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2 years ago

NEED HELP ASAP PLEASE ILL GIVE BRAINLIEST!!!

Mathematics

1 answer:

spayn [35]2 years ago

8 0

Answer:

the answer is in the picture

Step-by-step explanation:

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Please help me with number ten and twelve

jasenka [17]

Answer: $93,\ 34^{\circ}$

Step-by-step explanation:

Given

For figure (i) both triangles are congruent

$\therefore 8x-27=4x+33\\\Rightarrow 8x-4x=33+27\\\Rightarrow 4x=60\\\Rightarrow x=15$

$\Rightarrow PQ=8x-27\\\Rightarrow PQ=93$

For figure (ii)

Two triangles WXZ and WZY are congruent

$\therefore 9x-28=5x-8\\\Rightarrow 4x=20\\\Rightarrow x=5^{\circ}\\\Rightarrow \angle YWX=2(9x-28)\\\Rightarrow \angle YWX=2(17)\\\Rightarrow \angle YWX=34^{\circ}$

5 0

2 years ago

Can somebody help me. Will Mark brainliest.

Alex17521 [72]

I think the answer is 43.2 degrees

5 0

3 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

Whats 2x + y = 10 5x - y = 18 in elimination method

Naya [18.7K]

2x + y = 10
5x - y = 18

7x = 28

x = 4

2(4) + y = 10

8 + y = 10

y = 2

(4,2)

8 0

3 years ago

What's the square root of 0.4 repeating

avanturin [10]

The answer depends on knowing that 0.4444.... is equal to 4/9;
<span>the square root of 4/9 is 2/3 = 0.6666....</span>

6 0

3 years ago