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3 years ago

What is the answer to this following question 11(a+6)

Mathematics

2 answers:

Ipatiy [6.2K]3 years ago

7 0

11a+66 hope this helped

r-ruslan [8.4K]3 years ago

5 0

11(a+6) = 11(a)+11(6) = 11a+66

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I NEEED HELP!!!!!<br> find the equation of the line containing (5, -2) and has a slope of -1

mojhsa [17]

I think it is y=-5x+12

8 0

2 years ago

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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.

3 0

3 years ago

A rectangular prism with a volume of 222 cubic units is filled with cubes with side lengths of \dfrac14 4 1 start fraction, 1,

skad [1K]

The correct question is
<span>A rectangular prism with a volume of 2 cubic units is filled with cubes with side lengths of 1/4 unit. How many cubes does it take to fill the prism?

we know that
[volume of the cube]=b</span>³
b=1/4 unit
[volume of the cube]=(1/4)³------> 1/64 unit³

if 1 cube has a volume of ----------------> (1/64) unit³
x cubes--------------------------------> 2 unit³
x=2/(1/64)-----> x=2*64----> x=128 cubes

the answer is
128 cubes

8 0

3 years ago

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2x-3= 0 and 3x-1= 0 solved please, I'LL GIVE BRAINLIEST

drek231 [11]

Answer:

1)x=1.5

2)x=0.33

Step-by-step explanation:

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3 years ago

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A baseball team wants a pitcher no more 150 innings during a season. So far, the pitcher has pitched <img src="https://tex.z-dn.

butalik [34]

Answer:

C. 6 because any answer besides that is too low or too high.

Step-by-step explanation:

110.25 pitches so far- 150 at max. 5.75 pitches per game.

4: 4x5.75 = 23

110.25 + 23 = 133.25

5. 5x5.75 = 28.75

110.25 + 28.75 = 139

6: 6x5.75 = 34.5

110.25 + 34.5 = 144.75

7: 7x5.75 = 40.25

110.25 + 40.25 = 150.50 > which rounds up to 151

5 0

3 years ago