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pentagon [3]
4 years ago
12

Triangle A B C is congruent to triangle E F G.

Mathematics
2 answers:
ale4655 [162]4 years ago
7 0

Answer:

Angle C B A is congruent to angle G E F is not true.

Step-by-step explanation:

We have given, triangle ABC is congruent to triangle EFG.

We need to find which statement is not true about congruence of these two triangles from the given options.

We know that,

Triangle ABC is congruent to triangle EFG, means same shape and different size.

Option a)  Angle B A C is congruent to angle F E G., means angles are similar.

Option c)  Segment A B is congruent to segment E F, means sides are similar.

Option d)  Segment A C is congruent to segment E G, means sides are similar.

We, can see that all the three options are correct about the  triangle ABC is congruent to triangle EFG, because a triangle is congruent when a side, angle, side (SAS) conditions, are followed by given triangles, option b) is not follow. Therefore, Angle CBA is not congruent to angle GEF.

Elan Coil [88]4 years ago
3 0
Angle CBA is congruent to angle GEF
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f(0) = 440
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A student solves the following equation and
Phoenix [80]

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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