4 years ago

Anyone know how to solve this? If you can help, I really appreciate it!

1 answer:

4 0

Well if you know how to find the area and perimeters of the shapes, you would get a ratio like this. (Note: This is not one of them in the picture.) Ex. Is the picture above.

I hope this helped you.

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seraphim [82]

Answer:

I do lol

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

HELP!!! 20 POINTS!!! URGENT!!!

VashaNatasha [74]

We know that :

$\clubsuit$ $ln(A) - ln(B) = ln(\frac{A}{B})$

$\clubsuit$ $aln(x) = ln(x)^a$

$\clubsuit$ $ln(A) + ln(B) = ln(AB)$

Using above ideas we can solve the Problem :

⇒ $\frac{3}{8}ln(x + 3) = ln(x + 3)^\frac{3}{8}$

⇒ $ln(x - 3) - ln(x + 3)^\frac{3}{8} = ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}]$

⇒ $4ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}] = ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}]^4 = ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}]$

⇒ $\frac{1}{3}lnx + ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}] = ln(x)^\frac{1}{3} + ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}] = ln[\frac{\sqrt[3]{x}(x - 3)^4}{\sqrt{(x + 3)^{3}}}]$