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3 years ago

Determine whether theses lines are parallel, perpendicular or neither

Mathematics

1 answer:

belka [17]3 years ago

8 0

Answer:

Lines 2 and 3 are parallel with slope (-2/3) and different y intercepts, and both are perpendicular to line 1 (-1) / (3/2) = (-2/3)

Step-by-step explanation:

Parallel lines have the same slope

Perpendicular lines have negative reciprocal slopes

line 1: 6x - 4y = 2

line 1: 4y = 6x - 2

line 1: y = (3/2)x - 0.5

line 2: y = (-2/3)x - 6

line 3:3y = -2x + 4

line 3: y = (-2/3)x + 4/3

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3 years ago

Write a polynomial of lowest degree with zeros 1/3 (multiplicity 2) and -1/4 (multiplicity 1) and with f(0)=-3

Aleksandr-060686 [28]

Answer:

Step-by-step explanation:

y = C(x - 1/3)(x - 1/3)(x + 1/4)

y = C(x² - 2/3x + 1/9)(x + 1/4)

y = C(x³ - 2/3x² + 1/4x² + 1/9x - 1/6x + 1/36)

y = C(x³ - 5/12x² - 1/18x + 1/36)

now to account for the y intercept (x = 0)

-3 = C((0)³ - 5/12(0)² - 1/18(0) + 1/36)

-3 = C(1/36)

C = -108

distribute the C term across all variables.

y = -108x³ + 45x² + 6x - 3

we can verify our solution using a plotting calculator.

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3 years ago

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4 years ago

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Solve the differential equation dy/dx=x/49y. Find an implicit solution and put your answer in the following form: = constant. he

anygoal [31]

Answer:

The general solution of the differential equation is $\frac{49y^{2} }{2}-\frac{x^{2} }{2} = c_{3}$

The equation of the solution through the point (x,y)=(7,1) is $y=\frac{x}{7}$

The equation of the solution through the point (x,y)=(0,-3) is $\:y=-\frac{\sqrt{441+x^2}}{7}$

Step-by-step explanation:

This differential equation $\frac{dy}{dx}=\frac{x}{49y}$ is a separable first-order differential equation.

We know this because a first order differential equation (ODE) $y' =f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of x and y

$f(x,y)=p(x)\cdot h(y)$ where p(x) and h(y) are continuous functions. And this ODE is equal to $\frac{dy}{dx}=x\cdot \frac{1}{49y}$

To solve this differential equation we rewrite in this form:

$49y\cdot dy=x \cdot dx$

And next we integrate both sides

$\int\limits {49y} \, dy=\int\limits {x} \, dx$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1}\\\int\limits {49y} \, dy=\frac{49y^{2} }{2} + c_{1}$

$\int\limits {x} \, dx=\frac{x^{2} }{2} +c_{2}$

$\int\limits {49y} \, dy=\int\limits {x} \, dx\\\frac{49y^{2} }{2} + c_{1} =\frac{x^{2} }{2} +c_{2}$

We can subtract constants $c_{3}=c_{2}-c_{1}$

$\frac{49y^{2} }{2} =\frac{x^{2} }{2} +c_{3}$

An explicit solution is any solution that is given in the form $y=y(t)$ . That means that the only place that y actually shows up is once on the left side and only raised to the first power.

An implicit solution is any solution of the form $f(x,y)=g(x,y)$ which means that y and x are mixed (y is not expressed in terms of x only).

The general solution of this differential equation is:

$\frac{49y^{2} }{2}-\frac{x^{2} }{2} = c_{3}$

To find the equation of the solution through the point (x,y)=(7,1)

We find the value of the $c_{3}$ with the help of the point (x,y)=(7,1)

$\frac{49*1^2\:}{2}-\frac{7^2\:}{2}\:=\:c_3\\c_3 = 0$

Plug this into the general solution and then solve to get an explicit solution.

$\frac{49y^2\:}{2}-\frac{x^2\:}{2}\:=\:0$

$\mathrm{Add\:}\frac{x^2}{2}\mathrm{\:to\:both\:sides}\\\frac{49y^2}{2}-\frac{x^2}{2}+\frac{x^2}{2}=0+\frac{x^2}{2}\\Simplify\\\frac{49y^2}{2}=\frac{x^2}{2}\\\mathrm{Multiply\:both\:sides\:by\:}2\\\frac{2\cdot \:49y^2}{2}=\frac{2x^2}{2}\\Simplify\\9y^2=x^2\\\mathrm{Divide\:both\:sides\:by\:}49\\\frac{49y^2}{49}=\frac{x^2}{49}\\Simplify\\y^2=\frac{x^2}{49}\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$y=\frac{x}{7},\:y=-\frac{x}{7}$

We need to check the solutions by applying the initial conditions

With the first solution we get:

$y=\frac{x}{7}=\\1=\frac{7}{7}\\1=1\\$

With the second solution we get:

$\:y=-\frac{x}{7}\\1=-\frac{7}{7}\\1\neq -1$

Therefore the equation of the solution through the point (x,y)=(7,1) is $y=\frac{x}{7}$

To find the equation of the solution through the point (x,y)=(0,-3)

We find the value of the $c_{3}$ with the help of the point (x,y)=(0,-3)

$\frac{49*-3^2\:}{2}-\frac{0^2\:}{2}\:=\:c_3\\c_3 = \frac{441}{2}$

Plug this into the general solution and then solve to get an explicit solution.

$\frac{49y^2\:}{2}-\frac{x^2\:}{2}\:=\:\frac{441}{2}$

$y^2=\frac{441+x^2}{49}\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\y=\frac{\sqrt{441+x^2}}{7},\:y=-\frac{\sqrt{441+x^2}}{7}$

We need to check the solutions by applying the initial conditions

With the first solution we get:

$y=\frac{\sqrt{441+x^2}}{7}\\-3=\frac{\sqrt{441+0^2}}{7}\\-3\neq 3$

With the second solution we get:

$y=-\frac{\sqrt{441+x^2}}{7}\\-3=-\frac{\sqrt{441+0^2}}{7}\\-3=-3$

Therefore the equation of the solution through the point (x,y)=(0,-3) is $\:y=-\frac{\sqrt{441+x^2}}{7}$

4 0

3 years ago