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vazorg [7]
4 years ago
10

A square sheet of art paper has an area of 625 square inches. What is the minimum side length of an easel that supports the whol

e sheet of paper?
Mathematics
2 answers:
Mila [183]4 years ago
7 0
If the area of the paper is 625 sq in, the length of one side of this square is sqrt(625 sq in), or 25 inches.  That's the min. side length of an easel that is to support the whole sheet of paper along its lowest side.
svetlana [45]4 years ago
4 0

area of a square = side^2

 to find the length of a side using area find the square root of the area

sqrt(625) = 25

so the easel has to be at least 25 inches

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Which statement correctly describe the data shown in the scatter plot?
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Option C is correct, The scatter plot shows a positive association because y increases as x increases.

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3 years ago
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Pauline's average speed for this trip is 64 km/hr.

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the distance traveled for next 40 minutes at 75 km/hr speed =(75*\frac{2}{3})km=50 km and

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So, <u>the total distance traveled </u>=(58.67+50+18.78) km = 127.45 km

She spends 17 minutes eating lunch and buying gas.

So, <u>the total time taken for the trip</u> = (40+40+17+23)minutes = 120 minutes= 2 hours

Thus, Pauline's average speed for this trip = (Total distance / Total time) =(\frac{127.45}{2})km/hr = 63.725 km/hr =64 km/hr <em>(Rounded to the nearest km/hr)</em>

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3 years ago
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4 years ago
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so the sun is one of the foci, and we have an eccentricity of 0.97. So the ellipse will look more or less like the picture below.


\bf \textit{ellipse, horizontal major axis}\\\\\cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1\qquad \begin{cases}center\ ( h, k)\\vertices\ ( h\pm a,  k)\\c=\textit{distance from}\\\qquad \textit{center to foci}\\\qquad \sqrt{ a ^2- b ^2}\\eccentricity\quad e=\cfrac{c}{a}\end{cases}\\\\-------------------------------\\\\\begin{cases}h=0\\k=0\\a=35.88\\e=0.97\end{cases}\implies \cfrac{(x-0)^2}{35.88^2}+\cfrac{(y-0)^2}{b^2}=1\\\\-------------------------------


\bf e=\cfrac{c}{a}\implies 0.97=\cfrac{c}{35.88}\implies \boxed{34.8036=c} \\\\\\ c=\sqrt{a^2-b^2}\implies b=\sqrt{a^2-c^2}\implies b=\sqrt{35.88^2-34.8036^2} \\\\\\ \boxed{b\approx 8.72}\\\\ -------------------------------\\\\ \cfrac{(x-0)^2}{35.88^2}+\cfrac{(y-0)^2}{8.72^2}=1\implies \cfrac{x^2}{1287.3744}+\cfrac{y^2}{76.0384}=1

5 0
4 years ago
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