{y=2x+14<br> {−4x−y=4<br><br> y = ???

the size of the second application is 3.45 MB less than he first application. What is the size of the second application

mr_godi [17]

The size of the second application given the size of the first application and the expression ( x - 3.45 mb) for the size of the second application is 293.55 MB.

<h3>Equation</h3>

Let

Size of the first application = x

Size of the second application= x - 3.45 mb

For instance,

if the size of the first application is 297 MB

Size of the second application= x - 3.45 mb

= 297 MB - 3.45 MB

= 293.55 MB

Therefore, the size of the second application given the size of the first application and the expression for the size of the second application is 293.55 MB

Learn more about equation:

brainly.com/question/2972832

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6 0

2 years ago

Simplify the expression <br><br> c+2+c+c+4

professor190 [17]

Answer:

2+4+ccc

Step-by-step explanation:

7 0

2 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

I need help with with problem please thanks!

liraira [26]

40 is the correct answer. Hope this helps :)

5 0

3 years ago

Last Tuesday was silly hat day at Toby's school. 74 students wore a silly hat and 666 students did not. What percentage of the s

valina [46]

Answer:

10%

Step-by-step explanation:

The total population of students will be equal to the sum of both those who wore the silly hat and those who didn't hence

Total students= 74+666=740

Number of students who wore silly hat is given as 74 hence when expressed as a percentage of the total students' population we get that

percentage of the students wore a silly hat= $\frac {74\times 100}{740}=10\%$

Therefore, 10% of students wore a silly hat last Tuesday

4 0

3 years ago

{y=2x+14 {−4x−y=4 y = ???