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Degger [83]
3 years ago
10

1. 8q+ 6 ; q = 2

Mathematics
1 answer:
Alex17521 [72]3 years ago
6 0
1. 8q + 6 ; q = 2
  = 8 ( 2 ) + 6
   = 16 + 6
  = 22 option D.

2. 9 + 4 ( -3 )
  = 9 - 12
  = -3 option A

3. 7 ( 4 ) - 7
   = 28 - 7
   = 21
option B

4. 10 - 3.2 ( 2 )
 = 10 - 6.4
= 3.6 option B

5. 12 (1.5) + 4
= 18 + 4 
= 22 option D.

6. 5w + 10 = 40  
    5w = 30 
   w = 6 option A.

7. 4y - 12 = 60
    4y = 72
    y = 18 option B

8 . 10k + 5 = 65
     10k = 60
     k = 6 option A

9. 2b - 15 = 33
    2b = 48
     b = 24 option C

10. 3d + 18 = 21
      3d = 3
        d = 1 option B
     




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Step-by-step explanation:

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ArbitrLikvidat [17]

Answer:

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, c) x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right), y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right).

Step-by-step explanation:

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After some algebraic and trigonometric handling:

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1

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Where:

\frac{x}{4} = \cos t

\frac{y-1}{4} = \sin t

Finally,

x = 4\cdot \cos t

y = 1 + 4\cdot \sin t

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

c) x = 4\cdot \cos t'', y = 1 + 4\cdot \sin t''

Where:

4\cdot \cos t' = 0

1 + 4\cdot \sin t' = 5

The solution is t' = \frac{\pi}{2}

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x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right)

y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)

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Answer:

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Step-by-step explanation:

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Find the value of x in each case:
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Answer:

Step-by-step explanation:

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