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3 years ago

The perimeter of a rectangular field is 312 m.

Mathematics

1 answer:

ahrayia [7]3 years ago

7 0

Answer:

length = 194 (97+97)

Step-by-step explanation:

p= 312 width= 59

= 59×2-312

= 118 - 312

= 194

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Which equation is a direct variation function? y = 4 y = 1/3x+ 4 y = x y = x3

Aleks04 [339]

Answer:

y=4

Step-by-step explanation:

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2 years ago

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Brian needs to paint a logo using two right triangles. The dimensions of the logo are shown below. What is the difference betwee

solong [7]

Answer:

7.5 cm²

Step-by-step explanation:

Dimensions of the large ∆:

$base (b) = 3cm, height (h) = 9cm$

$Area = 0.5*b*h = 0.5*3*9 = 13.5 cm^2$

Dimensions of the small ∆:

$base (b) = 2cm, height (h) = 6cm$

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3 years ago

Assume y=x²+3x+6. what is the avg rate of change of y with respect to x as x changes from -4 to -2

AURORKA [14]

The average rate of change is 3

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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

What is the equation of the following line? Be sure to scroll down first to see all answer options.

Annette [7]

From the figure the given line passes through the points (0, 0) and (-4, 8).

Recall that the equation of a straight line is given by
$\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$

Thus, The equation of the given figure is given by
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8 0

3 years ago