<img src="https://tex.z-dn.net/?f=9%28x%2B1%29%20-4x%20%3D%205-%28x-3%29" id="TexFormula1" title="9(x+1) -4x = 5-(x-3)" alt="9(x

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morpeh [17]

3 years ago

+1) -4x = 5-(x-3)" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

nekit [7.7K]3 years ago

4 0

Answer:

$x=\frac{-1}{6}$

Step-by-step explanation:

$9(x+1) -4x = 5-(x-3)$

$9(x+1) -4x = 5-(x-3) \\$

$9x+9\cdot1-4x=5-\left(x-3\right)$

$9x+9-4x=5-\left(x-3\right)$

$\left(9x-4x\right)+9=5-\left(x-3\right)$

$5x+9=5-\left(x-3\right)$

$5x+9=5-x+3$

$5x+9=-x+\left(5+3\right)$

$5x+9=-x+8$

$5x+9+x=-x+8+x$

$5x+x+9=-x+8+x$

$6x+9=-x+8+x$

$6x+9=-x+x+8$

$6x+9=8$

$6x+9-9=8-9$

$6x=8-9$

$6x=-1$

$\frac{6x}{6}=\frac{-1}{6}$

$x=\frac{-1}{6}$

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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>

7 0

3 years ago

Can someone help me fill this out

Gwar [14]

Answer:

Step-by-step explanation:

4 0

3 years ago

WILL GET BRAILIEST FOR CORRECT ANSWER

Alex Ar [27]

Answer:

x = ± $\sqrt{20}$

Step-by-step explanation:

Given

x² = 20 ( take the square root of both sides )

x = ± $\sqrt{20}$

which can be simplified to

x = ± 2 $\sqrt{5}$

[ ± means plus or minus ]

5 0

3 years ago

It’s in the picture if u can help I will be thankful

Delicious77 [7]

Answer:

The length of LM,hypotenuse is 55.866ft

Step-by-step explanation:

Given that LN=39ft and MN=40ft so you can solve it using Pythogoras' Theorem:

side²+side² = hypo²

LN² + MN² = LM²

hypo² = 39²+40²

= 3121

LM = √3121

= 55.866ft (3d.p)

6 0

4 years ago

What is 15/8 as a mixed number

9966 [12]

It would be equal to 1 7/8

6 0

4 years ago

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