= 2^-16 * 3^10 *3^-8/ 2^12 * 2^28
= 2^(-16 - 12 + 28) * 3^(10 - 8)
= 2^0 * 3^2
= 9
answer
9
Answer:
31.9secs
6,183.3m
Step-by-step explanation:
Given the equation that models the height expressed as;
h(t ) = -4.9t²+313t+269
At the the max g=height, the velocity is zero
dh/dt = 0
dh/dt = -9,8t+313
0 = -9.8t + 313
9.8t = 313
t = 313/9.8
t = 31.94secs
Hence it takes the rocket 31.9secs to reach the max height
Get the max height
Recall that h(t ) = -4.9t²+313t+269
h(31.9) = -4.9(31.9)²+313(31.9)+269
h(31.9) = -4,070.44+9,984.7+269
h(31.9) = 6,183.3m
Hence the maximum height reached is 6,183.3m
Write and solve an equation of ratios:
5 *10^9 years 100 m
------------------- = --------------
5 * 10^3 yrs x
Reducing the first fraction,
100 m
10^6 = ----------
x
100 m
Solving for x, x = ------------- = 10^(-4) m, or [10^(-1)] * [10^(-3))]
10^6
this comes out to one tenth of 1 millimeter.
Answer:
y = -2.8x +69.4
Step-by-step explanation:
Let y represent units of inventory, and x represent days since the last replenishment. We are given points (x, y) = (3, 61) and (13, 33). The line through these points can be described using the 2-point form of the equation of a line:
... y -y1 = (y2-y1)/(x2 -x1)(x -x1)
Filling in the given point values, we have ...
... y -61 = (33 -61)/(13 -3)(x -3)
Simplifying and adding 61, we get ...
... y = -2.8x +69.4
