A) AB is where plane P and plane R intersect.
B) A, D, B are collinear (on the same line).
C) plane ADG (three points on the plane)
D) F, D, G, A are all on plane R
E) D lies on both planes.
The position function of a particle is given by:
![X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t](https://tex.z-dn.net/?f=X%5Cmleft%28t%5Cmright%29%3D%5Cfrac%7B2%7D%7B3%7Dt%5E3-%5Cfrac%7B9%7D%7B2%7Dt%5E2-18t)
The velocity function is the derivative of the position:
![\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V%28t%29%3D%5Cfrac%7B2%7D%7B3%7D%283t%5E2%29-%5Cfrac%7B9%7D%7B2%7D%282t%29-18%20%5C%5C%20%5Ctext%7BSimplifying%3A%7D%20%5C%5C%20V%28t%29%3D2t%5E2-9t-18%20%5Cend%7Bgathered%7D)
The particle will be at rest when the velocity is 0, thus we solve the equation:
![2t^2-9t-18=0](https://tex.z-dn.net/?f=2t%5E2-9t-18%3D0)
The coefficients of this equation are: a = 2, b = -9, c = -18
Solve by using the formula:
![t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Substituting:
![\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81-4%282%29%28-18%29%7D%7D%7B2%282%29%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81%2B144%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B225%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm15%7D%7B4%7D%20%5Cend%7Bgathered%7D)
We have two possible answers:
![\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B9%2B15%7D%7B4%7D%3D6%20%5C%5C%20t%3D%5Cfrac%7B9-15%7D%7B4%7D%3D-%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bgathered%7D)
We only accept the positive answer because the time cannot be negative.
Now calculate the position for t = 6:
Answer:
answer: obtuse
Step-by-step explanation:
Hope that helps!
A geometric shapes that can be used to create a logo that will fit in a measuring 15 cm by 15 cm is to use a square.
<h3>What are geometric shapes?</h3>
It should be noted that geometric shapes are the figure of area that is closed by a boundary.
In this case, a geometric shapes that. an be used to create a logo that will fit in a measuring 15 cm by 15 cm is to use a square.
This is because all the sides in a square are equal.
Learn more about shapes on:
brainly.com/question/25965491
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