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devlian [24]
2 years ago
6

A pool table is shown. On the table, a triangle is drawn to connect the cue ball, the eight ball, and the top right pocket. The

distance between the cue ball and eight ball is 8 inches, the distance between the eight ball and the pocket is 6 inches, and the distance between the cue ball and the pocket is unknown.
You are told that in a billiards shot, the cue ball was shot at the eight ball, which was 8 inches away. As a result, the eight ball rolled into a pocket, which was 6 inches away.


Knowing that the angle made with the path of the cue ball and the resulting path of the eight ball is larger than 90°, it can be determined that the original distance from the cue ball to the pocket was greater than ______ inches.
Ask for details
Mathematics
1 answer:
inna [77]2 years ago
6 0

Answer:

Its 10

Step-by-step explanation:

i just did it

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Check
algol [13]

Answer: Option A is the correct answer, y = x + 2

y - 3 = x-1

Step-by-step explanation:

6 0
2 years ago
Help please ASAP !!!!!!
Mashcka [7]
A ∩ B = {2, 4}                [intersection (∩) means the common terms]
7 0
3 years ago
What is the greatest common fact of 9 and 22
AlexFokin [52]

Answer:

1

Step-by-step explanation:

the greatest common factor of 9 and 22 is 1.

3 0
3 years ago
Suzette ran and biked for a total of 34 mi in 3.5 h. Her average running speed was 6 mph and her average biking speed was 12.5 m
sveticcg [70]

Answer:

The speed of an object is the rate of distance traveled over time. Suzette ran for 1.5 hours, and she biked for 2 hours.

Given that: Speed is calculated as: Make Distance the subject Average running speed = 6 mph for x hours implies that: Average biking speed = 12.5 mph for y hours implies that: So, the total distance is: The total distance is 34 miles. So, we have: The total time is 3.5h. So, we have: Make x the subject Substitute in Collect like terms Divide both sides by 6.5 Recall that: Hence, Suzette ran for 1.5 hours, and she biked for 2 hours.'

x=3.5-y

x-3.5-2

x=1.5

4 0
2 years ago
An object is launched vertically in the air at 24.5 meters per second from a 11-meter-tall platform. Using the projectile motion
ElenaW [278]
Check the picture below.

\bf \qquad \textit{initial velocity}\\\\
\begin{array}{llll}
\qquad \textit{in meters}\\\\
h(t) = -4.9t^2+v_ot+h_o
\end{array} 
\quad 
\begin{cases}
v_o=\textit{initial velocity of the object}\\
\qquad \qquad 25\\
h_o=\textit{initial height of the object}\\
\qquad \qquad 11\\
h=\textit{height of the object at "t" seconds}
\end{cases}
\\\\\\
h(t)=-4.9t^2+25t+11\\\\
-------------------------------\\\\

\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h(t) = &{{ -4.9}}x^2&{{ +25}}x&{{ +11}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left(\stackrel{\textit{how long it took}}{-\cfrac{25^2}{2(-4.9)}}~~,~~\stackrel{\textit{how high it went up}}{11-\cfrac{25^2}{4(-4.9)}}  \right)

6 0
3 years ago
Read 2 more answers
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