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3 years ago

14

WHAT ARE ALL PRIME NUMBERS X WHICH ARE

1" title="5\ \textless \ x\leq 19" alt="5\ \textless \ x\leq 19" align="absmiddle" class="latex-formula">
WHOEVER ANSWERS FIRST AND CORRECTLY IS BRAINLIEST

2 answers:

Anit [1.1K]3 years ago

7 0

The prime numbers that X could be are 7, 11, 13, 17, and 19.

Hope this helps; have a great day!

ella [17]3 years ago

5 0

Answer:

7, 11, 13, 17, 19

All of these numbers only have 1 and themselves as factors, and every other number can be written as a product of two or more primes, making the remaining number composite:

$6=2\cdot3\\8=2^3\\9=3^2\\10=2\cdot5\\12=2^2\cdot3\\14=2\cdot7\\15=3\cdot5\\16=2^4\\18=2\cdot3^2$

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Let S represent the number of randomly selected adults in a community surveyed to find someone with a certain genetic trait.

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3 years ago

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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

5=1 2/3x how do I solve for x?

LiRa [457]

You'll first Cross multiply: 5× 3x

that'll give you 15x.

15x=12(12 is a numerical constant).

Divide both sides by the the coefficient of X .That is 15x/15= 12/15

Therefore the answer is 12/15= x

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I believe the answer is B can anyone confirm please?

Marizza181 [45]

The answer is D, not B

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4 years ago