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3 years ago

5

you are lying 120 ft away from a tree that is 50 feet tall you look up at the top of the tree approxmaelty how far is your head

from the top of the tree in a straight line

1 answer:

Pepsi [2]3 years ago

5 0

Check the picture below.

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Number 8 please, I don't understand it

Nuetrik [128]

Lovin lemons sells 4 gallons for 24$ while sour co sells 4 gallons for 32 $

loving lemons is cheaper and ignore my grammer

7 0

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.

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

A 12 inch candle and an 18 inch Candle are lit at 6 p.m. The 12 inch candle burns 0.5 inches every hour. The 18 inch candle burn

Stella [2.4K]

18 - 2H = 12 - .5H
Combine the variables
18 - 2H + .5H = 12 - .5H + .5H
18 - 1.5H = 12
Combine the whole numbers
18 - 18 - 1.5H = 12 - 18
-1.5H = -6
Divide by -1.5
-1.5H/-1.5 = -6 / -1.5
H = 4
The answer is 4 hours

7 0

3 years ago

H<br> nlaced inside a cylinder as shown. The radius of the cone is half the radius of the cylinder.

Sophie [7]

Answer:

umm what or we doing?

Step-by-step explanation:

5 0

3 years ago

Carrie has 300 marbles 25 of the marbles are green 15% of the marbles are red and the rest of the marbles are blue how many blue

Rus_ich [418]

Answer:

230.

Step-by-step explanation:

$\: \: \: \: 300 - 25 - \frac{15}{100} \times 300 = \\ = 275 - 45 = \\ = 230$

Hope this helps!

4 0

3 years ago

Read 2 more answers