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Stells [14]
2 years ago
14

A ball has a mass of 3g and a volume of 6 cm3. It's density is what

Mathematics
2 answers:
Naily [24]2 years ago
8 0
<h3>Answer ↓</h3>

pls view the calculations section of this ans for more details.

<h3>Calculations ↓</h3>

\boxed{\\\begin{minipage}{3cm}\xrightarrow{density}\\$\displaystyle\frac{mass}{volume} \\ \end{minipage}}

Mass  = 3g

Volume = 6

3/6

1/2 <----- density

<h3>So the density is 1/2 mL³</h3>

<h3>It will float because it's density is less than the density of water</h3>

hope helpful ~

Dvinal [7]2 years ago
7 0

Answer:

float

Step-by-step explanation:

density = mass/voume

3g/6c.m. ^3

0.5g/c.m.^3 = 0.5g/mL^3

Since the density of water is about 1g/mL^3 and 0.5g/mL^3 is less than that, it will float

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Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ
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y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}

The arc length of the curve is

\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx

which has a value of about 5.99086.

Let f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}. Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

\ell_i=\dfrac{i-1}2

r_i=\dfrac i2

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These subintervals have midpoints given by

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

Over each subinterval, we approximate f(x) with the quadratic polynomial

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that the integral we want to find can be estimated as

\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that

\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6

so that the arc length is approximately

\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086

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Thus, all three are placed in the intersection of the both circle of the vein diagram.

To know more about the regular polygon, here

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