3. A model rocket is launched from a roof into a field. The path rocket can be modleed by the equation. y=-0.04x^2+8.3x+4.3, where x is the horizontal distance, in meters, from the starting point on the roof and the y is the height, in meters, of the rocket above the ground. How far horizontally from its strating point will the rocket land? Round to the nearest 100th meter. A- 208.02m B-416.03m C-0.52m D-208.19m
When the rocket lands, the height of the rocket above the ground will equal zero. Set y = 0 and solve the quadratic equation for x.
so is it C
I cannot give out the answer. You have to show the work how you got C.
well im guessing
no its D k
-0.04x^2 + 8.3x + 4.3 = 0 Divide throughout by -0.04 Then solve the quadratic equation. Or you can directly use the quadratic formula and solve for x. \[\Large x = \frac{ -b \pm \sqrt{b ^{2} - 4ac} }{ 2a }\]
so i divide
i got d well i was told D is tha rit
\[\Large x = \frac{ -8.3 \pm \sqrt{(8.3) ^{2} - 4(-0.04)(4.3)} }{ 2(-0.04) } = ?\]
hold on k let me put that in my calculater k
um.... can u jus giv me the answer i don't get it
I will get a warning from the moderator if I just give out the answer. But you should be able to do the above calculation. Quadratic equation occurs often in math and it is better to learn to solve it so you don't have to ask someone each time.
ok i'll try but nvm i'll close this question so i can put another up k
alright.
can u help me with that one
well you need to let y = 0 that means the reockat has no height, and solve for x easiest way seems to be the general quadratic formula x=−8.3±8.32−4×−0.04×4.3−−−−−−−−−−−−−−−−−−−√2×−0.04
thanx
no problem :)
but im going to close the question k
ok
u think u can help me wit the next one i puit up
sure
wait are you from mulberry middle?
no where is that
message me anyway k
haha sorry my collage is next to it and i know a girl named phebe.
ok
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