Express 1120 product of powers of its prime factors

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)

lawyer [11315]

Answer:

Cardiac output: $F=0.055 L\s$

Step-by-step explanation:

Details: The cardiac output is assessed using the dye dilution technique with 3 mg of dye.

Goal: To determine the cardiac output value.

Solution:

Cardiac output formula: $F=\frac{A}{\int\limits^T_0 {c(t)} \, dt}$ ---1

A = 3 mg

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

Utilize integration by parts

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

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

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

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

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

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

Insert the value into 1

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

Cardiac output: $F=0.055 L\s$

Consequently, Cardiac output: $F=0.055 L\s$

4 0

26 days ago

On a dell river, a boat will pass the Colby drawbridge and then the wave drawbridge opening times.There are 60 minutes in a hour

Svet_ta [11404]

The conversions yield 45/60 and 10/60 since the common denominator is established as 60, thus 3/4 of an hour equals 45 minutes and 1/6 of an hour translates to 10 minutes.

6 0

8 days ago

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PQ⊥PS , m∠QPR=7x−9, m∠RPS=4x+22<br> Find : m∠QPR

zzz [11052]

Given:

It is stated that

PQ ⊥ PS and

∠QPR = 7x-9

∠RPS = 4x+22

To find the measure of ∠QPR.

Formula

According to the problem PR sits between PQ and PS,

Consequently,

∠QPR+∠RPS = 90°

Now,

Inserting the values for ∠RPS and ∠QPR yields

$7x-9+4x+22 = 90$

or, $11x = 90-22+9$

or, $11x = 77$

or, $x = \frac{77}{11}$

or, $x = 7$

Substituting the value of $x = 7$ into ∠QPR gives us

∠QPR = $7(7)-9$

or, ∠QPR = $40^\circ$

So,

The measure of ∠QPR is 40°.

3 0

18 days ago

Sarah and her children went into a movie theater and she bought $35.50 worth of drinks and candies. Each drink costs $5 and each

AnnZ [11017]

Answer:

Sarah purchased 2 drinks and 6 candies.

Step-by-step explanation:

Let

x ----> the quantity of drinks Sarah bought.

y ----> the number of candies acquired by Sarah.

We know that

the total spent on drinks and candies was $35.50

therefore,

$5x+4.25y=35.50$ -----> equation A

She bought 3 times more candies compared to drinks.

thus,

$y=3x$ -----> equation B

To resolve the equations graphically

The solution lies at the intersection of the two graphs

utilizing a graphing tool

The result is the coordinate (2,6)

therefore,

Sarah bought 2 drinks and 6 candies.

4 0

1 month ago

The graph of the function f(x) = (x – 4)(x + 1) is shown below. On a coordinate plane, a parabola opens up. It goes through (neg

tester [10867]

Answer:

The function decreases across all real numbers x where x < 1.5.

Step-by-step explanation:

We have

$f(x)=(x-4)(x+1)$

$f(x)=x^2+x-4x-4$

$f(x)=x^2-3x-4$

This represents an upward-opening vertical parabola

The vertex denotes a minimum point

The vertex is located at (1.5,-6.25)

We know that

The function is decreasing within the interval ----> (-∞, 1.5) x < 1.5

This means----> the function is continuously decreasing for all real x values under 1.5

Conversely, the function is increasing within the interval ----> (1.5, ∞) x> 1.5

Thus, for all real x values greater than 1.5, the function is increasing

Check the attached figure for further clarification

Therefore

The true statement is

The function decreases across all real numbers x where x < 1.5.

5 0

1 month ago

Read 2 more answers