-4x+7x-14
3x-14>-18
3x>-4
X>-4/3
EXPLANATION
Let's see the facts:
The scale is---> 14 milimeters ----> 5 meters
Width = 42 milimeters
Applying the unitary method, the actual width of the pool is:

The width of the pool is 15m
As x approaches -inf f(x) -> -inf
and as x approaches inf, f(x) approaches +inf
Mark brainliest please
Answer:
The correct answer is:
Between 600 and 700 years (B)
Step-by-step explanation:
At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

First, let us calculate the decay constant (k)

Next, let us calculate the half-life as follows:

Therefore the half-life is between 600 and 700 years