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Basile [38]
3 years ago
15

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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0.037 in scientific notation
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\huge\boxed{3.7\ \times\ 10^{-2}}

First, you need to find where you can move the decimal point to so that there is only one non-zero digit to the left of it.

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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

<u>Step-by-step explanation:</u>

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

5) cos (t) = 1

Where on the Unit Circle does cos = 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}

<em>Hint: sin is only positive in Quadrants I and II</em>

\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)

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

\text{Answer: at}\  \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n

In degrees:    t = 120° + 360n  and  300° + 360n

4 0
3 years ago
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