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3 years ago

15 points please look at picture BC bisects ABD

Mathematics

2 answers:

BabaBlast [244]3 years ago

6 0

A: = any higher degree
b: = 180 degrees
c: = 230 degrees
d: 140 degrees

dybincka [34]3 years ago

3 0

Shii ione even know it’s A

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Lucia is wrapping packages that are in the shape of a triangular prism. The net of the prism is shown below: ( Help me answer th

solniwko [45]

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Step-by-step explanation:

Let's find the surface area of ONE prism and then multiply that result by 6 to obtain the final answer.

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3 years ago

In Exercises 5-7, find all the exact t-values for which the given statement is true,

bearhunter [10]

Answer: See Below

Step-by-step explanation:

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

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In radians: t = 0π + 2πn

In degrees: t = 0° + 360n

******************************************************************************

$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does $sin = \dfrac{1}{2}$

Hint: sin is only positive in Quadrants I and II

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees: t = 30° + 360n and 150° + 360n

******************************************************************************

$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

Hint: sin and cos are only opposite signs in Quadrants II and IV

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3 years ago