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3 years ago

Help!

Mathematics

1 answer:

dsp733 years ago

6 0

Your answer in scientific notation is 2.355 * 10^14

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Someone pleasee help me with this freaking IXL.

Norma-Jean [14]

The answer is 1761 1/2 (1761.5 in decimal form)

6 0

2 years ago

The vertices of polygon ABCD are at A(1, 1), B(2, 3), C(3, 2), and D(2, 1). ABCD is reflected across the x-axis and translated 2

Lana71 [14]

-- Reflecting across the x-axis makes all the x-coordinates the negative of
what they were before the reflection. The y-coordinates don't change.

-- Translating 2 units up makes all the y-coordinates 2 greater than
they were before the translation. The x-coordinates don't change.

You didn't give us a list of new coordinates, so there's nothing
to match with.

8 0

3 years ago

Plot (−2 3/4, −4 1/2) on the coordinate plane.

Makovka662 [10]

Because -2 3/4 is on the x axis, it is the diagonal one, -4 1/2 is on the y axis it’s vertical.

The first number is x, the second is y.

Brainliest answer please?

7 0

2 years ago

Turn 0.68 into a fraction in simplest form

sattari [20]

Since .68 is 68/100, you can simplify to 17/25.

7 0

2 years ago

Read 2 more answers

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.

4 0

2 years ago