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Oksanka [162]
3 years ago
13

Find the volume of the sphere .

Mathematics
1 answer:
mars1129 [50]3 years ago
8 0

Answer:

1436.8cm³

Step-by-step explanation:

Formula :

\frac{4}{3}  \times \pi \times r ^{3}

\frac{4}{3}  \times \pi \times 7^{3}

=1436.75504cm³

to 1dp

=1436.8cm³

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dalvyx [7]

Answer:

38 is the correct solution

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1 year ago
What is 1/7 divided by ? = 14
user100 [1]
First you multiply the numbers you do have to get the number your looking for 1/7•14=2 and 2divided by 1/7=14 so 2 would be your answer
4 0
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Y= 2x+1 Graph this line
maksim [4K]

Answer:

look up desmos graphing calculator! you're welcome x

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Harry earns an hourly rate of $15. For each additional hour over 40 hours, he earns 150% percent of his regular hourly rate. Las
irinina [24]

Answer:

51 hours

Step-by-step explanation:

$15 x 40 = $600

$15 x 1.50 = $22.5

$22.5 x 11 = $247.5

$600 + $247.5 = $847.5

40 + 11 = 51 hours

7 0
3 years ago
Read 2 more answers
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]

The left and right endpoints of the i-th subinterval, respectively, are

\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8

r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8

for 1\le i\le4, and the respective midpoints are

m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8

  • Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}

  • Midpoint rule

We approximate the area for each subinterval by

M_i=f(m_i)(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}

  • Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial p_i(x), where

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

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It so happens that the integral of p_i(x) reduces nicely to the form you're probably more familiar with,

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

Then the integral is approximately

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0
3 years ago
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