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:
Step-by-step explanation:
(cos 4a*cos 2a+sin 4a*sin 2a)/sin 4a
=[cos (4a-2a)]/sin 4a
=(cos 2a)/sin 4a
=(cos 2a) /(2 sin 2a cos 2a)
=1/(2 sin 2a)
=1/2 csc 2a
Answer:
48 member did not vote
Step-by-step explanation:
32 is 40% of 80
80-32=48
Answer:
x√5 = x square root 5
Step-by-step explanation:
√35x^5 ÷ √7x^3 = √(35x^5 / 7x^3)
√(35x^5 / 7x^3) = √(5x^5 / x^3)
√(5x^5 / x^3) = √5x^(5-3) = √5x²
√5x² = x√5
Answer:
The Answer: 50°
Step-by-step explanation:
