Answer:
proof below
Step-by-step explanation:
Remember that a number is even if it is expressed so n = 2k. It is odd if it is in the form 2k + 1 (k is just an integer)
Let's say we have to odd numbers, 2a + 1, and 2b + 1. We are after the sum of their squares, so we have (2a + 1)^2 + (2b + 1)^2. Now let's expand this;
(2a + 1)^2 + (2b + 1)^2 = 4a^2 + 4a + 4b + 4b^2 + 4b + 2
= 2(2a^2 + 2a + 2b^2 + 2b + 1)
Now the sum in the parenthesis, 2a^2 + 2a + 2b^2 + 2b + 1, is just another integer, which we can pose as k. Remember that 2 times any random integer, either odd or even, is always even. Therefore the sum of the squares of any two odd numbers is always even.
Step-by-step explanation:
what is THE partial product ? there is a mistake in your problem description.
it must read "what are valid partial products".
1262×3 has the following partial products :
1000×3 = 3000
200×3 = 600
60 × 3 = 180
2×3 = 6
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3786
so, D seems to be the best answer, because both numbers are partial products here.
A, B and C have each only one correct.
Use a Position time graph
Your answer would be X=8 and Y=11.
C because it has the x variables in the numerator. This is wrong because the x axis represents horizontal movement and slope is rise over run, not run over rise
