The answer is C. you can check the attached picture.
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:
brainly.com/question/11377865
#SPJ4
Answer:
Yes
Step-by-step explanation:
Multiplying an integer by 0.01 is the same as Dividing such as :
0.01 = 10^-2
100= 10
For instance :
A number, n = 5 is multiplied by 10^-2
5 * 0.01 = 0.05
Thus, divide the number, n by 100;
5 / 100 = 0.05
Hence, the proof.
For line B to AC: y - 6 = (1/3)(x - 4); y - 6 = (x/3) - (4/3); 3y - 18 = x - 4, so 3y - x = 14
For line A to BC: y - 6 = (-1)(x - 0); y - 6 = -x, so y + x = 6
Since these lines intersect at one point (the orthocenter), we can use simultaneous equations to solve for x and/or y:
(3y - x = 14) + (y + x = 6) => 4y = 20, y = +5; Substitute this into y + x = 6: 5 + x = 6, x = +1
<span>So the orthocenter is at coordinates (1,5), and the slopes of all three orthocenter lines are above.</span>
