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kotykmax [81]
3 years ago
10

On sports day at Lakewood middle school, gold, silver, and bronze medals were awarded to the top three finishers in each event M

iniature replicas of the medals were sold as souvenirs in booths around the field. For $26, attendees could purchase a package containing 5 gold, 4 silver, and 2 bronze replicas, or one containing 4 gold, 4 silver, and 4 bronze replicas, or one that had 3 gold, 4 silver, and 6 bronze replicas. Which system of equations may be used to find the price, in dollars, of each type of replica?
Mathematics
1 answer:
8090 [49]3 years ago
8 0

Answer:

5G + 4S + 2B = 26

2G + 2S + 2B = 13

3G + 4S + 6B = 26

Step-by-step explanation:

Given

Represent Gold with G, Silver with S and Bronze with B

Required

Determine the system of equations

The packages are represented as:

The first: 5G + 4S + 2B

The second: 4G + 4S + 4B

The third: 3G + 4S + 6B

Each of the package costs $26.

So, we have:

5G + 4S + 2B = 26 -- (1)

4G + 4S + 4B = 26 -- (2)

3G + 4S + 6B = 26 -- (3)

Divide both sides of (2) by 2

4G + 4S + 4B = 26

2G + 2S + 2B = 13

So, the system of equations are:

5G + 4S + 2B = 26

2G + 2S + 2B = 13

3G + 4S + 6B = 26

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\huge\bold{Given:}

Length of the base = 16 km.

Length of the hypotenuse = 34 km. \huge\bold{To\:find:}

✎ The length of the missing leg ''a".

\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}

The length of the missing leg "a" is\boxed{30\:km}.

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}

Using Pythagoras theorem, we have

({perpendicular})^{2}  +  ({base})^{2}  =  ({hypotenuse})^{2}  \\ ⇢ {a}^{2}  +  ({16 \: km})^{2}  =  ({34 \: km})^{2}  \\ ⇢ {a}^{2}   + 256 \:  {km}^{2}  = 1156 \:  {km}^{2}  \\ ⇢ {a}^{2}  = 1156 \:  {km}^{2}  - 256 \:  {km}^{2}  \\ ⇢ {a}^{2}  = 900 \:  {km}^{2}  \\ ⇢a \:  =  \sqrt{900  \: {km}^{2} }  \\ ⇢a =  \sqrt{30 \times 30 \:  {km}^{2} }  \\ ⇢a = 30 \: km

\sf\blue{Therefore,\:the\:length\:of\:the\:missing\:leg\:"a"\:is\:30\:km.}

\huge\bold{To\:verify :}

( {30 \: km})^{2}  +  ({16 \: km})^{2}  =(  {34 \: km})^{2}  \\ ⇝900 \:  {km}^{2}  + 256 \:  {km}^{2}  = 1156 \:  {km}^{2}  \\⇝1156 \:  {km}^{2}  = 1156 \:  {km}^{2}   \\ ⇝L.H.S.=R. H. S

Hence verified. ✔

\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘

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2 years ago
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