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2 years ago

9

Solve for x. x = ln 1 x =

1 answer:

VikaD [51]2 years ago

3 0

$x=\ln 1\\
x=0$

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Determine whether the improper integral converges or diverges, and find the value of each that converges.

Temka [501]

Answer:

It diverges.

Step-by-step explanation:

We are given the inetegral: $\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx$

$\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx=\int\limits^{\infty}_2 (\ln x)^2 d(\ln x)=\\\\=\lim_{t \to \infty} \int\limits^t_2 (\ln x)^2d(\ln x)=\lim_{t \to \infty} \frac{(\ln t)^3}{3} |^t_2=\infty-\frac{(\ln 2)^3}{3} =\infty$

So it is divergent.

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2 years ago

How do you wire 0.52 in words

Alexxx [7]

Answer: fifty-two hundredths

Step-by-step explanation:

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2 years ago

The sum of two numbers is 56 the difference is 4 find the two numbers

belka [17]

Answer:

30 and 26

Step-by-step explanation:

30 plus 26 is 55

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2 years ago

If ON= 5x - 7, LM = 4x+ 4, NM = x -7, and OL =3y-4, find the values of c and y for which LMNO MUST BE a parallelogram.

Sholpan [36]

Answer:

x=11, y=8/3

Step-by-step explanation:

Consider quadrilateral LMNO. If this quadrilateral MUST BE a parallelogram, then

LM=NO

and

LO=MN

Thus,

$\left\{\begin{array}{l}4x+4=5x-7\\ \\x-7=3y-4\end{array}\right.$

Solve this system of two equations. From the first equation:

$4x-5x=-7-4\\ \\-x=-11\\ \\x=11$

Substitute it into the second equation:

$11-7=3y-4\\ \\3y-4=4\\ \\3y=4+4\\ \\3y=8\\ \\y=\dfrac{8}{3}$

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2 years ago

I have the question in a screengrab. Thank you, whoever helps me.

nata0808 [166]

To simplify

$\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}$

we need to use the fact that

$\sqrt[4]{x^4}=|x|$

Why the absolute value? It's because $(-x)^4=(-1)^4x^4=x^4$ .

We start by rewriting as

$\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}$

$\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}$

Since $x\neq0$ , we have $\dfrac xx=1$ , and the above reduces to

$\sqrt[4]{\dfrac{3x^2y}{2^4y^4y}}$

Then we pull out any 4th powers under the radical, and simplify everything we can:

$\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}$

$\dfrac1{|2y|}\sqrt[4]{3x^2}$

where $y>0$ allows us to write $\dfrac yy=1$ , and this also means that $|y|=y$ . So we end up with

$\dfrac{\sqrt[4]{3x^2}}{2y}$

making the last option the correct answer.

7 0

2 years ago