Fernando M. answered • 09/30/16
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Guidance from a CPA for Accounting, Business, Finance, and Writing.
We start with the equation:
D = -16(t^2) + 144
We set D to "0" so that we can eliminate it as a viarble. Essentially, we're asking, if the antenna is at 0 feet, how long does it have to fall to get to that height?
If D = 0,
0 = -16(t^2) + 144
0 = -16(t^2) + 144
Divide both sides by 16
0 = -t^2 + 9
From here, having simplified the equation, we need to construct it into its quadratic expression to show us when it hits "0"
0 = -(t + 3)(t - 3)
This equation is true when t = 3 or -3, so it would take three seconds for the antenna to hit the ground. It wouldn't make sense for it to take -3 seconds.
The only issue here is, what requirements do the common core regulations also impose upon you? What else does common core require you to provide other than this explanation for the answer? Could you include that in the question? Thanks.
I apologize for the edit. My first answer gives you an imaginary number and it's wrong. This answer is better.
Hannah M.
d = -16(t^2) +144 is correct. Sorry I wrote it down wrong. No wonder I was having problems!
Thank you for your help!
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09/30/16
Fernando M.
Hi, The answer is written above, but were you asking for clarification? I know it's a bit of a skip from "0= -t^2 +9" to "0 = -(t + 3)(t - 3)," but the way I arrived at it is simply by trial and error. Better mathematicians than me can offer a better explanation, but essentially, when we have to simplify a quadratic equation, I approach it by asking, "in what way could I rewrite this equation to keep the same result?" So I just come up with some equation I think would fit, and see if the result of applying the FOIL method matches the original equation we started with. Using the FOIL method (multiplying the first, outer, inner, and last numbers together in that order), if I tried 0 = -(t + 3)(t - 3), first, multiply the "t" variables to get t^2, multiply the outer numbers to get -3t, the inner ones for 3t, and finally, the last digits; 3 x -3 = -9. So far, the equation now looks like this: -(t^2 - 3t + 3t -9)=0. The "3t" and "-3t" cancel each other out, so we're at -(t^2 - 9), and from here we distribute the negative sign (-), so t^2 becomes -t^2, and -9 becomes positive 9. Finally we arrive at the equation with which we started, -t^2 +9 = 0. This means our quadratic expression works. The number "3" is the solution to this equation, so if we plug it in we get: -(3^2) +9 = 0 -(3 x 3 = 9) + 9 = 0 -9 + 9 = 0 For the antenna to reach 0 feet in the air, beginning from a height of 144 feet up high, it has to travel "t" seconds, and in this case, "t" is 3. It needs to go for 3 seconds. I hope this clarifies it a bit!
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04/14/20
Fernando M.
09/30/16