To get a close estimate, we can round 49 up to 50 and 311 down to 300, obtaining an estimate of 50/300 = 1/6, or 0.1666... as a repeating decimal. That decimal approximation is a little less than one hundredth away from the actual decimal approximation of ≈ 0.1576
<span>Dawn was at 6 am.
Variables
a = distance from a to passing point
b = distance from b to passing point
c = speed of hiker 1
d = speed of hiker 2
x = number of hours prior to noon when dawn is
The first hiker travels for x hours to cover distance a, and the 2nd hiker then takes 9 hours to cover that same distance. This can be expressed as
a = cx = 9d
cx = 9d
x = 9d/c
The second hiker travels for x hours to cover distance b, and the 1st hiker then takes 4 hours to cover than same distance. Expressed as
b = dx = 4c
dx = 4c
x = 4c/d
We now have two expressions for x, set them equal to each other.
9d/c = 4c/d
Multiply both sides by d
9d^2/c = 4c
Divide both sides by c
9d^2/c^2 = 4
Interesting... Both sides are exact squares. Take the square root of both sides
3d/c = 2
d/c = 2/3
We now know the ratio of the speeds of the two hikers. Let's see what X is now.
x = 9d/c = 9*2/3 = 18/3 = 6
x = 4c/d = 4*3/2 = 12/2 = 6
Both expressions for x, claim x to be 6 hours. And 6 hours prior to noon is 6am.
We don't know the actual speeds of the two hikers, nor how far they actually walked. But we do know their relative speeds. And that's enough to figure out when dawn was.</span>
Answer:
what do u need help with?? :)
Step-by-step explanation:
Answer: x>-6
Explanation: -2x -6>-18
-2x >-12
-x>6
X>-6
Answer:
The answer to your question is t = 1.3 s
Step-by-step explanation:
Data
Equation h(t) = -4.9t² + v₀t + h₀
v₀ = 0 m/s
h₀ = 8 m
t = ?
h = 0 m
Process
1.- Substitute the values in the formula
0 = -4.9t² + 0t + 8
2.- Simplification
0 = -4.9t² + 8
3.- Solve for t
4.9t² = 8
t² = 8/4.9
t² = 1.63
4.- Result
t = 1.27 ≈ 1.3 s
