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:
A. (-7 -2)
Step-by-step explanation:
You can eliminate y by multiplying the first equation by 7 and subtracting 6 times the second equation:
7(-3x +6y) -6(5x +7y) = 7(9) -6(-49)
-21x +42y -30x -42y = 63 +294 . . . . eliminate parentheses
-51x = 357 . . . . . . . . collect terms
x = -7 . . . . . . . divide by -51. This matches answer choice A.
9514 1404 393
Answer:
multiply by 3 and divide by 2 to get 6x -y = 12.
Step-by-step explanation:
A standard form equation has mutually prime integers for coefficients, and the leading one is positive. We can get that here by multiplying by 3 (to eliminate the denominator) and dividing by 2 (to remove that common factor). The result is ...
4(3/2)x -(2/3)(3/2)y = 8(3/2)
6x -y = 12
Well lets solve for both.
6 + 2 + 4x
Combine like terms.
8 + 4x
Now solve the other one.
6 - (2 - 4x)
Distribute.
6 - 2 + 4x
Combine like terms.
4 + 4x.
These equations are not equivalent because instead of adding two, like in the first one, the second problem is subtracting. So you get 8 + 4x and 4 + 4x. I hope this helps love! :)
Answer:
a
d
A
c
c
Step-by-step explanation:
