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3 years ago

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Please help me with this question

1 answer:

azamat3 years ago

7 0

It is 90 your welcome

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Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

HELP PLZZZ I NEED ANSWERS

asambeis [7]

Answer:

D.

Step-by-step explanation:

y = (x - 5)^2 + 16

= x^2 - 5x - 5x + 25 + 16

= x^2 - 10x + 41

That corresponds with answer choice D.

Hope this helps!

3 0

3 years ago

an athlete completes a race in 55.72 seconds how many times greater is the digit in the tens place than the digit in the ones pl

Inessa05 [86]

7 is 5 greater than the 2

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3 years ago

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FrozenT [24]

Answer:

Step-by-step explanation:

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7 0

2 years ago

arrange the expressions in the correct sequence to rationalize the denominator of the expression -(2)/(\sqrt(x+y-2)-\sqrt(x+y+2)

cupoosta [38]

We have to rationalize the denominator:
$\frac{-2}{ \sqrt{x+y-2} - \sqrt{x+y+2} } = \\ \frac{-2}{ \sqrt{x+y-2} - \sqrt{x+y+2} }* \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2} }{ \sqrt{x+y-2}+ \sqrt{x+y+2} }= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2}) }{x+y-2-(x+y+2)}= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2}) }{x+y-2-x-y-2}= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2} }{-4}= \\ \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2} }{2}$

6 0

3 years ago

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