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andrew11 [14]
4 years ago
10

The first term of a geometric sequence is -2 and the common ratio is 3 what is the 12th term of the sequence

Mathematics
1 answer:
lina2011 [118]4 years ago
6 0

The n-th term of a geometric sequence with initial value a and common ratio r is given by a_n=a_1r^{n-1}.

We have a_1=-2 and r=3 and n=12

Substitute:

a_{12}=-2\cdot3^{12-1}=-2\cdot3^{11}=-2\cdot177147=-354294

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What is the range of the function f(x)=3|2x-4|+5
natita [175]
Y/f(x) is greater than or equal to 5
3 0
3 years ago
Help me please thank youu ❤️
Nata [24]

Answer: 5^6=15,625

Step-by-step explanation:

To solve this exercise it is important to remembet the definition Definition of a Logarithm:

If x >0 and, b>0,b\neq 1, then y = log_bx is equivalent to b^y=x (which is the exponential form).

Knowing this definition and having  the following logarithmic equation:

6=log_515,625

You can identify that:

b=5\\y=6\\x=15,625

Therefore, susbtituting values into b^y=x, you can rewrite this equation as an exponential equation. Then:

5^6=15,625

7 0
4 years ago
A red kangaroo jumps along at 40 km/h. At this rate how long would it take to jump 2 km?
Nookie1986 [14]
We have to calculate how would it take for a red kangaroo to jump 2 km, if he jumps at a rate of 40 km/h.
Formula for the velocity is: v = d / t, where d stays for the distance and t stays for the time.
Therefore: t = d / v
t = 2 km / 40 km/h
t = 0.05 h = 3 minutes
Answer:
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8 0
3 years ago
If michael jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time
spin [16.1K]
The formula for the vertical height, h, attained with a vertical take off speed of u is
u² - 2gh = 0
where
g = 9.8 m/s², acceleration due to gravity.

Because h = 1.29 m, therefore
u² = 2*9.8*1.39 = 27.244
u = 5.2196 m/s

Also, the time of flight, t, required when the vertical take-off speed is u is given by
ut  - 0.5gt² = 0.
Therefore
5.2196t - 0.5*9.8*t² = 0
5.2196t - 4.9t² = 0
t(5.2196 - 4.9t) = 0
t = 0, or t = 5.2196/4.9 = 1.0652 s

t = 0 corresponds to take-off.
t = 1.0652 s corresponds to landing.
The hang time is 1.0652 s.

Answer:
Take-off speed = 5.22 m/s (nearest hundredth)
Hang time  = 1.065 s (nearest thousandth)

7 0
3 years ago
If a series of rigid transformations maps ∠E onto ∠B where ∠E is congruent to ∠B, then which of the following statements is true
Svetlanka [38]

Answer:

C

Step-by-step explanation:

We know that there are 2 pairs of congruent angles, angles B and E along with the pair of right angles.

4 0
2 years ago
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