The total number of ways you can toss a coin 3 times is 8.
H H H
H H T
H T H
H T T
T H H
T H T
T T H
T T T
That should be 8 times.
H H T is the only toss that works
1/8 times
No idea because i don’t speak english
Yes it is a function Bc the x values each have a unique y value.
Answer:
a. 0.9 Hz b. 0.37 Hz
Step-by-step explanation:
The frequency of the simple pendulum f = (1/2π)√g/l where g = acceleration due to gravity and l = length of pendulum
a. Find the frequency of a pendulum whose length is 1 foot and where the gravitational field is approximately 32 ft/s2
To find f on Earth, g = 32 ft/s² and l = 1 ft
So, f = (1/2π)√(g/l)
f = (1/2π)√(32 ft/s²/1 ft)
f = (1/2π)√(32/s²)
f = (1/2π)(5.66 Hz)
f = 0.9 Hz
b. The strength of the gravitational field on the moon is about 1/6 as strong as on Earth.. Find the frequency of the same pendulum on the moon.
On the moon when acceleration due to gravity g' = g/6,
f = (1/2π)√(g'/l)
f = (1/2π)√(g/6l)
f = (1/2π)√[32 ft/s²/(6 × 1 ft)]
f = (1/2π)√(32/s²)/√6
f = (1/2π)(5.66 Hz)/√6
f = 0.9/√6 Hz
f = 0.37 Hz
Recall that the the vertical shifting of functions is associated with the numerical value that is added to the original one. A possitive value k will shift the function k units up, and a negativa value (or subtracting k) will move it vertically down k units.
If I understand correctly, you forgot to type an "x"in the original function f(x), which should read:
f(x) = 0.5 x
Correct?
The new graph is g(x) = 0.5 x - k which is obtained by shifting the original f(x) DOWN by 3 units.
In order to shift the graph of f(x) DOWN three units, we need to SUBTRACT 3 from f(x), leading to : 0.5 x - 3
Therefore, the value of "k"in the given g(x) = 0.5 x - k must be 3.
which agrees with the last option in your list (option D)
<br />a)By how much will your stopping distance increase if you increase your speed from 20 mi/h to 30 mi/h?<br />b)By how much will your stopping distance increase if you increase your speed from 50 mi/h to 60 mi/h? <br />c)An important driving guideline is that you leave one car length between your car and the car in front of you for every 10 mi/h of speed. Thus, you should leave three car lengths if you are driving at 30 mi/h and 5 car lengths if you are driving at 50 mi/h. Is this guideline consistent with the model f(x)?