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9966 [12]
3 years ago
7

Mr. Birdie has worked hard as a Math teacher all his

Mathematics
1 answer:
Nikitich [7]3 years ago
6 0

Answer:

You'll likely need assets worth 10 to 16 times your salary by the time you leave your job. A 45-year-old making $120,000 who hopes to retire at age 60, say, should already have nearly $700,000 set aside. (See the Retire Early calculator.) You can get by with less if you'll have other sources of income.

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The diameter of a circle is 10 ft find it’s circumference in terms of pi.
Vilka [71]

Answer:

10π ft.

Step-by-step explanation:

Circumference = diameter * π.

3 0
3 years ago
Please help!!
love history [14]

Subtract to find the weight she gained since birth.

32 - 8 = 24

Divide to see how many years passed.

24 / 4 = 6

Therefore, Darrel's baby sister is 6 years old.

Best of Luck!

7 0
3 years ago
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LeAnn wants to.giftwrap a present she got for her little brother. How many square inches of.giftwrap will be needed to cover a b
sergiy2304 [10]
In all you will need 48 square inches of gift wrap. 4*6= 24 and 24*2= 48
4 0
3 years ago
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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
What’s the slope of the line that passes through 5,10 and 7,12
Marrrta [24]

Slope of the line passes through (5,10) and (7,12) is 1.

Step-by-step explanation:

Given,

The two points are (5,10) and (7,12).

To find the slope passing through the given points.

Formula

The slope of the line passing through (x_{1} ,y_{1}) and (x_{2} ,y_{2}) is \frac{y_{2} -y_{1} }{x_{2} -x_{1} }

Now, putting x_{1} =5,y_{1}=10, x_{2}=7, y_{2} =12 we get,

Slope = \frac{12-10}{7-5} = \frac{2}{2} = 1

Hence,

Slope of the line passes through (5,10) and (7,12) is 1.

7 0
3 years ago
Read 2 more answers
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