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mojhsa [17]
2 years ago
13

Find the coordinates of the midpoint M of ST. Then find the distance between points S and T. Round the distance to the nearest

Mathematics
1 answer:
ololo11 [35]2 years ago
8 0

Answer:

Step-by-step explanation:

M is the midpoint

=>\left \{ {{x=\frac{6+7}{2} } \atop {y=\frac{-3-2}{2} }} \right

⇔\left \{ {x=\frac{13}{2} } \atop {y=\frac{-5}{2}}} \right.

⇒M(\frac{13}{2};\frac{-5}{2})

ST=\sqrt{(7-6) ^{2} +(-2+3)^{2}=\sqrt{2}=1.414

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Answer: im having the same problem

Step-by-step explanation:

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3 years ago
A small independent organic food store offers a variety of specialty coffees. To determine whether price has an impact on sales,
lorasvet [3.4K]

Answer:

b= -4.68

Step-by-step explanation:

Hello!

We have to study variables:

Y: Monthly coffe sales

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The estimate regression line is

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The slope of the estimated regression line is represented by b.

The formula I'll use to calculate it is:

b= [n∑xiyi -(∑xi)*(∑yi)]/ n∑xi²-(∑xi)²

To calculate b we need to do some auxiliary calculations:

∑xiyi= 4213.15

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∑yi= 545

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b= [10*4213.15-(87.47)*(545)]/ n883.3703-(87.47)²

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A: partial variation

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Answer:

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Derive the equation of the parabola with a focus at (3,1) and a directrix of y = 5
serg [7]
So hmmm  check the picture below

so... the vertex is "p" distance from the focus and the directrix, thus, the vertex is really half-way between both

in this case, 2 units up from the focus or 2 units down from the directrix, and thus it lands at 3,3

now, the "p" distance is 2, however, the directrix is up, the focus point is below it, the parabola opens towards the focus point, thus, the parabola is opening downwards, and the squared variable is the "x"

because the parabola opens downwards, "p" is negative, and thus, -2

now, let's plug all those fellows in then

\bf \begin{array}{llll}
(x-{{ h}})^2=4{{ p}}(y-{{ k}})\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-----------------------------\\\\

\begin{cases}
h=3\\
k=3\\
p=-2
\end{cases}\implies (x-3)^2=4(-2)(y-3)\implies (x-3)^2=-8(y-3)
\\\\\\
-\cfrac{(x-3)^2}{8}=y-3\implies \boxed{-\cfrac{1}{8}(x-3)^2+3=y}

5 0
3 years ago
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