2 years ago

Find the oth term of the geometric sequence 3, 15,75, ...

Mathematics

1 answer:

Zinaida [17]2 years ago

4 0

Answer:

375

Step-by-step explanation:

hope this helps

You might be interested in

7/9-2n+1/9 simplify the fraction please help!

koban [17]

Answer: -2*(9n-4)/9

Step-by-step explanation: The -2*(9n-4) is over the 9

7 0

2 years ago

What is $184.94 if the shipping 5%

pychu [463]

The anwser is $924.70

4 0

3 years ago

Read 2 more answers

The value of y varies directly with x. If x=4, then y=10. a. Find the constant of proportionality. b. Write an equation to repre

Lynna [10]

Answer:

aaaa

Step-by-step explanation:

In a short paragraph, describe how you would determine the amount of work needed to pull a toboggan the length of a football field. Be sure to include the equipment list and describe the measurements that you will take.

3 0

3 years ago

The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet

vladimir1956 [14]

$\bf s(t)=\sqrt{t^3+1}
\\\\\\
\cfrac{ds}{dt}=\cfrac{1}{2}(t^3+1)^{-\frac{1}{2}}\cdot 3t^2\implies \boxed{\cfrac{ds}{dt}=\cfrac{3t^2}{2\sqrt{t^3+1}}}\leftarrow v(t)
\\\\\\
\cfrac{d^2s}{dt^2}=\cfrac{6t(2\sqrt{t^3+1})-3t^2\left( \frac{3t^2}{\sqrt{t^3+1}} \right)}{(2\sqrt{t^3+1})^2}\implies 
\cfrac{d^2s}{dt^2}=\cfrac{ \frac{6t(2\sqrt{t^3+1})-1}{\sqrt{t^3+1}} }{4(t^3+1)}$

$\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies 
\boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\
-------------------------------\\\\a(2)=\cfrac{215~ft^2}{44~sec}$

8 0

2 years ago

Read 2 more answers

The volume of a cone with a height of 6 cm is 8 cubic centimeters. Which expression can be used to find r, the radius of

Salsk061 [2.6K]

The base of the cone has to be B-3(n6)

5 0

2 years ago