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3 years ago

Please help with these two questions ASAP, will give BRAINLIEST!

Mathematics

2 answers:

algol133 years ago

7 0

Let me try my best to help you out :) ❤️

A) if the area of the square is 81, than that will mean that the length of all the sides are 9. You find the Perimeter by adding up all of the sides. So that will mean you do 9+9+9+9 which equals 36.

B) The diagonal that I got for the square is 12.728. The formula to find the diagonal is d=a sqrt 2.

Hope this helps!

OlgaM077 [116]3 years ago

3 0

Step-by-step explanation:

write what you know

square- all sides are equal

area of a square

A = L x W

where A = 81

L = sqrt(81)

W = sqrt(81)

perimeter the length around the square

P = sum of all sides

you can find hypotenuse using pythagorean thm

c=sqrt( L^2 + W^2)

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The difference between two integers is 32.The ratio of these integers is 1:3.Find these integers

shutvik [7]

Answer:

16 and 48

Step-by-step explanation:

let the 2 integers be x and 3x ← ratio 1 : 3

Then

3x - x = 32 ← difference between the integers

2x = 32 ( divide both sides by 2 )

x = 16

and 3x = 3 × 16 = 48

Since magnitude of difference is 32

We can also express the difference as

x - 3x = 32

- 2x = 32 ( divide both sides by - 2 )

x = - 16

and 3x = 3 × - 16 = - 48

The 2 integers are 16 and 48 or - 16 and - 48

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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.

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3 years ago