Answer is 3 seconds
When the bullet reaches the ground, ground being x in graph (and here its s which is = 0)
s = -16t^2 + 48t
s = 0, solve for t
0 = -16t^2 + 48t
0 = t ( -16t + 48)
0 = 16t ( - t + 3)
now you have two equation
0 = 16t and 0 = -t +3 ( you can look at the graph line touches x twice)
0 = 16 t
0 = t ( you know its false, because time = 0)
You are left with
0 = -t + 3
t = 3
It takes 3 seconds for the bullet to return to the ground.
// Hope this helps.
Mean, or average, is solved by adding all the numbers and dividing by the number of numbers. The answer is 88.
Answer:
Present value = $4,122.4
Accumulated amount = $4,742
Step-by-step explanation:
Data provided in the question:
Amount at the Start of money flow = $1,000
Increase in amount is exponentially at the rate of 5% per year
Time = 4 years
Interest rate = 3.5% compounded continuously
Now,
Accumulated Value of the money flow = 
The present value of the money flow = 
= 
= ![1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4](https://tex.z-dn.net/?f=1000%5Cleft%20%5B%5Cfrac%7Be%5E%7B0.015t%7D%7D%7B0.015%7D%20%5Cright%20%5D_0%5E4)
= ![1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right]](https://tex.z-dn.net/?f=1000%5Ctimes%5Cleft%20%5B%5Cfrac%7Be%5E%7B0.015%284%29%7D%7D%7B0.015%7D%20-%5Cfrac%7Be%5E%7B0.015%280%29%7D%7D%7B0.015%7D%20%5Cright%5D)
= 1000 × [70.7891 - 66.6667]
= $4,122.4
Accumulated interest = 
= 
= $4,742
Answer:
The point of intersection is (6, 8). This indicates that in 6 days, asteroid A and B will collide at a point that is 8 million miles away from asteroid A's current position and 12 million miles away from asteroid B's current position.
Answer:
They lose about 2.79% in purchasing power.
Step-by-step explanation:
Whenever you're dealing with purchasing power and inflation, you need to carefully define what the reference is for any changes you might be talking about. Here, we take <em>purchasing power at the beginning of the year</em> as the reference. Since we don't know when the 6% year occurred relative to the year in which the saving balance was $200,000, we choose to deal primarily with percentages, rather than dollar amounts.
Each day, the account value is multiplied by (1 + 0.03/365), so at the end of the year the value is multiplied by about
... (1 +0.03/365)^365 ≈ 1.03045326
Something that had a cost of 1 at the beginning of the year will have a cost of 1.06 at the end of the year. A savings account value of 1 at the beginning of the year would purchase one whole item. At the end of the year, the value of the savings account will purchase ...
... 1.03045326 / 1.06 ≈ 0.9721 . . . items
That is, the loss of purchasing power is about ...
... 1 - 0.9721 = 2.79%
_____
If the account value is $200,000 at the beginning of the year in question, then the purchasing power <em>normalized to what it was at the beginning of the year</em> is now $194,425.14, about $5,574.85 less.
