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Svetllana [295]
3 years ago
9

Lupe bought 8 pounds of pretzels at a local wholesaler for $18. Her friend Charles bought 6 pounds of pretzels at the supermarke

t for $15. Charles thinks he got the better deal because $15 is less than $18
How much would Charles have spent at the supermarket if he had purchased the same amount of pretzels as Lupe bought at the wholesaler? Show your work explain your answer, in a paragraph
Mathematics
2 answers:
Novay_Z [31]3 years ago
3 0

Answer:

20

Step-by-step explanation:

Charle would have spent that much because one pound is 2.50

15 divided by 6=2.50

2.50x8=20

alexdok [17]3 years ago
3 0

Answer:

Charles would've spent $20 at the supermarket if he bought the same amount of pounds as Lupe.

Step-by-step explanation:

18÷8= 2.25

2.25 is the amount of money per pound at the wholesaler.

15÷6= 2.50

2.50 is the amount of money per pound at the supermarket.

2.50x8=  20

Charles would've spent $20 if he bought the same amount of pounds as Lupe. Since one pound cost 2.50 at the supermarket, 8 times 2.50 is 20. He would actually spend more at the supermarket than the wholesaler.

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<img src="https://tex.z-dn.net/?f=Simplify%3A%20%5Cfrac%7B%205%C3%97%2825%29%5E%7Bn%2B1%7D%20-%2025%20%C3%97%20%285%29%5E%7B2n%7
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\green{\large\underline{\sf{Solution-}}}

<u>Given expression is </u>

\rm :\longmapsto\:\dfrac{5 \times  {25}^{n + 1}  - 25 \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {25}^{n + 1} }

can be rewritten as

\rm \:  =  \: \dfrac{5 \times  { {(5}^{2} )}^{n + 1}  -  {5}^{2}  \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {( {5}^{2} )}^{n + 1} }

We know,

\purple{\rm :\longmapsto\:\boxed{\tt{  {( {x}^{m} )}^{n}  \: = \:   {x}^{mn}}}} \\

And

\purple{\rm :\longmapsto\:\boxed{\tt{ \:  \:   {x}^{m} \times  {x}^{n} =  {x}^{m + n} \: }}} \\

So, using this identity, we

\rm \:  =  \: \dfrac{5 \times  {5}^{2n + 2}  - {5}^{2n + 2} }{{5}^{2n + 3 + 1}  -  {5}^{2n + 2} }

\rm \:  =  \: \dfrac{{5}^{2n + 2 + 1}  - {5}^{2n + 2} }{{5}^{2n + 4}  -  {5}^{2n + 2} }

can be further rewritten as

\rm \:  =  \: \dfrac{{5}^{2n + 2 + 1}  - {5}^{2n + 2} }{{5}^{2n + 2 + 2}  -  {5}^{2n + 2} }

\rm \:  =  \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2}  - 1)}

\rm \:  =  \: \dfrac{4}{25 - 1}

\rm \:  =  \: \dfrac{4}{24}

\rm \:  =  \: \dfrac{1}{6}

<u>Hence, </u>

\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times  {25}^{n + 1}  - 25 \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {25}^{n + 1} }  =  \frac{1}{6} }}

4 0
3 years ago
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