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2 years ago

In the figure if angleACB = angleCDA, AC = 8 cm and AD= 3 cm, then BD is

Mathematics

1 answer:

tatiyna2 years ago

6 0

Answer:

AC = 8 cm,

AD = 3 cm and ∠ACB = ∠CDA

From figure,

∠CDA = 90°

∴ ∠ACB = ∠CDA = 90°

In right angled ∆ADC,

AC2 = AD2 + CD2

⇒ (8)2 = (3)2 + (CD)2

CD2 = 64 – 9 = 55

⇒ CD = √55 cm

In ∆CDB and ADC.

∠BDC = ∠AD [each 90°]

∠DBC = ∠DCA [each equal to 90°-∠A]

∴ ∠CDB ∼ ∆ADC

Then,

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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.

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3 years ago

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Solnce55 [7]

A Point is needed to define a circle Hope this helps!!!

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3 years ago

Each school day Lauren rides her

Licemer1 [7]

Answer:

21.41667

Step-by-step explanation:

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3 years ago

Read 2 more answers

rick jogged the same distance on tuesday snd friday, and 8 miles on sunday for a total of 20 miles for the week. solve to find t

saveliy_v [14]

Ricky jogged 6 miles on tuesday and 6 miles on friday

Solution:

Given that,

Rick jogged the same distance on tuesday and friday

Let "x" be the distance jooged on each tuesday and friday

He also jogged for 8 miles on sunday

Total of 20 miles for the week

Therefore, we frame a equation as,

total distance jogged = miles jogged on tuesday + miles jogged on friday + miles jogged on sunday

20 = x + x + 8

20 = 2x + 8

2x = 20 - 8

2x = 12

x = 6

Thus Ricky jogged 6 miles on tuesday and 6 miles on friday

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2 years ago

In the figure if angleACB = angleCDA, AC = 8 cm and AD= 3 cm, then BD is​

In the figure if angleACB = angleCDA, AC = 8 cm and AD= 3 cm, then BD is