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Alenkasestr [34]
3 years ago
5

Please help solve this, I’ve tried my best. My answers are a height of 6.7 using 30-60-90 as a last resort, and the glass volume

of 14 cubic inches

Mathematics
1 answer:
Neko [114]3 years ago
3 0

Answer:

A) To ensure the correct and exact straight height

B) 15 inches

C) 24.87 in^{3}

Step-by-step explanation:

Hello! Lets start!

First, it asks us to explain why the two bases must be parallel. In order to measure the correct and exact height, we must be sure that the bases are parallel, or else there would be a range of heights; in order to ensure that you are measuring a straight distance, they must be parallel!

Second, it asks us to identify the height of cone. To do this, we must make a slope! We can see that, for every time we go down 5 inches, we lose 1 inch of our base! Our slope in this situation would be -5. So, lets continue to see how long before the cone reaches its tip, in which the base would be equal to 0. At 5 inches, our base is 2 inches. At 10 inches, our base is 1 inch. At 15 inches, we do not have a base, as it equals 0! So therefore, our cone height is 15 inches.

Third, it asks us to identify the volume of the cup, which is also called a <em>circular truncated cone. </em>To do so, we can use the formula:

₁V = \frac{1}{3} \pi (r_{1}^2 + r_{1}r_{2} + r_2^2)h

Lets plug in our information!

V = \frac{1}{3} \pi (1.5^{2} + (1.5)(1) + 1^{2}) (5)\\V = 24.87

Hope this helps! Remember, math is fun!

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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}

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