Answer:
Step-by-step explanation:
The vertex form of an equation of a parabola y = ax² + bx + c:
(h, k) - vertex
We have the equation:
Substitute:
Finally:
Answer:
x=1/3
Step-by-step explanation:
A function f is given as
in the interval [0,1]
This function f being an algebraic polynomial is continuous in the interval [0,1] and also f is differntiable in the open interval (0,1)
Hence mean value theorem applies for f in the given interval
The value
Find derivative for f
Equate this to -5 to check mean value theorem
We find that 1/3 lies inside the interval (0,1)
<em>Greetings from Brasil...</em>
First degree equation. The variables are in the 1st member and the numbers in the 2nd member.
X + 6 - 2X = X - 24
X - 2X - X = - 24 - 6
- 2X = - 30
<h2>X = 15</h2>
Answer:
120.8 feet
Step-by-step explanation:
Answer and Step-by-step explanation:
(a) Given that x and y is even, we want to prove that xy is also even.
For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.
(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:
Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.
(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.
