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1 year ago

5

A snake is born measuring 5 inches long, and it grows 6 inches each month. Write an equation that shows the relationship between

the snake's age in months, x, and its length in inches, y.

1 answer:

Mkey [24]1 year ago

5 0

Answer:

y=6x

Step-by-step explanation:

it's because it grows every 6 inches

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Find M of angle ABC

pochemuha

Answer:

25

Step-by-step explanation:

$\sf Answer \begin{cases}\sf \rightarrowtail \:( 3x - 6) + (2x + 25) + (x + 11) = 180 \\ \sf \rightarrowtail \: 3x - 6 + 2x + 25 + x + 11 = 180 \\ \sf \rightarrowtail \: 3x + 2x + x - 6 + 25 + 11 = 180 \\ \sf \rightarrowtail \: 6x + 30 = 180 \\ \sf \rightarrowtail \: 6x = 180 - 30 \\ \sf \rightarrowtail \: 6x = 150 \\ \sf \rightarrowtail \: x = \frac{150}{6} \\ \sf \rightarrowtail \: x = 25\end{cases}$

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2 years ago

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2 years ago

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Help please work with answer

Basile [38]

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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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2 years ago

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mixas84 [53]

Answer:

1. Below 2. B. d = 8h

Step-by-step explanation:

The graph is a bit confusing. But basically, substitute x for the numbers they give you.

5x

5 · 1 = 5 = y

5 · 2 = 10 = y

5 · 3 = 15 = y

5 · 4 = 20 = y

7x

7 · 1 = 7 = y

7 · 2 = 14 = y

7 · 3 = 21 = y

7 · 4 = 28 = y

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2 years ago

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