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Nookie1986 [14]
3 years ago
8

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Mathematics
1 answer:
Igoryamba3 years ago
3 0

Answer:

<L = 65, <LJK = 50°

Step-by-step explanation:

note: ? is the angle name that is cut off in the picture

<?JL + <LJK = 180

130 + <LJK = 180

<LJK = 50

<LJK + <K + <J = 180

50 + 65 + <L = 180

115 + <L = 180

<L = 65

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avanturin [10]

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8 0
2 years ago
I need help in the two questions
Sholpan [36]
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3 0
3 years ago
I need help...like fast plezz
liubo4ka [24]

Answer:

80°

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6 0
2 years ago
The formula for any geometric sequence is an = a1 · rn - 1 , where a n represents the value of the n th term, a1 represents the
const2013 [10]

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3 0
2 years ago
Solve 2cos ²y -siny -1=0 for 0° ≤y≤360°
natima [27]

 

\displaystyle\bf\\2cos^2y -sin\,y -1=0~~~for~~0^o\leq y \leq360^o\\\\cos^2y=1-sin^2y\\\\2(1-sin^2y) -sin\,y -1=0\\\\2-2sin^2y-sin\,y-1=0\\\\-2sin^2y-sin\,y+2-1=0\\\\-2sin^2y-sin\,y+1=0~~~\Big|\times(-1)\\\\2sin^2y+sin\,y-1=0

.

\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}

 

 

5 0
3 years ago
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