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IgorC [24]
2 years ago
8

What is the 13th term of the geometric sequence 4, 12, 36, 108, ...?

Mathematics
1 answer:
bogdanovich [222]2 years ago
7 0

Answer: 2,125,764

Step-by-step explanation:

You can use the equation y=ar^n-1 to find the 13th term where a is the starting term, r is the common ratio, and n is the nth term

The next number is 12/4=3 times greater than the first. So if this sequence is a geometric one, then the next term must be 12*3=36. This is true, and the next term must be 36*3=108, which is also true. Therefore this sequence is a geometric one and its common ratio is 3. Therefore the equation is:

y=4*3^(n-1)

Substitute 13 for n:

y=4*3^(13-1)

y=4*3^12

y=2125764

Therefore, the 13th term of this geometric sequence is 2,125,764

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Natalka [10]

9514 1404 393

Answer:

  • 4/13 h ≈ 0.308 h
  • 22 9/13 km ≈ 22.7 km

Step-by-step explanation:

Mary and Amaruk are starting 35 -15 = 20 km apart. Their closing speed is 40 +25 = 65 km/h. The time it will take to close the gap between them is ...

  t = (20 km)/(65 km/h) = 20/65 h = 4/13 h ≈ 0.308 h

They will pass each other after 4/13 hours, about 18 minutes 28 seconds.

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In that time, Mary will have increased her distance from Hall Beach by ...

  d = (4/13 h)(25 km/h) = 100/13 km = 7 9/13 km

Her distance from Hall Beach when she passes Amaruk will be ...

  15 km + 7 9/13 km = 22 9/13 km ≈ 22.7 km . . . . . distance from Hall Beach

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3 years ago
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Step-by-step explanation:

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If y=34 what is y50+y/60
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I need answer Immediately pls!!!!!!
Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

n(U)=27

n(A)=7

n(B)=18

n(A'\cap B')=7

We know that,

n(A\cup B)=n(U)-n(A'\cap B')

n(A\cup B)=27-7

n(A\cup B)=20

Now,

n(A\cup B)=n(A)+n(B)-n(A\cap B)

20=7+18-n(A\cap B)

n(A\cap B)=7+18-20

n(A\cap B)=25-20

n(A\cap B)=5

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}

\text{Probability}=\dfrac{5}{27}

Therefore, the required probability is \dfrac{5}{27}.

3 0
3 years ago
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