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3 years ago

If x= -2 and y=8 than how do you solve this problem in its simplest form. 4x=y/xy-4y=

Mathematics

2 answers:

Reika [66]3 years ago

6 0

4(-2)=8/-2(8)-4(8)

-8=-4(8)-32

-8=32-32

-8=0

son4ous [18]3 years ago

3 0

Hi there!

So we know that:

x = -2

y = 8

And we have this equation to solve : 4x = $\frac{y}{xy - 4y}$

To solve the equation, you need to replace the letters "x" and "y" in the equation with their value :

4 x -2 = $\frac{8}{(-2x8) - (4x8)}$

Now all you need to do is calculate everything:

4 × -2 = -8

-2 × 8 = -16

4 × 8 = 32

-8 = $\frac{8}{-16 - 32}$

-16 + -32 = -48

-8 = $\frac{8}{-48}$

$\frac{8}{-48} \left \{ {{divided by 8 = 1} \atop {divided by 8 = -6}} \right. = \frac{1}{-6}$

-8 = - $\frac{1}{6}$

I'm not sure if this would be the simplest form you are looking for..

If there's anything, just let me know! :)

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2 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

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oksian1 [2.3K]

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Step-by-step explanation:

Given, A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.05.

Let $\overline{X}$ represents the sample mean GPA for each student.

Then, the probability that the random sample of 100 male students has a mean GPA greater than 3.42:

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hence, the required probability is 0.0548.

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3 years ago