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qwelly [4]
3 years ago
11

Helpppppp meeeeeeeeeeeee

Mathematics
1 answer:
denis23 [38]3 years ago
8 0

Answer:

81000 people went to Bonnaroo.

192000 people went to Lollapalooza.

Step-by-step explanation:

Given:

Total number of people attending festivals, N=273000

Number of people attending Lollapalooza is 30,000 more than twice the people attending Bonnaroo.

Unknown:

Number of people attending Lollapalooza and Bonnaroo.

Let the number of people attending Bonnaroo be x.

Therefore, the number of people attending Lollapalooza is equal to 30000+2x

Evaluate:

Sum of people attending Bonnaroo and Lollapalooza is equal to the total number of people attending the festivals. So,

x+30000+2x=N\\3x+30000=273000\\3x=273000-30000\\3x=243000\\x=\frac{243000}{3}=81000

Substitute and Solve:

Now, we have, x=81000.

Substitute 81000 for x and solve for the number of people attending both the festivals. This gives,

Number of people attending Bonnaroo = x=81000

Number of people attending Lollapalooza = 30000+2x=30000+2(81000)=30000+162000=192000

Therefore, 81000 people went to Bonnaroo and 192000 people went to Lollapalooza.

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The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
Lemur [1.5K]

Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

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Answer:

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At the old rate, the CO₂ concentration in 2100 will be 516 ppm.

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Time = 2100 - 2010 = 90 years

Increase in CO₂ = 90 yr × (1.9 ppm/1 yr) = 171 ppm

CO₂ in 2010 = 390 + 171 = 561 ppm

At the new rate, the CO₂ concentration in 2100 will be 561 ppm.

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