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fredd [130]
3 years ago
13

Cuál es el problema actual que están viviendo los emprendedores?

Mathematics
1 answer:
GarryVolchara [31]3 years ago
7 0

Answer:

Los emprendedores se enfrentan a muchos desafíos en el mundo empresarial ultracompetitivo de hoy. Afortunadamente, los emprendedores también tienen más recursos que nunca para abordar esos problemas.

Hoy en día, muchos emprendedores enfrentan los siguientes 10 desafíos. Quizás ya te hayas enfrentado a algunos de ellos. Siga leyendo para saber por qué existe cada desafío y para obtener soluciones y soluciones para que pueda operar su negocio de manera eficiente y exitosa.

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Help on thishwwhejjddjhdhddhdhdj
QveST [7]
The answer is D.
The first one says you add -4 each time in the sequence.
The second says you add -2 each time in the sequence.
The third says you multiply by -2 each time in the sequence.
The fourth says you multiply by -4 each time in the sequence.
-2*-4=8
8*-4=-32
5 0
3 years ago
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In ΔJKL, j = 93 cm, k = 82 cm and l=86 cm. Find the measure of ∠J to the nearest degree.
zhuklara [117]

Answer:

5

Step-by-step explanation:

Answer:

85t

Step-by-step explanation:

1. Volume of a cone: 1. V = (1/3)πr2h 2. Slant height of a cone: 1. s = √(r2 + h2) 3. Lateral surface area of a cone: 1. L = πrs = πr√(r2 + h2) 4. Base surface area of a cone (a circle): 1. B = πr2 5. Total surface area of a cone: 1. A = L + B = πrs + πr2 = πr(s + r) = πr(r + √(r2 + h2))

hope this helps

8 0
3 years ago
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What is the probability of rolling a number greater than or equal to 9 with the sum of two dice, given that at least one of the
olga55 [171]
Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
6 0
3 years ago
Joe has 5 dimes and 4 pennies. Jamal has 2 dollars, 4 dimes, and 5 pennies. Jimmy has 6 dollars and 4 dimes. They want to put th
sesenic [268]

Answer:

no they had $9.39

Step-by-step explanation:

i just added them all together

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3 years ago
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