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Natalija [7]
3 years ago
7

Help!

Mathematics
2 answers:
REY [17]3 years ago
6 0

Answer:

0.79 yw hop ethis helped.

Step-by-step explanation:

padilas [110]3 years ago
5 0

Answer:

0.390

Step-by-step explanation:

0.79×0.79×0.79×0.79

= 0.38950081

= 0.390

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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
What is the area of the shape below? Hint: use the formula: A =1/2 h (b1 + b2) *
Inga [223]

Answer:

B. 172.5 yd squared

Step-by-step explanation:

FIrst, we must find half of the height, which is 15/2 or 7.5.

Next we have to multiply 7.5 times (15+8) which is B.

5 0
3 years ago
Giving thanks and brainliest for best answer &lt;3<br> ::
satela [25.4K]
I can help if I see the problem
7 0
3 years ago
Read 2 more answers
Is algebra.
Dmitry [639]
Factor By Grouping
GCF
7 0
3 years ago
Joelle is working on a school report in a word processor. She wants to place an oversized image which is 30.51 \text{ cm}30.51 c
ehidna [41]

Answer:

the horizontal position for the left edge of the image is -4.46 cm.

Step-by-step explanation:

The horizontal midpoint of the page is:

21.59 / 2 = 10.79 cm

The horizontal midpoint of the image is given as:

30.51 / 2 = 15.25 cm

These midpoints must be on the same point in the axis.

By taking the leftmost edge of the paper to be point zero on the axis, then, the distance accommodated by the paper is 10.795 cm.  

The distance that goes beyond this leftmost edge is computed as;

This is on the negative side on the axis.

Thus the horizontal position for the left edge of the image is -4.46 cm.

3 0
3 years ago
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