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alexira [117]
4 years ago
15

Suppose a triangle has sides a, b, and c, and let theta be opposite the side of length a. If cos theta < 0, what must be true

?
Mathematics
1 answer:
katrin2010 [14]4 years ago
6 0
If \cos \theta\ \textless \ 0, then angle \theta is obtuse and triangle with sides a, b, c is obtuse triangle.
In an arbitrary triangle can be only one obtuse angle, and the side which lies opposite to the largest angle is the largest. Then since <span>angle \theta is opposite the side of length a</span>  you can conclude that a>c and a>b.



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(10x+8)°<br> (9x+18<br> x=8<br> (12x-10)
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Answer:

Step-by-step explanation:

(10x + 8)^0 = 1

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12x - 10

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(1) ( 90) ( 86)

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A piece of wire 40 cm long is to be cut into two pieces. One piece will be bent to form a circle and; the other will be bent to
Brums [2.3K]
<h3>The perimeter of circle  = 17.6 cm</h3><h3>Perimeter of square  = 22.4 cm</h3>

Step-by-step explanation:

The total length of the wire  = 40 cm

Let us assume the circumference of the circle  = k cm

So, the circumference of the square  = ( 40 - k) cm

Now,as circumference of circle = 2πR

⇒2πR = k cm    or, R  = (\frac{k}{2\pi})

Area  of the  circle = πR²  

or, A  = \pi (\frac{k}{2\pi} )^2  

 = \pi \times (\frac{k^2}{(2\pi)^2} )  = \frac{k^2}{4 \pi} \\\implies A =  \frac{k^2}{4 \pi}   ....... (1)

Similarly , perimeter of square  = 4 x Side

⇒ 4 x Side  =  ( 40 - k) cm    ⇒ S = (\frac{40 - k}{4} )

Area  of the  square = (Side)²    = (\frac{40 - k}{4} ) ^2

 Solving, we get:  A = \frac{1600 + k^2 - 80 k}{16}   ....... (1)

So, combining (1) and (2), total area A is

A = (\frac{k^2}{4 \pi} ) + \frac{k^2 + 1600 - 80 k }{16}

or, we get: A  = 0.07962 k²  + 0.0625 k² + 100 - 5 k

A  =  0.1422 k² - 5 k + 100

For axis of symmetry :  k = (\frac{-b}{2a} ) = (-\frac{-5}{2(0.1420})  = 17.6

⇒ k = 17.6 inches

So, the perimeter of circle  = 17.6 cm

Perimeter of square  = (40 - 17.6 )  = 22.4 cm

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Answer:

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