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1 year ago

6

A raku pot at 1,409 degrees fahrenheit. the artist fores a porcelain pot at 266 degrees fahrenheit less than 2 times the tempera

ture at which the raku pot is fired. you want to find the temperature at which the porcelain pot is fired

1 answer:

Alona [7]1 year ago

4 0

The temperature at which porcelain pot is fired is 2552° F.

Firstly forming the equation to find the temperature at which the porcelain pot is fired. Secondly we will solve the equation based on operators. Let us represent the required temperature by t.

t = 2×1409 - 266

Performing multiplication on Right Hand Side of the equation to find the value of t

t = 2818 - 266

Performing subtraction on Right Hand Side of the equation to find the value of t

t = 2552

Based on the temperature, the porcelain pot is fired at 2552° F.

Learn more about equations -

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$=8\sqrt{15}b^{\frac{7}{2}}$

Step-by-step explanation:

$\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}$

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$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}$

$2^{\frac{3}{2}+\frac{3}{2}}$

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$=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}$

$b^{\frac{3}{2}+2}$

$=b^{\frac{7}{2}}$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}$

$=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}$

$=8\sqrt{15}b^{\frac{7}{2}}$

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3 years ago