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Alinara
5 days ago
8

These tables of values represent continuous functions. For which function will the y-values be the greatest for very large value

s of x?
PlzzPlease help me bunnies <3
No monkey business plz. Or I will report

Mathematics
1 answer:
zzz [9K]5 days ago
7 0

It's option C.

The value escalates threefold compared to the last step.

All other function values increase by a constant amount, resulting in linear behavior

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f Khan had an army of 10,000 soldiers at the lowest level, how many men in total were under him in his organization? (b) If Khan
babunello [8423]

Cette question est incomplète, la question complète est;

Gengis Khan organisait ses hommes en groupes de 10 soldats dirigés par un « leader de 10 ». Dix « leaders de 10 » étaient sous un « leader de 100 ». Dix « leaders de 100 » étaient sous un « leader de 1 000 ».

(a) Si Khan avait une armée de 10 000 soldats au niveau le plus bas, combien d'hommes en tout étaient sous son commandement dans son organisation?

(b) Si Khan avait une armée de 5 763 soldats au niveau le plus bas, combien d'hommes en tout étaient sous son commandement dans son organisation?

On suppose que les groupes de 10 doivent contenir 10 si possible, mais qu'un groupe à chaque niveau peut avoir besoin de témoigner moins

Réponse:

a) = 11110 soldats

b = 6404 soldats

Explication étape par étape:

Étant donné que Khan a disposé ses hommes en groupes de 10 sous un « leader de 10 », dix « leaders de 10 » étaient supervisés par un « leader de 100 », et dix « leaders de 100 » étaient dirigés par un « leader de 1000 »

alors

a) Si Khan avait une armée de 10 000 soldats au niveau le plus bas, combien d'hommes au total étaient sous son organisation?

MAINTENANT

Nombre de soldats = 10 000

Rang le plus bas (soldat) = 10000

troisième plus haut (leader de 10) = 10000/10 = 1000

deuxième le plus élevé (leader de 100) = 1000/10 = 100

Le plus haut (Leader de 1000) = 100/10 = 10

donc le total = 10000 + 1000 + 100 + 10 = 11110 soldats

b) Si Khan avait une armée de 5 763 soldats au rang le plus bas, combien d'hommes au total étaient sous son commandement dans son organisation? On suppose que les groupes de 10 doivent contenir 10 si possible, mais qu'un groupe à chaque niveau peut avoir besoin de témoigner moins

MAINTENANT

Nombre de soldats = 5 763

Rang le plus bas (soldat) = 5763

troisième plus haut (leader de 10) = 5763/10 = 576.3 = 577 (nombre entier requis pour un humain)

deuxième le plus élevé (leader de 100) = 577/10 = 57.7 = 58

Le plus haut (Leader de 1000) = 58/10 = 5.8 = 6

donc le total = 5763 + 577 + 58 + 6 = 6404 soldats

8 0
11 days ago
In a state where tax is 5%, Jane bought a telescope and paid $4.55 in tax. What is the price of the telescope?
tester [8842]
The telescope’s cost is $91

4.55 divided by 5% (0.05) equals 91.
7 0
1 day ago
Read 2 more answers
Find a parametrization, using cos(t) and sin(t), of the following curve: The intersection of the plane y= 5 with the sphere
Svet_ta [9518]

Answer:

The parametrization of the specified curve is x = 10 cos(t) \ \,z = 10 sin(t)

Step-by-step explanation:

From the provided problem statement, we derive the function

x^2 + y^2 + z^2 = 125

When y = 5

x^2 + 5^2 +z^2 =125

x^2 + z^2 = 100

x^2 + z^2 = 10^2

Converting the above to a polar equation yields

x = 10 cos(t) \ \,z = 10 sin(t)

6 0
10 days ago
Mrs. Adams bought cans of pears and cans of mixed fruit.
Svet_ta [9518]

Answer: 5 Cans

Step-by-step explanation:

4 0
1 month ago
Consider randomly selecting a student at a large university. Let A be the event that the selected student has a Visa card, let B
Zina [9179]

Response:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) signifies the likelihood that a student with a visa card also possesses a MasterCard.

p(A/B) indicates the probability that a student with a MasterCard also has a visa card.

e. 0.35

f. 0.31

Detailed explanation:

a. p(AUBUC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)

        = 0.6 + 0.4 + 0.2 - 0.3 - 0.11 - 0.1 + 0.07 = 0.76

b. P(AnBnC') = P(AnB) - P(AnBnC)

        = 0.3 - 0.07 = 0.23

c. P(B/A) = P(AnB)/P(A)

        = 0.3/0.6 = 0.5

e. P((AnB)/C) = P((AnB)nC)/P(C)

        = P(AnBnC)/P(C)

        = 0.07/0.2 = 0.35

f. P((AUB)/C) = P((AUB)nC)/P(C)

        = (P(AnC) U P(BnC))/P(C)

        = (0.11 + 0.1)/0.2

        = 0.21/0.2 = 0.31

7 0
21 day ago
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