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:
Together they equal 180 l so you need to do 180-128 which equals 52. The answer is <D measures 52
Step-by-step explanation:
Answer:
-5x^2-2x
Step-by-step explanation:
When distributing, you multiply the term outside the brackets to all the terms in brackets.
If the term outside the bracket is negative, then when distributing/opening the brackets, the signs of the terms changes.
So in this prob. -x would multiply to both +5x and +2
-x*5x+-x*2
∴-5x^2-2x
120/4 is 30
You'll be able to put 4 nails thirty times.
You put them every 6 feet
So, you have 5 feet before running out.
