The quadrilateral ABCD is a rhombus with BD = BC. On the side

1 answer:

5 0

The halves of the rhombus separated by line BD are equilateral triangles.
∠BAD = 60°
so
∠ADM is 0.80*60° = 48°.

∠BMD is an exterior angle to ΔAMD, so is equal to the sum
∠BMD = ∠BAD + ∠ADM
= 60° +48° = 108°

∠BMD = 108°

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Answer:

We have function,

$y = 3 - 6 \sin {}^{} (2x + \frac{\pi}{2} )$

Standard Form of Sinusoid is

$y = - 6 \sin(2x + \frac{\pi}{2} ) + 3$

Which corresponds to

$y = a \sin(b(x + c)) + d$

where a is the amplitude

2pi/b is the period

c is phase shift

d is vertical shift or midline.

In the equation equation, we must factor out 2 so we get

$y = - 6(2(x + \frac{\pi}{4} )) + 3$

Also remeber a and b is always positive

So now let answer the questions.

a. The period is

$\frac{2\pi}{ |b| }$

$\frac{2\pi}{ |2| } = \pi$

So the period is pi radians.

b. Amplitude is

$| - 6| = 6$

Amplitude is 6.

c. Domain of a sinusoid is all reals. Here that stays the same. Range of a sinusoid is [-a+c, a-c]. Put the least number first, and the greatest next.

So using that rule, our range is [6+3, -6+3]= [9,-3] So our range is [-3,9].

D. Plug in 0 for x.

$3 - 6 \sin((2(0) + \frac{\pi}{2} )$