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LiRa [457]
3 years ago
12

What is the sum of the infinite geometric series with a1=42 and r=6/5?

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
8 0
This infinite sum is given by

          a     
S =  ------  but ONLY if r<1.  Here, r>1, so the infinite sum does not exist.    
        1-r                                     
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21 Last year Chelsea and Sonya took 236 pictures during Thanksgiving. This year Chelsea and Sonya each took the same amount of p
Leokris [45]

Let 'c' represent the number of pictures Chelsea took.

Let 's' represent the number of pictures Sonya took.

For last year's Thanksgiving, c + s = 236

For this year's Thanksgiving, let 'x' represent the number of photos taken in total. x = c + s, where c and s are two integers that are the same (c = s).

And we know that for both years, c + s + x = 500.

As we know that c + s is already 236 from last year, we can remove c + s from the equation in bold and replace it with 236 instead.

236 + x = 500.

Now we have to isolate the x term.

x = 500 - 236

x = 264.

We know that x = c + s, where c and s are the same, so we can just use one of the variables and double it (so you either get 2c or 2s - it doesn't matter which one you pick because they're both the same).

2c = 264

c = 132

c = s

s = 132.

Both took 132 pictures this year.

4 0
3 years ago
A rectangular, above-ground swimming
gregori [183]
The answer is A. 158 lbs
3 0
3 years ago
A group of five people are all working on the same mathematics problem. On the night before it is due, they call each other to d
UkoKoshka [18]

Answer:

Minimum number of calls = 10

Step-by-step explanation:

Lets name the five people as A,B,C,D and E.

<u>On the night before ,each person talks to every other person atleast once that means A would talk to B,C,D and E atleast once.</u>

Lets start with A. He would talk to other 4 people which means there would be 4 phone calls made.

Now lets take B. He can talk to A,C,D and E. But  A has already talked to C therefore to get minimum number of phone calls , B need not call A again. So he calls only C,D and E.

In case of C using similar logic he need to talk to only D and E.

For D , he talks to E alone.

E does not have to talk to anyone as he has already talked to everyone atleast once.

Total calls = 4 + 3 + 2 + 1

                    = 10

5 0
3 years ago
URGENT! If you get the correct answer, you will get a BRAINLIEST!
trasher [3.6K]

Answer:

x_{1} =  - 2 -  \sqrt{2}  \\ x_{2} = - 2 +  \sqrt{2}

5 0
3 years ago
Q3<br> Help pls.. urgent!
Komok [63]
Alright, so 3f-g=4 and f+2g=5. 
3f-g=4
f+2g=5
Multiplying the first equation by 2 and adding it to the second, we get 7f=13 and by dividing both sides by 7 we get f=13/7. Since f+2g=5, then we can plug 13/7 in for f to get 13/7+2g=5. Next, we subtract 13/7 from both sides to get 2g=3+1/7=22/7 (since 3*7=21 and 21+1=22). DIviding both sides by 2, we get 22/14=g. Plugging that into f/39g, we get (13/7)/(22*39/14)
= (13/7)/(858/14) 
= (13/7)*(14/858)
=182/6006
= 91/3003 (by dividing both numbers by 2)
= 13/429 (by dividing both numbers by 7)
 = 1/33 (by dividing both numbers by 13)
7 0
3 years ago
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