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
Answer: ...well can you be a little specific please¿
Step-by-step explanation:
Factor:
3x^2 + 27
= 3(x^2 + 9)
Answer is 3(x^2 + 9), when factored.
A) (3x + 9i)(x + 3i)
= (3x + 9i)(x + 3i)
= (3x)(x) + (3x)(3i) + (9i)(x) + (9i)(3i)
= 3x^2 + 9ix + 9ix + 27i^2
= 27i^2 + 18ix + 3x^2
B) (3x - 9i)(x + 3i)
= (3x + - 9i)(x + 3i)
= (3x)(x) + (3x)(3i) + ( - 9i)(x) + (- 9i)(3i)
= 3x^2 + 9ix - 9ix - 27i^2
= 27i^2 + 3x^2
C) (3x - 6i)(x + 21i)
= (3x + - 6i)(x + 21i)
= (3x)(x) + (3x)(21i) + (- 6i)(x) + ( -6i)(21i)
= 3x^2 + 63ix - 6ix - 126i^2
= - 126i^2 + 57ix + 3x^2
D) (3x - 9i)(x - 3i)
= (3x + - 9)(x + - 3)
= (3x)(x) + (3x)( - 3i) + (- 9)(x) + ( - 9)( - 3i)
= 3x^2 - 9ix - 9x + 27i
= 9ix + 3x^2 + 27i - 9x
Hope that helps!!!
Answer:
I will just choosé the second one
Because it is possible to compare sizes In one triangle rather than comparing with another one which we don't reàlly know of its sizes.
The lowest common multiple of 7 and 5 is 35, so you divide 365 by 35, the whole number is your answer,
the answer is 10