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valentinak56 [21]
3 years ago
8

What is the maximum number of real distinct roots that a cubic equation can have?

Mathematics
2 answers:
Hoochie [10]3 years ago
5 0
3 but in most cases it is 2 real roots and one root has a multiplicity of 2
gogolik [260]3 years ago
4 0
<span>an degree polynomial can have as many as n real roots.</span>
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Im solving variables i need to simplify -5x + 3 (x+5) = -3
victus00 [196]

Answer:

x=9

Step-by-step explanation:

Add similar elements: -5x+3x=-2x

-2x+15=-3

Subtract 15 from both sides

-2x + 15-15=-3-15

Simplify

-2x=-18

Divide both sides by -2

-21/-2 = -18/-2

Simplify

x=9

8 0
3 years ago
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The profit on a jacket can be found by the formula P=24x-125, where x is the number of jackets sold. What is the profit if 15 ja
horrorfan [7]
Your answer would be 235
5 0
3 years ago
Which equation shows the correct use of the Power of Products Property?
Vadim26 [7]
The power of products property states that for number enclosed in a bracket or parenthesis, if it is raised to a power, it must be multiplied to the power of the enclosed number no matter how different the base is. You cannot add it because it is not raised. You can only add it if they have the same base. But in this problem, you will just multiply it. The breakdown of the solution to this problem is shown below. So,
<span><span>• (2x⁵y²)³=(21x3x5*3y2*3) = 6x15y6</span><span>

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7 0
3 years ago
Pls help me I’m failing lolz:/
Marta_Voda [28]

Answer:

6= 4k/9

Step-by-step explanation:

RLLY HOPE THIS HELPS QUICK ANSWER TRYING Q.8

8 0
3 years ago
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La barbería El Caleño, tiene en promedio 120 clientes a la semana a
Luba_88 [7]

Queremos maximizar el precio de tal forma que los ingresos no disminuyan.

Ese maximo precio es: $14,040.6

Sabemos que actualmente el precio es:

p = $6,000

El número de clientes es:

C = 120

Actualmente los ingresos son el producto de esos dos números, es decir:

ingresos = $6,000*120 = $720,000

Ahora sabemos que por cada incremento de $700 en el precio, el número de clientes decrece en 10.

Entonces podemos escribir el número de clientes como una ecuación lineal.

C(p) = a*p + b

tal que tenemos dos puntos en esa linea:

($6,000, 120)

($6,700, 110)

La pendiente es:

a = \frac{110 - 120}{\$6,700 - \$6,000} = \frac{-10}{\$ 700}

Entonces tenemos:

C(p) = (-10/$700)*p + b

Sabemos que:

C($6,000) = 120 = (-10/$700)*$6,000 + b

                     120 = -85.71 + b

                     120 + 85.71 = b =

Entonces la ecuación lineal es:

C(p) = (-10/$700)*p + 205.71

Los ingresos serán dados por:

ingresos = C(p)*p = (-10/$700)*p^2 + 205.71*p

Y queremos maximizar p de tal forma que esto sea igual a lo que obtuvimos antes:

(-10/$700)*p^2 + 205.71*p = $720,000

Entonces debemos resolver la ecuación cuadratica:

(-10/$700)*p^2 + 205.71*p - $720,000 = 0.

Las soluciones son dadas por la formula de Bhaskara.

p = \frac{-205.71 \pm \sqrt{(205.71)^2 - 4*(-10/\$ 700)*\$ 720,000} }{2*(-10/\$ 700)} \\\\p = \frac{-205.71 \pm 195.45}{(-20/\$ 700)}

La solución de maximo valor es:

p = (-205.71 - 195.45)/(-20/$700) = $14,040.6

Sí quieres aprender más, puedes leer.

brainly.com/question/8926135

7 0
3 years ago
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