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3 years ago

Highest common factor of 72 and 96

Mathematics

1 answer:

Likurg_2 [28]3 years ago

4 0

Answer:

Step-by-step explanation:

Prime Factorise 72:

$72=2*36\\72=2*2*18\\72=2*2*2*9\\72=2*2*2*3*3\\\\72=2^3*3^2$

Prime factorise 96:

$96=2*48\\96=2*2*24\\96=2*2*2*12\\96=2*2*2*2*6\\96=2*2*2*2*2*3\\96=2^5*3$

Multiply the common factors of $2^3*3^2$ and $2^5*3$ :

$2^3*3=24$

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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

Crickets can jump with a vertical velocity of up to 14 ft/s. Which equation models the height of such a jump, in feet, after t s

Natasha_Volkova [10]

Acceleration is simplified by assuming it is the constant -g

a=-g we integrate this with respect to time to get v...

v=-gt+C where C is the initial velocity in this case 14ft/s so

v=-gt+14 integrate again to get the height function

h=(-gt^2)/2 +14t +C we are not given an initial height so C is 0

h(t)=14t-gt^2/2 letting g=32 and neatening up a bit...

h(t)=14t-16t^2

8 0

3 years ago

Read 2 more answers

Given the function f(x) = -5x2 + 3x, express the value of f(a+h) – f(x) in simplest<br> form.

chubhunter [2.5K]

Step-by-step explanation:

it comes to

$- 7x$