Please help right noww pleasee i will put brainliest

Mathematics

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

Answer:

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Piecewise function help

Vlad1618 [11]

Nothing is mentioned...how am I supposed to help you?

3 0

3 years ago

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago

Julia can finish a 20-mile bike ride in 1.2 hours. Katie can finish the same bike ride in 1.6 hours. How much faster does Julia

Nimfa-mama [501]

We know that, $speed = \frac{distance}{time}$

Julia can finish a 20-mile bike ride in 1.2 hours.

The distance Julia travels is 20 miles and the time she takes is 1.2 hours.

So, Julia's speed = $\frac{20}{1.2}$ = 16.67 mph

Katie can finish the same bike ride in 1.6 hours.

The distance Katie travels is 20 miles and the time she takes is 1.6 hours.

So, Katie's speed = $\frac{20}{1.6}$ = 12.5 mph

Now, to find how much faster Julia rides than Katie we subtract Katie's speed from Julia's speed.

So, 16.67 mph - 12.5 mph = 4.17 mph = 4.2 mph (approximately)

Thus, Julia rides 4.2 mph faster than Katie.

5 0

3 years ago

Read 2 more answers

Solve for x & y. 30x−42y=−6 and 5x−7y=−1

vladimir1956 [14]

Um.. Both of those equations are the exact same. Divide both sides of the first equation by 6. You get:

$\frac{30x - 42y}{6} = \frac{-6}{6} \\ 5x - 7y = -1$

That is the exact same as the second equation. This system has an infinite number of solutions. 5x - 7y = -1 is a line, so basically every point on that line is a solution to the system.

For example, $x = 0$ and $y = \frac{1}{7}$ would work, but so would $x = 1$ and $y = \frac{6}{7}$