 10^{-6}m^{3}$

$V_{n} = 6*10^{-18}*(2^{(n-1)})$

$10^{-6} = 6*10^{-18}*(2^{(n-1)})$

$\frac{10^{-6}}{6*10^{-18}} = 2^{(n-1)})$

Obs: $a^{b-c} = \frac{a^{b}}{a^{c}}$

$\frac{10^{12}}{6} = \frac{2^{n}}{2}$

$2^{n} = \frac{10^{12}}{3}$

Now, we have to apply these following logarithim proprierties to find the value of n:

$log a^{n} = n log a$

$log(\frac{a}{b}) = log a - log b$

$log 10^{n} = n$

log 2 = 0.30

log 3 = 0.48

$log 2^{n} = log(\frac{10^{12}}{3})$

$n log 2 = log(10^{12}) - log 3$

$0.30n = 12 - 0.48$

$0.30n = 11.52$

$n = 38.4$

Lets find $T_{38}$ and $T_{39}$ in the arithmetic sequence.

$T_{n} = 0.5n$

$T_{38} = 0.5*38 = 19$

$T_{39} = 0.5*39 = 19.5$

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$1.08*10^{8} km^{3} = 1.08*10^{17} m^{3} = 108*10^{15}m^{3}$

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log a*b = log a + log b

$log (2^{n}) = log(36 * 10^{33})$

$n log 2 = log 36 + log 10^{33}$

$0.30n = 1.56 + 33$

$0.30n = 34.56$

$n = \frac{34.56}{0.30}$

$n = 115.2$

$T_{115} = 0.5*115.2 = 57.6$

It will take 57.6 hours to for the volume to fill the entire earth

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