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Mariana [72]
3 years ago
15

make a conjecture, is there a value of N for which there could be a triangle with sides of length N, 2N, and 3N?

Mathematics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer: Conjecture: There is no triangle with side lengths N, 2N, and 3N (where N is a positive real number)

Proof:

We prove this by contradiction: Suppose there was an N for which we can construct a triangle with side lengths N, 2N, and 3N. We then apply the triangle inequalities tests. It must hold that:

N + 2N > 3N

3N > 3N

3 > 3

which is False, for any value of N. This means that the original choice of N is not possible. Since the inequality is False for any value of N, there cannot be any triangle with the given side lengths, thus proving our conjecture.

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Find the distance between the points
Tasya [4]

Answer:

5 units

Step-by-step explanation:

Calculate the distance d using the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = (2, 1) and (x₂, y₂ ) = (6, 4)

d = \sqrt{(6-2)^2+(4-1)^2}

   = \sqrt{4^2+3^2}

   = \sqrt{16+9}

    = \sqrt{25}

    = 5

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3 years ago
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The figure below shows a shaded region and a non-shaded region. Angles in the figure that appear to be right angles are right an
Naddika [18.5K]

Answer:

the shaded area is 26, the non-shaded area 16

Step-by-step explanation:

The red shaded lines had numbers as well so I add all of them up.

The non-shaded lines didn't have a number, but the bottom had the same length of the top line so therefore your answer is 26 for Shaded, and 16 for non-shaded.

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2 years ago
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Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t
Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1.f(x)=\frac{x^2+x-20}{x^2+4}

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345

3.h(x)=\frac{3x-5}{x^2-5x+7}

x^2-5x+7=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43

2.g(x)=\frac{x-17}{x^2+75}

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12

4.i(x)=\frac{x^2-9}{x-9}

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is  not continuous on the set of real numbers.

5.j(x)=\frac{4x^2-7x-65}{x^2+10}

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0

6.k(x)=\frac{x+1}{x^2+x+29}

x^2+x+29=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102

7.l(x)=\frac{5x-1}{x^2-9x+8}

x^2-9x+8=0

x^2-8x-x+8=0

x(x-8)-1(x-8)=0

(x-8)(x-1)=0

x=8,1

The function is not defined for x=8 and x=1

Hence, function l is not  defined for all real values.

8.m(x)=\frac{x^2+5x-24}{x^2+11}

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

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Answer:

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