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3 years ago

UOW and UOV as well would be nice but not required.

Mathematics

1 answer:

Allushta [10]3 years ago

5 0

Answer:

No-------Uow and Uov are------Tpu and Uov arent

Step-by-step explanation:

adjacent=having a common vertex and a common side.

they don't have the vertex in common

Hope this helps!

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What is the value of 36X negative Y over to when X equals three and Y equals -6

Rus_ich [418]

Note:

Your question is a little unclear. But, from my understanding, I assume you may be asking the evaluation of the expression 36x - y when x = 3 and y = -6.

so I will solve based on my assumption which anyways would clear your concept.

Answer:

The value of 36x - y when x = 3 and y = -6 will be: 114

Step-by-step explanation:

Given

Assuming the expression

36x - y

To Determine

Evaluate 36x - y when x = 3 and y = -6

Given the expression

$36x - y$

substitute x = 3 and y = -6 to evaluate the expression

$36x-y= 36(3) - (-6)$

$= 108+6$

$= 114$

Therefore, the value of 36x - y when x = 3 and y = -6 will be: 114

5 0

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.

6 0

3 years ago

Please help asap please

zhannawk [14.2K]

Answer:

take away 8 from 14 then see if the answer works for x by doing 28 - that answer

Step-by-step explanation:

6 0

3 years ago

Over which interval(s) is g(x) = x2 + x - 12 positive?

VikaD [51]

Answer:

B. x < -4 and x > 3

Step-by-step explanation:

Factor and set = to 0

$x^{2} + x - 12\\( x + 4)(x - 3)$ = 0

x = - 4 or x = 3 I call these critical values

The two numbers would divide a number line into 3 intervals. Pick a value in one of the intervals and put it in the original expression. If it makes the function positive, then all the values in that interval make the function positive. If the value you picked makes the function negative, then the values in the other intervals will make the function negative. Let's pick the value of 0 and substitute it into the function

We get $0^{2}$ + 0 - 12 = -12 which is not positive. Therefore, all the values between -4 and 3 will make the function negative. So, the values less than -4 or greater than 3 will make the function positive. Therefore, B is the correct answer.

Another way to do this problem is to graph the function and see where the graph is above the x-axis. But, sometimes it is not easy to graph the function.

4 0

3 years ago

If point M lies at (3,5) and it was reflected over the x- axis, what would be the points location?

Dmitry_Shevchenko [17]

Answer:

(3, -5)

Step-by-step explanation:

To reflect a point over the x axis you change the value to a negative.

6 0

3 years ago