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2 years ago

Write an equation for the nth term of the arithmetic sequence 8, 3, -2, -7,... Find the 9th term of the sequence.

Mathematics

1 answer:

vodka [1.7K]2 years ago

8 0

Answer: -32

Rule: -5

Explanation: 8-5=3 3-5=-2 (-2)-5=(-7) (-7)-5=(-12), etc.

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When Theresa got her puppy, it weighed 6 ponds less than 2/3 of its weight now. If the puppy weighed 4 pounds when she first rec

sesenic [268]

Answer:

15 lb

Step-by-step explanation:

let the current weight of the puppy be X, the initial weight of the puppy is 4lb or $\frac{2}{3}X-6\ lb$ , (6 lb less 2/3 X) .

#We equate the current weight to the initial weight to determine X:

$4\ lb=\frac{2}{3}X-6\ lb\\\\10 \ lb=\frac{2}{3}X\\\\X=15$

Hence, the current weight of the puppy is 15 lb

5 0

4 years ago

4) 240

Maksim231197 [3]

Answer: I belive it is 360 inches cubed. I could be wrong

Step-by-step explanation:

8 0

3 years ago

Solve for 8:<br> -0.2 = 8 +(-0.8)

Nookie1986 [14]

S=0.6
Step by step explanation: subtract the greater number by the smaller number. Plug in the decimal after.

6 0

3 years ago

To visit his grandmother, Luis takes a train 6.376.376, point, 37 kilometers and a car 5.455.455, point, 45 kilometers.

murzikaleks [220]

Answer:

Luis traveled 11.82 km in total.

Step-by-step explanation:

<em>The statement should say</em>: To visit his grandmother, Luis takes a train 6.37 kilometers and a car 5.45 kilometers. How many kilometers is Luis's journey in total?

Hence, to find the total distance traveled by Luis we must add the distances of the train (6.37 km) and the car (5.45 km):

$d = d_{t} + d_{c} = 6.37 km + 5.45 km = 11.82 km$

Therefore, Luis traveled 11.82 km in total.

I hope it helps you!

8 0

3 years ago

How do you use the limit comparison test on this particular series?<br>Calculus series tests

barxatty [35]

Compare $\dfrac1{\sqrt{n^2+1}}$ to $\dfrac1{\sqrt{n^2}}=\dfrac1n$ . Then in applying the LCT, we have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac1{\sqrt{n^2+1}}}{\frac1n}\right|=\lim_{n\to\infty}\frac n{\sqrt{n^2+1}}=1$

Because this limit is finite, both

$\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}$

and

$\displaystyle\sum_{n=1}^\infty\frac1n$

behave the same way. The second series diverges, so

$\displaystyle\sum_{n=0}^\infty\frac1{\sqrt{n^2+1}}=1+\sum_{n=1}^\infty\frac1n$

is divergent.

4 0

4 years ago