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frez [133]
3 years ago
12

What is the sum of two consecutive numbers is 141

Mathematics
1 answer:
dmitriy555 [2]3 years ago
6 0
Two consecutive numbers can be written as:
x , x+1

So their sum being 141
x + x +1 = 141

Combine and solve for c

2x = 140
x = 70

So two numbers are:

70,71
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An automotive part manufacturer regularly makes shipments of 400 parts to merks auto repair. On average, 14 of the 400 parts are
Radda [10]

Answer:

no matter how many parts are in the shipment, 3.5% of the parts will be defective.

hope this helped, if so hit that thanks button ;D

Step-by-step explanation:


4 0
3 years ago
Given ƒ(x) = 9x + 54, find x when ƒ(x) = 18. <br> A)−6 B)−4 C)3 D)8
Katen [24]

Answer:

B. -4

Step-by-step explanation:

We want to find x such that f(x)=18.

We need to therefore solve 9x+54=18 forx.

9x+54=18

Divide both sides by 9:

(9x+54)/9=18/9

9x/9+54/9=18/9

Simplify:

x+6=2

Subtract 6 on both sides:

x=2-6

Simplify

x=-4.

So f(-4) should have value 18.

Let's check.

f(x)=9x+54

f(-4)=9(-4)+54

f(-4)=-36+54

f(-4)=18

So the answer is B.

7 0
4 years ago
Read 2 more answers
I really need help!!!
Licemer1 [7]

Answer:

3

Step-by-step explanation

cause there are three rooms that need to be painted

8 0
3 years ago
Read 2 more answers
Who could help me please:)))I'll apreciate the answers:))))
Wittaler [7]
The answer in order is    b   a     c 

6 0
4 years ago
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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