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emmainna [20.7K]
3 years ago
12

I don’t understand...

Mathematics
2 answers:
Makovka662 [10]3 years ago
6 0

Hi there,

I have done 3-6. I think the first 2 are correct.

Sonbull [250]3 years ago
6 0
I have trouble with this tooooo
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Answer:

969

Step-by-step explanation:

173 +  212 + 185 +  197 + 202  = 969

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What is MEAN ABSOLUTE DEVIATION and how do you find it?
Fynjy0 [20]

Step-by-step explanation:

Step 1: Calculate the mean. Step 2: Calculate how far away each data point is from the mean using positive distances. These are called absolute deviations. Step 3: Add those deviations together. Step 4: Divide the sum by the number of data points.

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A ribbon is cut into n parts. The length of each part increases such that they form a geometric progression. SPM Given the lengt
emmainna [20.7K]

Answer:

Step-by-step explanation:

Let\ say\ r \ the \common\ ratio:\\a)a_4=r^2*a_2=9*a_2\Longrightarrow\ r=3\ (not\ -3\ since\ parts\ are\ positive)\\\\b)\\\displaystyle \sum_{i=1}^{n} a_i =a_1+a_2+a_3+...+a_n\\\\=a_1+a_1*r+a_1*r^2+a_1*r^3+...+a_1*r^{n-1}\\\\=a_1*(1+r+r^2+...+r^{n-1})\\\\=a_1*\dfrac{r^n-1}{r-1} \\\\147620=5*\frac{(3^n-1)}{3-1} \\\\3^n-1=\frac{2*147620}{5} \\\\3^n=59049 \\\\n*ln(3)=ln(59049)\\\\n=\dfrac{ln(59049)}{ln(3)} \\\\n=10\\\\a_n=5*3^{10-1}=5*19683=98415(cm)

5 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 :)

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