Simplify. 4x2+2xx2+x−28x2+4x3x2+10x+8

Mathematics

1 answer:

Tatiana [17]3 years ago

3 0

What’s the answer? I can’t divide anything if it’s just x. Example: 12x+24x+30 = 467. Then you can just divide the numbers after they are added up to see what x equals. 30 and 24x + 12x won’t go together because they have x, 30 doesn’t. Move your x’s to Texas.

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What type of triangle has an exterior angle that is obtuse at two vertices? Explain your reasoning.

navik [9.2K]

Answer:

A right triangle

Step-by-step explanation:

A right triangle would meet these characteristics. A triangle's inner angles always add up to be 180 degrees. A right triangle has one angle at 90 degrees meaning the other two angles need to be less than 90 degrees and sum up to be 90 degrees. This would indicate that both of these angles' exterior angles would be obtuse because they would be wider than 90 degrees.

5 0

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.