Answer:
smaller x is to the power of 2 3x is the larger.
Step-by-step explanation:
Answer:
4Joules
Step-by-step explanation:
According to Hooke's law which states that extension of an elastic material is directly proportional to the applied force provide that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
K is the elastic constant
e is the extension
If a spring exerts a force of 6 N when stretched 3 m beyond its natural length, its elastic constant 'k'
can be gotten using k = f/e where
F = 6N, e = 3m
K = 6N/3m
K = 2N/m
Work done on an elastic string is calculated using 1/2ke².
If the spring is stretched 2 m beyond its natural length, the work done on the spring will be;
1/2× 2× (2)²
= 4Joules
If your functions are
f(x) = 16·x
g(x) = 16·(1/2x)
The output values of g(x) are one-half the output values of f(x) for the same value of x.
If your functions are
f(x) = 16^x
g(x) = 16^(1/2x)
The output vaules of g(x) are the square root of the output values of f(x) for the same value of x.
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You seem to have left out the relevant operator symbol, so we don't really know your intent. Copy/paste operations will do that. Editing is sometimes needed.
Answer:
Step-by-step explanation:
D.All of the above
