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solniwko [45]
3 years ago
12

he parking garage of a mall is underground. Dr. Taylor parks her car 2 floors below ground level. She plans to walk to the 3rd f

loor of the mall from her car. How many floors will she travel from her car to the 3rd floor?
Mathematics
1 answer:
ad-work [718]3 years ago
4 0

Answer:

she will travel 5 floors from her car to the 3rd floor.

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\bf -------------------------------\\\\&#10;~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&&(~ 9 &,& 4~) &#10;%  (c,d)&#10;&&(~ -3 &,& -2~)&#10;\end{array}&#10;\\\\\\&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;d=\sqrt{(-3-9)^2+(-2-4)^2}\implies d=\sqrt{(-12)^2+(-6)^2}&#10;\\\\\\&#10;d=\sqrt{180}\qquad \qquad \qquad radius=\cfrac{\sqrt{180}}{2}

\bf -------------------------------\\\\&#10;\textit{equation of a circle}\\\\ &#10;(x- h)^2+(y- k)^2= r^2&#10;\qquad &#10;center~~(\stackrel{3}{ h},\stackrel{1}{ k})\qquad \qquad &#10;radius=\stackrel{\frac{\sqrt{180}}{2}}{ r}&#10;\\\\\\&#10;(x-3)^2+(y-1)^2=\left( \frac{\sqrt{180}}{2} \right)^2\implies (x-3)^2+(y-1)^2=\cfrac{(\sqrt{180})^2}{2^2}&#10;\\\\\\&#10;(x-3)^2+(y-1)^2=\cfrac{180}{4}\implies (x-3)^2+(y-1)^2=45
7 0
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In △RST, RW=27 cm.
sertanlavr [38]

Hello!

I just completed this test and if I'm correct in assuming that this is the same problem I did, then your answer would be 18 cm.


Hope this helps! :)

7 0
3 years ago
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