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timofeeve [1]
3 years ago
15

Does anyone know this

Mathematics
1 answer:
Aliun [14]3 years ago
3 0

Answer:

ns esta en ingles AAAHHHHHHHHHHHHHHHHHHHHHHH

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Find the sum of the infinite geometric series: 9 – 6 + 4 – …
balandron [24]
The defining characteristic of all geometric sequences is a common ratio which is a constant when dividing any term by the term preceding it.

In this case the common ratio is:  -6/9=4/-6=r=-2/3

An infinite series will have a sum when r^2<1, so in this case the sum will converge to an actual value because (-2/3)^(+oo) approaches zero.

The sum of any geometric sequence is:

s(n)=a(1-r^n)/(1-r), since we have a common ratio of -2/3 and we want to calculate an infinite series, ie, n approaches infinity, the sum becomes simply:

s(n)=a/(1-r)    (because (1-r^+oo) approaches 1 as n approaches +oo)

So our infinite sum is:

s(+oo)=9/(1--2/3)

s(+oo)=9/(1+2/3)

s(+oo)=9/(5/3)

s(+oo)=27/5

s(+oo)=54/10

s(+oo)=5.4
7 0
4 years ago
Chuck and Dana agree to meet in Chicago for the weekend. Chuck travels 252 miles in the same time that Dana travels 228 miles. I
vlada-n [284]

Answer:

Chuck's rate of travel = 63mph

Step-by-step explanation:

Given chuck travels 252 miles and dana travels 228 miles. Given that they both take same time to travel .

Let the travelling time be T.

Also given that the speed of chuck is 6mph greater than that of dana's.

let the speed of chuck be x. Now

Speed = \frac{distance }{time}

x= \frac{252}{T}

Now for Dana's speed

x-6= \frac{228}{T}

When we divide both the equations we get

\frac{x}{x-6}=\frac{252}{228}

\frac{x}{x-6}=\frac{63}{57}

x=63mph

7 0
3 years ago
Read 2 more answers
Arrange the following fractions in increasing order. 1/8, 2/10, 11/12, 3/4
Nitella [24]
Hlo mate :-

Solution :-

☆ 1/8 > 2/10 > 3/4 > 11/12 ans
4 0
3 years ago
Please help me answer question number 3.
scoray [572]
It’s A
six more than a number which is y- (y+6)
is 10.03
so y+6=10.03
5 0
4 years ago
Ute of Change
spin [16.1K]

Answer:

1

2

0.5

Step-by-step explanation:

8 0
3 years ago
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