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ExtremeBDS [4]
3 years ago
11

Simplify. StartStart Fraction StartFraction 16 Over a squared EndFraction plus StartFraction 24 Over ab EndFraction plus StartFr

action 9 Over b squared EndFraction OverOver StartFraction 16 Over a squared EndFraction minus StartFraction 9 Over b squared EndFraction End EndFraction Start StartFraction StartFraction 16 Over a squared EndFraction plus StartFraction 24 Over ab EndFraction plus StartFraction 9 Over b squared EndFraction OverOver StartFraction 16 Over a squared EndFraction minus StartFraction 9 Over b squared EndFraction End EndFraction equals nothing ​(Simplify your​ answer.)
Engineering
2 answers:
sertanlavr [38]3 years ago
5 0

Answer:

(3a + 4b)² / a²b²

Explanation:

This question is difficult to interpret, however the concept being tested in this question is the ability to simplify fractions with variables.

It is a basic rule that only like terms (as denominators) can be added and subtracted from each other, however if you need to add or subtract two fractions with different denominators, it is important to make sure you create a common denominator.

For example, in the beginning, the first fractions are:

16/a² + 24/ab + 9/b²

So it is important to make sure there is a common denominator between all 3 fractions. An easy way of doing this would be to multiply denominators together:

(16b²)/a²b² + (9a²)/a²b² + 24/ab

(9a²+16b²)/a²b² + 24/ab

So now it has gone from 3 fractions to 2, and then to 1:

(9a²+16b²)/a²b² + 24ab/a²b²

(9a² + 24ab + 16b²) / a²b²

Then all that needs to be done is simplify the numerator:

(3a + 4b)² / a²b²

inysia [295]3 years ago
5 0

Answer:

(3a + 4b)² / a²b²

Explanation:

This question is difficult to interpret, however the concept being tested in this question is the ability to simplify fractions with variables.

It is a basic rule that only like terms (as denominators) can be added and subtracted from each other, however if you need to add or subtract two fractions with different denominators, it is important to make sure you create a common denominator.

For example, in the beginning, the first fractions are:

16/a² + 24/ab + 9/b²

So it is important to make sure there is a common denominator between all 3 fractions. An easy way of doing this would be to multiply denominators together:

(16b²)/a²b² + (9a²)/a²b² + 24/ab

(9a²+16b²)/a²b² + 24/ab

So now it has gone from 3 fractions to 2, and then to 1:

(9a²+16b²)/a²b² + 24ab/a²b²

(9a² + 24ab + 16b²) / a²b²

Then all that needs to be done is simplify the numerator:

(3a + 4b)² / a²b²

Explanation:

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A sample of semiconductor has a cross sectional area of 1cm^2 and a thickness of 0.1cm.
pogonyaev

Answer:

a. 3.17*10¹⁹ b. 3.17*10¹⁴

Explanation:

a. Area A = 1cm²

Thickness h = 0.1cm

Energy of photon E = hf=hc/λ

Where f = frequency; c=speed of light; λ=wavelength = 6300Α=6300*10⁻¹⁰;

Planck's constant h = 6.626*10⁻³⁴ joule-seconds; speed of light c = 3*10⁸

Therefore E = (6.626*10⁻³⁴)*3*10⁸/6300*10⁻¹⁰=3.155*10¹⁹J

1Watt of light releases 3.17*10¹⁸ photons per second.

Volume of sample = Area * Thickness = 1*0.1=0.1cm³

Therefore, number of electron hole pairs that are generated per unit volume per unit time = 3.17*10¹⁸/0.1 = 3.17*10¹⁹photon/cm³-s

b. Steady state excess carrier concentration = 3.17*10¹⁹*10μs

=3.17*10¹⁹*10*10⁻⁶

=3.17*10¹⁴/cm³

6 0
3 years ago
A heat engine does 210 J of work per cycle while exhausting 440 J of waste heat. Part A What is the engine's thermal efficiency?
Oxana [17]

Answer:

The engine's thermal efficiency is 0.32

Explanation:

Thermal efficiency = work done ÷ quantity of heat supplied

Work done = 210 J

Quantity of heat supplied = work done + waste heat = 210 + 440 = 650 J

Thermal efficiency = 210 ÷ 650 = 0.32

6 0
4 years ago
If you are involved in a collision and your vehicle is blocking the flow of traffic, you should
zimovet [89]

Answer:

leave the vehicle where it is until the police arrive

Explanation:

4 0
3 years ago
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