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Lana71 [14]
4 years ago
6

Which pair of triangles can be proven congruent by the HL theorem

Mathematics
2 answers:
Lelu [443]4 years ago
7 0

Answer: Third pair

Step-by-step explanation:

  • The HL theorem says that if the hypotenuse and a leg of one right triangle are equal to the hypotenuse and a leg of another right triangle, then the triangles are said to be congruent.

From the given pictures only third pair of triangles have hypotenuse and one leg are equal.

Then by HL theorem, the triangles in third pair are congruent.

lys-0071 [83]4 years ago
4 0

third one is your answer

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Which expressions are equivalent to \dfrac{4^{-3}}{4^{-1}} 4 −1 4 −3 ​ start fraction, 4, start superscript, minus, 3, end super
Slav-nsk [51]

Answer:

\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}

\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}

Step-by-step explanation:

Given

\dfrac{4^{-3}}{4^{-1}}

Required

Choose equivalent expressions

Choosing the first answer:

\dfrac{4^{-3}}{4^{-1}}

Split expressions

4^{-3} * \frac{1}{4^{-1}}

Apply laws of indices: (a^{-b} = \frac{1}{a^b})

\frac{1}{4^3} * \frac{1}{4^{-1}}

Apply laws of indices: (a^{-b} = \frac{1}{a^b})

\frac{1}{4^3} * \frac{1}{1/4}

\frac{1}{4^3} * \frac{4^1}{1}

\frac{4^1}{4^3}

Hence:

\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}

Choosing the second:

\dfrac{4^{-3}}{4^{-1}}

Apply law of indices: (\frac{a^m}{a^n} = a^{m-n})

So,

\dfrac{4^{-3}}{4^{-1}} = 4^{-3-(-1)}

\dfrac{4^{-3}}{4^{-1}} = 4^{-3+1)}

\dfrac{4^{-3}}{4^{-1}} = 4^{-2}

Apply law of indices: (a^{-b} = \frac{1}{a^b})

So:

\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}

4 0
3 years ago
Please help anyone will give Points
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Simply try each value using long division.

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Long division 5238/4=1309.5 decimal here, so 4 isn't an answer

5:

This is also very easy, simply check if the number ends in 0 or 5. Since 5,238 does not, 5 doesn't work either.

6:

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9:

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10:

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Step-by-step explanation:

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Step-by-step explanation:

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