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Ilia_Sergeevich [38]
3 years ago
7

Throwing a paper ball across the room follows a/an _________ pattern.

Mathematics
2 answers:
ipn [44]3 years ago
8 0

Answer:

B) Quadratic is the option your looking for.

ehidna [41]3 years ago
3 0

Answer:

B. Quadratic

Step-by-step explanation:

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12w - 27 = -34 + 14w
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the answer is in fact 7/2

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3 years ago
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*Please Help!* (Easy question) 20 POINTS
mariarad [96]
Yes - they have to be similar.  Since they are using the same line as the hypotenuse, the ratio of the other two sides are in the same ratio (ie, the slope of the line), the 3 inner angles will be the same.  Thus, the triangles will be similar.
7 0
3 years ago
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PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!
zmey [24]

Step-by-step explanation:

underneath the x axis is the minuses so you draw across from -4

4 0
3 years ago
Find the area of the shaded region. Round your answer to the nearest tenth.
Alex
Check the picture below on the left-side.

we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.

now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.

so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.

\bf \textit{area of a sector of a circle}\\\\
A_x=\cfrac{\theta \pi r^2}{360}\quad 
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad degrees\\
------\\
r=6\\
\theta =60
\end{cases}\implies A_x=\cfrac{60\cdot \pi \cdot 6^2}{360}\implies \boxed{A_x=6\pi} \\\\
-------------------------------\\\\

\bf \textit{area of a segment of a circle}\\\\
A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta )  \right]
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad degrees\\
------\\
r=6\\
\theta =120
\end{cases}

\bf A_y=\cfrac{6^2}{2}\left[\cfrac{\pi\cdot 120 }{180}~-~sin(120^o )  \right]
\\\\\\
A_y=18\left[\cfrac{2\pi }{3}~-~\cfrac{\sqrt{3}}{2} \right]\implies \boxed{A_y=12\pi -9\sqrt{3}}\\\\
-------------------------------\\\\
\textit{shaded area}\qquad \stackrel{A_x}{6\pi }~~+~~\stackrel{A_y}{12\pi -9\sqrt{3}}\implies 18\pi -9\sqrt{3}

7 0
4 years ago
Let U={q, r, s, t, u, V, W, X, Y, Z},
trapecia [35]

Answer:

Ooh is theis unions I did it but

3 0
2 years ago
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