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3 years ago

Aboard is 4 ft 8 in long how long is the board in meters

Mathematics

1 answer:

andreev551 [17]3 years ago

3 0

If a board is 4ft 8 in long than it is 1.4224 meters long

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Simplify the exponential expression. (−8) 2 =

sweet-ann [11.9K]

Answer:

-16

Step-by-step explanation:

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3 years ago

CAN SOMEONE HELP ME

Xelga [282]

Answer:

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2 years ago

Match each equation with a diagram:

Flura [38]

Answer:

a= 3

b=3

c=3

d=3

Step-by-step explanation:

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3 years ago

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What is the average speed for the interval t=1 hour to t=3 hours

inysia [295]

Answer:

7.5 miles per hour

Step-by-step explanation:

Average speed over the interval [1, 3] = (distance at 3 hrs -distance at 1 hr)/(3 hrs - 1 hr)

From the graph:

Distance at 3 hrs = 30 => (3, 30)

Distance at 1 hr = 15 => (1, 15)

Average speed = (30 - 15)/(3 - 1)

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= 7.5

Therefore, average speed = 7.5 miles per hour

5 0

3 years ago

Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y

DiKsa [7]

The area of the surface is given exactly by the integral,

$\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx$

We have

$y(x)=\dfrac15x^5\implies y'(x)=x^4$

so the area is

$\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx$

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the $i$ -th subinterval are, respectively,

$\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2$

$r_i=\dfrac{5-0}{10}i=\dfrac i2$

with midpoint

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

with $1\le i\le10$ .

Over each subinterval, we interpolate $f(x)=\sqrt{1+x^8}$ with the quadratic polynomial,

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

Then

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that the latter integral reduces significantly to

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$

which is about 651.918, so that the area is approximately $651.918\pi\approx\boxed{2048}$ .

Compare this to actual value of the integral, which is closer to 1967.

4 0

3 years ago