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2 years ago

The figure shows ST

Mathematics

2 answers:

Sidana [21]2 years ago

8 0

Answer:

A,B,E,F those the answer i have EOC review packet done

Step-by-step explanation:

musickatia [10]2 years ago

4 0

Answer:

Step-by-step explanation:

it's the correct answer because they are the same

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What is m=3 and the line goes through the point (2,4) in point - slope form

Inga [223]

Answer:

y-4=3(x-2)

Step-by-step explanation:

y-y1=m(x-x1)

6 0

3 years ago

True or False? A circle could be circumscribed about the quadrilateral below.

KengaRu [80]

The answer is false your welcome

6 0

3 years ago

Read 2 more answers

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

Simplify x^2+3x-10/x^2-25

Natasha_Volkova [10]

Answer:

3x -10 / -25

Step-by-step explanation:

cancel off the x²

6 0

3 years ago

Kyle held a balloon 8ft off the ground above his head and Ava is standing 15 feet directly to g the right of Kyle how far is Ava

ryzh [129]

Answer:

Step-by-step explanation:

Pythagorean theory baby! Okay so the Pythagorean theory is a² + b² = c² so lets draw a triangle and label our sides. So we know 8² + 15² = c² so I'm going to find the square root of all of our numbers. We get 64 + 225 = c² so then add 64 + 225 = 289 then I take the square root of 289 which equals 17 . This answer also makes sense because out hypotenuse should always be bigger then the legs.

I'm pretty sure thats right, Hope this helps.

5 0

3 years ago