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In-s [12.5K]
3 years ago
6

The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 42 and the common rati

o is three fourths. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.
Mathematics
2 answers:
Scilla [17]3 years ago
8 0


Since this is the formula for the infinite geometric series with
a1 = 42
r = 3/4

Solution will be

sum = 42 / 1-3/4
sum = 42 / 0.25
sum = 168

Therefore, the total is 168






JulsSmile [24]3 years ago
4 0

Answer: the answer is D


Step-by-step explanation: We can immediately eliminate options A and B because they aren’t even formatted properly. The common ratio(3/4) ALWAYS comes after a_1 which in this case was 42. That leaves us with C and D. Now, for series to be divergent, the common ratio(number in parentheses) HAS to be GREATER than one. Since this equation has 3/4 or .75 in the parentheses it is NOT divergent. That leaves us with answer choice D, and the answer is 168. My friend above explains how to get to that answer. Thanks and please rate.


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3 years ago
Use Theorem 7.4.1. THEOREM 7.4.1 Derivatives of Transforms If F(s) = ℒ{f(t)} and n = 1, 2, 3, . . . , then ℒ{tnf(t)} = (−1)n dn
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L\left(te^{2t }sin3t\right)=\frac{6s-12}{(s^2-4s+13)^2}.

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If F(s)= L{f(t)}

Then L\left\{(t^nf(t)\right\}=(-1)^n\frac{\mathrm{d^n}F(s)}{\mathrm{d^n}s}

L\left\{te^{2t}sin3t\right\}

f(t)=e^{2t}sin3t

L\left\{e^{at}sinbt\right\}=\frac{b}{(s-a)^2+b^2}

Therefore,L\left\{e^{2t}sin3t\right\}=\frac{3}{(s-2)^2+(3)^2}

L\left\{e^{2t}sin3t\right\}=\frac{3}{s^2-4s+13}

L\left\{te^{2t}sin3t\right}=-\frac{\mathrm{d}F(s)}{\mathrm{d}s}

=-\frac{\mathrm{d}e^{2t}sin3t}{\mathrm{d}s}

L\left\{te^{2t}sin3t\right\}

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7 0
4 years ago
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svlad2 [7]

Answer:

Lo primero que debemos saber, es que la gráfica de pastel nos permite dividir a una población, la cual sería el 100%, en porcentajes menores que representan algo.

En este caso tenemos los porcentajes 20%, 50%, 10%  (notar que no suman 100%, entonces nos falta algún porcentaje)

Bueno, como sabemos el gráfico (el cual es un círculo) representa el 100%.

Y en un círculo, tenemos un ángulo de 360°

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x = 50%

Si tomamos el cociente entre las ecuaciones:

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resolviendo para x, obtenemos:

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Para los otros porcentajes hacemos lo mismo.

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8 0
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