An object moves along the x-axis, its position at each time t> 0 given by \"x(t) = 81/2*t^2-3/4*t^4\" Determine the time inte rval(s), if any, during which the object slows down.
1 answer:
You can say an object slow down when the acceleration is negative which can be observed on a negative slope during the graph of acceleration, from a position equation lets derive 2 times to attain the acceleration equation <span>x(t) = 81/2*t^2-3/4*t^4 V(t) = 81X - 3t^3 A(t) = 81 - 9t^2 </span>81 - 9t^2 < 0<span> t^2 > 9-3 > t > 3 </span>
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My answer is A. I hope it hepls
-3(6)+10=4(-2) -18+10=-8 -8=-8 Yes
Answer:
x=10
B=98
Step-by-step explanation:
It will equal 82 in the end.
9×10-8
90-8
82
~
To find B I took 180 and subtracted 82 to get 98
Hope this helps ^-^
It’s b!!!! …… (need 20 characters)