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3 years ago

Find distance and midpoint between two points,

Mathematics

1 answer:

IrinaVladis [17]3 years ago

8 0

Answer:

P(Px, Py) and Q(Qx, Qy)

Midpoint

M((Px + Qx)/2, (Py + Qy)/2)

Distance

d = √((Px - Qx)^2 + (Py - Qy)^2)

For example

(A)

Midpoint

M((-5 + 3)/2, (2 + (-4))/2) = M(-1, -1)

Distance

d = √((-5 - 3)^2 + (2 - (-4))^2) = √(64 + 36) = 10

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2 years ago

Worker leafcutter ants can measure 0.5 inches. Name a decimal that is equivalent to 0.5. Remember not use add spaces to your ans

Sergeeva-Olga [200]

Answer:

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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: $F=0.055 L\s$

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2 years ago

If the height of the cylinder is 3 inches, what is the volume of the cylinder? Write your answer in terms of p.

azamat

Formula for Volume of Cylinder you need the radius

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3 years ago

Read 2 more answers