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3 years ago

Find the equation of the line that is perpendicular to 3x-4y=5

Mathematics

1 answer:

Kaylis [27]3 years ago

7 0

Answer:

4x + 3y = 0

Step-by-step explanation:

A line perpendicular to one written in standard form can be found by swapping the x- and y-coefficients and negating one of them. Usually, you want the leading (x) coefficient to be positive, so here we can negate that one. This process give ...

4x + 3y = 0

The constant on the right can be chosen to make the line go through a desired point. Here, no point is specified, so we have set it to zero.

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Select the two values of x that are roots of this equation

Jet001 [13]

<h2>Hello!</h2>

The answers are:

B. $\frac{-3-\sqrt{29}}{2}$

C. $\frac{-3+\sqrt{29}}{2}$

We can use the quadratic equation to find the two values of x that are roots of the given equation. We must remember that most of the quadratic equations have two roots, however, we could find quadratic equations with just one root or even with no roots, at least in the real numbers.

Quadratic equation:

$\frac{-b+-\sqrt{b^{2}-4ac} }{2a}$

So,

From the given equation we have:

$a=1\\b=3\\c=-5$

Substituting it into the quadratic equation to find the roots, we have:

$\frac{-b+-\sqrt{b^{2}-4ac} }{2a}=\frac{-3+-\sqrt{3^{2}-4*1*-5} }{2*1}\\\\\frac{-3+-\sqrt{3^{2}-4*1*-5} }{2*1}=\frac{-3+-\sqrt{9+20} }{2}\\\\\frac{-3+-\sqrt{29}}{2}$

So,

$x_{1}=\frac{-3-\sqrt{29}}{2}\\\\x_{2}=\frac{-3+\sqrt{29}}{2}$

Hence, the correct options are B and C.

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4 years ago

Ya yeet answer dis plzz

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He would keep 3/8 for himself. Hope it helps! :)

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Answer:

y=x+ -5

i think

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Find the equation of the line that is perpendicular to 3x-4y=5​

Find the equation of the line that is perpendicular to 3x-4y=5