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babunello [35]
3 years ago
10

Find the 7th term in the sequence 1/3,1,3,9

Mathematics
2 answers:
Setler79 [48]3 years ago
8 0
Y = 3x

Proof:

3(1) = 3

3(3) = 9


5. 9(3) = 27

6. 27(3) = 81

7. 81(3) = 243


The 7th term is 243.
Eduardwww [97]3 years ago
6 0
So peosjdjnds. Sksjsjsnsnejdd
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Step-by-step explanation:

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PLEASE HELP!!! WILL MARK BRAINLIEST!! THX! A golfer hits a ball with an initial velocity of 32.7 m/s from the ground. Find the f
OLEGan [10]

Answer:

See below

Step-by-step explanation:

<u>First Problem</u>

The ball hits the ground when h(t)=0, therefore:

h(t)=-4.9t^2+v_0t+h_0

0=-4.9t^2+32.7t

0=t(-4.9t+32.7)

t=0 and t=\frac{32.7}{4.9}\approx6.67

Since the ball is in the air before it hits the ground, t=6.67 (seconds) is the more appropriate choice.

<u>Second Problem</u>

The maximum height of the ball is determined when t=-\frac{b}{2a}, therefore:

t=-\frac{b}{2a}

t=-\frac{32.7}{2(-4.9)}

t=-\frac{32.7}{-9.8}

t\approx3.34

This means that the height of the ball is at its maximum after 3.34 seconds:

h(t)=-4.9t^2+32.7t

h(3.34)=-4.9(3.34)^2+32.7(3.34)

h(3.34)\approx54.55

Thus, the answer is 54.55 (meters).

<u>Third Problem</u>

Refer to the second problem

<u>Fourth Problem</u>

<u />h(t)=-4.9t^2+32.7t<u />

<u />h(4.3)=-4.9(4.3)^2+32.7(4.3)<u />

<u />h(4.3)\approx50.01<u />

<u />

Therefore, the height of the ball after 4.3 seconds is 50.01 (meters).

<u>Fifth Problem</u>

The ball will be 24 meters off the ground when h(t)=24, therefore:

h(t)=-4.9t^2+32.7t

24=-4.9t^2+32.7t

0=-4.9t^2+32.7t-24

t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

t=\frac{-32.7\pm\sqrt{(32.7)^2-4(-4.9)(-24)}}{2(-4.9)}

t_1\approx0.84

t_2\approx5.83

Therefore, the ball will be 24 meters off the ground after 0.84 (seconds) and 5.83 (seconds)

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2 years ago
Write (-6, 3) and (4, 3) in slope form
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Answer:

y-3=0 (x+6)

Step-by-step explanation:

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Subtract 1/3x from the left side of the equation and subtract it from the right side of the equation. So it would be y = 4 - 1/3x
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