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Brrunno [24]
1 year ago
10

Lim x-> 0 sqrt x+4 -2 / x

Mathematics
1 answer:
Wittaler [7]1 year ago
3 0

Greetings from Brasil...

Here we have an indeterminacy 0/0

We can change variable or rationalize.......

Let's rationalize

{[√(X + 4) - 2]/X} · {[√(X + 4) + 2]/[√(X + 4) + 2]} = 1/[√(X + 4) + 2]

So, the limit will be 1/4

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PLEASE HELP!!! I WILL RATE BRAINLIEST TO THE FASTEST AND CORRECT ANSWER!!!
Andrej [43]
Not easy:

- 4 < x^4 + 4x^2 ---> 0 < x^4+4x^2+4 = (x^2+2)^2, so it is always positive. Always satisfied. Curious ...

Now the other one:

x^4+4x^2<21 --> x^4+4x^2-21 < 0 --> (x^2+2)^2 - 25 < 0

(x^2+2)^2<25 ->>> x^2+2 < 5 ---> x^2 <3,

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3 years ago
A) Function<br> B) Not a function
svetlana [45]

Answer:

b

Step-by-step explanation:

parabola's aren't functions

6 0
3 years ago
Which expression is equivalent to<br> logbase12 ((1/2)/8w)?
Lunna [17]

\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ log_a(xy)\implies log_a(x)+log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm of rationals} \\\\ log_a\left( \frac{x}{y}\right)\implies log_a(x)-log_a(y) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_{12}\left( \cfrac{~~\frac{1}{2}~~}{8w} \right)\implies \begin{array}{llll} \log_{12}\left( \frac{1}{2} \right)&-&\log_{12}(8w)\\\\ \log_{12}(1)-\log_{12}(2)&-&[\log_{12}(8)+\log_{12}(w)] \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \log_{12}(1)-\log_{12}(2)-\log_{12}(8)-\log_{12}(w)~\hfill

7 0
3 years ago
Tina and maryann each play tennis. Maryann has won 5 more games than Tina. Is it possible for tina to have won games if the sum
Soloha48 [4]

Answer:

Tina won 5 games more than maryann, and the sum of their games is 29.. 29-5/2= 12

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4 0
3 years ago
Read 2 more answers
Suppose you choose a marble from a bag containing 3 red marbles, 5 white marbles, and 4 blue marbles. You return the first marbl
Veseljchak [2.6K]

Answer:

P(red and blue) = 0.1667

Step-by-step explanation:

Let A represent choosing a red marble, there are 3 red marbles

Let B represent choosing a blue marble, there are 4 blue marbles

There are 12 total marbles

Then...

P(A) = 3/12

P(B) = 4/12

choosing the first one, then replacing means the first choice has no effect on the probabilities of the second choice, so the situation is independent.  

When calculating the probability of independent events, you multiply the probabilities together.  There are 2 scenarios where we can get a red and blue marble..

Choosing a red, then a blue marble, the probability is

(3/12)( 4/12) = 12/144 = 1/12

Choosing a blue, then a red marble, the probability is

(4/12)(3/12) = 12/144 = 1/12

So we have a

(1/12) + (1/12) = 2/12 = 1/6 chance for that to happen, to

P(A and B) = 1/6 = 0.1667

6 0
3 years ago
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