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2 years ago

A 5.0g of salt was weighed by Victoria as 5.1g. What is the percentage error?

Mathematics

1 answer:

Aleksandr-060686 [28]2 years ago

7 0

qwerty blablablablabal

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3 years ago

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

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2 years ago

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neonofarm [45]

Answer:

Is this a question or is this spamming numbers ?

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2 years ago

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A 5.0g of salt was weighed by Victoria as 5.1g. What is the percentage error?​

A 5.0g of salt was weighed by Victoria as 5.1g. What is the percentage error?