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3 years ago

What are the values of a, b, and c in the quadratic equation –2x2 + 4x – 3 = 0?

Mathematics

2 answers:

Marina86 [1]3 years ago

7 0

Answer:

a=-2

b=4

c=-3

Step-by-step explanation:

You just compare -2x^2+4x-3=0 to

ax^2+bx+c=0

There is really no work here.

Korvikt [17]3 years ago

3 0

For this case we have that by definition, a quadratic equation is of the form:

$ax ^ 2 + bx + c = 0$

The roots are given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

We have the following equation:

$-2x ^ 2 + 4x-3 = 0$

So:

$a = -2\\b = 4\\c = -3$

Answer:

$a = -2\\b = 4\\c = -3$

You might be interested in

(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

Aloiza [94]

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:

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

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

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

3 0

3 years ago

Please help i really need help

lukranit [14]

9514 1404 393

Answer:

17/99

Step-by-step explanation:

Replace the digits 23 in your example with the digits 17 and you have your answer:

$0.\overline{17}=\dfrac{17}{99}$

_____

In general, a 2-digit repeat will have 99 as its denominator. If the digits are a multiple of 3 or 11, then the fraction can be reduced. 17 is prime, so the fraction cannot be reduced.

3 0

3 years ago

PLEASE HELP ITS OVERDUE PLEASE I NEED HELP!!

IgorLugansk [536]

Answer:

First question: Choice A

second question:Choice D

Third question: Choice c

Fourth question: Choise b

5th Question: choice b

5 0

3 years ago