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4 years ago

if you draw 2 equilateral triangles are congruent and share a side. what polygon is formed? Is it a regular polygon?

Mathematics

2 answers:

Jobisdone [24]4 years ago

8 0

Answer:

The polygon formed is rhombus which is regular polygon.

Step-by-step explanation:

Given that if we draw 2 equilateral triangles are congruent and share a side AC as shoen in figure then

we have to tell what polygon is formed it is regular or not.

As two equilateral triangles are congruent it means also the sides of two equilateral triangle are equal.

Also the angles are of 60° which implies the angle formed at intersecting the diagonals at right angle i.e of 90°

Hence, all the properties of rhombus satisfied which implies

The polygon formed is rhombus which is regular polygon.

Annette [7]4 years ago

3 0

A rhombus is formed when you draw 2 equilateral triangles that are congruent and share a side. A rhombus is a rugular polygon.

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4 years ago

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Find the largest prime divisor of 76^2 − 1.

saveliy_v [14]

Answer:

Step-by-step explanation:

This is a difference of squares and therefore can be written as (76-1)(76+1)=(75)(77).

So lets look at the factors....

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3 years ago

Which expression is equivalent to (4g3h2k4)3

RSB [31]

Answer:

d) $8g^{6}h^{4} k^{12} - (h^{25} k^{15} )$

$\frac{(4g^{3} h^{2}k^{4} )^{3} }{8g^{3}h^{2} } - (h^{5} k^{3} )^{5}$ $= 8g^{6}h^{4} k^{12} - (h^{25} k^{15} )$

Step-by-step explanation:

Explanation

Given expression

= $\frac{(4g^{3} h^{2}k^{4} )^{3} }{8g^{3}h^{2} } - (h^{5} k^{3} )^{5}$

By using

(ab)ⁿ = aⁿbⁿ

$\frac{a^{m} }{a^{n} } = a^{m-n}$

= $\frac{(4)^{3} g^{9} h^{6}k^{12} ) }{8g^{3}h^{2} } - (h^{5} k^{3} )^{5}$

After simplification , we get

$= 8g^{9}g^{-3} h^{6} h^{-2} k^{12} - (h^{5} k^{3} )^{5}$

$= 8g^{9-3}h^{6-2} k^{12} - (h^{5} k^{3} )^{5}$

$= 8g^{6}h^{4} k^{12} - (h^{25} k^{15} )$

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3 years ago

Problems for you to try.

Alecsey [184]

Step-by-step explanation:

8⁴.... I hope it's correct

HAVE A GOOD DAY!

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3 years ago