The answer is
<span>a) 1000=-16t^2+1700, implies t² = -700 /-16, and t= 6.61s
b) </span><span>970= -16t^2+1700, </span><span>implies t² = -730 /-16, and t=6.75s
c)
reasonable domain of h
h is polynomial function, so its domain is R, (all real number)
its range
the inverse of h is h^-1 = sqrt (1700- t / 16), and its domain is </span>
<span><span><span>1700- t / 16>=0, so t <1700,
the range of h is I= ]-infinity, 1700]</span> </span> </span>
Answer:
2. y intercept is at 0 because it needs to go through the origin if it is proportional
3. y intercept is positive the line crosses above 0. when y intercept negative the line crosses below 0
4. y intercept is 0
5. y intercept is 1.8. it represents the time remaining as its starting timer.
6. (0, 1). y intercept is 1
7. y intercept is -4
8. y intercept is 0
9. 80m
The answer is 4. If you multiply 4 • 4 = 16 • 4 = 64 • 4 = 256.
Answer
simplified is log5 (2)
the 5 is suposed to be the subscript one.
Answer:
10
Step-by-step explanation:
Ground level is where h = 0, so solve the equation ...
h(x) = 0
-5(x -4)^2 +180 = 0 . . . . substitute for h(x)
(x -4)^2 = 36 . . . . . . . . . . divide by -5, add 36
x -4 = 6 . . . . . . . . . . . . . . positive square root*
x = 10 . . . . . . add 4
The object will hit the ground 10 seconds after launch.
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* The negative square root also gives an answer that satisfies the equation, but is not in the practical domain. That answer would be x = -2. The equation is only useful for time at and after the launch time: x ≥ 0.
