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3 years ago

The sum of two numbers is 52. One number is 3 times as large as the other number. Which is the larger number

Mathematics

1 answer:

creativ13 [48]3 years ago

4 0

The two number are 39 and 13

Solution:

Let the two numbers be "a" and "b"

Let the larger number be "a" and the smaller number be "b"

Given that, sum of two numbers is 52

a + b = 52 ---------- eqn 1

One number is 3 times as large as the other number

Larger number = 3 times smaller number

a = 3b -------- eqn 2

Let us solve eqn 1and eqn 2

Substitute eqn 2 in eqn 1

3b + b = 52

4b = 52

b = 13

Substitute b = 13 in eqn 2

a = 3(13)

a = 39

Thus the two number are 39 and 13

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2 years ago

A mail carrier can deliver mail to 36 houses in 30 minutes. Mark wants to determine how many houses the carrier can deliver mail

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Answer:

Mark’s method is correct, because even though it is impossible to deliver mail to 1.2 houses in a minute, 1.2 represents the unit rate of houses per minute.

Step-by-step explanation:

Let's rule out the options given one by one.

Option 1: Mark’s method is wrong, because it is impossible to deliver mail to 1.2 houses in a minute. The carrier can only deliver to a whole number of houses.

This is not true, because 1.2 is a rate, it just means that the carrier can finish 1 house in a minute, and also have time to get started delivering to the second house. The total number of houses has to be a whole number, but the rate doesn't.

Option 2: Mark’s method is wrong, because it is impossible to deliver mail for 7.5 minutes. The carrier can only deliver mail for a whole number of minutes.

This option is also incorrect. Delivering mail for 7.5 minutes just means 7 minutes and 30 seconds, which is definitely possible.

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This option is correct. 1.2 is the rate of which the carrier delivers mail.

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4 0

3 years ago

Read 2 more answers

Miranda found out that there are 20 drops in 11 mL of medicine. She measured 15 drops of medicine into a measuring cup. How many

andreyandreev [35.5K]

Answer: Her measuring cup has 8.25 mL of medicine.

Step-by-step explanation:

As per given,

In 11 mL of medicine, there are 20 drops.

That means quantity in 1 drop = $\dfrac{11}{20}$ mL

Now , the quantity in 15 drops = 15 x quantity in 1 drop

= $15\times\dfrac{11}{20}$

$=\dfrac{33}{4}$

= 8.25 mL

Hence, her measuring cup has 8.25 mL of medicine.

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3 years ago

What is the quotient of Negative StartFraction 3 over 8 EndFraction and Negative one-third?

nadezda [96]

Answer:

Step-by-step explanation:

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3 years ago

How to find the limit

Ludmilka [50]

$\displaystyle\lim_{n\to\infty}\left(k!+\frac{(k+1)!}{1!}+\cdots+\frac{(k+n)!}{n!}\right)=\lim_{n\to\infty}\dfrac{\displaystyle\sum_{i=0}^n\frac{(k+i)!}{i!}}{n^{k+1}}=\lim_{n\to\infty}\frac{a_n}{b_n}$

By the Stolz-Cesaro theorem, this limit exists if

$\displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$

also exists, and the limits would be equal. The theorem requires that $b_n$ be strictly monotone and divergent, which is the case since $k\in\mathbb N$ .

You have

$a_{n+1}-a_n=\displaystyle\sum_{i=0}^{n+1}\frac{(k+i)!}{i!}-\sum_{i=0}^n\frac{(k+i)!}{i!}=\frac{(k+n+1)!}{(n+1)!}$

so we're left with computing

$\displaystyle\lim_{n\to\infty}\frac{(k+n+1)!}{(n+1)!\left((n+1)^{k+1}-n^{k+1}\right)}$

This can be done with the help of Stirling's approximation, which says that for large $n$ , $n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n$ . By this reasoning our limit is

$\displaystyle\lim_{n\to\infty}\frac{\sqrt{2\pi(k+n+1)}\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\sqrt{2\pi(n+1)}\left(\dfrac{n+1}e\right)^{n+1}\left((n+1)^{k+1}-n^{k+1}\right)}$

Let's examine this limit in parts. First,

$\dfrac{\sqrt{2\pi(k+n+1)}}{\sqrt{2\pi(n+1)}}=\sqrt{\dfrac{k+n+1}{n+1}}=\sqrt{1+\dfrac k{n+1}}$

As $n\to\infty$ , this term approaches 1.

Next,

$\dfrac{\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\left(\dfrac{n+1}e\right)^{n+1}}=(k+n+1)^k\left(\dfrac{k+n+1}{n+1}\right)^{n+1}=e^{-k}(k+n+1)^k\left(1+\dfrac k{n+1}\right)^{n+1}$

The term on the right approaches $e^k$ , cancelling the $e^{-k}$ . So we're left with

$\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}$

Expand the numerator and denominator, and just examine the first few leading terms and their coefficients.

$\displaystyle\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{n^{k+1}+(k+1)n^k+\cdots+1+n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}$

Divide through the numerator and denominator by $n^k$ :

$\dfrac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}=\dfrac{1+\cdots+\left(\frac{k+1}n\right)^k}{(k+1)+\cdots+\frac1{n^k}}$

So you can see that, by comparison, we have

$\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\lim_{n\to\infty}\frac1{k+1}=\frac1{k+1}$

so this is the value of the limit.

8 0

3 years ago