The first claim,
"If 2<em>n</em> + 4 is even, then <em>n</em> is even"
is false; as a counterexample, consider <em>n</em> = 1, which is odd, yet 2•1 + 4 = 6 is even.
The second claim,
"If <em>n</em> is even, then (<em>n</em> + 3)² is odd"
is true. This is because
(<em>n</em> + 3)² = <em>n</em> ² + 6<em>n</em> + 9
<em>n</em> ² + 6<em>n</em> is even because <em>n</em> is even. 9 is odd. The sum of an even and odd integer is odd.
Answer:
i would say it is b because its more likely to happen
Step-by-step explanation:
This = - 8 * 1/2 * sqrt2*sqrt32
= -4 * sqrt64
= -4 * 8
= -32 (answer) Its B
By working backwards,
<span>you get 76 as your answer because you have to remove the rightmost digit, right? so the three-digit number is </span>
<span>76x </span>
<span>(x stands for the rightmost digit, because we don't know what the rightmost digit is) </span>
<span>we are told we have to add six to the three digit number to get 76x, so we have to subtract 6 from 76x: </span>
<span>76x - 6 = ??? </span>
<span>since we assume that the original two-digit number is NOT 76, and any number from 760 to 765 inclusive minus 6 is logically in the 750-range, we can conclude that the original two-digit number is 75. </span>
Answer:
Step-by-step explanation:
