Answer:
t = 2.28 s
Step-by-step explanation:
h = 105 - 9t - 16t ^ 2
0 ft = 105 ft - 9t -16^t
To find the roots of a quadratic function we have to use the Bhaskara formula
, the roots will give us the time it takes to reach zero height
ax^2 + bx + c = 0
-16^t - 9t + 105 ft = 0 ft
a = -16 b = -9 c = 105
t1 = (-b + √ b^2 - 4ac)/2a
t2 =(-b - √ b^2 - 4ac)/2a
t1 = (9 + √(-9^2 - (4 * (-16) * 105)))/2 * (-16)
t1 = (9 + √(-81 + 6720))/ -32
t1 = (9 + √6639)/ -32
t1 = (9 + 81.84)/ -32
t1 = 90.84 / -32
t1 = -2.83 s
t2 = (9 - √(-9^2 - (4 * (-16) * 105)))/2 * (-16)
t2 = (9 - √(-81 + 6720))/ -32
t2 = (9 - √6639)/ -32
t2 = (9 - 81.84)/ -32
t2 = -72.84 / -32
t2 = 2.28 s
we have two possible values, we are only going to take the positive one, beacause we are talking about time
t2 = 2.28 s
Answer: I think it’s great but you should make a stronger opinion and always have back up evidence
Step-by-step explanation:
Answer:
The second one
Step-by-step explanation:
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Answer:
Distance of JK = 15 unit
Step-by-step explanation:
Given:
J(4,8)
K(-1,-2)
Find:
Distance of JK
Computation:
Distance = √(x₂-x₁)² + (y₂-y₁)²
Distance of JK = √(-1-4)² + (-2-8)²
Distance of JK = √25 + 100
Distance of JK = √125
Distance of JK = 15 unit
N
- -76=116
5
If the question gives you another variable, just switch it out with n
