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3 years ago

Are all integers rational

Mathematics

2 answers:

Ne4ueva [31]3 years ago

8 0

Answer: Every integer is a rational number, since each integer n can be written in the form n/1.

Lubov Fominskaja [6]3 years ago

5 0

All integers are rational, including zero

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The expression 35 exponet 2 is equivalent to the product of two perfect squares (each greater than 1). You can find the two

raketka [301]

Answer:

Step-by-step explanation:

7 0

3 years ago

Use slopes and y-intercepts to determine if the lines 10x+3y=−3 and 5x−4y=−3 are parallel.

kvasek [131]

Answer:

They are not parallel

Step-by-step explanation:

original equation

10x + 3y = -3

subtract 10x

3y = -10x - 3

divide by 3

y = -10/3x - 1

original equation

5x - 4y = -3

subtract 5x

-4y = -5x-3

divide by -4

y = 5/4x + 3/4

the slopes are not equal to each other (5/4x and -10/3x) so they are not parallel

4 0

1 year ago

Please help me with this sum

barxatty [35]

Answers:

781 × 8 = 6248

231 × 7 = 1617

421 × 6 = 2526

531 × 5 = 2655

681 × 2 = 1362

951 × 4 = 3804

741 × 3 = 2223

361 × 3 = 1083

931 × 7 = 6517

821 × 2 = 1642

8 0

2 years 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

which container hold the most? a. cylinder with a radius of 5 in and a height of 15 in b. sphere with a radius 6in c. come with

stepan [7]

Vcylinder=hpir^2 Vsphere=(4/3)pir^3 Vcone=(1/3)hpir^2 Vcylinder=15*pi*5^2=375pi in^3 Vsphere=(4/3)*pi*6^3=288pi in^3 Vcone=(1/3)*15*pi*8^2=320pi in^3 greatest is Vcylinder at 375pi in^3 answer is A (cylinder)

8 0

3 years ago