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2 years ago

15

If 5 and 12 are the two smallest values in a Pythagorean triple, what is the largest value, c, in the triple?

2 answers:

ohaa [14]2 years ago

6 0

13 would be the largest value in the triple.

s344n2d4d5 [400]2 years ago

6 0

<h3>Answer: 13</h3>

=============================================

Work Shown:

a^2 + b^2 = c^2

c = sqrt( a^2 + b^2 )

c = sqrt( 5^2 + 12^2 )

c = sqrt( 25 + 144 )

c = sqrt( 169 )

c = 13

---------------

Extra info (optional section)

The perimeter is 5+12+13 = 30 units and the area is base*height/2 = 5*12/2 = 60/2 = 30 square units. In this triangle type, the area and perimeter are the same value (different units though of course). This concept and particular triangle is discussed in the video titled "Superhero triangles" from Numberphile (timestamp around 1:23 of the video).

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Hatshy [7]

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3 years ago

Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)

vagabundo [1.1K]

Answer:

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And the ordered pairs in the format $(x, y)$

I don't know if it is log of base 3 or 10, but I will assume it is 3.

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So the ordered pair will be $(\frac{1}{3}, -1 )$

For $(1, a_{1} )$

$x=1$

$y=a_{1}$

$y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0$

So the ordered pair will be $(1, 0 )$

For $(3, a_{2} )$

$x=3$

$y=a_{2}$

$y = log_{3}x\\y = log_{3}3\\y = 1$

So the ordered pair will be $(3, 1 )$

For $(9, a_{3} )$

$x=9$

$y=a_{3}$

$y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2$

So the ordered pair will be $(9, 2 )$

For $(27, a_{4} )$

$x=27$

$y=a_{4}$

$y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3$

So the ordered pair will be $(27, 3 )$

For $(81, a_{5} )$

$x=81$

$y=a_{5}$

$y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4$

So the ordered pair will be $(81, 4 )$

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3 years ago

Triangle ABC is an equilateral triangle. How many lines of symmetry does triangle ABC have?

xxTIMURxx [149]

1 because it can't be reflected over more than one line to be the same

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2 years ago

How do I solve this?

Sonbull [250]

The question is somewhat poorly posed because the equation doesn't involve <em>θ</em> at all. I assume the author meant to use <em>x</em>.

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By definition of secant and cosecant,

1/cos(<em>x</em>) = 1/sin(<em>x</em>)

Multiply both sides by sin(<em>x</em>) :

sin(<em>x</em>)/cos(<em>x</em>) = sin(<em>x</em>)/sin(<em>x</em>)

As long as sin(<em>x</em>) ≠ 0, this reduces to

sin(<em>x</em>)/cos(<em>x</em>) = 1

By definition of tangent,

tan(<em>x</em>) = 1

Solve for <em>x</em> :

<em>x</em> = arctan(1) + <em>nπ</em>

<em>x</em> = <em>π</em>/4 + <em>nπ</em>

(where <em>n</em> is any integer)

In the interval 0 ≤ <em>x</em> ≤ 2<em>π</em>, you get 2 solutions when <em>n</em> = 0 and <em>n</em> = 1 of

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Answer:

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Step-by-step explanation:

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