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iren [92.7K]
3 years ago
14

Identify a counterexample to disprove n3 ≤ 3n2, where n is a real number.

Mathematics
2 answers:
Semmy [17]3 years ago
8 0
n^3\le3n^2\iff n^3-3n^2\le0\iff n^2(n-3)\le0

The quantity on the left hand side will be positive for any n>3, so suppose consider n=4.

Now,

4^3=64

but

3(4)^2=48

and 64\not\le48
dem82 [27]3 years ago
5 0
n^3-3n^2\ \leq   \ 0    \text{Move all terms to one side} 

n^2(n-3) \leq0   \text{Factor out the common term} \  {n}^{2} 

n(n-3) = 0  \text{when n=0,3} 

⇒ \text{From the values of} \ n^2 \text{ above, we have these 3 intervals to test} 

⇒ n \leq0 

⇒ 0 \leq   n \leq  3 

⇒ n \geq  3

⇒ \text{Pick a test point for each interval} 


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Anettt [7]
Hey There!

Remember, first find the gcf of 4,3 in this problem it is 12.

Steps to solve

1.6\frac{1}{4} - 3\frac{1}{3}

2.6\frac{1}{4} - 3 \frac{1}{3} = 2\frac{11}{12}

So, The answer is <span>2\frac{11}{12}</span>

Hope This Helps :)
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UNO [17]

1st number: 0

2nd number: 17

Answer

Step-by-step explanation:

3x+2y=34 x is the first number, y is the second.

1/2x+2y=34 multiply each number by two to get rid of the integer by the variable x.

x+4y=68 solve for x.

x=68-4y add this into the first equation to solve for variable y.

3(68-4y)+2y=34 solve for y.

204-10y=34

-10y= -170

y=17

now to solve for x

x= 68-4(17)

x= 0

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Answer:

Please show us the questions and will help further this discussion.


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3 years ago
12-2(n-1)=-18<br><br> aa help plzz
Luda [366]

Answer:

n=16

Step-by-step explanation:

Let's solve your equation step-by-step.

12−2(n−1)=−18

Step 1: Simplify both sides of the equation.

12−2(n−1)=−18

12+(−2)(n)+(−2)(−1)=−18(Distribute)

12+−2n+2=−18

(−2n)+(12+2)=−18(Combine Like Terms)

−2n+14=−18

−2n+14=−18

Step 2: Subtract 14 from both sides.

−2n+14−14=−18−14

−2n=−32

Step 3: Divide both sides by -2.

−2n /−2  =  −32 /−2

n=16

8 0
3 years ago
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