All categories

2 years ago

What is the circumference of a circle with a radius of 20 units

Mathematics

1 answer:

Kaylis [27]2 years ago

5 0

Answer:

125.66

Step-by-step explanation:

C = 2πr = 2·π·20 = 125.66371

You might be interested in

What is 19/20 as a terminating decimal

natka813 [3]

$\frac{19}{20} = 0.95$

3 0

3 years ago

Read 2 more answers

Find x, to the nearest tenth of a centimetre

kondor19780726 [428]

Answer:

In the picture above

Step-by-step explanation:

I think it's a clear explanation.

5 0

3 years ago

Simplify the numerical expression.

laiz [17]

Answer:

Step-by-step explanation:

13-2x 5

13-5= 8

2x 8

5 0

3 years ago

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: $F=0.055 L\s$

4 0

3 years ago

G a tanker has developed a leak in pristine bay and has spilled a considerable amount of oil. the good news is that oil-eating b

alexandr402 [8]

$V = \pi r^2 h \\ \\ \frac{dV}{dt} = 2\pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt} \\ \\ -5 =2 \pi (500)(.01)\frac{dr}{dt} + \pi (500^2)(-.001) \\ \\ \frac{dr}{dt} = \frac{250 \pi -5}{10 \pi} = \frac{50 \pi - 1}{2 \pi}$

7 0

3 years ago