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Delvig [45]
3 years ago
6

What is cubic inches of aquarium that is 3/4 full

Mathematics
1 answer:
Alex17521 [72]3 years ago
6 0
First, find the volume of the full tank 
V= 16*8.5*10.5
V=1428 cubic inches
Then, just find 3/4 of the volume
V2= 1428 * (3/4)
V2= 1071 cubic inches
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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>

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i don’t know for sure but i just figured it out....

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Simplify 8-(-5) - 4(-7).<br> О-63<br> оо ООО
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If \frac{3+\sqrt{8} }{3-\sqrt{8} } +\frac{3-\sqrt{8} }{3+\sqrt{8} }=a+b\sqrt{2}, the value of a and b is given as a = 34 and b = 0

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables or numbers.

\frac{3+\sqrt{8} }{3-\sqrt{8} } +\frac{3-\sqrt{8} }{3+\sqrt{8} }\\ \\\frac{9+6\sqrt{8}+8+9-6\sqrt{8}+8  }{(3-\sqrt{8} )(3+\sqrt{8} )} \\\\\frac{34}{9-8} =34+0\sqrt{2}

If \frac{3+\sqrt{8} }{3-\sqrt{8} } +\frac{3-\sqrt{8} }{3+\sqrt{8} }=a+b\sqrt{2}, the value of a and b is given as a = 34 and b = 0

Find out more on equation at: brainly.com/question/2972832

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Rashid [163]
Hola! yo te ayudaria, pero no se en que. No tienes el problema para que yo lo resuelva. Lo siento si mi espanol no es bueno, estoy aprendiendo como escribir lo. Sorry-
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