Please help, the subject is math

Find the distance between (2, 3) and (–4, 4). If necessary, round to the nearest tenth.

devlian [24]

Answer:

√37

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Match the integrals with the type of coordinates which make them the easiest to do.

WARRIOR [948]

Answer:

1. C. cylindrical coordinates

2 A. spherical coordinates

3. A. spherical coordinates

4. D. Cartesian coordinates

5 B. polar coordinates

Step-by-step explanation:

USE THE BOUNDARY INTERVALS TO IDENTIFY

1. ∭E dV where E is:

x^2 + y^2 + z^2<= 4, x>= 0, y>= 0, z>= 0 -- This is A CYLINDRICAL COORDINATES SINCE x>= 0, y>= 0, z>= 0

2. ∭E z^2 dV where E is:

-2 <= z <= 2,1 <= x^ 2 + y^2 <= 2 This is A SPHERICAL COORDINATES

3. ∭E z dV where E is:

1 <= x <= 2, 3<= y <= 4,5 <= z <= 6 -- This is A SPHERICAL COORDINATES

4. ∫10∫y^20 1/x dx ---- This is A CARTESIAN COORDINATES

5. ∬D 1/x^2 + y^2 dA where D is: x^2 + y^2 <=4 This is A POLAR COORDINATES

6 0

3 years ago

Which of the following expressions can be used to calculate the area of a square with side lengths 5 inches?

NemiM [27]

Step-by-step explanation:

5x5

because 5^2 equals the area

4 0

3 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

Geometry: please only comment if you know the answer I need help with this please

stich3 [128]

Answer:

Step-by-step explanation:

8 0

3 years ago