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3 years ago

Please help me solve this equation I'm in a hurry and have to leave soon!!!

2 answers:

5 0

Hey there!

So we know that x = 2, now we have to multiply the numbers 4 and 2 . . .

4 * 2 = 8

Now multiply the numbers 6 and 2 . . .

6 * 2 = 12

Now add them together . . .

12 + 8 = 20 <----

Your answer is 20.

4x + 6x = 20

Hope this helps you.
Have a great day! :)

zhuklara [117]3 years ago

3 0

Hi there!

To solve this problem, we need to plug in the values of x into the equation.

Since x=2....

4x + 6x
= 4*2 + 6*2
= 8 + 12
= 20

So, your answer is 20.

Hope this helps!

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2 years ago

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3 years ago

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The circumference of a circle is 29.83 cm. What are the radius and diameter, respectively? Use 3.14 for pi.

nasty-shy [4]

Answer:

Diameter: 9.5

Radius: 4.75

Step-by-step explanation:

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2 years ago

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.

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3 years ago

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Troyanec [42]

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2 years ago

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