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Stella [2.4K]
3 years ago
14

How do you do this question?

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
5 0

Step-by-step explanation:

y = 3 + 8x^(³/₂), 0 ≤ x ≤ 1

dy/dx = 12√x

Arc length is:

s = ∫ ds

s = ∫₀¹ √(1 + (dy/dx)²) dx

s = ∫₀¹ √(1 + (12√x)²) dx

s = ∫₀¹ √(1 + 144x) dx

If u = 1 + 144x, then du = 144 dx.

s = 1/144 ∫ √u du

s = 1/144 (⅔ u^(³/₂))

s = 1/216 u^(³/₂)

Substitute back:

s = 1/216 (1 + 144x)^(³/₂)

Evaluate between x=0 and x=1.

s = [1/216 (1 + 144)^(³/₂)] − [1/216 (1 + 0)^(³/₂)]

s = 1/216 (145)^(³/₂) − 1/216

s = (145√145 − 1) / 216

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Answer:  195.5 or 12 7/32

Step-by-step explanation:

There is no letter tetha in the table so I use α instead. However it is not sence to final result.

The expression is:

(sinα+cosα)/(cosα*(1-cosα))

Lets divide the nominator and denominator by cosα

(sinα/cosα+cosα/cosα)/(cosα*(1-cosα)/cosα)= (tanα+1)/(1-cosα)=

=(8/15+1)/(1-cosα)= 23/(15*(1-cosα))    (1)

As known cos²α=1-sin²α   (divide by cos²α both sides of equation)

cos²a/cos²α=1/cos²α-sin²α/cos²α

1=1/cos²α-tg²α

1/cos²α=1+tg²α

cos²α=1/(1+tg²α)

cosα=sqrt(1/(1+tg²α))= +-sqrt(1/(1+64/225))=+-sqrt(225/(225+64))=

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Substitute in (1) cosα  by (2):

1st use cosα=15/17

1) 23/(15*(1-cosα)) =23/(15*(1-15/17))= 23*17/2=195.5

2-nd use cosα=-15/17

2)23/(15*(1-cosα)) =23/(15*(1+15/17))= 23*17/32=12 7/32

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What’s the value of x?
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A population of rabbits doubles evey 4 months. if the population starts out with 8 rabbits how many rabbits will there be in 1 y
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the factor for x^2 + 5x - 24 is

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Step-by-step explanation:

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