Answer:
148°
Step-by-step explanation:
it is very close to the 150 line and the option closest to 150 is 148
We have that
<span>Circle 1: center (8, 5) and radius 6
</span><span>Circle 2: center (−2, 1) and radius 2
we know that
the equation of a circle is
(x-h)</span>²+(y-k)²=r²
for the circle 1---------> (x-8)²+(y-5)²=36
for the circle 2---------> (x+2)²+(y-1)²=4
using a graph tool
see the attached figure
Part A)<span>What transformations can be applied to Circle 1 to prove that the circles are similar?
we know that
r1/r2---------> 6/2------> 3
</span><span>
to prove that the circle 1 and circle 2 are similar, the radius of circle 1 </span>must be divided by 3 and translate the center of the circle 1 (10) units to the left and (4) units down
<span>
the answer part A) is
</span>
the radius of circle 1 must be divided by 3 and translate the center of the circle 1 (10) units to the left and (4) units down
Part B) <span>What scale factor does the dilation from Circle 1 to Circle 2 have?
the answer Part B) is
the scale factor is (3/1)</span>
Answer:
What is it you need help with?
Step-by-step explanation:
Which one
Quadratic trapezoids have only one pair of parallel sides
<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>