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mrs_skeptik [129]

3 years ago

15

A basket weighs 12 oz. A watermelon weighing 7 lb 9 oz is added to the basket. What is the total weight of the basket and waterm

elon? (Give the answer in pounds and ounces.)

1 answer:

SSSSS [86.1K]3 years ago

3 0

7 lbs is 112 ounces. 112 plus 9 is 121. 121 plus 12 is 133. The total weight of the basket and the watermelon is 133 ounces. I'm not sure how you need the answer so I'll leave it in a couple different forms.

Ounces- 133
Pounds- 8.312 (rounds to the nearest tenth, 8.3)
Pounds and ounces- 8 pounds and 5 ounces

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Answer:

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

Step-by-step explanation:

Given that,

The given curve between the specified points is

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

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Using given equation

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On differentiating w.r.to y

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

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

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

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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 $\dfrac{21}{16}$

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