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e-lub [12.9K]
3 years ago
10

What's the measure of Z1 if Z CBD = 75° and ZABC = 135°?

Mathematics
2 answers:
svet-max [94.6K]3 years ago
6 0

Answer:

Brainliest goes to me!

Step-by-step explanation:

angle abc = 135 degrees

part of it is angle 1 and the other part is angle cbd

<abc (135) = cbd (75) + <1

angle 1 = 60 degrees

Paladinen [302]3 years ago
4 0

Answer:

60°

Step-by-step explanation:

∠ABC-∠CBD=∠1

135-75

=60

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rect = 10 \times 8  =80 \\ tria = 0.5 \times 4 \times 5 = 10 \\ the \: total \: area \: is \: 80 + 10 = 90

I hope that is useful for you

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The figure shown below is composed of a semicircle and a non-overlapping equilateral triangle and contains a hole that is also c
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Check the picture below.

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now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.

\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}

\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01

3 0
4 years ago
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