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3 years ago

6 (Picture) CONVERGENT AND DIVERGENT SERIES PLEASE HELP!!

Mathematics

1 answer:

V125BC [204]3 years ago

3 0

Answer:

The progression is Option A. convergent.

Step-by-step explanation:

The given series is( $-\frac{8}{5}+\frac{32}{25}-\frac{128}{125}+......$ )

= $(-\frac{8}{5})(1-\frac{4}{5}+(\frac{4}{5})^{2}-.....)$

Now we can write this series as $\sum_{n=0}^{n=\oe}(-\frac{8}{5})(-\frac{4}{5})^{n}$

In this expression common ration is 4/5= 0.8

As we know that in geometric progression if common factor is less than one then the progression converges.

Therefore we can say that this progression is convergent.

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Kareem lives 4/10of a mile from the mall . write equivalent fractions that show what fraction of a mile karen lives from the mal

Slav-nsk [51]

You just have to simplify it by two 2/5 8/20

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3 years ago

What is the equation of this line in slope-intercept form?

KatRina [158]

Answer: B

Step-by-step explanation:

The slope-intercept is usually written as y = mx + b form, where m is the slope, and b is the y-intercept. The y-intercept is the y value when x is equal to 0.

We can first eliminate C, because the y-intercept of the graph is 1.

To solve for slope, which is the m, choose two points, (-1, 5) and (1, -3).

Slope = (y2 - y1) / (x2 - x1)

= (-3 - 5) / (1 - (-1))

= (-3 - 5) / (1 + 1)

= -8/2

= -4

Plug in the slope and y-intercept, we get y = -4x + 1

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3 years ago

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77julia77 [94]

Answer:

Explanation:

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3 years ago

Compute: 3/8 divided -0.25

Makovka662 [10]

Answer:

-1.5

Step-by-step explanation:

3/8 is 0.375 in decimal form

3 0

2 years ago

Read 2 more answers

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

3 years ago