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2 years ago

Drag each number to show whether or not it is a solution to the inequality shown.

Mathematics

1 answer:

Pani-rosa [81]2 years ago

5 0

Answer:

The numbers that work are: -15, 5, 3.5, and 4 3/4

Step-by-step explanation:

Subtract 5 from both sides of the equation. Which will give you 15≥3x. Then you divide both sides by 3 which will give you 5≥x.

20≥5+3x

-5 -5

15≥ 3x

÷ 3 ÷ 3

5 ≥ x

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garri49 [273]

Answer:

(-2, 1)

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2 years ago

Find the solution of the initial value problem dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

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$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

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$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

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2 years ago

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iogann1982 [59]

Answer:

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hopefully this helps!

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