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Snowcat [4.5K]
3 years ago
6

A church rings its bells every 15 minutes, the school rings its bells every 20 minutes and the day care center rings its bells e

very 25 minutes. If they all ring their bells at noon on the same day, at what time will they next all ring their bells together
Mathematics
1 answer:
ankoles [38]3 years ago
7 0

Answer:

As few as just over 345 minutes (23×15) or as many as just under 375 minutes (25×15).

Imagine a simpler problem: the bell has rung just two times since Ms. Johnson went into her office. How long has Ms. Johnson been in her office? It could be almost as short as just 15 minutes (1×15), if Ms. Johnson went into her office just before the bell rang the first time, and the bell has just rung again for the second time.

Or it could be almost as long as 45 minutes (3×15), if Ms. Johnson went into her office just after the bells rang, and then 15 minutes later the bells rang for the first time, and then 15 minutes after that the bells rang for the second time, and now it’s been 15 minutes after that.

So if the bells have run two times since Ms. Johnson went into her office, she could have been there between 15 minutes and 45 minutes. The same logic applies to the case where the bells have rung 24 times—it could have been any duration between 345 and 375 minutes since the moment we started paying attention to the bells!

Step-by-step explanation:

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35/8 + 10/3 = simplest form
Nimfa-mama [501]

Answer:

185/24

Step-by-step explanation:

Convert both to fractions with the same denominator first.

35/8= 105/24

10/3= 80/24

105/24+80/24

= 185/24

6 0
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What is the y-coordinate of point T? Write a decimal coordinate.
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Where is the point t in question
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How many days are there from march 15 to september 15 inclusive?
Natasha2012 [34]

March 15 - March 31 = 17 days

April = 30days

May = 31 days

June = 30 days

July = 31 days

August = 31 days

Sept 1 - Sept 15 = 15 days


 total = 185 days

6 0
3 years ago
(a) Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the powe
Fofino [41]

Answer:

a) \mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!}  ...}

b)  See Below for proper explanation

Step-by-step explanation:

a) The objective here  is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.

The function is e^x + 3 \ cos \ x

The expansion is of  e^x is e^x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ...

The expansion of cos x is cos \ x = 1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...

Therefore; e^x + 3 \ cos \ x  = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...]

e^x + 3 \ cos \ x  = 4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} + \dfrac{x^3}{3!}+ ...

Thus, the first three terms of the above series are:

\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!}  ...}

b)

The series for e^x + 3 \ cos \ x is \sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!} +  3 \sum \limits^{\infty}_{x=0} ( -1 )^x  \dfrac{x^{2x}}{(2n)!}

let consider the series; \sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!}

|\frac{a_x+1}{a_x}| = | \frac{x^{n+1}}{(n+1)!} * \frac{n!}{x^x}| = |\frac{x}{(n+1)}| \to 0 \ as \ n \to \infty

Thus it converges for all value of x

Let also consider the series \sum \limits^{\infty}_{x=0}(-1)^x\dfrac{x^{2n}}{(2n)!}

It also converges for all values of x

7 0
3 years ago
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