C. 1.64
The object will fall back to the ground when s(t) = 0.
Setting s(t) equal to 0, we get:
0 = -16t^2 + 11t + 25
Rearranging and factoring the equation, we get:
16t^2 - 11t - 25 = 0
(4t+5)(-4t-5) = 0
This gives us t = -5/4 or t = 5/4. Since time cannot be negative in this context, the object will fall back to the ground after t = 5/4 = 1.25 seconds, which is equivalent to 1.64 seconds.
Therefore, the correct answer is C. 1.64.
An object is launched vertically upward from a 25 foot platform and is modeled by the function s(t)=−16t^2+11t+25
The object will fall back to the ground after
A. 0.95
B. −1.64
C. 1.64
D. −0.95
7 answers
seconds because blank cannot be a solution.
-0.95
-1.64
1.64
0.95
-0.95
-1.64
1.64
0.95
0.95
The object will fall back to the ground after t = 1.64 seconds.
Therefore, the correct answer is D. 0.95.
The object will fall back to the ground after t = 1.64 seconds.
Therefore, the correct answer is D. 0.95.
An object is launched vertically upward from a 25 foot platform and is modeled by the function s(t)=−16t^2+11t+25
The object will fall back to the ground after
A. 0.95
B. −1.64
C. 1.64
D. −0.95
seconds because blank cannot be a solution.
E. -0.95
F. -1.64
G. 1.64
H. 0.95
The object will fall back to the ground after
A. 0.95
B. −1.64
C. 1.64
D. −0.95
seconds because blank cannot be a solution.
E. -0.95
F. -1.64
G. 1.64
H. 0.95
To find out when the object falls back to the ground, we set the function s(t) equal to 0 and solve for t:
0 = -16t^2 + 11t + 25
This is a quadratic equation which can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -16, b = 11, and c = 25. Substituting these values into the formula, we get:
t = (-11 ± √(11^2 - 4 * (-16) * 25)) / (2 * -16)
t = (-11 ± √(121 + 1600)) / -32
t = (-11 ± √1721) / -32
This results in two potential solutions:
t = (-11 + √1721) / -32 ≈ 0.95 seconds
t = (-11 - √1721) / -32 ≈ -1.64 seconds
Since time cannot be negative in this context, the object will fall back to the ground after approximately 0.95 seconds.
Therefore, the correct answer is A. 0.95.
0 = -16t^2 + 11t + 25
This is a quadratic equation which can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -16, b = 11, and c = 25. Substituting these values into the formula, we get:
t = (-11 ± √(11^2 - 4 * (-16) * 25)) / (2 * -16)
t = (-11 ± √(121 + 1600)) / -32
t = (-11 ± √1721) / -32
This results in two potential solutions:
t = (-11 + √1721) / -32 ≈ 0.95 seconds
t = (-11 - √1721) / -32 ≈ -1.64 seconds
Since time cannot be negative in this context, the object will fall back to the ground after approximately 0.95 seconds.
Therefore, the correct answer is A. 0.95.
theres two part of the question what are the answers
My apologies for the oversight. The correct answers for the two options should be:
E. -0.95
H. 0.95
Thank you for pointing that out.
E. -0.95
H. 0.95
Thank you for pointing that out.
1 answer