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nordsb [41]
3 years ago
12

A submarine out of death of 2167 feet ascends to a depth of 609ft. How far did the submarine ascend?

Mathematics
1 answer:
Leya [2.2K]3 years ago
8 0

Answer:

A submarine was situated 800 feet below sea level. If it ascends 250 feet, what is its new ... Roman Civilization began in 508 B.C. and ended in 476 A.D. How long did Roman Civilization last? 476-(-508) = 476+508 = 984 ...

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Calculate the limit values:
Nataliya [291]
A) This particular limit is of the indeterminate form,
\frac{ \infty }{ \infty }
if we plug in infinity directly, though it is not a number just to check.

If a limit is in this form, we apply L'Hopital's Rule.

's
Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' }
So we take the derivatives and obtain,

Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1}

Still it is of the same indeterminate form, so we apply the rule again,

Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 2 }{2x}

This simplifies to,

Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 1 }{x} = 0

b) This limit is also of the indeterminate form,

\frac{0}{0}
we still apply the L'Hopital's Rule,

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' }

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (x) }{1 }

When we plug in zero now we obtain,

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (0) }{1 } = \frac{1}{1} = 1
c) This also in the same indeterminate form

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' }

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x }

It is still of that indeterminate form so we apply the rule again, to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 }

Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 }

This gives us;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } =\frac{ (4(1) )}{ 2 }=2

d) Lim_ {x \rightarrow +\infty }\sqrt{x^2+2x}-x

For this kind of question we need to rationalize the radical function, to obtain;

Lim_ {x \rightarrow +\infty }\frac{2x}{\sqrt{x^2+2x}+x}

We now divide both the numerator and denominator by x, to obtain,

Lim_ {x \rightarrow +\infty }\frac{2}{\sqrt{1+\frac{2}{x}}+1}

This simplifies to,

=\frac{2}{\sqrt{1+0}+1}=1
5 0
3 years ago
Answer question number 5 only.
irinina [24]

Answer:

θ = π<em>n</em>

<em />

Step-by-step explanation:

6 0
3 years ago
. Solve for the missing value.
choli [55]

Answer:

1. f= 9

2. y= 10

3. h= 2.5

4. q= 10

5. m= 6

Step-by-step explanation:

1. you divide both sides by 3 to get f by itself

2. you divide both sides with 6

3. you subtract because it is the opposite of addition. you could also write 5/2 but that’s not simplified enough

4. 10 x 4 = 40 so 10 is the answer

5. you subtract both sides by 6 and get 24. 24 divided by 4 equals 6

4 0
3 years ago
Read 2 more answers
How do i solve this? i need help
Korvikt [17]

A (103)(109)

=103*109

=(10*10*10)*(10*10*10*10*10*10*10*10*10)

=10*10*10*10*10*10*10*10*10*10*10*10

=1012

=1000000000000


c.1e+23

3 0
4 years ago
Andrew withdrew $332.55 from his savings account. He now has at most $529.71. How Much cash did he originally have?
Art [367]
Andrew originally had $197.16
7 0
3 years ago
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