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3 years ago

I'M BEGGING, SOMEONE PLEASE HELP ME WITH THIS!!!! I must get this done TODAY!!

Mathematics

1 answer:

zalisa [80]3 years ago

8 0

A)
π = 3.1
√3 = 1.7
2√3 = √12 = 3.4 (twice of √3)
√5 = 2.2

b)
ordered from least to greatest are:
√3 , √5, π, 2√3

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Scrieți multipli lui 7 mai mari decât 10 și mai mici decât 20

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The answer is here:

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2 years ago

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1 year ago

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2 years ago

What is the simplified form of the following expression? Assume y=0 ^3 sqrt 12x^2/16y

pochemuha

For this case we must simplify the following expression:

$\sqrt [3] {\frac {12x ^ 2} {16y}}$

We rewrite the expression as:

$\sqrt[3]{\frac{4(3x^2)}{4(4y)}}=\\\sqrt[3]{\frac{4(3x^2)}{4(4y)}}=\\\frac{\sqrt[3]{3x^2}}{\sqrt[3]{4y}}=$

We multiply the numerator and denominator by:

$(\sqrt[3]{4y})^2:\\\frac{\sqrt[3]{3x^2}*(\sqrt[3]{4y})^2}{\sqrt[3]{4y}*(\sqrt[3]{4y})^2}=$

We use the rule of power $a ^ n * a ^ m = a ^ {n + m}$ in the denominator:

$\frac{\sqrt[3]{3x^2}*(\sqrt[3]{4y})^2}{(\sqrt[3]{4y})^3}=\\\frac{\sqrt[3]{3x^2}*(\sqrt[3]{4y})^2}{4y}=$

Move the exponent within the radical:

$\frac{\sqrt[3]{3x^2}*(\sqrt[3]{16y^2}}{4y}=\\\frac{\sqrt[3]{3x^2}*(\sqrt[3]{2^3*(2y^2)}}{4y}=$

$\frac{2\sqrt[3]{3x^2}*(\sqrt[3]{(2y^2)}}{4y}=\\\frac{2\sqrt[3]{6x^2*y^2}}{4y}=$

$\frac{\sqrt[3]{6x^2*y^2}}{2y}$

Answer:

$\frac{\sqrt[3]{6x^2*y^2}}{2y}$

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3 years ago

Read 2 more answers