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ioda
3 years ago
8

What is a counterexample for the conjecture? Any number that is divisible by four is also divisible by eight

Mathematics
1 answer:
Cloud [144]3 years ago
3 0
The value 12 is a good counterexample. It is divisible by 4 (since 12/4 = 3 is a whole number with no remainders or decimal portions) but 12 is not divisible by 8 (note how 12/8 = 1 remainder 4 = 1.5)

To find other examples, add 8 to 12 to get 20. Then add on 8 more to get 28. This pattern continues on forever. 
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Need help with simplifying this question ​
Digiron [165]

{3}^{ - 3}

hope this helps :)

5 0
3 years ago
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3x^2-5x+5y-4+12x <br> WILL GIVE BRAINLY
qwelly [4]
3x^2+7x+5y-4 I believe
4 0
2 years ago
Solve by quadratic formula. x squared plus x plus 1 = 0
Tasya [4]

Remember that the quadratic formula is: \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.


a stands for the coefficient of the first term, the one associated with the x^2, b stands for the coefficient of the second term, the one associated with x, and c stands for the value of the constant.


In x^2 + x + 1 = 0, our a-value is 1, our b-value is also 1, and our c-value is also 1.


Thus, when we plug in our values into our formula, we get the answer:

\dfrac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \dfrac{-1 \pm \sqrt{-3}}{2}


Simplifying this answer using i = \sqrt{-1}, we get our final answer of:

\boxed{x = \dfrac{-1 \pm i\sqrt{3}}{2}}

5 0
3 years ago
Find two consecutive odd integers such that 28 more than the lesser is three times the greater.
Elan Coil [88]

Answer:

lesser odd integer =11

Greater odd integer=13

Step-by-step explanation:

let the consecutive odd integers be x,x+2

and  x<x+2

3(x+2)=x+28

3x+6=x+28

3x-x=28-6

2x=22

x=11

x+2=11+3=13

8 0
3 years ago
Let r = ⟨7, 1⟩ and s = ⟨–12, –3⟩. What is r – s?
Zigmanuir [339]
The answer is <19,4> because when subtracting a negative it becomes a postitive
8 0
3 years ago
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