Answer:
line
x-intercept | 0
f-intercept | 0
normal vector | (-3072/sqrt(9437185), 1/sqrt(9437185))≈(-1., 0.000325521)
slope | 3072
curvature | 0
Step-by-step explanation:
If you would like to solve (n^3 - n^4) - (3n^3 - 7n^4), you can do this using the following steps:
(n^3 - n^4) - (3n^3 - 7n^4) = <span>n^3 - n^4 - 3n^3 + 7n^4 = n^3 - 3n^3 - n^4 + 7n^4 = -2n^3 + 6n^4
</span>
The correct result would be <span>-2n^3 + 6n^4.</span>
Okay so 165 kids
10 for 1 instructor
just divide 165 by 10 ?
Answer:
maximum height is 4.058 metres
Time in air = 0.033 second
Step-by-step explanation:
Given that the equation height h
h = -212t^2 + 7t + 4
What is the toy's maximum height?
Let us assume that the equation is a perfect parabola
Time t at Maximum height will be
t = -b/2a
Where b = 7 and a = - 212
t = -7/ - 212 ×2
t = 7/ 424 = 0.0165s
Substitute t in the main equation
h = - 212(7/424)^2 + 7(7/424) + 4
h = - 0.05778 + 0.115567 + 4
h = 4.058 metres
Therefore the maximum height is 4.058 metres
How long is the toy in the air?
The object will go up and return to the ground.
At ground level, h = 0
-212t^2 + 7t + 4 = 0
212t^2 - 7t - 4 = 0
You can factorize the above equation and pick the positive time t since time can't be negative
Or
Since we have assumed that it's a perfect parabola,
Total time in air = (-b/2a) × 2
Time in air = 0.0165 × 2 = 0.033 s
