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MrRissso [65]
2 years ago
15

This is Grade 9 Mathematics so I kinda stuck here so, if somebody can help me Please sin30⁰+csc60⁰​

Mathematics
2 answers:
mamaluj [8]2 years ago
6 0

\huge\mathrm{Answer࿐}

  • \sin(30)  +  \csc(60)

  • \dfrac{1}{2}  +  \dfrac{2}{ \sqrt{3} }

  • \dfrac{ \sqrt{3 } + 4 }{2 \sqrt{3} }

To remove √3 from denominator, we multiply both numerator as well as denominator with √3

  • \dfrac{ \sqrt{3 } + 4 }{2 \sqrt{3} }  \times  \dfrac{ \sqrt{3} }{ \sqrt{3} }

  • \dfrac{3 + 4 \sqrt{3} }{6}

____________________________

\mathrm{ \#TeeNForeveR}

PSYCHO15rus [73]2 years ago
4 0

Answer:

give he brainlets never mind me

Step-by-step explanation:

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Which expression is equivalent to -<br>60x20, 24<br>30x10 12?​
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Answer:

30x10 12

Step-by-step explanation:

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Given: mAngleEDF = 120°; mAngleADB = (3x)°; mAngleBDC = (2x)° Prove: x = 24 3 lines are shown. A line with points E, D, C inters
N76 [4]

Answer:

" Vertical angles are congruent " ⇒ 2nd answer

Step-by-step explanation:

* <em>Look to the attached figure </em>

- There are three lines intersected at point D

- We need to find the missing in step 3

∵ Line FA intersects line EC at point D

- The angles formed when two lines cross each other are called

 vertical angles

- Vertical angles are congruent (vertical angles theorem)

∴ ∠ADC and ∠FDE are vertical angles

∵ Vertical angles are congruent

∴ ∠EDF ≅ ∠ADC

∴ m∠EDF ≅ m∠ADC

∵ m∠EDF = 120° ⇒ given

∵ m∠ADC = m∠ADB + m∠BDC

∴ m∠ADB + m∠BDC = 120°

∵ m∠ADB = (3x)° ⇒ given

∵ m∠BDC = (2x)° ⇒ given

∴ 3x + 2x = 120 ⇒ add like terms

∴ 5x = 120 ⇒ divide both sides by 5

∴ x = 24

Column (1)                                                     Column (2)

m∠EDF = 120°                                               given

m∠ADB = 3 x                                                 given

m∠BDC = 2 x                                                 given

∠EDF and ∠ADC are vertical angles           defin. of vert. ∠s

∠EDF is congruent to ∠ADC                        vertical angles are      

                                                                        congruent  

m∠ADC = m∠ADB + m∠BDC                        angle add. post.

m∠EDF = m∠ADC                                          defin. of cong.

m∠EDF = m∠ADB + m∠BDC                         substitution

120° = 3 x + 2 x                                               substitution

120 = 5 x                                                         addition

x = 24                                                              division  

∴ The missing reason is " vertical angles are congruent "

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 theorem

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When we equate this quadratic polynomial with 0 , it becomes a quadratic equation.

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