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AnnZ [28]
2 years ago
5

Find x step by step (10 points)

Mathematics
2 answers:
zhannawk [14.2K]2 years ago
8 0
<h2><u>Answer:</u></h2>

\boxed{\left \{ {{\sqrt{(n-m)(m+n)}} \atop {-\sqrt{(n-m)(m+n)}}} \right.}

<h2><u>Solution Steps:</u></h2><h3><em>- Steps using the quadratic formula - </em></h3>

______________________________

<h3>1.)  <u>Subtract n² from both sides:</u></h3>
  • n^2-n^2=0
  • m^2-n^2=m^2-n^2

<h3>2.) <u>Rewrite:</u></h3>

<em>This equation is in standard form: ax² + bx² + c = 0. Substitute 1 for a, 0 for b, and (m-n) (m+n) for c in the quadratic formula, </em>\frac{-b\frac{+}{}\sqrt{b^2-4ac}  }{2a}:

  • x^2+m^2-n^2=0

<em>    </em>2.a)  <u>Turns into:</u>

  • x=\frac{0\frac{+}{}\sqrt{0^2-4(m-n)(m+n)}}{2}

<h3>3.) <u>Square 0:</u></h3>
  • 0^2=0 <em>(Also means it Cancels out.)</em>

<h3>4.)  <u>Take the square root of −4(m−n)(m+n):</u></h3>

<em>This just means combine them by squaring. </em>

  • -4^2=2<em> (Half it, don't square it.)</em>
  • (m-n)(m+n)^2=(n^2-m^2)

<em>    </em>4a.)<em>  </em><u>Our equation should look like this now:</u>

  • x=\frac{0\frac{+}{}2\sqrt{(n^2-m^2)}}{2}

<h3><u>5.)  </u><u>Solve the equation using </u><u>±</u><u>:</u></h3>

<em>Solve the equation </em>x=\frac{0\frac{+}{}2\sqrt{(n^2-m^2)}}{2}<em>:</em>

<em>    </em>5a.) <em> </em><u><em>Solve the equation when </em></u><u><em>± </em></u><u><em>is plus:</em></u>

  • x=\sqrt{(n-m)(m+n)}

   5b.) <em> </em><u><em>Solve the equation when </em></u><u><em>± </em></u><u><em>is minus:</em></u>

  • x=-\sqrt{(n-m)(m+n)}

______________________________

Hope this helps!

If you have any questions, or need help with anything else, feel free to ask! I'm happy to help!

- TotallyNotTrillex -

AleksandrR [38]2 years ago
4 0

Answer:

\sqrt{(n-m)(n+m)} & -\sqrt{(n-m)(n+m)}

Step-by-step explanation:

1) Subtract m^{2} from both sides. This should leave you with x^{2}=n^{2}-m^{2}.

2) Square root both sides. This should leave you with x=\sqrt{n^2-m^2} & x=-\sqrt{n^2-m^2}.

<em>You can stop here if this is what the problem is asking for. However, it is not fully simplified.</em>

<em />

3) Factor the equation. This should leave you with \sqrt{(n-m)(n+m)} & -\sqrt{(n-m)(n+m)}.

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If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?
kondaur [170]

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

Here, x_1=-8, y_1=-6, x_2=-4, y_2=-14

$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)

$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)

$P(x, y) =(-6, -10)

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Here, x_1=-6, y_1=-10, x_2=-8, y_2=-6

r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}

r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}

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The standard form of the equation of a circle is

(x-a)^{2}+(y-b)^{2}=r^{2}, where (a, b) are center and r is the radius.

Here, center = (–6, –10) and r=\sqrt{20}

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(x+6)^{2}+(y+10)^{2}=20

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

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