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Afina-wow [57]
3 years ago
15

A membership to Movie Night Movie Club costs $10, plus $2 per movie. y = 2x +10. Use x-values: 0, 5, and 10.

Mathematics
2 answers:
Bumek [7]3 years ago
8 0

Answer:

$10, $20, $30

Step-by-step explanation:

So, the equation you have set up already, y = 2x + 10, models the cost of the membership and movies, with y as the total cost, and x as the number of movies. Here we can just plug in the x values to get the total costs. If x is 0, we get $10 total, if it is 5, we get $20 total, and if it is 10, we get $30 total.

siniylev [52]3 years ago
3 0

Answer:

ten, twenty , thirty

Step-by-step explanation:

So, the equation you have set up already, y = 2x + 10, models the cost of the membership and movies, with y as the total cost, and x as the number of movies. Here we can just plug in the x values to get the total costs. If x is zero, we get ten dollars total, if it is five, we get twenty dollars total, and if it is ten, we get thirty dollars total.

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Tamiku [17]

Answer:

Step-by-step explanation:

15. \frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx}   \\

16.  \frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }

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

=> The clause is correct

17. Do the same (16)

18. \frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx

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\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\

=> The clause is correct

20. Do the same (18)

21. Left = \frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA}  } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}

Right = cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA}   \\

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

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