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3 years ago

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Can someone help me with this problem.

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Amiraneli [1.4K]3 years ago

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Answer:

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borishaifa [10]

(x^a)(x^b)=x^(a+b)

(ab)(cd)=(a)(b)(c)(d)

x^-m=1/(x^m)

(3y^-4)(2y^-4)=
(3)(y^-4)(2)(y^-4)=
(6)(y^-8)=
6/(y^8)

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3 years ago

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To the nearest tenth, find the perimeter of ABC with vertices A (-2,-2) B (0,5) and C (3,1)

Pavlova-9 [17]

the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.

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$\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill$

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3 years ago

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Troyanec [42]

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Step-by-step explanation:

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Range: (-∞, 2)∪ (2, ∞)

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