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14 votes
100 points please help me and answer right!

100 points please help me and answer right!-example-1

2 Answers

6 votes
  • y=t²-82t+160

#a

Solve for t

  • t²-82t+160=0
  • t²-80t-2t+160=0
  • t(t-80)-2(t-80)=0
  • (t-2)(t-80)=0
  • t=2,80

The journey began at 2s and vanished at 80s

#b

Let's check for vertex

x coordinate

  • -b/2a
  • 82/2(1)
  • 41

Solving for y

  • y=41²-82(41)+160
  • y=-1521

Vertex at (41,-1521)

Time=41s

User Andrey Smelik
by
8.7k points
13 votes

Answer: Scroll down to find out :)

Explanation:

a) To solve t^2 - 82t + 160 = 0, we can factor it.

I got (t - 2)(t - 80) = 0, so t is either 2 or 80. So it begins in 2 secs and ends in 80 secs.

b) To find the vertex, we need to do (-b/2a), (substituting -b/2a in the equation). That gives us (82/2), (41^2 - 82*41 + 160) = (41, -1521)

So, the vertex is (41, -1521)

P.S. I tried my best lol, sorry if it isn't right.

User Deleplace
by
8.2k points

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