All categories

3 years ago

3vfef e2fc1d,My ex the rcrcr

Advanced Placement (AP)

1 answer:

Mariulka [41]3 years ago

3 0

Answer: ok

Explanation:

You might be interested in

What is the tundra’s permanently frozen layer of ground called?

jek_recluse [69]

Answer:

permafrost

Explanation:

The arctic soil is so cold that the ground beneath the tundra surface remains frozen all year. This permanently frozen ground is called permafrost. Each summer, when the sun warms the tundra surface, the top few inches of soil thaw.

7 0

3 years ago

Read 2 more answers

Which of the following statements is correct?

padilas [110]

Earth's orbit is not a perfect circle. It is elliptical, or slightly oval-shaped. This means there is one point in the orbit where Earth is closest to the Sun, and another where Earth is farthest from the Sun. The answer is the Earth’s orbit is elliptical.

3 0

3 years ago

Read 2 more answers

How do I get free points ?

SpyIntel [72]

Answer:

you answer questions

Explanation:

4 0

3 years ago

Read 2 more answers

Carolyn was enjoying an evening walk around her neighborhood pond. As she walked over the wooden bridge, she heard a loud splash

marissa [1.9K]

Answer:

parasympathetic nervous system

Explanation:

parasympathetic nervous system is the answer because she is getting nervous and scared about what is going to happen. I hope this helps. HAVE A GREAT DAY!

6 0

3 years ago

Read 2 more answers

Please solve problem below

Tanya [424]

(a) This is a Bernoulli equation:

$\dfrac{dy}{dx} + \dfrac{x}{2(x^2+1)} y = x^3y^5$

$\implies y^{-5} \dfrac{dy}{dx} + \dfrac{x}{2(x^2+1)} y^{-4} = x^3$

Substitute $z=y^{-4}$ and $\frac{dz}{dx} = -4y^{-5}\frac{dy}{dx}$ to transform the ODE to

$-\dfrac14 \dfrac{dz}{dx} + \dfrac{x}{2(x^2+1)} z = x^3$

$\implies \dfrac{dz}{dx} - \dfrac{2x}{x^2+1} z = -4x^3$

which is now linear in $z$ . Using the integrating factor method, the I.F. is

$\mu = \displaystyle \exp\left(\int -\frac{x}{x^2+1} \, dx\right) = \exp\left(-\frac12 \ln(1+x^2)\right) = \dfrac1{\sqrt{1+x^2}}$

Distribute $\mu$ on both sides to get a derivative of a product on the left side.

$\dfrac1{\sqrt{1+x^2}} \dfrac{dz}{dx} - \dfrac{2x}{(x^2+1)^{3/2}} z = -\dfrac{4x^3}{\sqrt{1+x^2}}$

$\implies \dfrac{d}{dx} \left(\dfrac1{\sqrt{1+x^2}} z\right) = -\dfrac{4x^3}{\sqrt{1+x^2}}$

Integrate both sides (the integral on the right can be done by parts) to get

$\displaystyle \frac1{\sqrt{1+x^2}} z = -4 \int \frac{x^3}{\sqrt{1+x^2}} \, dx = -\frac43 (x^2-2) \sqrt{1+x^2} + C$

Solve for $z$ .

$\displaystyle \frac1{\sqrt{1+x^2}} z = -\frac43 (x^2-2) \sqrt{1+x^2} + C$

$\implies z = -\dfrac43 (x^2-2) (1+x^2) + C \sqrt{1+x^2}$

Solve for $y$ .

$\dfrac1{y^4} = -\dfrac43 (x^2-2) (1+x^2) + C \sqrt{1+x^2}$

$\implies \boxed{y^4 = -\dfrac3{4(x^2-2) (1+x^2) + C \sqrt{1+x^2}}}$

You could go on to solve explicitly for $y$ if you like.

(b) This is also a Bernoulli equation:

$x^2y' + 2xy - y^3 = 0$

$\implies x^2 y^{-3} y' + 2xy^{-2} = 1$

Substitute $z=y^{-2}$ and $z' = -2y^{-3}y'$ .

$-\dfrac{x^2}2 z' + 2xz = 1$

$\implies z' - \dfrac4x z = -\dfrac2{x^2}$

Now repeat the method from (a) to solve for $y$ .

$\mu = \exp\left(-\displaystyle \int \frac4x \, dx\right) = \dfrac1{x^4}$

$\implies \dfrac1{x^4} z' - \dfrac4{x^5} z = -\dfrac2{x^6}$

$\implies \left(\dfrac1{x^4} z\right)' = -\dfrac2{x^6}$

$\displaystyle \implies \dfrac1{x^4} z = -2 \int \frac{dx}{x^6}$

$\implies \dfrac1{x^4} z = \dfrac2{5x^5} + C$

$\implies z = \dfrac2{5x} + Cx^4$

$\implies \dfrac1{y^2} = \dfrac2{5x} + Cx^4$

$\implies \boxed{y^2 = \dfrac{5x}{2 + Cx^5}}$

3 0

2 years ago