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Natasha2012 [34]

2 years ago

9

Water leaks through a roof and collects in a bucket the table shows how many cups of water collect in the bucket for different n

umbers of hours

1 answer:

Crank2 years ago

7 0

Answer: 1/5 pie = y

Step-by-step explanation:

You might be interested in

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

where $m_i$ is the midpoint of the $i$ -th subinterval,

$m_i=\dfrac{\ell_i+r_i}2$

Then the integral $I$ is equal to the sum of the integrals of each interpolation over the corresponding $i$ -th subinterval.

$I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt$

It's easy to show that

$\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)$

so that the value of the overall integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}$

4 0

3 years ago

How many different ways can 5 students line up in a cafteria

lina2011 [118]

Ans: 120 different ways

7 0

2 years ago

Solve the following inequality for x. -5x<-15x+40

melisa1 [442]

Answer:

x is less than 4

or

x is greater than 4

Step-by-step explanation:

-5x < -15x + 40 -5x < -15x + 40

+15x +15x -40 -40

10x < 40 -40 - 5x < -15x

÷10 ÷10 +5x +5x

× < 4 -40 < -10x

÷10 ÷10

4 < ×

5 0

3 years ago

Find all critical points of the given plane autonomous system. (Enter your answers as a comma-separated list.) x′ = x 12 − x − 1

Andru [333]

Answer:

the critical points are (0,0) , (0, 20), (12, 0) , (4,16)

Step-by-step explanation:

To consider the autonomous system

$x' =x (12 -x - \dfrac{1}{2})$

$y' = y( 20 -y - x)$

The critical points of the above system can be derived by replacing x' = o and y' = 0.

i.e.

$x' =x (12 -x - \dfrac{y}{2}) = 0$

$\dfrac{x}{2} (24 -2x - y) = 0$

x = 0 or 24 - 2x - y = 0 ----- (1)

Also

$y' = y( 20-y-x) = 0$

y( 20 -y - x) = 0

y = 0 or 20 - y - x = 0 ----- (2)

By solving (1) and (2);

we get x = 4 and y = 16

Suppose x = 0 from (2)

y = 20

Also;

if y = 0 from (1)

x = 12

Thus, the critical points are (0,0) , (0, 20), (12, 0) , (4,16)

6 0

2 years ago

−9x+y= \,\,71 71 -x-y= −x−y= \,\,-1 −1 \mathbf{Add}\text{ to eliminate }\mathbf{y.}Add to eliminate y. \mathbf{Subtract}\text{ t

Rama09 [41]

Answer:

<h3>Add to eliminate y.</h3>

Step-by-step explanation:

Given the equations

-9x + y = 71.... 1

-x - y = -1 .... 2

Using elimination method we can eliminate y first since they have the same coefficient.

To do that we will need to add up both equations. We are adding because the coefficient of y in both equations has different signs. If they have the same sign, we would have subtracted.

<em>Hence the correct option in this case is to add to eliminate y.</em>

8 0

2 years ago