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.
I am pretty sure the answer is 180
The correct answer would be 16/24. You would need to multiply both the denominator and numerator by 4.
You need to find the total distance and the total time. The total distance is two times the distance to school, 2d.s=d/t=>t=d/sWhere s is speed, d is distance and t is time. So the time taken to get to school is:d/s1and the time taken to get back is:d/s2so the total time is:d/s1+d/s2=d(1/s1+1/s2)So, as speed = distance/time, divide the total distance by the total time:2d/d(1/s1+1/s2)=2/(1/s1+1/s2)=2/(1/20+1/40)=80/3≈26.7km/hr
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