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3 years ago

Please need help lots of math

Mathematics

1 answer:

Darya [45]3 years ago

4 0

I will help you do your math

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What is the domain and range of the function y = 1/x-1<br> show solutions pls

katovenus [111]

Answer:

(1,0)

D: 1

R: 0

Step-by-step explanation:

y=1/x-1

y=1/1-1

y=1-1

y=0

8 0

3 years ago

Rapunzel started counting flowers in a meadow. 5 of the first 15 were yellow. If this trend continues and she counts 6 more flow

kotegsom [21]

Answer:

Step-by-step explanation:

5 of 15. is 1/3

1/3 of 6 is equal to 2

6 0

3 years ago

Read 2 more answers

Cynthia is planning a party. for entertainment, she has designed a game that involves spinning two spinners. if the sum of the n

vesna_86 [32]

<span>The correct statement of the problem is attached

</span><span>Spinner A: 4 parts, 3/4 is 1, 1/4 is 3.----------1 1 1 3
Spinner B: 3 parts, 1/3 is 5, 1/3 is 7, 1/3 is 9--------5 7 9

</span><span>The combinations of the two spinners are
1 and 5
1 and 7
1 and 9
3 and 5
3 and 7
3 and 9

</span>cases get you thrown in the pool
1 and 5
1 and 7
3 and 5

calculation of the probabilities
1 and 5--------3/4*1/3 = 1/4
1 and 7--------3/4*1/3 = 1/4
3 and 5--------1/4*1/3 = 1/12
1/4+1/4+1/12=7/12-----------58.33%

probabilities that the guests will be thrown in the pool--------58.33%

5 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

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