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3 years ago

(-5)^2 divided by 5^4

Mathematics

2 answers:

Pani-rosa [81]3 years ago

4 0

Answer:

Step-by-step explanation:

divide

Nikitich [7]3 years ago

3 0

Answer:

0.04

Step-by-step explanation:

........ i just divided it

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Ted and Maggie solved the following equation, (3x - 2)(x + 5) = 0. Their

VLD [36.1K]

Answer:

maggie

Step-by-step explanation:

3 0

2 years ago

What is the answer of square root of -24 in complex numbers

Sergio039 [100]

With $\sqrt{-1}=i$ , we have

$\sqrt{-24}=\sqrt{(-1)(24)}=\sqrt{-1}\sqrt{24}=i\sqrt{24}$

Since $24=2^3\cdot3$ , we can further write this as

$i\sqrt{24}=i\sqrt{2^2\cdot6}=2i\sqrt6$

3 0

3 years ago

Which equation results from taking the square root of both sides of (x + 9)^2= 25?

MrRa [10]

Answer:

$x=-4,-15$

Step-by-step explanation:

$(x+9)^{2}=25 \\(x+9)=\pm5\\x=-9\pm5\\x=-4, -15$

6 0

3 years ago

Read 2 more answers

Salsk061 [2.6K]

Answer:

10+8=8+5+5

Step-by-step explanation:

4 0

3 years ago

Calculus 2 Master needed, evaluate the indefinite integral of: <img src="https://tex.z-dn.net/?f=%5Cint%5C%28%20%28lnx%29%5E2%7D

viva [34]

Answer:

$\int (\ln(x))^2dx=x(\ln(x)^2-2\ln(x)+2)+C$

Step-by-step explanation:

So we have the indefinite integral:

$\int (\ln(x))^2dx$

This is the same thing as:

$=\int 1\cdot (\ln(x))^2dx$

So, let's do integration by parts.

Let u be (ln(x))². And let dv be (1)dx. Therefore:

$u=(\ln(x))^2\\\text{Find du. Use the chain rule.}\\\frac{du}{dx}=2(\ln(x))\cdot\frac{1}{x}$

Simplify:

$du=\frac{2\ln(x)}{x}dx$

And:

$dv=(1)dx\\v=x$

Therefore:

$\int (\ln(x))^2dx=x\ln(x)^2-\int(x)(\frac{2\ln(x)}{x})dx$

The x cancel:

$=x\ln(x)^2-\int2\ln(x)dx$

Move the 2 to the front:

$=x\ln(x)^2-2\int\ln(x)dx$

(I'm not exactly sure how you got what you got. Perhaps you differentiated incorrectly?)

Now, let's use integrations by parts again for the integral. Similarly, let's put a 1 in front:

$=x\ln(x)^2-2\int 1\cdot\ln(x)dx$

Let u be ln(x) and let dv be (1)dx. Thus:

$u=\ln(x)\\du=\frac{1}{x}dx$

And:

$dv=(1)dx\\v=x$

So:

$=x\ln(x)^2-2(x\ln(x)-\int (x)\frac{1}{x}dx)$

Simplify the integral:

$=x\ln(x)^2-2(x\ln(x)-\int (1)dx)$

Evaluate:

$=x\ln(x)^2-2(x\ln(x)-x)$

Now, we just have to simplify :)

Distribute the -2:

$=x\ln(x)^2-2x\ln(x)+2x$

And if preferred, we can factor out a x:

$=x(\ln(x)^2-2\ln(x)+2)$

And, of course, don't forget about the constant of integration!

$=x(\ln(x)^2-2\ln(x)+2)+C$

And we are done :)

8 0

3 years ago