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4 years ago

Please use the image below in order to find out the answer to the following question:

Mathematics

1 answer:

vesna_86 [32]4 years ago

7 0

<u>Given</u>:

The measure of ∠EGF is 51°

We need to determine the measure of ∠EHF

<u>Measure of ∠EHF:</u>

By inscribed angle theorem, we know that, "if two inscribed angles of a circle intercept the same arc then the angles are congruent".

Applying the theorem, we have that the two inscribed angles of a circle intercepting the same arc are ∠G and ∠H

Then, the two angles are congruent.

Thus, we have;

∠G ≅ ∠H

Therefore, the measure of ∠G = ∠H = 51°

Hence, the measure of ∠EHF = 51°

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Conference A and conference B each have 10 football teams. The table shows the number of football games that each team in the tw

babymother [125]

Answer with explanation:

1.Mean of a Data set

$=\frac{\text{Sum of all the observation}}{\text{total number of observation}}$

$=\frac{10+9+11+8+10+12+2+7+8+0+12+10+8+11+3+12+6+7+4+3}{20}\\\\=\frac{153}{20}\\\\=7.65$

2. Range = Maximum value in two conferences - Minimum value in two conference

= 12 - 0

=12

The mean number of wins for teams from the two conferences is 7.65.

The range in wins is 12.

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3 years ago

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mrs_skeptik [129]

Answer : Y=-15/7

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3 years ago

14. What is the rate if the distance traveled is 360 miles for 6 hours?

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Answer:

Step-by-step explanation:

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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.

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3 years ago

1+1 equals 2 plus 3 get brainlist if u answer first

dusya [7]

Cool thanks bro I appreciate it

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3 years ago

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