5. The blue one is y >= 5x - 7
and green is y >= -0.65x + 3
6. y <= (4/5)x
y >= -(5/4)x
Answer:127 pages of paper
Step-by-step explanation:
18 times 7
Acute: anything less than 90 but greater than 0
so, m<CAD
right: angles at exactly 90 degrees, so m<AEC,
and obtuse angles are anything greater than 90 but less than 180 degrees,
so CDA (you can tell by the picture that it is bigger than 90- 90 degrees is exactly perpendicular)
If it costs 3n+5, you have to plug 7 in for n and solve. 3(7)+5=26 so $26
<span> sin20 * sin40 * sin60 * sin80
since sin 60 = </span><span> √3/2
</span>√3/<span>2 (sin 20 * sin 40 * sin 80)
</span>√3/<span>2 (sin 20) [sin 40 * sin 80]
</span>
Using identity: <span>sin A sin B = (1/2) [ cos(A - B) - cos(A + B) ]
</span>√3/<span>2 (sin 20) (1 / 2) [cos 40 - cos 120]
</span>√3/4<span> (sin 20) [cos 40 + cos 60]
</span>
Since cos 60 = 1/2:
√3/4<span> (sin 20) [cos 40 + (1/2)]
</span>√3/4 (sin 20)(cos 40) + √3/8<span> (sin 20)
</span>
Using identity: <span> sin A cos B = 1/2 [ sin(A + B) + sin(A - B) ]
</span>√<span>3/4 (1 / 2) [sin 60 + sin (-20)] + </span>√3/8<span> (sin 20)
</span>
Since sin 60 = √3/<span>2
</span>√3/8 [√3/2 - sin 20] + √3/8 (sin 20)
3/16 - √3/8 sin 20 + √3/8<span> sin 20
</span>
Cancelling out the 2 terms:
3/16
Therefore, sin20 * sin40 * sin60 * <span>sin80 = 3/16</span>
