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Aleksandr [31]
3 years ago
7

Audra's paper airplane is 13.8 centimeters long. Jordan's paper airplane is 9/10 centimeters shorter than Audra's. How long is J

ordan's paper airplane?
Mathematics
1 answer:
pychu [463]3 years ago
5 0

If Audra's airplane is 13.8 centimeters long, and Jordans airplane is 9/10 centimeters shorter than Audra's. First multiply 13.8 with 9/10, which equals 12.42, next subtract 13.8 and 12.42 which equals 1.38. So Jordan's airplane is I believe 1.38 centimeters long.

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Make d the subject of the formula h=d/3+2
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A student solves the following equation and
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Answer:

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Step-by-step explanation:

Considering the expression

\frac{3}{a+2}-6\cdot \frac{a}{-4+a^2}=\frac{1}{a-2}

\frac{3}{a+2}-\frac{6a}{-4+a^2}=\frac{1}{a-2}

\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)

\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)

\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)

as

  • \frac{3}{a+2}\left(a+2\right)\left(a-2\right):\quad 3\left(a-2\right)
  • -\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right):\quad -6a
  • \frac{1}{a-2}\left(a+2\right)\left(a-2\right):\quad a+2

so equation becomes

3\left(a-2\right)-6a=a+2  

-3a-6=a+2

-3a-6+6=a+2+6

-4a=8

\mathrm{Divide\:both\:sides\:by\:}-4

\frac{-4a}{-4}=\frac{8}{-4}

a=-2

\mathrm{Verify\:Solutions}

\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}

\mathrm{Solve\:}\:a+2=0:\quad a=-2

\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2

\mathrm{Solve\:}\:a-2=0:\quad a=2

So the following points are undefined

a=-2,\:a=2

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

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