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5 days ago

Given f(x) = log2x, which of the functions below represents g(x) resulting from reflecting the graph of f(x) in the x-axis and s

hifting left by 2 units?

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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)

lawyer [12517]

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

2 months ago

In the spinner below, the large wedges are twice the size of the smaller ones. What is true about the probability of landing on

tester [12383]

Response:

Detailed explanation:

I would choose option c as the spinner has a chance to land on any area at any moment. I wish I could provide a definitive answer.

6 0

2 months ago

Read 2 more answers

An oblique prism has trapezoidal bases and a vertical height of 10 units. An oblique trapezoidal prism is shown. The trapezoid h

Svet_ta [12734]

Response:

Volume of the trapezoidal prism = 15x^2 cubic units

Detailed explanation:

First, let’s calculate the area of the trapezoidal bases.

The lengths of the parallel sides are x and 2x, averaging to 1.5x.

The height stands at x

Consequently, area of the trapezoidal base comes out to be 1.5x * x = 1.5x^2

The volume of this prism is computed as area of the base multiplied by height

(the length is not a factor, but height certainly is)

Thus, 1.5x^2 * 10 yields 15x^2

5 0

2 months ago