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3 years ago

Look at the function below. f(x)=−x2−2x+8

Mathematics

1 answer:

enyata [817]3 years ago

7 0

Answer:

f(x) = -1(x+1)² + 9

Step-by-step explanation:

f(x) = -x² - 2x + 8

This is a down-opening parabola. Put the equation into vertex form. The y-coordinate of the vertex is the maximum.

factor out the leading coefficient

f(x) = -1(x²+2x) + 8

Complete the square

coefficient of x term: 2

divide in half: 1

square it: 1²

use 1² to complete the square:

f(x) = -1(x²+2x+1²) + 1² + 8

f(x) = -1(x+1)² + 9

vertex (-1,9)

maximum value of f(x) = 9

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The automatic opening device of a military cargo parachute has been designed to open when the parachute is 200 m above the groun

kirill115 [55]

Step-by-step answer:

Given:

mean, mu = 200 m

standard deviation, sigma = 30 m

sample size, N = 5

Maximum deviation for no damage, D = 100 m

Solution:

Z-score for maximum deviation

= (D-mu)/sigma

= (100-200)/30

= -10/3

From normal distribution tables, the probability of right tail with

Z= - 10/3

is 0.9995709, which represents the probability that the parachute will open at 100m or more.

Thus, by the multiplication rule, the probability that all five parachutes will ALL open at 100m or more is the product of the individual probabilities, i.e.

P(all five safe) = 0.9995709^5 = 0.9978565

So there is an approximately 1-0.9978565 = 0.214% probability that at least one of the five parachutes will open below 100m

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4 years ago

Angle measure for 225

Alex73 [517]

Answer:

3.92699 radians

Step-by-step explanation:

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3 years ago

Jamie purchased a DVD that was on sale for 15% off. The sales tax in her county is 5%. Let y represent the original price of the

Mamont248 [21]

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3 years ago

Which value would make the statement true? 2/? is greater than 4/12 ? A.4 B.6 C.12 D.24

Liula [17]

Answer:

A.4

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.

3 0

3 years ago