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AlekseyPX
3 years ago
9

There is no smallest positive rational number because, if there were, then it could be divided by two to get a smaller one. EXPL

AINED
Mathematics
1 answer:
Gennadij [26K]3 years ago
3 0

Answer:

Since a/2⁽ⁿ ⁺ ¹⁾b <  a/2ⁿb,  we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.

Step-by-step explanation:

Let a/b be the rational number in its simplest form. If we divide a/b by 2, we get another rational number a/2b. a/2b < a/b. If we divide a/2b we have a/2b ÷ 2 = a/4b = a/2²b. So, for a given rational number a/b divided by 2, n times, we have our new number c = a/2ⁿb where n ≥ 1

Since \lim_{n \to \infty} \frac{a}{2^{n}b } = a/(2^∞)b = a/b × 1/∞ = a/b × 0 = 0, the sequence converges.

Now for each successive division by 2, a/2⁽ⁿ ⁺ ¹⁾b <  a/2ⁿb and

a/2⁽ⁿ ⁺ ¹⁾b/a/2ⁿb = 1/2, so the next number is always half the previous number.

So, we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.

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P/3 - 2 provide an algebraic expression for p divided by 3 plus 2
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\frac{p}{3}+2
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Joanne's Dress Shop received an invoice dated July 25 for $1,400, with terms of 2/10, 1/15, n/60. On August 8, Joanne's Dress Sh
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The actual amount that should be credited is $757.58.

<u>Solution:</u>

Given that,

2% Discount if paid in 10 days.

1% Discount if paid in 15 days, (but greater that 10 days)

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That is,

\bold{1400\times0.98 = 1372} will close the acount on days 1-10

\bold{1400\times0.99 = 1386} will close the acount on days 11-15

And August 8 is Day 15 and falls under the 1% discount rule.

Therefore, we have to divide the partial payment by the complement of the discount rate.

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5 0
4 years ago
Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and t
WINSTONCH [101]

Answer:

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.  

How long will it take for this population to grow to a hundred rodents? To a thousand rodents?

Step-by-step explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}

Seperate the differential equation and solve for the constant C.

\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2}  }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}

You have 100 rodents when:

100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months

You have 1000 rodents when:

1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months

7 0
3 years ago
Pls help ASAP!! Urgent
Mama L [17]

A= 1/2Bh

A= 1/2 (10.2 x 8)

A= 1/2 (81.6)

A=81.6 / 2

A= 40.8


5 0
3 years ago
What’s the correct answer ???
Svetlanka [38]
The correct answer is the second one
7 0
3 years ago
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