if two fractions have the same denominator but different new Mary which fraction is greater? Give an example

Mathematics

1 answer:

german4 years ago

5 0

It would be the one with the greater numerator 6/6 would be greater than 5/6

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Can someone help me plz<br> (-5x^4y^2)^3

yan [13]

(-5x⁴y²)³

To avoid confusion, we can separate each variable and raise it to the 3rd power.

-5³ * (x⁴)³ * (y²)³

-5³ = -5 * -5 * -5 = -125
(x⁴)³ = x⁴*³ = x¹²
(y²)³ = y²*³ = y⁶

<span>(-5x⁴y²)³ = -125x</span>¹²y⁶

5 0

4 years ago

The increase in length of an aluminum rod is twice the increase in length of an Invar rod with only a third of the temperature i

cluponka [151]

Answer:

the ratio of lengths of the two rods, Aluminum to Invar is 11.27

Step-by-step explanation:

coefficient of linear expansion of aluminum, $\alpha _{Al} = 23 \times 10^{-6} /K$

Coefficient of linear expansion of Invar, $\alpha _{Iv} = 1.2 \times 10^{-6}/K$

Linear thermal expansion is given as;

$\Delta L = L_0 \times \alpha\times \Delta T\\\\where;\\\\L_0 \ is \ the \ original \ length \ of \ the \ metal\\\\\Delta L \ is \ the \ increase \ in \ length$

The increase in length of Invar is given as;

$\Delta L_{Iv} = L_0_{Iv} \times \alpha _{Iv}\times \Delta T_{Iv}$

The increase in length of the Aluminum;

$\Delta L_{ Al} = L_0_{Al} \times \alpha _{Al} \times \Delta T_{Al}\\\\from \ the\ given \ question, \ the \ relationship \ between \ the \ rods \ is \ given \ as\\\\ L_0_{Al} \times \alpha _{Al} \times \frac{1}{3} \Delta T_{Iv}= 2( L_0_{Iv} \times \alpha _{Iv} \times \Delta T_{Iv})\\\\ L_0_{Al} \times \alpha _{Al} \times \Delta T_{Iv}= 6( L_0_{Iv} \times \alpha _{Iv} \times \Delta T_{Iv})\\\\ L_0_{Al} \times \alpha _{Al} \times \Delta T_{Iv} = 6L_0_{Iv} \times 6\alpha _{Iv} \times 6 \Delta T_{Iv}\\\\$

$\frac{L_0_{Al}}{6L_0_{Iv} } = \frac{6\alpha _{Iv} \ \times \ 6 \Delta T_{Iv}}{\alpha _{Al} \ \times \ \Delta T_{Iv}} \\\\\frac{L_0_{Al}}{6L_0_{Iv} } = \frac{6\alpha _{Iv} \ \times \ 6}{\alpha _{Al} \ } \\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{\alpha _{Iv} }{\alpha _{Al} } )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{1.2 \times 10^{-6} }{23\times 10^{-6} } )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{1.2}{23} )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = \frac{259.2}{23} \\\\\frac{L_0_{Al}}{L_0_{Iv} } = 11.27$