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.
Given expression:
2x(x + 7)-(3x +1);
To distribute implies to spread a process equally in an expression. Often times, the distributive property is used for expressions inside parentheses;
2x(x + 7)-(3x +1)
Distribute 2x over the first parentheses;
the negative sign distributes over the second part
= (2x² + 14x) - 3x -1
= 2x² + 14x -3x -1
Add like terms;
= 2x² + 11x -1
This expression 2x² + 11x -1 is equal to the given one
6x = 16 +14
6x = 30
30 / 6 =5
x= 5
1+1 = 2 that is the answer
Yes. As long as the shape has 4 sides, it's a quadrilateral.