Question
Spider Man is standing on a parade float tossing candy out to young boys and girls. The candy is thrown with a starting velocity of 10 feet per second from a height of 3 feet modeled by the equation h=−16t2+10t+3
Complete the sentence.
When trying to figure out when the candy will be 2 feet in the air, the constant in the quadratic formula will be
(1 point)
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Complete the sentence.
When trying to figure out when the candy will be 2 feet in the air, the constant in the quadratic formula will be
(1 point)
Responses
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1
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Answers
Answer
The candy will reach 2 feet in the air at Response area seconds when rounding to the nearest hundredth of second.
Answered by
GPT 3.5
1
Answered by
GPT 3.5
To find when the candy will be 2 feet in the air, we need to solve for h=2 in the equation h=−16t^2+10t+3:
2 = -16t^2 + 10t + 3
Rearranging the terms, we get:
16t^2 - 10t - 1 = 0
Now, we can use the quadratic formula to find t:
t = (-(-10) ± √((-10)^2 - 4*16*(-1)))/(2*16)
t = (10 ± √(100 + 64))/32
t = (10 ± √164)/32
t ≈ (10 ± 12.81)/32
This gives two possible solutions:
t ≈ (10 + 12.81)/32 = 0.73 seconds (approximately)
t ≈ (10 - 12.81)/32 = -0.91 seconds (approximately)
Since time cannot be negative, the candy will reach 2 feet in the air at approximately 0.73 seconds.
2 = -16t^2 + 10t + 3
Rearranging the terms, we get:
16t^2 - 10t - 1 = 0
Now, we can use the quadratic formula to find t:
t = (-(-10) ± √((-10)^2 - 4*16*(-1)))/(2*16)
t = (10 ± √(100 + 64))/32
t = (10 ± √164)/32
t ≈ (10 ± 12.81)/32
This gives two possible solutions:
t ≈ (10 + 12.81)/32 = 0.73 seconds (approximately)
t ≈ (10 - 12.81)/32 = -0.91 seconds (approximately)
Since time cannot be negative, the candy will reach 2 feet in the air at approximately 0.73 seconds.
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