All categories

3 years ago

Solving Systems Using Elimination.

Mathematics

1 answer:

LUCKY_DIMON [66]3 years ago

3 0

Answer:

no solution

i think idk

You might be interested in

6- 6x^2- 3x = 0 solve

VARVARA [1.3K]

Answer:

The solutions are $x=\frac{3)+\sqrt{153} }{-12}$ , $x=\frac{3)-\sqrt{153} }{-12}$

If you need to have them in a decimal form rounded to 3 decimals, the answers would be $x=-1.281$ , $x=0.781$

Step-by-step explanation:

We are given the equation $6-6x^2-3x=0$

The first thing that can be done is to rearrange the terms so that they are in descending order of power. This makes it easier to work with

This gives us $-6x^2-3x+6=0$

To find the solutions to this equation, we can use the quadratic formula, which will tell us that the solutions are equal to $\frac{-b+\sqrt{b^2-4ac} }{2a} ,\frac{-b-\sqrt{b^2-4ac} }{2a}$

From our equation, we can gather that $a=-6, b=-3, c=6$

Now that we have our variables, we can plug them into the quadratic formula to get our solutions.

$\frac{-(-3)+\sqrt{(-3)^2-4(-6)(6)} }{2(-6)}\\\\\frac{3)+\sqrt{9+144} }{-12}\\\\\frac{3)+\sqrt{153} }{-12}$

And now for our second equation

$\frac{-(-3)-\sqrt{(-3)^2-4(-6)(6)} }{2(-6)}\\\\\frac{3)-\sqrt{9+144} }{-12}\\\\\frac{3)-\sqrt{153} }{-12}$

3 0

3 years ago

How are finding the volume of a rectangular prism and a triangular prism different?

Norma-Jean [14]

Answer:

Step-by-step explanation:

A triangular prism has its bases in a triangle shape and a rectangular prism has its bases in rectangle shape.

4 0

3 years ago

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv$

which has a value of 0, since each component integral is 0:

$\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0$

4 0

3 years ago

Can someone solve for me pls!!!!!

g100num [7]

Answer:

$\frac{7}{15}$

Step-by-step explanation:

Let 1 unit be 0.4666...

16 units = $0.4666...*16=7.4666...$

15 units = $7.4666...-0.4666...$

15 units = $7$

1 unit = $\frac{7}{15}$

8 0

3 years ago

Linear Inequality Systems Graphically

faltersainse [42]

Answer:

SNITCHY SNATCH

Step-by-step explanation:

I actually don't know how to show how to graph it. Sorry. But I can give a few coordinates. For y < 1/3x + 3: (-1,2 2/3) (0,3) (1,3 1/3). For y > -2/3x - 3: (-1,-2 1/3) (0,-3) (1,-3 2/3).

6 0

3 years ago