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}.
Learn more about Intermediate Value Theorem on:
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Answer:
Option 1
Figure Length (feet) Width (feet)
small rectangle 14 6
large rectangle 20 7
Figure Base (feet) Height (feet)
triangle 6 6
Option 2
Figure Length (feet) Width (feet)
small rectangle 6 7
large rectangle 14 13
Figure Base (feet) Height (feet)
triangle 6 6
Step-by-step explanation:
Answer:
0.90
Step-by-step explanation:
hope this helps
Solution:
we are given that
A six sided number cube has faces with the numbers 1 through 6 marked on them.
we have been asked to find the probability that a number less than 2 will occur on one toss of the number cube.
Since a number less than 2 is only one and that is "1" and total number of possible outcome is 6.
and as we know that probability is given using the formula
Substitute the values we get
Hence the required probability is 1/6.
