All categories

3 years ago

The difference of two numbers is 13. The sum of two numbers is 75

Mathematics

1 answer:

NARA [144]3 years ago

8 0

Answer:

The numbers are 31 and 44.

Step-by-step explanation:

Given that the difference of two numbers is 13, and the sum of two numbers is 75, to determine what these numbers are, the following calculations must be performed:

(75 - 13) / 2 = X

62/2 = X

31 = X

31 + (31 + 13) = X

31 + 44 = X

75 = X

Therefore, the numbers are 31 and 44.

You might be interested in

Idk how to do this. Please help?

Soloha48 [4]

Answer:

Is that algebra??

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

I really need help Please thank you

natulia [17]

Answer:

Step-by-step explanation:

the median is 85, the mean is 85, the mode is 80.

it can't be A, B, or C. so it must be D.

4 0

3 years ago

Solve for the value of r.

Katyanochek1 [597]

Answer:

r=14

Step-by-step explanation:

7r+6=8r-8

14=r

r=14

7 0

3 years ago

Read 2 more answers

Dan earns £332 over the course of a five-day week. How much is that per day?

puteri [66]

Answer:

£66.4

Step-by-step explanation:

£332/5

6 0

3 years ago

A leprechaun places a magic penny under a girls pillow. The next night there are 2 magic pennies under her pillow. The following

kiruha [24]

Answer:

After 47 days she will have more than 90 trillion pennies.

Step-by-step explanation:

At the beginning there was 1 penny. At the second day the amount of pennies under the pillow became 2.

The amount of pennies doubled each day. So the series is,

$1,2,4,8,16,32,.....$

This series is in geometric progression.

As the pennies from each of the previous days are not being stored away until more pennies magically appear so the sum of series will be,

$S_n=\dfrac{a(r^n-1)}{r-1}$

where,

a = initial term = 1,

r = common ratio = 2,

As we have find the number of days that would elapse before she has a total of more than 90 trillion, so

$\Rightarrow 90\times 10^{12}\le \dfrac{1(2^n-1)}{2-1}$

$\Rightarrow 90\times 10^{12}\le \dfrac{2^n-1}{1}$

$\Rightarrow 90\times 10^{12}\le 2^n-1$

$\Rightarrow 2^n\ge 90\times 10^{12}+1$

$\Rightarrow \log 2^n\ge \log (90\times 10^{12}+1)$

$\Rightarrow n\times \log 2\ge \log (90\times 10^{12}+1)$

$\Rightarrow n \ge \dfrac{\log (90\times 10^{12}+1)}{\log 2}$

$\Rightarrow n \ge 46.4$

$\Rightarrow n\approx 47$

8 0

3 years ago