V(t) = 2t^2 - 3t - 3
for an acceleration a(t),
a(t) = d (v(t)) / d t
a(t) = 4t - 3
at t=2
a(2) = 8 - 3
a(2) = 5 feet/s^2
(9,30)(18,30)....when u have 2 sets of points and there y value is the same, then u have a horizontal line with a slope of 0
answer : A...it has a slope of 0 because the change in the y values
Answer:
SAS
Step-by-step explanation:
It doesnt show the hypotenuse so its no HL
It doesnt show 2 angles so its not SAA or ASA
It doesnt show three sides so its not SSS
The little line indicates that the sides are congruent. They both have a right angle and the bottom lines are right on top of each other so its SAS
Answer:
- x = ±√3, and they are actual solutions
- x = 3, but it is an extraneous solution
Step-by-step explanation:
The method often recommended for solving an equation of this sort is to multiply by the product of the denominators, then solve the resulting polynomial equation. When you do that, you get ...
... x^2(6x -18) = (2x -6)(9)
... 6x^2(x -3) -18(x -3) = 0
...6(x -3)(x^2 -3) = 0
... x = 3, x = ±√3
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Alternatively, you can subtract the right side of the equation and collect terms to get ...
... x^2/(2(x -3)) - 9/(6(x -3)) = 0
... (1/2)(x^2 -3)/(x -3) = 0
Here, the solution will be values of x that make the numerator zero:
... x = ±√3
_____
So, the actual solutions are x = ±3, and x = 3 is an extraneous solution. The value x=3 is actually excluded from the domain of the original equation, because the equation is undefined at that point.
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<em>Comment on the graph</em>
For the graph, we have rewritten the equation so it is of the form f(x)=0. The graphing program is able to highlight zero crossings, so this is a convenient form. When the equation is multiplied as described above, the resulting cubic has an extra zero-crossing at x=3 (blue curve). This is the extraneous solution.
The answer is -14 ........