Let 3<em>n</em> + 1 denote the "number" in question. The claim is that
(3<em>n</em> + 1)² = 3<em>m</em> + 1
for some integer <em>m</em>.
Now,
(3<em>n</em> + 1)² = (3<em>n</em>)² + 2 (3<em>n</em>) + 1²
… = 9<em>n</em>² + 6<em>n</em> + 1
… = 3<em>n</em> (3<em>n</em> + 2) + 1
… = 3<em>m</em> + 1
where we take <em>m</em> = <em>n</em> (3<em>n</em> + 2).
Answer:
Your answer is B
Step-by-step explanation:
6.14 Checkpoint: Systems of Linear Inequalities
I took the test.
the answer is x is greater then -7
to find this add 4 to each side
then divide by -2.
since you divided both sides by a negative the symbol flips and you get x is greater then -7
Answer:
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First step to factor the trinomial, multiply the constant -28 with the leading coefficient 6.
So, -28 * 6 = - 168.
Next step is to breakdown -168 into two multiples so that their addition will result the coefficient of w= - 13.
So, -168 = -21 * 8.
And addition of - 21 and 8 will give - 13.
So, next step is to replace -13x with -21x + 8. Therefore,
6x² - 21x + 8x - 28 =0
(6x² - 21x) + (8x - 28 )=0 Make the group of two terms.
3x (2x - 7) + 4 (2x - 7) = 0 Take out the common factor from each group.
(2x - 7)(3x + 4) = 0 Again take out the common factor (2x -7).
So, 2x - 7 =0, 3x + 4 = 0 Set both the factor equal to 0.
2x = 7 , 3x = -4
So, 
Hope this helps you !