(PLEASE HELP)simplify 2 with an exponent of 5 over 3 with an exponent of 2 and the equation has an exponent of 4
1 answer:
Assuming you mean this as your problem I hope I can assist you.
So just apply the exponent rule to start, which is (a/b)^c = a^c/b^c
This translates to....
Now apply the exponent rule (a^b)^c = a^bc
Which translates to 3^2 * 4 which is refined to 3^8
So now we have....
Just apply the exponent rule (a^b)^c
So we get 2^5 * 4 which refines to 2^20
Therefore our final answer is 2^20/3^8
So just apply the exponent rule to start, which is (a/b)^c = a^c/b^c
This translates to....
Now apply the exponent rule (a^b)^c = a^bc
Which translates to 3^2 * 4 which is refined to 3^8
So now we have....
Just apply the exponent rule (a^b)^c
So we get 2^5 * 4 which refines to 2^20
Therefore our final answer is 2^20/3^8
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Answer:
Part a)
It takes 1.57 seconds for an object to fall a distance of 2 feet
Part b) see the explanation
Part c)
Step-by-step explanation:
Let
f(d) -----> the time in seconds it takes for an object to fall
d -----> distance in feet
we have
Part a): Determine f(2) and explain what it represents
we know that
f(2) represent the time in seconds it takes for an object to fall a distance of 2 feet
For
substitute in the function above and solve for f(2)
therefore
It takes 1.57 seconds for an object to fall a distance of 2 feet
Part b) The imbrium basin is the largest basin on the moon. A reasonable domain for the height above the lowest point in the basin is given by {d|0 ≤ d ≤ 3805774}
What does this tell you about the basin?
The height of the basin is greater than or equal to 0 ft and less than or equal to 3,805,774 ft
so
The maximum height of the basin is 3,805,774 ft
Part c) How long would it take a rock to drop from the rim to the bottom of the basin?
we know that
The distance from the rim to the bottom of the basin is equal to the maximum height of the basin
so
substitute the value of d in the function f(d)