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3 years ago

7

Please help, its due tonight

1 answer:

NikAS [45]3 years ago

3 0

Answer:

AS14 = 1107.4

Step-by-step explanation:

Hopefully this helps

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Solve the system of linear equations by elimination.<br> 5 - 2y = -8<br> 4x – 4y= 8

Alenkasestr [34]

Answer:

x = 17/2

y = 13/2

Step-by-step explanation:

Hope this helps!

6 0

2 years ago

A student used the graph below to identify the minimum/maximum. The student said the graph had a maximum of 1. Explain the stude

11Alexandr11 [23.1K]

Answer: Answer is in the step.

Step-by-step explanation:

The student made a mistake by identifying the maximum point by the x coordinate value of the vertex. Minimum or maximum points are determine using the y coordinate value.The student can determine the difference between a maximum and minimum by identifying the y coordinate of the vertex.

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3 years ago

In Exercise, simplify the expression. <br> eln(2x+1)

Galina-37 [17]

Answer:

The simplified expression is 2x + 1

Step-by-step explanation:

We are given the following in the question:

$e^{ln(2x+1)}$

We have to simplify this expression.

The logarithmic function is the inverse of the exponential function. Thus, we can write

$e^{ln(2x+1)}\\=2x + 1$

Thus, the simplified expression is 2x + 1

8 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

You are walking from home to a grocery store you stop for a rest after tofit miles the grocery store is actually 3/4 miles from

svp [43]

Answer:

no I don't know I am sorry for this question

8 0

3 years ago