Which of the following cannot be the lengths of the three sides of a triangle? A 3.4.5 B 6.7.10 C 8.38 D 8.7.15

Mathematics

1 answer:

lesantik [10]3 years ago

3 0

Answer:

Step-by-step explanation:

For three line segments to be able to form any triangle you must be able to take any two sides, add their length and this sum be greater than the remaining side.

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$We\ know:\ i=\sqrt{-1}\to i^2=-1$

$\dfrac{2+4i}{3i}=\dfrac{2+4i}{3i}\cdot\dfrac{3i}{3i}=\dfrac{6i+12i^2}{9i^2}=\dfrac{6i+12(-1)}{9(-1)}\\\\=\dfrac{6i-12}{-9}=\dfrac{2i-4}{-3}=\dfrac{4}{3}-\dfrac{2}{3}i$

$\dfrac{3+2i}{4+i}=\dfrac{3+2i}{4+i}\cdot\dfrac{4-i}{4-i}=\dfrac{(3+2i)(4-i)}{4^2-i^2}\\\\=\dfrac{(3)(4)+(3)(-i)+(2i)(4)+(2i)(-i)}{16-(-1)}=\dfrac{12-3i+8i-2i^2}{16+1}\\\\=\dfrac{12+5i-2(-1)}{17}=\dfrac{12+5i+2}{17}=\dfrac{14+5i}{17}=\dfrac{14}{17}+\dfrac{5}{17}i$