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Vesna [10]
3 years ago
8

Help me please!! Joe went to a fast food restaurant and bought 4 orders of fries and 2 burgers for $14. Becca went to the same f

ast food restaurant and bought 3 orders of fries and one burger for $9.
Question:

Write a system of equations that models this situation. Let x represent orders of fries and y represent burgers.
Mathematics
1 answer:
pashok25 [27]3 years ago
6 0
Here are the equations that are represented by the scenarios:

4x + 2y = $14 (Joe)
3x + y = $9 (Becca) or y = 9 - 3x

To find the cost of each burger and fry you will solve the second equation for y and substitute it in for the y in the first equation.  This will put everything in terms of x, the price of a fry.

4x + 2(9 - 3x) = $14
4x + 18 - 6x = 14
-2x +18 = 14
     -18     -18
    <u>-2x</u> = <u>-4</u>
    -2      -2
     x = 2

Fry cost $2, and burgers cost (9 - 3 x 2) or $3.

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Simplify: [5×(25)^n+1 - 25 × (5)^2n]/[5×(5)^2n+3 - (25)^n+1​]​
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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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