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3 years ago

What is the perimeter P of △DEF to the nearest whole number?

Mathematics

1 answer:

inna [77]3 years ago

3 0

Answer:

Can you please tell me what is the area of the given triangle?

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4.5 times 6-12 divided by 3

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Answer:5

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$\frac{(2a+1)(a+5)}{(a+1)^2(a+3)} hope this helps
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3 years ago

Item 16 A sphere has a radius of 8 centimeters. A second sphere has a radius of 2 centimeters. What is the difference of the vol

Zigmanuir [339]

Answer:

$672\pi \text{ cm}^3$ .

Step-by-step explanation:

We have been given that a sphere has a radius of 8 centimeters. A second sphere has a radius of 2 centimeters. We are asked to find the difference of the volumes of the spheres.

We will use volume formula of sphere to solve our given problem.

$\text{Volume of sphere}=\frac{4}{3}\pi r^3$ , where r is radius of sphere.

The difference of volumes would be volume of larger sphere minus volume of smaller sphere.

$\text{Difference of volumes}=\frac{4}{3}\pi(\text{8 cm})^3-\frac{4}{3}\pi(\text{2 cm})^3$

$\text{Difference of volumes}=\frac{4}{3}\pi(512)\text{ cm}^3-\frac{4}{3}\pi(8)\text{ cm}^3$

$\text{Difference of volumes}=\frac{4}{3}\pi(512-8)\text{ cm}^3$

$\text{Difference of volumes}=4\pi(168)\text{ cm}^3$

$\text{Difference of volumes}=672\pi\text{ cm}^3$

Therefore, the difference between volumes of the spheres is $672\pi \text{ cm}^3$ .

3 0

3 years ago

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