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2 years ago

8

If DEF and MNO are two triangles such that DE=MN and EF=NO, which of the following would be sufficient to prove that triangles a

re congruent

1 answer:

Delvig [45]2 years ago

3 0

Answer:

D. ∠E ≅ ∠N

Step-by-step explanation:

The pair of sides meet at vertex E in ∆DEF and at vertex N in ∆MNO. Since the sides that make up angles E and N are shown congruent, it is sufficient to show ...

∠E ≅ ∠N

Then the SAS congruence postulate can be claimed.

__

<em>Additional comment</em>

The alternative is to show DF ≅ MO. That would allow you to claim SSS congruence. That is not an answer choice.

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Answer: $y(x) = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\$

==========================================================

Explanation:

The given differential equation (DE) is

$y'-\frac{7}{x}y = \frac{y^3}{x^8}\\\\$

Which is the same as

$y'-\frac{7}{x}y = \frac{1}{x^8}y^3\\\\$

This 2nd DE is in the form $y' + P(x)y = Q(x)y^n$

where

$P(x) = -\frac{7}{x}\\\\Q(x) = \frac{1}{x^8}\\\\n = 3$

As the instructions state, we'll use the substitution $u = y^{1-n}$

We specifically use $u = y^{1-n} = y^{1-3} = y^{-2}$

-----------------

After making the substitution, we'll end up with this form

$\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\$

Plugging in the items mentioned, we get:

$\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\\frac{du}{dx} + (1-3)*\frac{-7}{x}u = (1-3)\frac{1}{x^8}\\\\\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\$

We can see that we have a new P(x) and Q(x)

$P(x) = \frac{14}{x}\\\\Q(x) = -\frac{2}{x^8}$

-------------------

To solve the linear DE $\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\$ , we'll need the integrating factor which I'll call m

$m(x) = e^{\int P(x) dx} = e^{\int \frac{14}{x}dx} = e^{14\ln(x)}$

$m(x) = e^{\ln(x^{14})} = x^{14}$

We will multiply both sides of the linear DE by this m(x) integrating factor to help with further integration down the road.

$\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\m(x)*\left(\frac{du}{dx} + \frac{14}{x}u\right) = m(x)*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + x^{14}*\frac{14}{x}u = x^{14}*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + 14x^{13}*u = -2x^6\\\\\left(x^{14}*u\right)' = -2x^6\\\\$

It might help to think of the product rule being done in reverse.

Now we can integrate both sides to solve for u

$\left(x^{14}*u\right)' = -2x^6\\\\\displaystyle \int\left(x^{14}*u\right)'dx = \int -2x^6 dx\\\\\displaystyle x^{14}*u = \frac{-2x^7}{7}+C\\\\\displaystyle u = x^{-14}*\left(\frac{-2x^7}{7}+C\right)\\\\\displaystyle u = x^{-14}*\frac{-2x^7}{7}+Cx^{-14}\\\\\displaystyle u = \frac{-2x^{-7}}{7}+Cx^{-14}\\\\$

$u = \frac{-2}{7x^7} + \frac{C}{x^{14}}\\\\u = \frac{-2}{7x^7}*\frac{x^7}{x^7} + \frac{C}{x^{14}}*\frac{7}{7}\\\\u = \frac{-2x^7}{7x^{14}} + \frac{7C}{7x^{14}}\\\\u = \frac{-2x^7+7C}{7x^{14}}\\\\$

Unfortunately, this isn't the last step. We still need to find y.

Recall that we found $u = y^{-2}\\\\$

So,

$u = \frac{-2x^7+7C}{7x^{14}}\\\\y^{-2} = \frac{-2x^7+7C}{7x^{14}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7C}$

We're told that y(1) = 1. This means plugging x = 1 leads to the output y = 1. So the RHS of the last equation should lead to 1. We'll plug x = 1 into that RHS, set the result equal to 1 and solve for C

$\frac{7*1^{14}}{-2*1^7+7C} = 1\\\\\frac{7}{-2+7C} = 1\\\\7 = -2+7C\\\\7+2 = 7C\\\\7C = 9\\\\C = \frac{9}{7}$

So,

$y^{2} = \frac{7x^{14}}{-2x^7+7C}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7*\frac{9}{7}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+9}\\\\y = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\$

We go with the positive version of the root because y(1) is positive, which must mean y(x) is positive for all x in the domain.

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