So set up an equation!

7* n/4 > 7

So trying to get the variable alone...

n/4 > (7/7) aka 1
n> 1*4
n> 4

Try it out!

7* 6/4 = 42/4 = 21/2= 10 and 1/2

It is greater than 7

7 0

3 years ago

Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.

Mama L [17]

Multiplying both sides by $y^2$ gives

$xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1$

so that substituting $v=y^3$ and hence $\frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx}$ gives the linear ODE,

$\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1$

Now multiply both sides by $3x^2$ to get

$x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2$

so that the left side condenses into the derivative of a product.

$\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2$

Integrate both sides, then solve for $v$ , then for $y$ :

$x^3v=\displaystyle\int3x^2\,\mathrm dx$

$x^3v=x^3+C$

$v=1+\dfrac C{x^3}$

$y^3=1+\dfrac C{x^3}$

$\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}$

6 0

2 years ago

the sum of 43.9 and a number is 49.65, as shown below l. what number should go in the box to complete the addition problem

torisob [31]

5.75 is the answer.
If you subtract 43.9 FROM 49.65, that's what you get.

7 0

2 years ago

What is the vertex of the quadratic function y=3x^2-12x+8

pashok25 [27]

Answer:

the vertex of 3x^2-12+8 is (2,-4)

4 0

2 years ago

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