All categories

2 years ago

Which two fractions are equivelent to 1 over 4

Mathematics

1 answer:

Serhud [2]2 years ago

8 0

Answer:

2/8 and 3/12

Step-by-step explanation:

You multiply both the numerator and denominator by the same number to get equivalent fractions.

1 x 2

4 x 2

= 2/8

and

1 x 3

4 x 3

= 3/12

You might be interested in

Please answer the question in the picture please its urgent 200 pts plus brainly!!!!

OverLord2011 [107]

Answer:

The answers B.

Step-by-step explanation:

6 0

2 years ago

use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r

Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt$

If we ask that $f$ has continuous $(n+1)$ th derivative we can apply the mean value theorem for integrals. Then, there exists $c$ between $a$ and $x$ such that

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x$

Hence,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .$

Thus,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

and the Taylor theorem with Lagrange remainder is

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ .

Step-by-step explanation:

5 0

2 years ago

Write the roman numeral of the Olympics Games that follow the 28th games

bogdanovich [222]

The Olympics Games that follow the 28th games will be "XXIX Games."

The Olympic Games that follow the 28th Olympic games is the 29th Olympic Games.

29 in Roman numerals is:

= XXVIII

An X in Roman numerals is 10 so the first two Xs are to signify "20."

An "IX" in Roman numerals is "9" which will then take the number to 29.

It will therefore be the "XXIX Olympic Games."

In conclusion, the number 29 is XXIX in Roman numerals so Olympic Games after the 28th games will be "XXIX Games."

<em>Find out more at brainly.com/question/11381536.</em>

7 0

1 year ago

The U.S. RDA for protein for a girl is 46 g. If a piece of friedchicken contains 23 g, how many pieces would the girl have to ea

Nataliya [291]

The U.S RDA for protein is 46g for girl

If a piece of fried chicken contains 23g

To find the pieces of chicken she is to eat to satisfy the U.S RDA requirement,

The number of pieces of chicken is

$\text{Number of pieces}=\frac{U.S.\text{ RDA g protein required for a girl}}{g\text{ of protein in a piece of chicken}}$

The number of pieces of chicken is

$\begin{gathered} \text{Number of pieces of chicken}=\frac{46}{23}=2\text{ pieces of chicken} \\ \text{Number of pieces of chicken}=2\text{ pieces of chicken} \end{gathered}$

Hence, the right option is B

6 0

9 months ago

Solve y + (-18) = -12

natali 33 [55]

Answer:

Y = 6 I’m pretty sure if its not 6 it’s -6

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers