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vaieri [72.5K]
3 years ago
15

Help me with the last part at the bottom please it says x =

Mathematics
1 answer:
lukranit [14]3 years ago
7 0

Answer:

Step-by-step explanation:

x = 0.8 + 0.2y

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The rate (In mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function 110I
Ksju [112]

Answer:

P is maximum at I = 2

Step-by-step explanation:

Here is the complete question

The rate (in mg carbon/m³/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 100I/(I² + I + 4) where I is the light intensity (measured in thousands of foot candles). For what light intensity P is a maximum?

To find the value of I at which P is maximum, we differentiate P with respect to I and equate it to zero.

So, dP/dI =  d[100I/(I² + I + 4)]/dI

= [(I² + I + 4)d(100I)/dI - 100Id(I² + I + 4)/dI]/(I² + I + 4)²

= [(I² + I + 4)100 - 100I(2I + 1)]/(I² + I + 4)²

= [100I² + 100I + 400 - 200I² - 100I]/(I² + I + 4)²

= [-100I² + 400]/(I² + I + 4)²

=  -100[I² - 4]/(I² + I + 4)²

Since dP/dI = 0,  -100[I² - 4]/(I² + I + 4)² = 0 ⇒ I² - 4 = 0 ⇒ I² = 4 ⇒ I = ±√4

I = ±2

Since I cannot be negative, we ignore the minus sign

To determine if this is a maximum point, we differentiate dP/dI. So,

d(dP/dI)/dI = d²P/dI² = d[-100[I² - 4]/(I² + I + 4)²]/dI

= [(I² + I + 4)²d(-100[I² - 4])/dI - (-100[I² - 4])d(I² + I + 4)²/dt]/[(I² + I + 4)²]²

= [(I² + I + 4)²(-200I) + 100[I² - 4]) × (2I + 1) × 2(I² + I + 4)]/(I² + I + 4)⁴

= [-200I(I² + I + 4)² + 200[I² - 4])(2I + 1)(I² + I + 4)]/(I² + I + 4)⁴

= [-200(I² + I + 4)[I(I² + I + 4) - [I² - 4])(2I + 1)]]/(I² + I + 4)⁴

= [-200(I² + I + 4)[I³ + I² + 4I - I² + 4])(2I + 1)]]/(I² + I + 4)⁴

= [-200(I² + I + 4)[I³ + 4I + 8])(2I + 1)]]/(I² + I + 4)⁴

Substituting I = 2 into d²P/dI², we have

= [-200(2² + 2 + 4)[2³ + 4(2) + 8])(2(2) + 1)]]/(2² + 2 + 4)⁴

= [-200(4 + 2 + 4)[8 + 8 + 8])(4 + 1)]]/(4 + 2 + 4)⁴

= [-200(10)[24](5)]]/(10)⁴

= -240000/10⁴

= -24

Since d²P/dI² = -24 < 0 at I = 2,  this shows that it I = 2 is a maximum point.

So, P is maximum at I = 2

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2 years ago
Dinkle Company purchased equipment for $50,000. The equipment has an estimated residual value of $5,000 and an expected useful l
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Answer:

Year 1

As per company book deprecation: (Equipment cost-Salvage value)/Life time = (50,000 - 5,000)/10 = 45,000/10 = $4,500

As per income tax depreciation = Cost * MACRS rate for year 1 = 50,000 * 20% = $10,000

Therefore, difference in year 1 will be: $10,000 - $4,500 = $5,500

Year 1

As per company book deprecation: (Equipment cost-Salvage value)/Life time = (50,000 - 5,000)/10 = 45,000/10 = $4,500

As per income tax depreciation = Cost * MACRS rate for year 2 = 50,000 * 32% = $16,000

Therefore, difference in year 1 will be: $16,000 - $4,500 = $11,500

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Between 1910 and 1930, what was the approximate increase in us homes with electricity
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nearly 68 percent of American homes were electrified. But, if you don't count farms, about 85 percent of Americans had electricity by the end of the 1920s.

And there's the rub. Rural communities, including farms, were far slower in adopting electricity than the rest of the country. Of the roughly 6.3 million American farms in 1922, only about 3% had electricity. It wasn't until 1935 that the U.S. government addressed this huge rural vs urban electric divide with the formation of the Rural Electrification Administration.

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Since ALL the data from the weekdays are lower than the weekends, that means the fewer cars were sold during the week than on weekends.
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b

Step-by-step explanation:

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