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3 years ago

Sally ran 2 % miles in 12 hour. What is her speed in miles per hour?

Mathematics

1 answer:

andreev551 [17]3 years ago

7 0

Answer:

Step-by-step explanation:

if you're saying she ran .2 miles in 12 hours

then she ran .016 miles per hour

if you're saying she ran 2 miles in 12 hours,

then she ran 1/3 mile per hour or 0.3333 miles per hour

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You multiply a number by 3 subtract 6 then add 2 the results is 20 what's the number

e-lub [12.9K]

You multiply a number by 3 subtract 6 then add 2 the results is 20

what's the number ?

Let the number = x

multiply a number by 3

so, it becomes 3x

subtract 6 , so, 3x - 6

then add 2

so, 3x - 6 + 2

The result will be 20

So,

3x - 6 + 2 = 20

solve for x

3x - 4 = 20

Add 4 to both sides

3x - 4 + 4 = 20 + 4

3x = 24

Divide both sides by 3

3x/3 = 24/3

So,

x = 8

5 0

1 year 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

3 over 8 what does that =

inn [45]

Answer:

.375

Step-by-step explanation:

divide 3 by 8

7 0

3 years ago

Read 2 more answers

Tell whether the angle on the circle is 1/4,1/2,3/4 or i full turn clockwise ir counterclockwise

Komok [63]

Their is no circle.........

6 0

3 years ago

Jon's weight loss for each week of the month is 4lbs., 3.5 lbs. and 2 lbs. He gained 4.5 lbs. the last week. If Jon originally w

aalyn [17]

Since he gained 4.5 lbs but he lost 4 + 3.5 + 2 = 9.5 lbs, in total, he lost 5 lbs.
hope it helps!
hit the thank you button!

4 0

3 years ago