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Alborosie
2 years ago
7

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

Mathematics
2 answers:
MatroZZZ [7]2 years ago
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)
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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
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2 years ago
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5 0
3 years ago
Solve the equation by dividing both sides by the appropriate number.<br><br><br> 5=18* N
mamaluj [8]

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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
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