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3 years ago

I NEED HELP ON THIS WKSH ASAP 19 POINTS!!!!!!!!

Mathematics

1 answer:

melisa1 [442]3 years ago

8 0

Answer:

PART A: Venti has the best unit rate because it is the cheapest per oz.

PART B: Venti has the best unit rate because it is the cheapest per oz. Since we are viewing the unit rate for oz, you divide the cost over the amount of oz. This will give you the unit rate and for us, the worst unit rate was the Tall cup and the best unit rate was the Venti cup.

Step-by-step explanation:

PART A

Step 1: Figure out what is the unit rate for each cup size

Tall -> 3.55/12 = 0.30 per oz

Grande -> 4.25/16 = 0.27 per oz

Venti -> 4.55/20 = 0.23 per oz

Answer: Venti has the best unit rate because it is the cheapest per oz.

PART B

Venti has the best unit rate because it is the cheapest per oz. Since we are viewing the unit rate for oz, you divide the cost over the amount of oz. This will give you the unit rate and for us, the worst unit rate was the Tall cup and the best unit rate was the Venti cup.

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Step-by-step explanation:

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Tasya [4]

Answer:

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Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

$\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}$

$\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}$

$\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}$

$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$

Thus,

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