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3 years ago

Can u please help me with this question?

Mathematics

1 answer:

Masteriza [31]3 years ago

3 0

Answer:

B) $y=\frac{-2}{3}x-\frac{-13}{3}$

Step-by-step explanation:

2x +3y = 5

3y = -2x + 5

Divide both sides by 3

$\frac{3y}{3}=\frac{-2x}{3}+\frac{5}{3}\\\\y =\frac{-2}{3}x+\frac{5}{3}$

Parallel lines have same slope.

m = (-2/3) ; (-2 , -3)

y - y1 = m(x -x1)

$y-[-3]=\frac{-2}{3}(x-[-2])\\\\y + 3 = \frac{-2}{3}(x +2)\\\\y + 3 =\frac{-2}{3}x + 2*\frac{-2}{3}\\\\y+3=\frac{-2}{3}x-\frac{4}{3}\\\\y=\frac{-2}{3}x-\frac{4}{3}-3\\\\y=\frac{-2}{3}x-\frac{4}{3}-\frac{3*3}{1*3}\\\\y=\frac{-2}{3}x-\frac{4}{3}-\frac{9}{3}\\\\y=\frac{-2}{3}x-\frac{-13}{3}$

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Help me show work please

masha68 [24]

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

(x₁, y₁) - given point

$\bullet\ (2,\ 7),\ m=-4\\\\y-7=-4(x-2)\\\\\bullet(-6,\ -2),\ m=3\\\\y-(-2)=3(x-(-6))\to y+2=3(x+6)\\\\\bullet(12,\ 5),\ m=-3\\\\y-5=-3(x-12)\\\\\bullet(7,\ -6),\ m=\dfrac{1}{2}\\\\y-(-6)=\dfrac{1}{2}(x-7)\to y+6=\dfrac{1}{2}(x-7)\\\\\bullet(4,\ -5),\ m=6\\\\y-(-5)=6(x-4)\to y=y+5=6(x-4)\\\\\bullet(-8,\ 2),\ m=-\dfrac{3}{4}\\\\y-2=-\dfrac{3}{4}(x-(-8))\to y-2=-\dfrac{3}{4}(x+8)$

8 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

Find the volume of the triangular prism described below.

Luda [366]

The volume of the prism 121.6 cube in.

Step-by-step explanation:

Given,

The base is right angle triangle whose two sides are equal.

So,

Base (b) = 8 in

Another side of right angle triangle (l) = 8 in

Height (h) of the prism = 3.8 in

To find the volume of the prism.

Formula:

V = $\frac{1}{2}$ bhl

Now,

V = $\frac{1}{2}$ ×8×8×3.8 cube in

= 121.6 cube in.

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3 years ago

HURRY!!!!!!! Which symbol replaces ? to make the statement true? 16÷4+12 ? 4+24÷2 A.< B.= C.>

satela [25.4K]

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JulijaS [17]

Answer:

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Step-by-step explanation:

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