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kobusy [5.1K]
4 years ago
13

I need help learning how to solve this problem please.

Mathematics
1 answer:
Radda [10]4 years ago
8 0

we have 4 gals, Stephanie, Andrea, Emily and  Becca.

btw the end point should be (-6, -7) and the start point is (8, 6).

so, if we know the distance between those two points, we can just cut it in 4 equal pieces and each gal will be driving a piece.

now, Emily's turn is after Andrea, BUT, we had first Stephenie driving ¼ of the way, then Andrea driving ¼ of the way too, but after ¼+¼, Emily comes on, but ¼+¼ is really 1/2, so when Emily starts, is really half-way through, or the midpoint of that distance.


\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\stackrel{\textit{Hometown}}{(\stackrel{x_1}{8}~,~\stackrel{y_1}{6})}\qquad
\stackrel{\textit{San Antonio}}{(\stackrel{x_2}{-6}~,~\stackrel{y_2}{-7})}
\qquad
\left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-6+8}{2}~~,~~\cfrac{-7+6}{2} \right)\implies \left(\cfrac{2}{2}~,~\cfrac{-1}{2} \right)\implies \stackrel{\textit{Emily starts off}}{\left(1~,-\frac{1}{2} \right)}


Emily's turn ends ¼ of the way later, which will be pretty much half-way in between  (1, -½) and (-6, -7), so is really the midpoint of those two.


\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
(\stackrel{x_1}{1}~,~\stackrel{y_1}{-\frac{1}{2}})\qquad
\stackrel{\textit{San Antonio}}{(\stackrel{x_2}{-6}~,~\stackrel{y_2}{-7})}
\\\\\\
\left( \cfrac{-6+1}{2}~~,~~\cfrac{-7-\frac{1}{2}}{2} \right)\implies \left( \cfrac{-5}{2}~,~\cfrac{~\frac{-14-1}{2}~}{2} \right)\implies \left( \cfrac{-5}{2}~,~\cfrac{~\frac{-15}{2}~}{2} \right)
\\\\\\
\left( \cfrac{-5}{2}~,~\cfrac{-15}{4} \right)\implies \left( -2\frac{1}{2}~,~-3\frac{3}{4} \right)

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Then the sum is

<em>u</em> + <em>v</em> = (4 cos(45°) + 2 cos(135°)) <em>x</em> + (4 sin(45°) + 2 sin(135°)) <em>y</em>

and its magnitude is

||<em>u</em> + <em>v</em>|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)

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… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))

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… = √(20 + 16 cos(90°))

… = √20 = 2√5

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Answer and Step-by-step explanation

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The upper quartile is the median of upper half of the data.

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The upper half of the data is 9,10,10,12,12,14,15

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