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AnnZ [28]
4 years ago
12

Given that triangle ABC is a right triangle, select a set of possible side length measures.

Mathematics
2 answers:
ad-work [718]4 years ago
5 0

You didn't supply any rules or constraints so.... 1 in., 1 in., and 1 in.

You can use the Pythagorean Theorem to check if the side lengths are appropriate.

a^2 + b^2 = c^2

1^2 + 1^2 = 1^2 –this is true.

Sindrei [870]4 years ago
3 0

There are an infinite number of sets of side-length measures that can make a right triangle.  It would take me quite a long time to write down all of them, and then an infinite length of time to upload them so you could see them.  Also, it would cost me a bundle to push all that data through on my wifi plan, and you wouldn't have enough space in your phone or computer to download them.

The whole idea is that a set that works MUST follow this rule:

<em>The square of the biggest one</em>

<em>is equal to</em>

<em>(the square of the 2nd one) + (the square of the last one) .</em>

Any set of 3 numbers that fits that rule can be the lengths of the sides of a right triangle.  

This was discovered by a math pro named Pythagoras, around the year 500.

==>  BC

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3 years ago
3^x= 3*2^x solve this equation​
kompoz [17]

In the equation

3^x = 3\cdot 2^x

divide both sides by 2^x to get

\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3

Take the base-3/2 logarithm of both sides:

\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}

Alternatively, you can divide both sides by 3^x:

\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13

Then take the base-2/3 logarith of both sides to get

\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}

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7 0
2 years ago
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serious [3.7K]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
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Verdich [7]

Answer:

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4 0
3 years ago
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