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Anon25 [30]
3 years ago
15

Rationalize the denominator and simplify.

Mathematics
1 answer:
cestrela7 [59]3 years ago
4 0
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3151018

——————————
 
     \mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}


Multiply and divide by the conjugate of the denominator, which is  \mathsf{4\sqrt{6}+3\sqrt{2}:}

     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}


Expand those products and eliminate the brackets:
 
     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}

     \mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}} 

     \mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}} 

     \mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

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The height of a triangle is 5 m less than its base. The area of the triangle is 42 m².
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A = (1/2) b * h
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42 = (1/2) * b * (b - 5)
42 = (1/2) * b^2 - 5b
multiply both sides by 2 to clear the fraction (1/2)
2(42) = 2(1/2) *b^2 - 5b
84 = b^2 -5b
Since this is a quadratic equation, subtract 84 from both sides so that it is set = to zero.
b^2 - 5b - 84 = 0
Now factor.
(b - 12)(b + 5) = 0
b - 12 = 0; b = 12
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Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)


Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)


Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1, y = 1, z = 0

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