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3 years ago

Solve the equation -9v-9=-27 for v.

Mathematics

2 answers:

Nikolay [14]3 years ago

4 0

Answer:

v=2

Step-by-step explanation:

-9v-9=-27

add nine to both sides

-9v-9+9=-27+9

simplify

-9v=-18

divide both sides by -9

-9v/-9 -18/-9

simplify y=2

hope this helps :) have a nice day !!

Vadim26 [7]3 years ago

3 0

Answer:

v=2

Step-by-step explanation:

-9v-9=-27

add nine to both sides

-9v-9+9=-27+9

simplify

-9v=-18

divide both sides by -9

-9v/-9 -18/-9

simplify y=2

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What is the solution to the inequality below x^2>81

dem82 [27]

Answer:

x>9

Step-by-step explanation:

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4 years ago

If A and B are independent events, P(A) = 0.25, and P(B) = 0.3, what is P(AB)?

Marrrta [24]

Answer:

$P(A) = 0.25, P(B= 0.3$

And if we want to find $P(A \cap B)$ we can use this formula from the definition of independent events :

$P(A \cap B) =P(A) *P(B) = 0.25*0.3= 0.075$

And the best option would be:

$P(A \cap B) =0.075$

Step-by-step explanation:

For this case we have the following events A and B and we also have the probabilities for each one given:

$P(A) = 0.25, P(B= 0.3$

And if we want to find $P(A \cap B)$ we can use this formula from the definition of independent events :

$P(A \cap B) =P(A) *P(B) = 0.25*0.3= 0.075$

And the best option would be:

$P(A \cap B) =0.075$

3 0

3 years ago

Alice wants to fence in a rectangular pay area for her rabbits. The length of the area should be at least 20ft, and the distance

dlinn [17]

This looks like common core which is never straight forward.It seems as though they are asking for the perimeter and not area,since they used feet for the distance around instead of sq.ft.

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3 years ago

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.