Verdadero o falso a) al reducir a comun denominador las fracciones se sustituyen por otras equivalentes

Mathematics

1 answer:

alexgriva [62]3 years ago

3 0

Answer:

Verdadero

Step-by-step explanation:

Verdadero, cuando cambias dos fracciones para darles un denominador comun la fraccion entera se sustituye para otra que tiene el mismo valor numerico que la original. Por ejemplo tenemos una fraccion 2/6 en la cual tenemos que cambiarle el denominador a uno comun con otra fraccion que es 18. Entonces sustituyemos la fraccion por la siguente fraccion 6/18. Esta nueva fraccion es equivalente a la original y los dos tienen el mismo valor numerico...

2 / 6 = 0.333

6 / 18 = 0.333

You might be interested in

What are partial products

skad [1K]

The product of one term of a multiplicand and one term of its multiplier

7 0

3 years ago

Read 2 more answers

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$