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Fynjy0 [20]
3 years ago
10

Question question question question helppppppppp Thanks!

Mathematics
1 answer:
Vadim26 [7]3 years ago
5 0

Answer:

\frac{5}{12}   \:  \:  \:  \frac{5}{8}  \:  \:  \:  \frac{5}{7}  \:  \:  \:  \frac{5}{6}

Step-by-step explanation:

The trick to this is looking at the denominator. If the numerators are the same for all fractions you are comparing then things are going to be a bit easier for comparing them. The one that has the biggest denominator is small and the one with the smaller denominator is a bigger fraction.

hope this makes sense

:)

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Can you guys solve this please step by step thank you so much x + 18 = 2x + 9
Bezzdna [24]

Answer:

x=9

Step-by-step explanation:

x+18=2x+9

-x       -x

18=x+9

-9     -9

9=x

x=9

8 0
3 years ago
Read 2 more answers
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
Write the equation of the line for a line that passes through (4, 4) and (1, -2).
pshichka [43]

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6 0
2 years ago
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Pls help with this question
Brrunno [24]
I wish i could help but i dont know the answer
7 0
3 years ago
10.3 divided by 1.4 =what
bekas [8.4K]

Answer:

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Step-by-step explanation:

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