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AleksandrR [38]
3 years ago
8

Write 54 + 63 as the product of the GCF of 54 and 63 and another sum

Mathematics
2 answers:
levacccp [35]3 years ago
8 0

Answer:

9*(6+7)

Step-by-step explanation:

First, we have to find the Greatest Common Factor (GCF), to do this we have to see all the factors of 54 and 63 and find the greatest factor that they have in common.

Factors of 54

1,2,3,6,9,18,27,54

Factors of 63

1,3,7,9,21,63

The GCF is 9 because is the greatest factor that is common to both numbers.

Now we have to divide 54/9 and 63/9

54/9 = 6

63/9 = 7

So now we can write the product of the GCF and another sum:

9*(6+7)

<em>We can prove this by solving both expressions:</em>

<em>54+63 = 9*(6+7)</em>

<em>117 = 9*13</em>

<em>117 = 117 </em>

<em>The results are equal so we prove it is right.</em>

PolarNik [594]3 years ago
5 0
The GCF: 54: 1,2,3,6,9,18,27,54 63: 1,3,7,9,21,63 The common factors: 1,3,9 Greatest one: 9 Divided: 54 divide 9= 6 63 divided 9= 7 54+63= 9x(6+7)
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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
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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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3 years ago
Geometry math question please help
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A square is a figure with four equal sides and four right angles.In a square the diagonals bisect at right angles .

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So it is not a necessary condition for a parallelogram to be a square .

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Answer:

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