Answer:
a. The rock's velocity is 
and the acceleration is 
b. It takes 22.5 seconds to reach the highest point.
c. The rock goes up to 405 m.
d. It reach half its maximum height when time is 6.59 s or 38.41 s.
e. The rock is aloft for 45 seconds.
Step-by-step explanation:
- Velocity is defined as the rate of change of position or the rate of displacement.
- Acceleration is defined as the rate of change of velocity.
a.
The rock's velocity is the derivative of the height function 
The rock's acceleration is the derivative of the velocity function 
b. The rock will reach its highest point when the velocity becomes zero.
It takes 22.5 seconds to reach the highest point.
c. The rock reach its highest point when t = 22.5 s
Thus
So the rock goes up to 405 m.
d. The maximum height is 405 m. So the half of its maximum height = 
To find the time it reach half its maximum height, we need to solve
For a quadratic equation of the form 
the solutions are
It reach half its maximum height when time is 6.59 s or 38.41 s.
e. It is aloft until s(t) = 0 again
The rock is aloft for 45 seconds.
Answer:
a)
a1 = log(1) = 0 (2⁰ = 1)
a2 = log(2) = 1 (2¹ = 2)
a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849
a4 = log(4) = 2 (2² = 4)
a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322
a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)
a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807
a8 = log(8) = 3 (2³ = 8)
a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )
a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322
b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:
log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that
a1 = 1
a2 = 2
a3 = 3
a4 = 4
a5 = 5
a6 = 6
a7 = 7
a8 = 8
a9 = 9
a10 = 10
I hope this works for you!!
Did somebody say you're supposed to draw the graph of the equation ?
Is that the assignment ?
OK. Just like every other equation you need to graph, get it in the
standard form, where 'y' is all alone on one side, and everything else
is on the other side. When you do that, you'll be able to spot the slope
and y-intercept of the line, or get some points, or whatever you want.
4y + 12 = 0
Subtract 12 from each side: 4y = -12
Divide each side by 4: y = -3
There's the equation you can handle.
The y-intercept is -3, and the slope is zero.
Would you like some points ? OK. Pick a couple of values for 'x',
and calculate the value of 'y' for each one:
The first value I picked for 'x': x = 72
The equation is y=-3, so when x=72, y=-3. The point is (72, -3)
The second value I picked for 'x' is: x = 1
The equation is y=-3, so when x=1, y=-3. The second point is (1, -3).
The third value I picked for 'x' is 4 billion.
The equation is y=-3, so when x=4 billion, y=-3. The third point is (1, -3).
Do you see what's going on here ? Your original equation didn't even
have 'x' in it, so we could tell right away that when the graph is drawn,
the value of 'y' at every point can't depend on 'x'.
When we simplified the equation and got it in standard form, we found that
the slope of the graph is zero. That means the graph doesn't rise or fall ...
it's just a horizontal line. Sure enough, the height of points on the line
doesn't depend on 'x'. The value of 'y' at every point on the line is -3 .
Answer:
3 x^2 - x + -1
Step-by-step explanation:
Simplify the following:
-(4 x - 2 x^2 - 3) + x^2 + 3 x - 4
Factor -1 out of -2 x^2 + 4 x - 3:
--(2 x^2 - 4 x + 3) + x^2 + 3 x - 4
(-1)^2 = 1:
2 x^2 - 4 x + 3 + x^2 + 3 x - 4
Grouping like terms, 2 x^2 + x^2 + 3 x - 4 x - 4 + 3 = (x^2 + 2 x^2) + (3 x - 4 x) + (-4 + 3):
(x^2 + 2 x^2) + (3 x - 4 x) + (-4 + 3)
x^2 + 2 x^2 = 3 x^2:
3 x^2 + (3 x - 4 x) + (-4 + 3)
3 x - 4 x = -x:
3 x^2 + -x + (-4 + 3)
3 - 4 = -1:
Answer: 3 x^2 - x + -1
Answer:
where is the picture of the graph beautiful
