44% of what number is 25?

Mathematics

2 answers:

Sunny_sXe [5.5K]2 years ago

4 0

Hi

x is 100%
25 is 44%

x/100 = 25/44
44x = 25·(100)
44x = 2500
x = 2500/44
x = 56.82

44% of 56.82 is 25

Oduvanchick [21]2 years ago

4 0

We set up a proportion: 25/?= 44/100

Cross multiply: 44*?= 25*100
⇒ ?= 25*100/44= 56 9/11

44% of 56 9/11 is 25.

You might be interested in

Multiply and simplify<br><br><img src="https://tex.z-dn.net/?f=35%20%5Ctimes%20%20%5Cfrac%7B7%7D%7B10%7D%20" id="TexFormula1" ti

Kobotan [32]

We can convert 7/10 to a decimal first.

7/10 = 0.7

Now we can multiply this problem easier.

35 x 0.7 = 24.5

6 0

2 years ago

Two boats start their journey from the same point A and travel along directions AC and AD, as shown below

adell [148]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Last year, Shen opened an investment account with $7800. At the end of the year, the amount in the account had decreased by 22.5

maks197457 [2]

Answer:

The decrease was $1755.

At the end of the year, he had $6045 in his account.

Step-by-step explanation:

Start with $7800.

Decrease of 22.5% of $7800.

22.5% of $7800 =

= $1755

The decrease was $1755.

$7800 - $1755 = $6045

At the end of the year, he had $6045 in his account.

4 0

2 years ago

Can anyone explain how to do mathematical induction. The question is below. Brainliest awarded for the right answer. Thanks.

igomit [66]

The sum of the triangular numbers is $\frac{n(n+1)(n+2)}{6}$

According to the question, the series of triangular numbers is given as in the form of the number of dots constituting the equilateral triangles i.e.
1, 3, 6, 10, 15 . . . . .n

<h3>what is triangular numbers?</h3>

triangular number are the sequence and series of the numbers and each number represents and constitute in visualization of series of equilateral triangle.

given series in the figure
1, 3, 6, 10, 15, . . . . . . ., n
each number represents the number of dots containing in triangle
now the series can be given by
$S = \frac{n(n+1)}{2}$ for n = 1, 2, 3, 4,. . . . .n
now , according to the question the sum of the series can be given as
⇒ ∑S
⇒ ∑ $\frac{n^2+n}{2}$
⇒ $\frac{1}{2}[\sum n^2+\sum n]\\\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}] \\\frac{n(n+1)}{4}[\frac{(2n+1)}{3}+1] \\\frac{n(n+1)}{4}[\frac{(2n+1+3)}{3}] \\\frac{n(n+1)}{4}[\frac{(2n+4)}{3}] \\\frac{n(n+1)}{4}2[\frac{(n+2)}{3}]\\\frac{n(n+1)(n+2)}{6}$