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irakobra [83]
3 years ago
8

What is the nth term to this sequence 18,16,14,12

Mathematics
2 answers:
densk [106]3 years ago
6 0

What is the nth term to this sequence 18,16,14,12 ?

Answer: 4

Explanation:

1. 18

2. 16

3. 14

4. 14

5. 12

6. 10

7. 8

8. 6

9. 4

Ratling [72]3 years ago
5 0

Answer:

Hey!

The nth term for your sequence is -2!

Step-by-step explanation:

18 - 2 = 16

16 - 2 =14

And so on! (the nth term of a sequence is basically the differences between the different terms!)

HOPE THIS HELPS!!

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Answer:

Triangle Q R P is shown. The length of Q R is 5 and the length of R P is 6. Angle Q R P is 40 degrees.

Step-by-step explanation:

The trigonometric formula refers the two sides length of the triangle and it also consists of included angle to find out the area

A = \frac{1}{2} ab sin C

QPR contains two sides and the included angle

XYZ has one side and the two angles

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Answer:

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You need a 60% alcohol solution. On hand, you have a 480 mL of a 75% alcohol mixture. How much pure water will you need to add t
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120 mL

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If the 480 mL has 75% concentration, then 360 mL are alcohol. To turn the 360 into only 60%, you need to increase the amount of water.

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A high school basketball coach wants to see whether there is a linear relationship between player height, x , and the number of
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HELP ME PLEASE
solmaris [256]

Answer:

The initial mass of the sample was 16 mg.

The mass after 5 weeks will be about 0.0372 mg.

Step-by-step explanation:

We can write an exponential function to model the situation.

Let the initial amount be A. The standard exponential function is given by:

P(t)=A(r)^t

Where r is the rate of growth/decay.

Since the half-life of Palladium-100 is four days, r = 1/2. We will also substitute t/4 for t to to represent one cycle every four days. Therefore:

\displaystyle P(t)=A\Big(\frac{1}{2}\Big)^{t/4}

After 12 days, a sample of Palladium-100 has been reduced to a mass of two milligrams.

Therefore, when x = 12, P(x) = 2. By substitution:

\displaystyle 2=A\Big(\frac{1}{2}\Big)^{12/4}

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\displaystyle 2=A\Big(\frac{1}{2}\Big)^3

Simplify:

\displaystyle 2=A\Big(\frac{1}{8}\Big)

Thus, the initial mass of the sample was:

A=16\text{ mg}

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