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11Alexandr11 [23.1K]
4 years ago
5

Given that x is a number there is not negative Kelvin conjectured that x(to the 2nd power) > x+1

Mathematics
2 answers:
Sav [38]4 years ago
4 0

The way this is written, you cannot use -1/2 or -2. Is that correct? Both of them are minus. So the only things you have going for you is 0 and 5

x^2 > x + 1  when x = 5

5^2 > 5 + 1 and his conjecture is true.

25 > 6

==============

What about 0?

0^2 = 0   so

0 > 0 + 1 and the conjecture is false.


Murrr4er [49]4 years ago
3 0

Answer:

5

Step-by-step explanation:

Given inequality,

x^2 > x + 1

For x=-\frac{1}{2}

(-\frac{1}{2})^2 > -\frac{1}{2} + 1

\implies \frac{1}{4} > \frac{1}{2} ( False )

For x=-2

(-2)^2 > -2 + 1

\implies 4 > 1 ( true )

For x=0

(0)^2 > 0+ 1

\implies 0 > 1 ( False )

For x=5

(5)^2 > 5 + 1

\implies 25 > 6 ( True )

Thus, x = -2 and 5 are the solutions of the given inequality,

But, negative numbers are not allowed.

Hence, x = 5 is a counterexample to kelvins conjecture.

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\boxed{x^{2} + 2x - 8}

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\begin{array}{lll}\textbf{Steps} & \textbf{Problem: }(x - 2)(x + 4) & \\\textbf{1. List variables} & a = x - 2 & \\ & b = x & \\ & c = 4 &\\\textbf{2. Distribute (x - 2)} & (x -2)(x + 4)\\ & = (x - 2)(x) + (x - 2)(4)\\\textbf{3. Distribute x and 4} & x^{2} -2x + 4x - 8\\\textbf{4. Combine like terms}& x^{2} + 2x - 8\\\end{array}\\\text{The area of the updated skatepark will be }\boxed{\mathbf{ x^{2} + 2x - 8}}

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The dimensions of a rectangle are √50a^3b^2 and √200a^3. What is the students error?
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The student incorrectly simplified 30ab\sqrt{2a}+20a\sqrt{2a} .

Thus, option D is correct.

<u></u>

Step-by-step explanation:

The formula to determine the Perimeter of a rectangle of width w and length l is expressed as:

P = 2l + 2w

In other words, the perimeter can be determined by multiplying the length and width by 2 and adding the result.

In our case, the dimensions of a rectangle are \sqrt{50a^3b^2}  and  \sqrt{200a^3}\:\:.

Here is the student's solution:

2\sqrt{50a^3b^2}+2\sqrt{200a^3}=2\cdot 5ab\sqrt{2a}+2\cdot 10a\sqrt{2a}

                                 =10ab\sqrt{2a}+20a\sqrt{2a}

                                 =30ab\sqrt{2a}          

The student made an error in calculating 30ab\sqrt{2a}+20a\sqrt{2a} , because 30ab\sqrt{2a}+20a\sqrt{2a} are not like terms.

Hence, 30ab\sqrt{2a}+20a\sqrt{2a} can not be simplified to 30ab\sqrt{2a}

Therefore, the student incorrectly simplified 30ab\sqrt{2a}+20a\sqrt{2a} .

Thus, option D is correct.

<u></u>

<u></u>

<u>Here is the correct Solution:</u>

2\sqrt{50a^3b^2}+2\sqrt{200a^3}=2\cdot 5ab\sqrt{2a}+2\cdot 10a\sqrt{2a}

                                 =10ab\sqrt{2a}+20a\sqrt{2a}

7 0
3 years ago
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