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Rina8888 [55]
3 years ago
14

G(n) = 9n² + 7n³ g'(n) = ? ​

Mathematics
2 answers:
Olin [163]3 years ago
4 0
<h2>♪Answer : </h2>

»g(n) = 9n² + 7n³

»g'(n) = 9(2)n + 7(3)n²

»g'(n) = 18n + 21n²✅

serg [7]3 years ago
3 0

Answer:

g'(n) = 3n(6 + 7n)

Step-by-step explanation:

g(n) = 9n² + 7n³

g'(n) = 18n + 21n²

g'(n) = 3n(6 + 7n)

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Is it possible to solve the differential equation
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No, and solutions is is not valid.

Step-by-step explanation:

Solving y'(x)= y(x)a|x| by parts yields y(x)=e^{\frac{a}{2}x^{2}  } is the solutions, for y(0)=0 (initial value), it yields e^{0}=0 which is not valid. Note a is assumed to be constant.

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Simplify the following expression:<br><br> 7472
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7^2

Step-by-step explanation:

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What is the smallest integer greater than 92
mylen [45]

After 92, we have the following integers,

93,94,95,...

The smallest of these integers is 93

The smallest integer after 92 is 93

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4 years ago
John rode 4 kilometers on his bike his sister sally rode 3400 meters on her bike who rode the farthest and how much farther did
emmasim [6.3K]
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7 0
4 years ago
Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordl
LenaWriter [7]

Answer:

a) 0.0498

b) 0.1489

c) 0.1818

Step-by-step explanation:

Given:

Number of telephones = 6+6+6= 18

6 cellular, 6 cordless, and 6 corded.

a) Probability that all the cordless phones are among the first twelve to be serviced:

12 are selected from 18 telephones, possible number of ways of selection = ¹⁸C₁₂

Then 6 cordless telephones are serviced, the remaining telephones are: 12 - 6 = 6.

The possible ways of selecting thr remaining 6 telephones = ¹²C₆

Probability of servicing all cordless phones among the first twelve:

= (⁶C₆) (⁶C₁₂) / (¹⁸C₁₂)

= \frac{1 * 924}{18564}

= 0.0498

b) Probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced:

Here,

One type must be serviced first

The 6 remaining to be serviced can be a combination of the remaining two types.

Since there a 3 ways to select one type to be serviced, the probability will be:

= 3 [(⁶C₁)(⁶C₅) + (⁶C₂)(⁶C₄) + (⁶C₃)(⁶C₃) + (⁶C₄)(⁶C₂) + (⁶C₅)(⁶C₁)] / ¹⁸C₁₂

= \frac{3 * [(6)(6) + (15)(15) + (20)(20) + (15)(15) + (6)(6)]}{18564}

= \frac{2766}{18564}

= 0.1489

c) probability that two phones of each type are among the first six:

(⁶C₂)³/¹⁸C₆

\frac{3375}{18564}

=0.1818

5 0
4 years ago
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