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2 years ago

Numbers with the same absolute value are equal or equally distant from zero?

2 answers:

3 0

Equally distant is the answer, I believe. Because absolute value is the same whether it's negative or positive. So they would be equally distant

Llana [10]2 years ago

3 0

Equally distant. If -4 is 4 spaces away from 0, so is 4

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Parallel, orthogonal, or neither.

zzz [600]

Vectors are orthogonal, when dot product $u\cdot v=0$

$u\cdot v=\langle-1,3\rangle\cdot\langle-9,-3\rangle=(-1)\cdot(-9)+3\cdot(-3)=9-9=0$

Vectors $u$ and $v$ are orthogonal.

8 0

2 years ago

Prove that: (a2 - b2)3 + (b2-c2)3+ (c2-a2)3 = 3 (a+b) (b+c) (c+a) (a-b) (b-c) (c-a).

Vanyuwa [196]

$L=(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=(*)\\\\(a^2-b^2)^3=a^6-3a^4b^2+3a^2b^4-b^6\\\\(b^2-c^2)^3=b^6-3b^4c^2+3b^3c^4-c^6\\\\(c^2-a^2)^3=c^6-3c^4a^2+3c^2a^4-a^6\\\\(*)=a^6-3a^4b^2+3a^2b^4-b^6+b^6-3b^4c^2+3b^2c^4-c^6+c^6-\dots\\\dots-3c^4a^2+3c^2a^4-a^6\\\\=-3a^4b^2+3a^2b^4-3b^4c^2+3b^2c^4-3c^4a^2+3c^2a^4\\\\=3(-a^4b^2+a^2b^4-b^4c^2+b^2c^4-a^2c^4+a^4c^2)$

$R=3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)\\\\=3(a^2-b^2)(b^2-c^2)(c^2-a^2)\\\\=3(a^2b^2-a^2c^2-b^4+b^2c^2)(c^2-a^2)\\\\=3(a^2b^2c^2-a^4b^2-a^2c^4+a^4c^2-b^4c^2+a^2b^4+b^2c^4-a^2b^2c^2)\\\\=3(-a^4b^2+a^2b^4-b^4c^2+b^2c^4-a^2c^4+a^4c^2)\\\\L=R$

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2 years ago

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schepotkina [342]

Answer:

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2 years ago

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Step-by-step explanation:

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Answer:

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