2 years ago

I didnt mean to write this question so I guess Ill just write this down because Ive got to put something.

Mathematics

1 answer:

jasenka [17]2 years ago

6 0

Answer:

Thx

Step-by-step explanation:

so its free points right?

You might be interested in

If Figure E is multiplied by a scale factor of 2.5, reflected over the y-axis, and translated 1 unit right and 4 units up, which

EastWind [94]

Answer: A

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Is this right ? if not please help me !

sergejj [24]

Answer:

You are correct!!! :)

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Which missing numbers make the following expressions equivalent?

Vladimir79 [104]

The answer is A. 24; 40

3 0

3 years ago

Read 2 more answers

What is 12-(16)= lollolll

Ierofanga [76]

Answer:

-4

Step-by-step explanation:

pls give brainliest

8 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 :

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

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

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

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

Option 3 is the Answer

8 0

3 years ago