<span>8(R + 6) - 2R,
8R + 48 - 2R,
The result is 6R + 48</span>
You just subtract 7 from both sides, this way you isolate f and its calculation is very easy
Answer:
155.3 m
Step-by-step explanation:
<u>We have:</u>
θ: is the angle measured from the vertical = 75 °
v = is the speed of the submarine = 10 m/s
t = 1 min
The deep of the front end of the submarine at the end of a 1-minute drive is given by the following ratio:
<u>Where:</u>
d: is the distance traveled by the submarine in 1 minute
x: is the deep to find
The distance, d, is:
Now, the deep is:

Therefore, the deep of the front end of the submarine at the end of a 1-minute drive is 155.3 m.
I hope it helps you!
Answer:
Option A:
x^2 + (y - 2)^2 = 9
Step-by-step explanation:
We know that the equation for a circle centered in the point (a, b) and of radius R is given by:
(x - a)^2 + (y - b)^2 = R^2
So the first thing we need to find is the center of the circle.
We can see that the center is at:
x = 0
y = 2
Then the center is at the point (0, 2)
Now we want our circle to pass through point 2, located at a distance of 2 units from the radius of the first circle.
So the distance between the center and point 2 is 2 units plus the radius of the smaller circle:
And the radius of the smaller circle is one unit.
Then, the radius of a circle centered at (0, 2) that passes through point 2 is:
R = 1 + 2 = 3
Then we have a circle centered at (0, 2) and of radius R = 3
Replacing these in the equation for a circle we get:
(x - 0)^2 + (y - 2)^2 = 3^2
x^2 + (y - 2)^2 = 9
The correct option is A
Answer:
It takes 1 second for the tape to reach the ground.
Equation to use: ![y(t)=16-\frac{1}{2} g\,t^2[/tex with acceleration due to gravity "g" = 32ft/s^2]Step-by-step explanation:This is an object moving vertically under the action of the acceleration of gravity (32 ft/s^2), with a starting position of 16 feet, and with NO initial velocity (drops from the roof).The equation that describes the"y" position of the object as a function of time (t) will be written as:[tex]y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2](https://tex.z-dn.net/?f=y%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5B%2Ftex%20with%20acceleration%20due%20to%20gravity%20%22g%22%20%3D%2032ft%2Fs%5E2%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EStep-by-step%20explanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EThis%20is%20an%20object%20moving%20vertically%20under%20the%20action%20of%20the%20acceleration%20of%20gravity%20%2832%20ft%2Fs%5E2%29%2C%20with%20a%20starting%20position%20of%2016%20feet%2C%20and%20with%20NO%20initial%20velocity%20%28drops%20from%20the%20roof%29.%3C%2Fp%3E%3Cp%3EThe%20equation%20that%20describes%20the%22y%22%20%20position%20of%20the%20object%20as%20a%20function%20of%20time%20%28t%29%20will%20be%20written%20as%3A%3C%2Fp%3E%3Cp%3E%5Btex%5Dy%28t%29%20%3D%20y_0%2B%20v_0%2At-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2)
As explain above, the initial position
s 16 ft, there is no initial velocity, so
, and the acceleration of gravity is 32 ft/s^2, and should be considered negative [as pointing down in the y-direction], so the equation simplifies to:

In order to find the tima it takes it to hit the ground, we simple solve the equation for t when y(t) = 0 (the tape has reached the ground (zero height in the y-direction):

We select the positive time (+1 second) which is what makes physical sense, since a negative value in time would mean time before the tape was dropped.
So the answer is: It takes 1 second for the tape to reach the ground.