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lisabon 2012 [21]
3 years ago
14

A sporting goods store is having a 15% off sale on all items. Which functions can be used to find the sale price of an item that

has an original price of x? You may choose more than one correct answer.
ƒ(x) = x - .15x
Sale = Original - 15
ƒ(x) = 1.15x
Sale = Original - .15(Original)
Mathematics
2 answers:
AURORKA [14]3 years ago
5 0
F(x) = x - 0.15x and Sale = Original - 0.15(original)
oksano4ka [1.4K]3 years ago
4 0

According to the question, the store is giving 15% off sales on all items. That means there will be a reduction of 15% in sale price of each item.

Let x represent the original price of an item.

15% of the original price=\frac{15}{100} \times x=0.15x

The sale price of an item in the store will now reduce by 0.15x. This means that;

Sale\: Price =Original-0.15(Original)

But x is the original price, therefore we can substitutex, wherever we see original. This gives us;

Sale\: Price=x-0.15x

Since sale price is expressed in terms of x, we can also write it as a function of x. That is,

f(x)=x-0.15x

Therefore the correct answers are

f(x)=x-0.15x

and

Sale\: Price=Original-0.15(Original)



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Answer:(a) getting the equation:

We'll assume that the altitude is represented by the x-axis and that the boiling point is represented by the y-axis.

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Simplify: [5×(25)^n+1 - 25 × (5)^2n]/[5×(5)^2n+3 - (25)^n+1​]
blagie [28]

\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} }}

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