So lets try to prove it,
So let's consider the function f(x) = x^2.
Since f(x) is a polynomial, then it is continuous on the interval (- infinity, + infinity).
Using the Intermediate Value Theorem,
it would be enough to show that at some point a f(x) is less than 2 and at some point b f(x) is greater than 2. For example, let a = 0 and b = 3.
Therefore, f(0) = 0, which is less than 2, and f(3) = 9, which is greater than 2. Applying IVT to f(x) = x^2 on the interval [0,3}.
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Answer:
All of the points lie on the graph except for the coordinates (-2,1) and (1,5)
Step-by-step explanation:
we know that
The least
common multiple (LCM) of two or more numbers is the least
number, other than zero, that is a multiple of all the numbers.
In this problem, we are looking for the least common multiple of and
So
Multiply by
therefore
On the day after today, she will do both on the same day
the answer is
Answer: Company b
Step-by-step explanation:
because company B is only 5percent it is better and can pay in 8 years