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3 years ago

What is the 8th term of this geometric sequence? 5,-10,20,-40

1 answer:

3 0

You are multiply by -2.
term 1: 5
term 2: -10
terms 3: 20
term 4: -40
term 5: 80
term 6: -160
term 7: 320
term 8: -640
So, the 8th term is -640. Hope this helps, please mark brainliest and have an amazing day!

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How to write 246,305 in expanded notation

gregori [183]

Answer:

200 000 + 40 000 + 6 000 + 300 + 5

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3 years ago

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What's the numerator for the following rational<br> expression?<br> j/k + 1/k = ?/k

lys-0071 [83]

Answer:

j+1

Step-by-step explanation:

j/k + 1/k = ?/k

Since the denominator is the same, we can add the numerators

j/k + 1/k = ?/k

(j+1)/k = ?/k

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3 years ago

You can build a dog house in 6 hours. And you work 8 hours how long would it take you to build 10 dog house.

Alexus [3.1K]

Seven days and a half

For you work 8 hours a day
It takes you 6 hours to per each
You wanna make 10
10 by 6 is 60 on 8 = 7.5

Best luck!

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3 years ago

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(PLEASE HELP!!)

AlexFokin [52]

Answer:

X= 8

Y= 130

(8, 130)

Step-by-step explanation:

I used a graphing calculator to see where the two equations intersected.

Hope this helped :)

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3 years ago

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If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

4 years ago