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MAXImum [283]
3 years ago
8

Simplify the expression

Mathematics
2 answers:
Dafna11 [192]3 years ago
7 0
\bf \left( \cfrac{2}{5m^5} \right)^{-4}\implies \left( \cfrac{5m^5}{2} \right)^{4}\implies \cfrac{5^4m^{5\cdot 4}}{2^4}\implies \cfrac{625m^{20}}{16}
BaLLatris [955]3 years ago
6 0
The answer is 625m^20/16
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Find the greatest common factor and the least common multiple of 16 and 20. The prime factorizations of each number are given.
kykrilka [37]

Answer:

GCF 4, LCM 80

Step-by-step explanation:

Since both 16 and 20 have two 2s as factor, their greatest common factor is 4.

The LCM is found by multiplying all of the remaining factors by the LCM:

16 still has (2 x 2), 20 still has (5), times the GCF (4)  so 2 x 2 x 5 x 4 = 80.

5 0
2 years ago
Ram has 35 gel pens.. please help this is for kahn academy
disa [49]

Answer:

After the 35 gel pens??

Step-by-step explanation:

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3 years ago
PLS HELP ASAP!!!<br> domain &amp; range of g(x)?
Morgarella [4.7K]
Domain: (-infinity, +infinity)
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6 0
3 years ago
Help plzz (30) points for both of the question-plz help - this is life or dead answer
Elodia [21]

Answer:

The answer is B) 11.

Step-by-step explanation:

We can solve this by making an expression to represent the situation. Let x = the number of student books in the box (what we're looking for):

2.3x + 4.3 + 0.4 = 30

2.3x + 4.7 = 30

2.3x + 4.7 - 4.7 = 30 - 4.7

2.3x = 25.3

2.3x/2.3 = 25.3/2.3

x = 11

8 0
3 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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