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kykrilka [37]
3 years ago
14

Solve by using the quadratic formula. 3psquared+7p+2=0

Mathematics
2 answers:
Rainbow [258]3 years ago
8 0

Answer:

\large\boxed{x=-2\ or\ x=-\dfrac{1}{3}}

Step-by-step explanation:

The quadratic formula of a quadratic equation

ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

We have the equation:

3p^2+7p+2=0\to a=3,\ b=7,\ c=2

Substitute:

b^2-4ac=7^2-4(3)(2)=49-24=25\\\\\sqrt{b^2-4ac}=\sqrt{25}=5\\\\x_1=\dfrac{-7-5}{2(3)}=\dfrac{-12}{6}=-2\\\\x_2=\dfrac{-7+5}{2(3)}=\dfrac{-2}{6}=-\dfrac{1}{3}

Serhud [2]3 years ago
6 0

Answer:

-1/3,-2

Step-by-step explanation:

The quadratic formula is [-b±√(b^2-4ac)]/2a.  In this equations 3p^2 is a, 7p is b, and 2 is c.

Then just plug in the numbers.

[-7±√((-7^2)-4(3)(2)]/2(3)

[-7±√(25)]/6

(-7+5)/6  and  (-7-5)/6

-1/3 and -2 are the answers, if you plug these numbers into the original equation, you find that they equal 0 which means that they work.

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(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul
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Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

Integrating, we have:

=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

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F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

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Answer:

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Step-by-step explanation:

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f(3) = -3

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