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valentina_108 [34]
3 years ago
13

.

Mathematics
2 answers:
shusha [124]3 years ago
7 0

Answer:

-36.

Step-by-step explanation:

First term a1 = 21.

Common difference d = 18 - 21 = -3   (15-18 = -3)

nth term an = a1 + (n - 1)d

So 20th term = 21 + (20-1) -3

= 21 - 3 * 19

= 21 - 57

= -36.

mixas84 [53]3 years ago
4 0

Answer:

-36

Step-by-step explanation:

21

18

15

12

9

6

3

0

-3

-6

-9

-12

-15

-18

-21

-24

-27

-30

-33

-36

please mark brainliest

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27.539 rounded to the nearest tenth rounded to the nearest tenth
WINSTONCH [101]

Answer:

27.5

Step-by-step explanation:

tens    ones          .       tenths      hundredths       thousandths

2          7                .           5             3                          9

Rounding to the nearest tenth means we look at the hundredth

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Answer:

a)    tan (157.5) = \frac{1-cos 315}{sin315}

b)

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c)

      sin^{2} (157.5) = \frac{1-cos (315) }{2}

d)

  cos 330° = 1- 2 sin² (165°)

       

         

Step-by-step explanation:

<u><em>Step(i):-</em></u>

By using trigonometry formulas

a)

cos2∝  = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ =  2 cos² ∝/2

cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}

b)

cos2∝  = 1- 2 sin² ∝

cos∝  = 1- 2 sin² ∝/2

sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

<u><em>Step(i):-</em></u>

Given

              tan\alpha = \frac{sin\alpha }{cos\alpha }

          we know that trigonometry formulas

        sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

         1- cos∝ =  2 sin² ∝/2

      Given

         tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }

put ∝ = 315

      tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }

     multiply with ' 2 sin (∝/2) both numerator and denominator

        tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2}  }{2sin(\frac{315}{2} cos(\frac{315}{2}) }

Apply formulas

 sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

  1- cos∝ =  2 sin² ∝/2

now we get

 tan (157.5) = \frac{1-cos 315}{sin315}

       

b)

          sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 330° above formula

             sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}

            sin^{2} (165) = \frac{1-cos (330) }{2}

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c )

         sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 315° above formula

             sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}

            sin^{2} (157.5) = \frac{1-cos (315) }{2}

           

d)

     cos∝  = 1- 2 sin² ∝/2

   put      ∝ = 330°

       cos 330 = 1 - 2sin^{2} (\frac{330}{2} )

      cos 330° = 1- 2 sin² (165°)

3 0
3 years ago
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