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3 years ago

What is the missing value in the equation __ x 1\10 = 0.026

Mathematics

2 answers:

AleksAgata [21]3 years ago

5 0

0.26 * 1/10 = 0.026

because 1/10 = 0.1
0.26 * 0.1 = 0.026

Ostrovityanka [42]3 years ago

3 0

.26 X 1/10 = 0.026

Hope this helps =)

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What's 5/8 in its simplest form<br><br> I think i"m dum Ф∧Ф ≥∧≤

OlgaM077 [116]

Answer:

It is already in simplest form :)

Step-by-step explanation:

5/8=5/8

I hope this helps :)

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3 years ago

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A small junior college had an enrollment of 7,843 students in the year 1990 and 8,793 students in the year 2000. Let x=0 represe

abruzzese [7]

Correct answer is: (0,7843) and (10,8793)

Solution:-

Given that a junior college has an enrollment of 7843 students in 1990 and 8793 students in year 2000.

We have to write this data as (x,y) .

Where x= years after 1990 and y=number of students enrolled.

Since in 1990, 7843 students enrolled, x = 1990-1990=0

And y=7843.

Hence one ordered pair is (0,7843).

Let us find the years after 1990 for 2000 = 2000-1990 =10

Hence another ordered pair is (10,8793).

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3 years ago

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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.

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3 years ago

Pls help! Asap brainliest to helpful answers

riadik2000 [5.3K]

Answer:

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Step-by-step explanation:

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