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2 years ago

Which equation has a slope of -2 and passes through the point (1,-6)?

Mathematics

1 answer:

Naily [24]2 years ago

7 0

Answer:

y=-2x-4

Step-by-step explanation:

Given that,

Slope, m = -2

It passes through the point (1,-6).

We need to find the equation of line that passes throgh the given point. It can be calculated as:

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

Put y₁ = -6, x₁ = 1 and m = -2

$y-(-6)=(-2)(x-1)\\\\y+6=-2x+2\\\\y+2x+6-2=0\\\\y=-2x-4$

Hence, the equation of line is y=-2x-4.

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6x+3(4x-2)=5(2x-8) the answer is -17/4 but can someone explain how to get that answer? plz help lol

Sergeu [11.5K]

Answer:

Step-by-step explanation:

$6x + 3(4x - 2) = 5(2x - 8)$

Expand the terms in the bracket and simplify

That's

$6x + 12x - 6 = 10x - 40 \\ 18x - 6 = 10x - 40$

Using the subtraction property subtract 10x from both sides of the equation

$18x - 10x - 6 = 10x - 10x - 40 \\ 8x - 6 = - 40$

<u>Add 6 to both sides of the equation</u>

$8x + 6 - 6 = - 40 + 6 \\ 8x = - 34$

<u>Divide both sides by 8</u>

$\frac{8x}{8} = - \frac{34}{8}$

<u>Reduce the fraction with 2</u>

We have the final answer as

$x = - \frac{17}{4}$

Hope this helps you

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2 years ago

HELP PLZ WILL GIVE YOU BRAINIEST

slamgirl [31]

Answer:

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Step-by-step explanation:

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2 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

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Answer:

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Step-by-step explanation:

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