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2 years ago

Help me pleaseeee it’s algebra 2

Mathematics

1 answer:

aivan3 [116]2 years ago

4 0

The answer is yes. I already answered this lol

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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

I need help on these two problems

sattari [20]

50. since n_1/4<1 the equation is not true. it has no solution.
48. same thing since 45 <= z the equation is not true.

6 0

3 years ago

Please answer this multiple choice question for 27 points and brainliest!!

storchak [24]

Answer:

Step-by-step explanation:

The equation will be in the form

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (0, 5) and (x₂, y₂ ) = (2.5, 0) ← 2 points on the line

m = $\frac{0-5}{2.5-0}$ = - 2

The line crosses the y- axis at (0, 5) ⇒ c = 5

y = - 2x + 5 or

y = 5 - 2x → A

5 0

3 years ago

Someone help me solve this it's easy i just don't understand simple math

Lady_Fox [76]

Answer:

The equation — E = mc2 — means "energy equals mass times the speed of light squared." It shows that energy (E) and mass (m) are interchangeable; they are different forms of the same thing. If mass is somehow totally converted into energy, it also shows how much energy would reside inside that mass: quite a lot.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find the solution of 2/3x+1/4=-4/3

wolverine [178]

These questions can be daunting at first, but they're pretty simple to solve.

First, we need to establish a common denominator. We have 2 / 3, 1 / 4, and
-4 / 3. The least common denominator we can get is by multiplying 4 and 3 together to get 12. So we will change the denominator as follows;

2 / 3, 1 / 4, -4 / 3 = 8 / 12, 3 / 12, -16 / 12

Now we can put these back into the equation.

8/12x + 3/12 = -16/12
8x + 3 = -16

It's simple math from here on out, but I'll show the process. What we can basically do now is take away the denominator because it doesn't matter now that it's common.

Subtract 3 from both sides. Now we have 8x = -19

Dividing by 8 on both sides of the equation will get you your answer.

x = -19/8

Hope this helps!

6 0

3 years ago