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3 years ago

Which sequence of transformations produces

Mathematics

1 answer:

inysia [295]3 years ago

8 0

Answer: (A)

Step-by-step explanation:

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What is the solution of 5sin2θ=3 for 0≤θ≤2π? Please help.

grigory [225]

Answer:

C) $\theta\approx\{0.3218,1.249,3.463,4.3908\}$

Step-by-step explanation:

$5\sin2\theta=3;\: 0\leq\theta\leq 2\pi\\\\\sin2\theta=\frac{3}{5}\\ \\2\theta\approx\{0.6436,2.498,6.926,8.7816\}\\\\\theta\approx\{0.3218,1.249,3.463,4.3908\}$

7 0

2 years ago

If the fdic has an insurance fund of 67.8 billion and must use 7.6% of it to cover several failed banks, approximately how much

erastovalidia [21]

If you're using 7.6 percent, then you're not using 92.4 percent (100-7.6)

To find 92.4 percent of 67.8 billion do:
67,800,000,000*.924=<span>62,647,200,000 </span>

5 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

Help me with this thank you

nikklg [1K]

It’s a frequency table, you find the middle number from the prices of mail received and you would multiply it. Not sure what with but if you search on Google Frequency tables there should be a good explanation.

Hope this helped you!

6 0

4 years ago

The area of a rectangular painting is 5/8 square yard. The width is 3/4 yard. What is the length of the painting?

vfiekz [6]

Answer:

The answer is

<h2> $\frac{5}{6} \: \: \: \: yards \\$ </h2>

Step-by-step explanation:

Area of a rectangle = length × width

Since we are finding the length of the rectangle

$length = \frac{area}{width} \\$

From the question

area = 5/8 square yards

width = 3/4 yard

So we have

$length = \frac{5}{8} \div \frac{3}{4} \\ = \frac{5}{8} \times \frac{4}{3} \\ = \frac{5}{2} \times \frac{1}{3}$

We have the final answer as

$\frac{5}{6} \: \: \: \: yards \\$

Hope this helps you

3 0

3 years ago