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Klio2033 [76]
3 years ago
12

Two parallel lines are crossed by a transversal. Horizontal and parallel lines y and z are cut by transversal x. At the intersec

tion of lines y and x, the top right angle is (2 k + 11) degrees. At the intersection of lines z and x, the bottom left angle is 131 degrees. What is the value of k? k = 9 k = 20 k = 60 k = 71
Mathematics
2 answers:
BigorU [14]3 years ago
7 0

Answer:

60

Step-by-step explanation:

ruslelena [56]3 years ago
3 0

Those two angles have the same measure, so you have

2k+11=131 \iff 2k=120 \iff k=60

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Answer:

The answer is 7.2368421053 .

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How to prove this???
swat32
\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3)  \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 


go to right side now

4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)

use x^3 - y^3 = (x-y)(x^2 + xy + y^2) and x^3 + y^3 = x^2 - xy + y^2

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow  4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad  \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)

so \sin^2 A + \cos^2 A = 1

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
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3 years ago
There are 20 classrooms in the high school with 30 students in each class. How many students are in the school?​
Sophie [7]
<h2>Required Answer:</h2>

\Large{\boxed{\tt{600 \:  Students}}}

<h2 /><h2>Question:</h2>

  • There are 20 classrooms in the high school with 30 students in each class. How many students are in the school?
<h2>To find:</h2>
  • No. of students
<h2>Given:</h2>
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<h2>Answer:</h2>
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best of luck! ^_^

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Answer:

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Step-by-step explanation:

If you replace the value of c with 130, you can multiply 13 with 130 to get your answer.

13 x 130 = 1690

Or, you can multiply 13 with 13 and add the 0 onto the end. This makes the same answer. To add a zero onto the end of a number multiply it by ten.

169 x 10 = 1690

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