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lisabon 2012 [21]
3 years ago
14

a sequence starts a 200 and 30 is subtracted each time 200,170,140 what are the first two numbers in the sequence that are kess

then zero
Mathematics
2 answers:
Snowcat [4.5K]3 years ago
5 0
<h3>Answer:  -10 and -40</h3>

===============================================================

Explanation:

a = 200 = first term

d = -30 = common difference

Tn = nth term

Tn = a + d(n-1)

Tn = 200 + (-30)(n-1)

Tn = 200 - 30n + 30

Tn = -30n + 230

Set Tn less than 0 and isolate n

Tn < 0

-30n + 230 < 0

230 < 30n

30n > 230

n > 230/30

n > 7.667 approximately

Rounding up to the nearest whole number gets us n \ge 8

So Tn starts to turn negative when n = 8

We can see that,

Tn = -30n + 230

T7 = -30*7 + 230

T7 = 20

and

Tn = -30n + 230

T8 = -30*8 + 230

T8 = -10 is the 8th term

and lastly

Tn = -30n + 230

T9 = -30*9 + 230

T9 = -40 is the ninth term

Or once you determine that T7 = 20, you subtract 30 from it to get 20-30 = -10 which is the value of T8. Then T9 = -40 because -10-30 = -40.

ikadub [295]3 years ago
4 0

Answer:

- 10

- 40

Step-by-step explanation:

By the 7th term you should be pretty close to 0. Let's show that.

a1 = 200

n = 7

d = - 30

t7 = a1 + (n - 1)*d

t7 = 200 + (7 -1)*-30

t7 = 200 + 6*-30

t7 = 200 - 180

t7 = 20

This is the last term that is positive. when you take 30 away from t7 you are going to be in negative territory.

t8 = 200 + (8-1) * - 30

t8 = 200 + 7 * - 30

t8 = 200 - 210

t8 = - 10

Now the 9th term

t9 = 200 + (9 - 1)*-30

t9 = 200 + 8 * - 30

t9 = 200 - 240

t9 = - 40

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A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter
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At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

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which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

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There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

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