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2 years ago

Help please!

Mathematics

1 answer:

Charra [1.4K]2 years ago

5 0

Answer:

Step-by-step explanation:

The period is the x distance for one full cycle.

The 1st peak is at 1 and the second peak at 15

The period is 15 - 1 = 14

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In a company, 37‰ of the 1500 employees are female. how many are male?

Mashcka [7]

Answer:

945 males

Step-by-step explanation:

1500*37/100= 555 female

1500-555 = 945 male

Because 1500 is the total number so it is eual a 100% and you do the cross multiply. You will get 555 for female. And you use the total to subtract female employees and you will get male employees number.

7 0

3 years ago

Integrate the following. ∫<img src="https://tex.z-dn.net/?f=5x%5E%7B4%7D%20dx" id="TexFormula1" title="5x^{4} dx" alt="5x^{4} dx

Vadim26 [7]

Answer:

$\displaystyle D) {x}^{5} + \rm C$

Step-by-step explanation:

we would like to integrate the following Integral:

$\displaystyle \int 5 {x}^{4} \, dx$

well, to get the constant we can consider the following Integration rule:

$\displaystyle \int c{x} ^{n} \, dx = c\int {x}^{n} \, dx$

therefore,

$\displaystyle 5\int {x}^{4} \, dx$

recall exponent integration rule:

$\displaystyle \int {x} ^{n} \, dx = \frac{ {x}^{n + 1} }{n + 1}$

so let,

$n = 4$

Thus integrate:

$\displaystyle = 5\left( \frac{ {x}^{4+ 1} }{4 + 1} \right)$

simplify addition:

$\displaystyle = 5\left( \frac{ {x}^{5} }{5} \right)$

reduce fraction:

$\displaystyle = {x}^{5}$

finally we of course have to add the constant of integration:

$\displaystyle \boxed{ {x}^{5} + \rm C}$

hence,

our answer is D)

7 0

2 years ago

Read 2 more answers

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

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