All categories

2 years ago

Using a common denominator to write an equivalent fraction for 3/7?

2 answers:

5 0

I'm not sure exactly what you are trying to ask, but you can multiply the numerator and denominator by any number and it will stay the same value fraction, as long as you are multiplying them both by the same number.

Nutka1998 [239]2 years ago

4 0

6/14

Multiply both 3 and 7 by 2, this gives you equivalent fractions (there are more possible equivalent equations of course)

You might be interested in

Write a quadratic function f whose zeros are 4 and 3 .

Pani-rosa [81]

F(x) = (x-4)(x-3)

if you put x=4, it makes f(x) 0. if you put x=3 it makes f(x) 0

5 0

2 years ago

Plz help me on this question

zhuklara [117]

I cant see it please send another pic of it. closer to the screen.

5 0

3 years ago

Solve the equation by dividing both sides by the appropriate number.<br><br><br> 5=18* N

mamaluj [8]

Ben h484b8 jrh7br u3b37b u3v7 I Yea yebd7 yeve7 ueve7

6 0

3 years ago

Read 2 more answers

“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

$I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx$

Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

$I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx$

Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

$I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx$

Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

$\implies I=-\dfrac{(t+1)^4}{4t^4}+C$

4 0

3 years ago

What is 24% of 1350 I will give 75 points

nika2105 [10]

<h3>Answer:</h3><h2>324.</h2><h3>Step-by-step explanation:</h3>

To find 24% of 1,350, you must multiply 24% by 1,350.

To do this problem, we must turn the numbers to fractions.

Twenty-Four hundredths, 24 / 100 is your fraction for 24%.

Put 1,350 over 1 since 1,350 is a whole number.

So:

24 / 100 x 1,350 / 1.

1350 x 24 = 32400, 100 x 1 = 100.

Your answer is 32,400 / 100.

Get rid of the 2 zeroes in 400 and 100.

It should be 324 / 1.

Since 1 is just the number 1, the answer is 324.

If you have any questions, feel free to comment below.

Merry Christmas!

8 0

3 years ago

Read 2 more answers