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3 years ago

Please solve and explain steps for this problem.

Mathematics

1 answer:

Helen [10]3 years ago

7 0

This means Square root (^1/2). this is kind of hard to read. anyways. 32^(1/2) separates into 4^(1/2) and 8^(1/2), which further separates into 4^(1/2) and 2^(1/2), root 4 becomes 2, their are two root 4's, so you get 2 x 2, and you are left with 2^(1/2) now why go to all this trouble. because now you can multiply the 4 you created (2x2) times, the 7... giving you 28*2^(1/2) now subtract it from the the other one with root 2. -5*2^(1/2), giving you 23*2^(1/2)-...idk what that last bit is. if its a odd number then this is the end of the problem, if you can get it to root 2. then do that and simplify.

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Step-by-step explanation:

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Match the expressions with their equivalent simplified expressions.

Tasya [4]

Answer:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

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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Step-by-step explanation:

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Five friends take a maths test

gulaghasi [49]

Answer:

Adam scored 60, Brandon scored 66, Chen scored 74, Damion scored 75, and Erica scored 75.

Step-by-step explanation:

Since five friends took the maths test, and Adam, Brandon, and Chen together together scored 200 marks; Brandon, Chen and Damion together scored 215; Chen, Damion and Erica together scored 224; and Damion and Erica scored more than Chen; While the five of them together scored 350 marks, to determine what are their individual scores the following calculations must be performed:

Adam + Brandon + Chen = 200

Damion + Erica = 150

Brandon + Chen + Damion = 215

Adam + Erica = 135

Chen + Damion + Erica = 224

Adam + Brandon = 126

Adam + Brandon = 126 + Chen = 200

Chen = 200 - 126

Chen = 74

Damion and Erica scored more than Chen

Chen + Damion + Erica = 224

74 + Damion + Erica = 224

Damion + Erica = 150

Damion = 75

Erica = 75

Brandon + Chen + Damion = 215

Brandon + 74 + 75 = 215

Brandon = 215 - 74 - 75

Brandon = 66

Adam = 350 - 75 - 75 - 74 - 66

Adam = 60

Therefore, Adam scored 60, Brandon scored 66, Chen scored 74, Damion scored 75, and Erica scored 75.

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2 years ago