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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The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c

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Answer:

The maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

Step-by-step explanation:

The area of a rectangle with length $L$ and width $W$ is $A= L\cdot W$ so the differential of <em>A</em> is

$dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W$

$\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L$ so

$dA=W\Delta L+L \Delta W$

We know that each error is at most 0.1 cm, we have $|\Delta L|\leq 0.1$ , $|\Delta W|\leq 0.1$ . To find the maximum error in the calculated area of the rectangle we take $\Delta L = 0.1$ , $\Delta W = 0.1$ and $L=55$ , $W=49$ . This gives