All categories

3 years ago

Whats the domain and range of the relation

Mathematics

1 answer:

hodyreva [135]3 years ago

5 0

Answer:

See below in bold.

Step-by-step explanation:

The domain is the list of x values and the range is the list of y values.

Reading from the graph: The domain is {-3.5, -2, 0, 3.5}.

The range is {3, 1.5, -0.5, }.

You might be interested in

The sum of the measures of two angles is 180°.

Degger [83]

A should be here :brainly.com/question/2121320

5 0

2 years ago

Read 2 more answers

A teacher and a doctor each have 45 books. If 4/5 f the teacher's books and 2/3 of the doctor's books are novels, how many more

Anarel [89]

Answer:

3/5

Step-by-step explanation:

8 0

3 years ago

Solve each word problem using a system of equations. Use substitution or elimination.

nasty-shy [4]

Answer:

Step-by-step explanation:

x + 3y = 24.....multiply by -1

5x + 3y = 36

------------------

-x - 3y = -24 (result of multiplying by -1)

5x + 3y = 36

------------------add

4x = 12

x = 12/4

x = 3 <=======

x + 3y = 24

3 + 3y = 24

3y = 24 - 3

3y = 21

y = 21/3

y = 7 <========

4 0

3 years ago

Can someone help me. Please

Alex Ar [27]

The answer for this question is c.

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