Find the common ratio for the geometric sequence defined by the formula: an=40(2–√)n−1

Mathematics

1 answer:

son4ous [18]3 years ago

8 0

B is the answer..........

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Use the three steps to solve the problem.

Mazyrski [523]

Let's assume a the age of Joe
b the age of James
a=b+10 (1)
2*(a+8)=3(b+8) (2)

==> 2*(b+10+8)=3b+24
==>2b+36=3b+24
==>b=12
==>a=12+10=22
Joe is 22 years's old
James is 12 years's old.

5 0

3 years ago

Answeeeerr please ????

valentinak56 [21]

It is the first option.

4 0

2 years ago

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Pls help LOL it’s pretty easy but like I don’t have time

MakcuM [25]

Answer:because it’s just reversed

Step-by-step explanation:

3 0

2 years ago

1!+2(2!)+3(3!)+...+n(n!)=(n+1)!-1

anyanavicka [17]

Step-by-step explanation:

OK, let's assume it this way:

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!=(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+...+((n-1)n!-n!)=(2!-1!)+(3!-2!)+(4!-3!)+

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!=(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+...+((n-1)n!-n!)=(2!-1!)+(3!-2!)+(4!-3!)+...+(n+1)!-n!=(n+1)!-1!=(n+1)!-1

and boom problem solved

7 0

1 year ago

Find the distance between the given points: (2, -2) and (-4, -7)

miv72 [106K]

Formula: sqr((x2-x1)^2 + (y2-y1)^2)
sqr((-4-2)^2 + (-7+2)^2)
sqr(-6)^2 + (-5)^2)
sqr(36)+(25)) = sqr(61)
The answer is A. sqr(61)

3 0

2 years ago

Read 2 more answers