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3 years ago

Which pairs of polygons are congruent?

Mathematics

2 answers:

andriy [413]3 years ago

8 0

Pair 1 and pair 3

The two polygons in pair 2's perimeter aren't the same, thus they aren't congruent with each other, the same goes for pair 4

Alla [95]3 years ago

6 0

Answer: Pair 1 and Pair 4

Step-by-step explanation:

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Inversely proportional:

y = constant/x, or y*x = constant. Let's see which give the *same* product:

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3 years ago

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MrRissso [65]

Answer:

1) $\sqrt{-48}=4\sqrt{3} \ i$

2) $\sqrt{-63}=3\sqrt{7} \ i$

3) $\sqrt{-72}=6\sqrt{2} \ i$

4) $\sqrt{-24}=4\sqrt{6} \ i$

5) $\sqrt{-84}=2\sqrt{21} \ i$

6) $\sqrt{-99}=3\sqrt{11} \ i$

Step-by-step explanation:

We need to simply the following :

Note: For all questions given we know that $\sqrt{-1}=i$ using it in all the given questions.

1) $\sqrt{-48}$

Factors of 48 are: 2x2x2x2x3

$\sqrt{-48}\\=\sqrt{48}\sqrt{-1} \\=\sqrt{2\times2\times2\times2\times3}\ i\\= \sqrt{2^2\times2^2\times3}\ i\\=2\times2\sqrt{3} \ i \\=4\sqrt{3} \ i$

2) $\sqrt{-63}$

Factors of 63 are: 3x3x7

$\sqrt{-63}\\=\sqrt{63}\sqrt{-1} \\=\sqrt{3\times3\times7}\ i\\= \sqrt{3^2\times7}\ i\\=3\sqrt{7} \ i$

3) $\sqrt{-72}$

Factors of 72 are: 2x2x2x3x3

$\sqrt{-72}\\=\sqrt{72}\sqrt{-1} \\=\sqrt{2\times2\times2\times3\times3}\ i\\= \sqrt{2^2\times3^2\times2}\ i\\=2\times3\sqrt{2} \ i \\=6\sqrt{2} \ i$

4) $\sqrt{-24}$

Factors of 24 are: 2x2x2x3

$\sqrt{-24}\\=\sqrt{24}\sqrt{-1} \\=\sqrt{2\times2\times2\times3}\ i\\= \sqrt{2^2\times2\times3}\ i\\=2\sqrt{2\times3} \ i \\=2\sqrt{6} \ i$

5) $\sqrt{-84}$

Factors of 84 are: 2x2x3x7

$\sqrt{-84}\\=\sqrt{84}\sqrt{-1} \\=\sqrt{2\times2\times3\times7}\ i\\= \sqrt{2^2\times3\times7}\ i\\=2\sqrt{21} \ i$

6) $\sqrt{-99}\\$

Prime factors of 99 are: 3x3x11

$\sqrt{-99}\\=\sqrt{99}\sqrt{-1} \\=\sqrt{3\times3\times11}\ i\\= \sqrt{3^2\times11}\ i\\=3\sqrt{11} \ i \\=3\sqrt{11} \ i$

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3 years ago