Prove that if m + n and n + p are even integers, where m, n, and p are integers, then m + p is even.
m=2k-n, p=2l-n
Let m+n and n+p be even integers, thus m+n=2k and n+p=2l by definition of even
m+p= 2k-n + 2l-n substitution
= 2k+2l-2n
=2 (k+l-n)
=2x, where x=k+l-n ∈Z (integers)
Hence, m+p is even by direct proof.
Answer:
2x³+17x²+25x-50
Step-by-step explanation:
Standard Form: All your numbers and variables in order of the highest exponent.
(2x²+7x-10)(x+5)
We want to distribute x and 5 to every single value.
2x³+7x²-10x+10x²+35x-50
Simplify.
2x³+17x²+25x-50
This is standard form because the highest values are on the left (x³ is larger than x²).
Well Okay, 1.70 x 6 + .90x, Your pretty much taking 1.70 multiply by 6 then adding .90 for every page after.
Answer: 3
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Step-by-step explanation: because of the decimal you have to place a zero regardless and the other 2 zeros you can’t do anything with them. Hope this helps!
Answer:
D
Step-by-step explanation:
For some side lengths to create a triangle, any two of the side lengths must have an added length together of less than the third leg. Otherwise, the vertices would simply not be able to be connected by the lines. In a), 3+5=8, but since it is not more, the triangle will not be able to be formed (it would basically just be a line.) In B, 7+8< 16. In C, there is the same situation as with the first example (4+15=19). Finally, in the last one, 5+7>12, and all of the side lengths work out. Therefore, the correct answer is D. Hope this helps!
