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katen-ka-za [31]
4 years ago
9

If a pier is 200 meters long and each board is 80 centimeters wide, how many boards are in the pier?

Mathematics
2 answers:
VladimirAG [237]4 years ago
8 0


Well we know that 1 meter = 100 centimeters

so 80cm = 0.80m

so now that we now how many meters are the boards we could divide;

200 ÷ 0.80 = 250

Your final answer is;

250 boards

Luba_88 [7]4 years ago
6 0
<span>250. 100 CM in 1 Meter. That leaves you with with .8 meters a board, 200/ .8 leaves you with 250</span>
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Step-by-step explanation:

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On a coordinate grid, Ming's house is located 2 blocks to the right and blocks up from (0. 0). Joe's house is located 3 blocks t
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If o<span>n a coordinate grid, Ming's house is located 2 blocks to the right and (I suppose also 2) blocks up from (0. 0), then the ordered pair that describes the location of Ming's house is (2,2) (moving right 2 units from some point you have to add 2 units to the x-coordinate of the point from which you are moving and moving up 2 units from this point you have to add 2 units to y-coordinate of the point from which you are moving up).
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3 0
3 years ago
What is the value of r of the geometric series n=11.3(0.8)n-1
jeyben [28]

the value of r of the geometric series n=11.3(0.8)n-1

an=11.3(0.8)^{n-1}

General formula for nth term of any geometric series is   a_n=a_1(r)^{n-1}

Here 'r' is the common ratio

a_1 is the first term of the series

Now we compare the given formula with general formula

Compare  an=11.3(0.8)^{n-1}  with  a_n=a_1(r)^{n-1}

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6 0
3 years ago
Given:
Goryan [66]

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a.Converge

b.Diverge

Step-by-step explanation:

We are given that

A_n=\frac{6n}{-4n+9}

\lim_{n\rightarrow \infty}A_n=\lim_{n\rightarrow \infty}\frac{6n}{-4n+9}=\frac{6n}{n(-4+\frac{9}{n})}

\lim_{n\rightarrow \infty}\frac{6}{-4+\frac{9}{n}}=\frac{6}{-4}=\frac{-3}{2}

Because \frac{9}{\infty}=0

Hence, sequence An converges to \frac{-3}{2} when n tends to infinity or - infinity.

\sum_{n=1}^{\infty}An=\sum_{n=1}^{\infty}\frac{6n}{-4n+9}=diverges

Necessary condition for a series to converge :

an\rightarrow 0 when n tends to infinity .

But A_n tends to \frac{-3}{2}when n tends to infinity.

Therefore, given series is divergent.

5 0
3 years ago
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