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3 years ago

14

Show that each fn is bounded on A, but the pointwise limit f of the sequence is not bounded on A. Does (fn) converge uniformly t

o f on A?

1 answer:

ladessa [460]3 years ago

4 0

Answer:

It seems your question is not complete, but i will asume it is like the one in the image.

Step-by-step explanation:

Answer at the image

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gtnhenbr [62]

Answer:

D is the correct answer

Step-by-step explanation:

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2 years ago

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2 years ago

Simplify this expression, you are y years old. Your sister is 3 years older than you. Your uncle is 4 times older than your sist

uysha [10]

Answer:The simplified expression that represents the uncle's age is

4y + 12

Step-by-step explanation:

Let y represent your age.

Your sister is 3 years older than you. This means that your sister's age is y + 3

Your uncle is 4 times older than your sister. It means that the age of your uncle would be

4(y + 3)

The simplified expression that represents your uncle's age would be

4y + 12

8 0

3 years ago

**WILL MARK BRAINLIEST AND THANK YOU**

Alex777 [14]

0, 4, -8

use x(x-4)(x+8)=0

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3 years ago

A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

GuDViN [60]

Answer:

1. $\frac{dy}{dt}=ky$

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

$\frac{dy}{dt}$ =Number of bacteria per unit time

$\frac{dy}{dt}\proportional y$

$\frac{dy}{dt}=ky$

Where k=Proportionality constant

2. $\frac{dy}{y}=kdt$ ,y'(0)=100

Integrating on both sides then, we get

$lny=kt+C$

We have y(0)=200

Substitute the values then , we get

$ln 200=k(0)+C$

$C=ln 200$

Substitute the value of C then we get

$ln y=kt+ln 200$

$ln y-ln200=kt$

$ln\frac{y}{200}=kt$

$\frac{y}{200}=e^{kt}$

$y=200e^{kt}$

Differentiate w.r.t

$y'=200ke^{kt}$

Substitute the given condition then, we get

$100=200ke^{0}=200 \;because \;e^0=1$

$k=\frac{100}{200}=\frac{1}{2}$

$y=200e^{\frac{t}{2}}$

Substitute t=2

Then, we get $y=200e^{\frac{2}{2}}=200e$

$y=200(2.718)=543.6=543.6$

e=2.718

Hence, the number of bacteria after 2 hours=543.6

4 0

3 years ago