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2 years ago

I need help with this

Mathematics

1 answer:

ycow [4]2 years ago

5 0

Answer:

it looses 200 in value for the first year

Step-by-step explanation:

I am in middle school first year.I have no idea how to solve this but i hope it helps

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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

2 years ago

Suppose a normal distribution has a mean of 266 and a standard deviation of

Elis [28]

Answer:

I think the answer is P(x>178)

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Am trying to find the value of r<br> 1/r + 2/1-r = 4/r^2

Mashcka [7]

<span>1/r + 2/1-r = 4/r^2

1-r+2r/r(1-r)=4/r^2
(1+r)/r(1-r)=4/r^2 cancle r both side

1+r/1-r=4/r

cross multiply
r+r^2=4-4r
r^2+4r+r-4=0
r^2+5r-4=0
r^2+4r+r-4=0
solve it for r factor it...

</span>

6 0

2 years ago

If the angles of a traingle are in the ratio 2:3:4 , Find Them

DedPeter [7]

Answer:

40°, 60°, and 80°

Step-by-step explanation:

We know that the sum of the angles of a triangle is equal to 180°.

We can use this equation to solve for these angles:

180 = 2x + 3x + 4x

180 = 9x

20 = x

Then substitute the solution in for x to solve for the angles:

2(20) = 40°

3(20) = 60°

4(20) = 80°

Therefore, the angles are 40°, 60°, and 80°.

8 0

2 years ago

Since opening night, attendance at Play A has increased steadily, while attendance at Play B first rose and then fell. Equations

nexus9112 [7]

Answer: option d is correct .

Step-by-step explanation:

In order to find the same attendance we set both equation equal

8x+191 = -x^2+26x+126

simplifying equation ,we get

x^2 -18x+65=0

(x-5) (x-13) = 0

x =5 or x =13

therefore on 5th day and 13th day both plays attendance is same

and it is obtained by plugging x = 5 in and x =13 in either of the equation.

for x = 5 ,

y = 8(5) +191 = 40+191 = 231

for x = 13

y = 8(13)+191

= 104+191

= 295

therefore option d is correct

4 0

3 years ago

I need help with this ​

I need help with this