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7nadin3 [17]
3 years ago
9

A husband and wife celebrate their birthdays on the same day of the month, but in different months. every year, their birthdays

fall on the same day of the week. if the husband's birthday is in March and the wife's birthday is later in the year, find the month of the wife's birthday
Mathematics
2 answers:
irina [24]3 years ago
8 0

Answer:

My B-day is on March!

Step-by-step explanation:

pychu [463]3 years ago
3 0
Husband is in march
wife is after march
we have
b-days are same day
ex march 3 and june 3
also, they fall on the same day of the week
so continuing ex. march 3 wed and june 3 wed

so we just look at the calllendar for days that fall on the same day of the week and have the same nuber and are after march
basically, the easy way to find it is to look at the months and see which months that are after march start on the same day
ex. this year march 1 is on a tuesday
so find other months that start on tuesday
answer is September and December

so the Wife's birthday could be in September or December
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Kipish [7]

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44°

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7 0
3 years ago
Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ
SOVA2 [1]

y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}

The arc length of the curve is

\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx

which has a value of about 5.99086.

Let f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}. Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

\ell_i=\dfrac{i-1}2

r_i=\dfrac i2

with 1\le i\le10.

These subintervals have midpoints given by

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

Over each subinterval, we approximate f(x) with the quadratic polynomial

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\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 the integral we want to find can be estimated as

\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that

\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6

so that the arc length is approximately

\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086

5 0
3 years ago
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