Answer:
Step-by-step explanation:
6
Base case: if <em>n</em> = 1, then
1² - 1 = 0
which is even.
Induction hypothesis: assume the statement is true for <em>n</em> = <em>k</em>, namely that <em>k</em> ² - <em>k</em> is even. This means that <em>k</em> ² - <em>k</em> = 2<em>m</em> for some integer <em>m</em>.
Induction step: show that the assumption implies (<em>k</em> + 1)² - (<em>k</em> + 1) is also even. We have
(<em>k</em> + 1)² - (<em>k</em> + 1) = <em>k</em> ² + 2<em>k</em> + 1 - <em>k</em> - 1
… = (<em>k</em> ² - <em>k</em>) + 2<em>k</em>
… = 2<em>m</em> + 2<em>k</em>
… = 2 (<em>m</em> + <em>k</em>)
which is clearly even. QED
1 Remove parentheses.
2(5c−7)≥c−3
2 Expand.
10c−14 ≥ c −3
3 Add 14 to both sides.
10c-14+14≥ c − 3+14
4 Simplify c−3+14 to c+11
10c≥c+11
5 Subtract c from both sides.
10c−c≥11
6 Simplify 10c−c to 9c
9c≥11
7 Divide both sides by 9.
c ≥11/9
Hello!
Your answer will be $150! Hope this helps! :D
