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svetlana [45]
3 years ago
12

(8 x 10^ -4) x (7.2 x 10 ^5)

Mathematics
1 answer:
antoniya [11.8K]3 years ago
3 0

Answer:

576

Step-by-step explanation:

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Ok who wants 37 points! Just say anything so you can get it.... :)
mestny [16]

Answer:

hello governor

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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
A drink costs 3 dollars and a sandwich costs 5 dollars. You buy 10 items (some combination of drinks and sandwiches) and you spe
andrew-mc [135]

See https://web2.0calc.com/questions/can-someone-help-please_7.

7 0
3 years ago
Find the approximate side length of a square game board with an area of 186in2
Arturiano [62]
13.6in is the answer
8 0
3 years ago
Read 2 more answers
joe and beth both have landscaping businesses. Joe charges a flat fee of $15 to mow your yard and $8.00 for each tree he trims.
nalin [4]
15+8=23. 50+6=56
23+8=31. 56+6=62

Two Trees
5 0
3 years ago
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