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3 years ago

Two lines are perpendicular if the of their А and Lline are also perpendicular.

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

8 0

Answer:

u can mark the other person brainliest now. ur welcome

Step-by-step explanation:

lions [1.4K]3 years ago

3 0

If a line is perpendicular to one of the

two given parallel lines then it is also perpendicular to the other line.

As we know that by the theorem of perpendicular transversal theorem that in the plane, if a transverse is perpendicular to one or two parallel lines, then it is perpendicular to the third line. ... Since, they do not touch or interact with each other they are said to be parallel to each other.

<h3>Thnx for marking me as brainlist</h3>

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Allisa [31]

B) is the correct because the definition of Isosceles indicates to the two sides that are congruent to each other.
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4 years ago

3) In a study of 600 community college students, 200 of them indicated that they are part-time students. What percent of the stu

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Answer: 33.3333333...%

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3 years ago

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Neporo4naja [7]

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Step-by-step explanation:

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3 years ago

Write an inequality that compares-6, 3, and 5

olga_2 [115]

Is that 6 or negative 6? I'll do both situations, as I can't tell (sorry!)

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3 years ago

X ^ (2) y '' - 7xy '+ 16y = 0, y1 = x ^ 4

AfilCa [17]

Given a solution $y_1(x)=x^4$ , we can attempt to find another via reduction of order of the form $y_2(x)=x^4v(x)$ . This has derivatives

${y_2}'=4x^3v+x^4v'$
${y_2}''=12x^2v+8x^3v'+x^4v''$

Substituting into the ODE yields

$x^2(x^4v''+8x^3v'+12x^2v)-7x(x^4v'+4x^3v)+16x^4v=0$
$x^6v''+(8x^5-7x^5)v'+(12x^4-28x^4+16x^4)v=0$
$x^6v''+x^5v'=0$

Now letting $u(x)=v'(x)$ , so that $u'(x)=v''(x)$ , you end up with the ODE linear in $u$

$x^6u'+x^5u=0$

Assuming $x\neq0$ (which is reasonable, since $x=0$ is a singular point), you can divide through by $x^5$ and end up with

$xu'+u=(xu)'=0$

and integrating both sides with respect to $x$ gives

$xu=C_1\implies u=\dfrac{C_1}x$

Back-substitute to solve for $v$ :

$v'=\dfrac{C_1}x\implies v=C_1\ln|x|+C_2$

and again to solve for $y$ :

$y=x^4v\implies \dfrac y{x^4}=C_1\ln|x|+C_2$
$\implies y=C_1\underbrace{x^4\ln|x|}_{y_2}+C_2\underbrace{x^4}_{y_1}$

Alternatively, you can solve this ODE from scratch by employing the Euler substitution (which works because this equation is of the Cauchy-Euler type), $t=\ln x$ . You'll arrive at the same solution, but it doesn't hurt to know there's more than one way to solve this.

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4 years ago