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Kitty [74]
3 years ago
12

Write the equation the line in slope-intercept form. Slope is 3, and (1, 5) is on the line. m = ? To find b (the y-intercept), s

ubstitute ? in for m, ? in for x and ? in for y into y=mx+ b. b = ? The equation should be y= ? x + ?
Mathematics
1 answer:
Katen [24]3 years ago
4 0

Answer:

The value of b is 2  and The equation of line is y = 3 x + 2 .  

Step-by-step explanation:

Given as :

The slope of line is 3

The points through which line passes is ( x , y ) = ( 1 , 5 )

Now,

The equation of line in slope-intercept form

y = m x + c

Where x and y are the points on axis

m is the slope of line

c is the constant term

∵ Line y = m x + b , passes through point ( 1 , 5 ) having slope m = 3

So ,  5 = 3 × 1 + b

Or,   5 =  3 + b

∴  b = 5 - 3

I.e b = 2

So, The equation can be written as

y = 3 x + 2

Hence The value of b is 2  and The equation of line is y = 3 x + 2 .  Answer

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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

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