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3 years ago

Please HELP HELP HELP ME

Mathematics

2 answers:

kompoz [17]3 years ago

7 0

The answer is A my friend.

Keith_Richards [23]3 years ago

7 0

The answer is 6squareroot5 so the 3rd one that one should be correct

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A 24-ounce box of cornflakes costs $4.59

Nikitich [7]

what is the question?

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3 years ago

12(x-2)+3x=1/2(x+6) +2

garri49 [273]

Answer: I think your answer is 2 ( please let me know if this is wrong ) !

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please answer it I need your help

balu736 [363]

Answer:

Step-by-step explanation:

Range-find the largest number and subtract the smallest number from it

96-74

4 0

3 years ago

Read 2 more answers

Add and round to the correct number of significant digits: 13.8 + 60 70 73 73.8 74

LenaWriter [7]

When you add all the number together you get 364.6 round it = 365
Because 6 is > 5

4 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