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Novay_Z [31]
2 years ago
15

What is the discount on a pair of shoes with a regular price of $48?

Mathematics
1 answer:
Korolek [52]2 years ago
8 0
The discounts on a pair of shoes with a regular price of $48 is 24
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<h2>h = 4 cm</h2><h2 />

Step-by-step explanation:

V = π r² h

where r = 3 cm

V = 36π cm³

solve for h

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36π = π 3² h

h = <u> 36π </u>

      π 3²

h = 4 cm

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A black line is shorter than a white line.the white line is shorter than a gray line. Is the black line longer or shorter than t
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The operations manager of a plant that manufactures tires wishes to compare the actual inner diameter of two grades of tires, ea
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Step-by-step explanation:

given that the operations manager of a plant that manufactures tires wishes to compare the actual inner diameter of two grades of tires, each of which has a nominal value of 575 millimeters. A sample of five tires of each grade is selected, and the results representing the inner diameters of the tires, ranked from smallest to largest, are as follows:

Grade X

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3 years ago
The distance from the tip of a slice of pizza to the crust is 7 inches.
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radius

Step-by-step explanation:

4 0
2 years ago
<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
Katen [24]

\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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