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3 years ago

If a+b+c=16 and a^2+b^2+c^2=90 then find the value a^3+b^3+c^3-3abc

1 answer:

3 0

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\\(16)^2=90+2(ab+bc+ca)\\256-90=2(ab+bc+ca)\\166/2=ab+bc+ca\\83=ab+bc+ca\\a^3-b^3-c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\16(90-(83))\\16(7)\\=112$

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The distance from the top of her head to the floor is 6 feet 2 inches.

Step-by-step explanation:

In his case Juana's height is given to us with two kinds of units, feet and inches, in order to make our solution easyer we will transform her height to only inches. In 1 feet we have 12 inches, so we need to take the part of her height that is given in feet and multiply it by 12. We have:

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3 years ago

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