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Ivahew [28]
2 years ago
15

Victoria kicked a soccer ball laying on the ground. It was in the air for 5 seconds before it hit the ground. While the soccer b

all was in the air, it reached a height of approximately 15 feet. Assuming that the soccer ball height (in feet) is a function of time (in seconds), what is the domain in the context of this problem?​
Mathematics
1 answer:
Irina-Kira [14]2 years ago
5 0

Answer:

the Domain would be represented by Time in this problem, since time can only be positive, the domain would start at zero seconds and end at 5 seconds when the soccer ball hits the ground.

Step-by-step explanation:

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Lemur [1.5K]

Answer:

10 sqrt(3)/3 =x

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

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3 years ago
The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units
ad-work [718]
Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&A&(~ 2 &,& 8~) 
%  (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&A&(~ 2 &,& 8~) 
%  (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}

so, add AC + AB + CB, and that's the perimeter of the triangle.

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Read 2 more answers
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