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Flura [38]
3 years ago
13

A soccer field is a rectangle 90 meters wide and 120 meters long. The coach asks the players to run from one corner to the corne

r diagonally across the field. How far do the players run?
Mathematics
2 answers:
daser333 [38]3 years ago
6 0

Answer:

<h2>150 metres </h2>

Step-by-step explanation:

We can imagine the rectangle, when cut diagonally it forms a Right angled Triangle

Using this we can find the diagonal i.e hypotenuse.

From the question we know that,

Length = 120 metres

Width = 90 metres

So we can find the diagonal by keeping it as x using the pythogorean theorem.

{120}^{2}  +  {90}^{2}  =  {x}^{2}  \\ 14400 + 8100 =  {x}^{2}  \\ 22500 =  {x}^{2}  \\ x =  \sqrt{22500}  \\ x = 150

prohojiy [21]3 years ago
3 0

Answer:

150 meters

Step-by-step explanation:

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The scale from a square park to a drawing of the park is 5 miles to 1 miles. The actual park has an area of 1,600 m×2 what is th
Flauer [41]

The user corrected that the scale of a drawing of a park reads: 5 miles to 1 cm , and we know that the park measures 1,600 square meters (user insisted that this measure is given in square meters and not square miles).

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then meters square will be equivalents to:

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now, since 5 miles are represented by 1 cm, then 25 square miles will be represented by 1 square cm

and therefore 0.00061776 square miles will be the equivalent to:

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Given that f(x) = x + 3<br> a) Find f(2)
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{ \qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

In the above question, it is given that :

\qquad \sf  \boxed{ \sf f(x) =  \frac{x + 3}{2} }

A.) Find f(2) :

\qquad \sf  \dashrightarrow \: f(2) =  \dfrac{2 + 3}{2}

\qquad \sf  \dashrightarrow \: f(2) =  \dfrac{5}{2}

or

\qquad \sf  \dashrightarrow \: f(2) = 0.5

B.) Find { \sf {f}^{-1}(x) } :

\qquad \sf  \dashrightarrow \: let \: y = f (x)

so, we can write it as :

\qquad \sf  \dashrightarrow \: y =  \dfrac{x + 3}{2}

\qquad \sf  \dashrightarrow \: 2y = x + 3

\qquad \sf  \dashrightarrow \: x = 2y - 3

Now, put x = { \sf {f}^{-1}(x) }, and y = x and we will get our required inverse function ~

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C.) Find { \sf {f}^{-1}(12) } :

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(12) = 2(12)- 3

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(12) = 36- 3

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(12) = 33

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