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irga5000 [103]
4 years ago
8

What must be true to find the sum of a infinite sequence?

Mathematics
1 answer:
Georgia [21]4 years ago
3 0

Answer:

The series must be decreasing, and the limit as the n-th term goes to infinity should be 0.

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2. Suppose (220)_b and (251)_b are the squares of consecutive integers. Determine b.
Andrei [34K]

Let x and x+1 be these consecutive integers. For some base b, we can write

(x_{10})^2=220_b=2b^2+2b

((x+1)_{10})^2=251_b=2b^2+5b+1

If

x^2=2b^2+2b

then

x^2+1=2b^2+2b+1

Now,

x^2+2x+1=2b^2+5b+1

2x+2b^2+2b+1=2b^2+5b+1

\implies2x=3b

Squaring both sides gives

4x^2=9b^2

and so

4(2b^2+2b)=9b^2

\implies8b^2+8b=9b^2

\implies b^2-8b=0\implies\boxed{b=8}

(because the base has to be non-zero)

3 0
3 years ago
Fifteen tickets cost $193.75. What is the average cost of each ticket?
julsineya [31]

Answer:

about $12.92

Step-by-step explanation:

could be wrong though

8 0
3 years ago
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Help <br> Fjfjdjxjcjfjfjdjdjjdjdjd
White raven [17]

Answer:

a or b

Step-by-step explanation:

3 0
3 years ago
Solve the following inequality.
Luden [163]
First simplify.
-\frac{2}{3}x \ \textgreater \  \mathrm{Multiply\:fractions}:\ \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  -\frac{2x}{3}

\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)} \ \textgreater \ \ \textgreater \  \frac{2x}{3}\ \textless \ -8

\mathrm{Multiply\:both\:sides\:by\:}3 \ \textgreater \  \frac{3\cdot \:2x}{3}\ \textless \ 3\left(-8\right) \ \textgreater \  Simplify \ \textgreater \  2x\ \textless \ -24

\mathrm{Divide\:both\:sides\:by\:}2 \ \textgreater \  \frac{2x}{2}\ \textless \ \frac{-24}{2} \ \textgreater \  Simplify \ \textgreater \  x\ \textless \ -12

Therefore\; x\ \textless \ -12 \;is\;our\;answer!

Hope this helps!
7 0
3 years ago
Duncan shares 20 percent of all of his sales commissions with his personal assistant. If Duncan earns x dollars in commissions a
grandymaker [24]

Answer:

it b

Step-by-step explanation:

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