The law of an object moving with constant acceleration is

Where
is space,
is time,
is the initial position,
is the initial velocity and
is the acceleration.
In this case, if we choose a reference grid with the vertical axis pointing upwards, the acceleration of gravity will point downwards (and thus be negative). The initial position is zero, because the rocket is on the ground, and the initial velocity is 100 (positive because pointing upwards).
So, its law is

(I changed
for
since the rocket is moving vertically, so its position is actually its height. Also, g is the acceleration due to gravity).
The rocket hits the ground if its height is zero, so if we set
we have

Solving for t, we have either t=0, or

The solution t=0 means that at the beginning the rocket is on the ground. So, we're interested in the other solution. Considering that g is about 32.2 feet/s^2, we have

Answer:
The length of x in the simplest radical form with a rational denominator will be:
Step-by-step explanation:
Given
hypotenuse = 5
angle Ф = 60°
To determine
x = ?
Using the trigonometric ratio
cos Ф = adjacent / hypotenuse
here
Ф = 60°
adjacent of 60° = x
hypotenuse = 5
so substituting Ф = 60°, adjacent = x and hypotenuse = 5 in the equation
cos Ф = adjacent / hypotenuse
so


switch sides

Multiply both sides by 5

Simplify

Therefore, the length of x in the simplest radical form with a rational denominator will be:
<h3>
Answer: Choice C. 4*sqrt(6)</h3>
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Explanation:
Each cube has a side length of 4. Placed together like this, the total horizontal side combines to 4+8 = 8. This is the segment HP as shown in the diagram below. I've also added point Q to form triangle HPQ. This is a right triangle so we can find the hypotenuse QH
Use the pythagorean theorem to find QH
a^2 + b^2 = c^2
(HP)^2 + (PQ)^2 = (QH)^2
8^2 + 4^2 = (QH)^2
(QH)^2 = 64 + 16
(QH)^2 = 80
QH = sqrt(80)
Now we use segment QH to find the length of segment EH. Focus on triangle HQE, which is also a right triangle (right angle at point Q). Use the pythagorean theorem again
a^2 + b^2 = c^2
(QH)^2 + (QE)^2 = (EH)^2
(EH)^2 = (QH)^2 + (QE)^2
(EH)^2 = (sqrt(80))^2 + (4)^2
(EH)^2 = 80 + 16
(EH)^2 = 96
EH = sqrt(96)
EH = sqrt(16*6)
EH = sqrt(16)*sqrt(6)
EH = 4*sqrt(6), showing the answer is choice C
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A shortcut is to use the space diagonal formula. As the name suggests, a space diagonal is one that goes through the solid space (rather than stay entirely on a single face; which you could possibly refer to as a planar diagonal or face diagonal).
The space diagonal formula is
d = sqrt(a^2+b^2+c^2)
which is effectively the 3D version of the pythagorean theorem, or a variant of such.
We have a = HP = 8, b = PQ = 4, and c = QE = 4 which leads to...
d = sqrt(a^2+b^2+c^2)
d = sqrt(8^2+4^2+4^2)
d = sqrt(96)
d = sqrt(16*6)
d = sqrt(16)*sqrt(6)
d = 4*sqrt(6), we get the same answer as before
The space diagonal formula being "pythagorean" in nature isn't a coincidence. Repeated uses of the pythagorean theorem is exactly why this is.
Answer:
Step-by-step explanation:
The inequality will be split into two
It is know that, if a<b<c
Then a<b and b<c
-8<2x-4<4
Apply that to this
Then,
-8<2x-4. Equation 1
Also,
2x-4<4 equation 2
Solving equation 1
-8<2x-4
Add 4 to both side of the equation
-8+4<2x-4+4
-4<2x
Divide both sides by 2
-4/2<2x/2
-2<x
Note, if a is less than b, then, b is greater than a, e.g. 4 is less than 10, this implies 10 is greater than 4
Therefore,
-2<x
Then, x greater than -2
Equation 2
2x-4<4
Add 4 to both side of the inequalities
2x-4+4<4+4
2x<8
Divide both side by 2
Then,
2x/2<8/2
x<4
Therefore x is between -2 and +4.
Check attachment for graphical solution